Introduction
The idea of observing low-frequency gravitational waves (GWs) via their influence on radio pulsar arrival times first appears in work of Sazhin [1 ] . Subsequently Detweiler [2 ] described the improvements obtained by cross-correlating data from multiple pulsars. Such detectors, called “pulsar timing arrays” (PTAs), have been studied in detail by many authors [3 , 4 , 5 ] . Currently, a number of groups are working to detect gravitational waves in the nano-Hz regime with these techniques [6 , 7 , 8 ] .
The basic idea is as follows. Select a pulsar which is a very stable clock, emitting pulses at uniform intervals in its rest frame. Determine the arrival times t t t [9 , 10 ] ). In the absence of GWs, those pulses would have uniform time offsets from each other, whereas in the presence of a GW, the offsets Δ τ \Delta\tau Δ τ timing residuals Δ τ ( t ) \Delta\tau(t) Δ τ ( t )
In 1983, Hellings and Downs [11 ] predicted an important property of pulsar timing residuals, which is central to their detection with PTAs. The timing residuals from different pulsars are correlated, in a way that depends upon the angle γ \gamma γ [12 ] . To observe this, one averages the product of the timing residual time-series over years or decades to form a correlation
ρ = Δ τ 1 ( t ) Δ τ 2 ( t ) ‾ , (1) \rho= \overline{\Delta\tau_1(t)\Delta\tau_2(t)}, \tag*{(1)} ρ = Δ τ 1 ( t ) Δ τ 2 ( t ) , (1) where the overbar denotes a time average, and 1 and 2 label the pulsars. The expected value of ρ \rho ρ
The function which describes the behavior of the average or expected correlation ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩ γ \gamma γ μ u ( γ ) \mu_{\mathrm{u}}(\gamma) μ u ( γ ) (5) and illustrated in Figure 1 . On the average, the expected correlation is largest when the two pulsars are along the same line of sight, and (for example) vanishes when the two pulsars are separated by about 50°. A convincing PTA detection of GWs is expected to show this pattern of correlation among the different possible pulsar pairs [13 ] .
Figure 1. The Hellings-Downs correlation mean ⟨ ρ ⟩ = μ u ( γ ) \langle\rho\rangle= \mu_u(\gamma) ⟨ ρ ⟩ = μ u ( γ ) Δ ρ = σ u ( γ ) \Delta\rho= \sigma_u(\gamma) Δ ρ = σ u ( γ ) (5) and (7) . The dashed lines show a ± 1 σ u \pm1\sigma_u ± 1 σ u
Here, we ask: how close to this expectation would the observed correlations lie, in the absence of experimental noise or errors? Simulations have demonstrated that fluctuations away from the mean can be large [14 , 15 ] . These fluctuations away from the mean have two contributions: a “pulsar variance” and a “cosmic variance.” The first, which can in principle be measured and removed, reflects the variations in correlation between pairs of pulsars at different sky locations, but separated by the same angle γ \gamma γ
Previous work on estimating the variance relies on “Gaussian ensemble” techniques which were developed for the detection of audio-frequency GW stochastic backgrounds with interferometric detectors [16 , 17 ] . These methods were extended to PTAs in [18 , 19 ] , and used to estimate the variances in [20 , 21 , 22 ] . In contrast, our calculations use a finite collection of discrete sources and smoothly transition from one source to many. Similar models have been employed in the context of audio-frequency GW stochastic backgrounds [23 ] . If we let the density of sources approach infinity while keeping the mean-squared GW amplitude at Earth fixed, then our discrete source models reproduce the results of the Gaussian ensemble (see Appendix § 8 § 9.2 ).
One of the most important contributions of this paper is a straightforward method to separately identify the pulsar and the cosmic variance contributions to the total variance. The distinction is relevant, because observations with access to enough low-noise pulsars, uniformly distributed around the sky, could reduce the impact of pulsar variance as much as desired, by averaging the Hellings-Downs correlation over many pairs of pulsars separated by the same angle. However, the effects of cosmic variance cannot be removed. Our calculational technique to distinguish these effects corresponds closely to experimental practice. To compute the cosmic variance, we average the correlation over pulsar pairs separated by the same angle, before the first and second moments of the correlation are calculated with ensemble averages. The same technique can also be applied to the traditional Gaussian ensemble (see Appendix § 8 § 11 ).
Simulations modeling sources in the nearest few hundred Mpc to Earth suggest that these can cause significant deviations from isotropy [15 , 24 ] . It has been argued that these will not lead to substantial variations from the Hellings-Downs curve [13 , 25 ] . Our findings tend to support this conclusion for the mean of the correlation, but not for the variance, which is also an observable quantity. In this sense, the predicted variance of the Hellings-Downs curve contains additional information about the nature of the sources.
A brief outline of the paper is as follows. In Sec. II, we present and discuss pulsar variance for a single unpolarized point source, neglecting pulsar terms. In Sec. III, we show how those single-point results can be used to estimate the variance for a large collection of point sources, based on a simple model for the ensemble of sources, also including pulsar terms. As a simple example of the calculation technique (unrelated to PTAs) we calculate the expected value and variance of the GW energy density. In Sec. III A, we consider point sources which are all radiating at the same GW frequency, resulting in confusion noise, and in Sec. III B, we consider point sources radiating at distinct frequencies. Finally, in § 3 , we show how these variances can be cleanly split into pulsar variance and cosmic variance. This is followed by a short Conclusion.
The paper relies on detailed calculations given in a series of technical appendices, which are referenced throughout. In Appendix § 2.1 , we calculate the mean and variance of the Hellings-Downs correlation for a general GW point source and explicitly demonstrate that the “pulsar average” over three variables is equivalent to the “GW sky direction” average over two variables. As an example, including Earth terms only, we calculate the mean and variance for a binary inspiral GW source model (which is polarized if the orbit is not face-on or face-off) and its population average.
In Appendix § 2.2 , we examine the effects of the pulsar terms on the mean and variance of the Hellings-Downs correlation. These do not affect the mean, but for sources with slow evolution they increase the variance by a factor of four. As an example, we calculate the mean and variance for a nonevolving (single frequency) binary inspiral GW source model, and its population average.
In Appendix § 8 , we employ the standard Gaussian ensemble from [17 ] . We first estimate the importance of the pulsar terms in the Hellings-Downs correlation. We then compute the total variance (one pulsar pair) and the cosmic variance (many pulsar pairs) of the Hellings and Downs correlation for the Gaussian ensemble. We also compute the variance of the GW energy density. These Gaussian ensemble results are limits of our discrete-source confusion-noise ensemble results from Secs. III A and IV. Finally, we derive the cosmic variance as a harmonic sum.
Appendix D is the first of three in which we carry out the integrals needed to obtain the second moments of the Hellings-Downs correlation. Appendix D does the unpolarized case corresponding to the “I” term in the Stokes parameters. Appendix E does the polarized case corresponding to the “Q” Stokes parameter, and Appendix F does the cross term.
Appendix G calculates the Hellings-Downs two-point function μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) μ ~ 2 ( γ ) \widetilde{\mu}^{2}(\gamma) μ 2 ( γ ) β \beta β
Notes on terminology and notation:
(i) The term “Hellings-Downs correlation” refers to correlations among pulsar timing residuals, which might or might not follow the “Hellings-Downs curve” μ u ( γ ) \mu_{u}(\gamma) μ u ( γ )
(ii) Correlations between pulsar timing residuals or redshifts are denoted by ρ \rho ρ μ \mu μ σ 2 \sigma^{2} σ 2 ρ \rho ρ
(iii) In contrast, the expressions μ u ( γ ) \mu_{u}(\gamma) μ u ( γ ) σ u 2 ( γ ) \sigma_{u}^{2}(\gamma) σ u 2 ( γ ) σ p 2 ( γ ) \sigma_{p}^{2}(\gamma) σ p 2 ( γ ) σ c 2 ( γ ) \sigma_{c}^{2}(\gamma) σ c 2 ( γ ) μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) μ ~ 2 ( γ ) \widetilde{\mu}^{2}(\gamma) μ 2 ( γ ) (203) , (D37), (E8), (F2), (G5), and (G11).
(iv) Angle brackets ⟨ Q ⟩ \langle Q\rangle ⟨ Q ⟩ Q Q Q ⟨ 1 ⟩ = 1 \langle1\rangle=1 ⟨ 1 ⟩ = 1
(v) Three-vectors are indicated in boldface x \mathbf{x} x a a a b b b c c c d d d
(vi) Indices A , A ′ , … A,A',\ldots A , A ′ , … × \times ×
(vii) We use fixed Cartesian coordinates centered at the SSB with x ^ , y ^ , z ^ \hat{\mathbf{x}},\hat{\mathbf{y}},\hat{\mathbf{z}} x ^ , y ^ , z ^ t t t
(viii) Unit vectors on the two-sphere are written Ω \mathbf{\Omega} Ω
(ix) The angles θ ∈ [ 0 , π ] \theta\in[0,\pi] θ ∈ [ 0 , π ] ϕ ∈ [ 0 , 2 π ) \phi\in[0,2\pi) ϕ ∈ [ 0 , 2 π ) θ \theta θ z z z ϕ \phi ϕ x x x x x x y y y
(x) Indices j j j k k k ℓ \ell ℓ m m m 1 , 2 , … , N 1,2,\ldots,N 1 , 2 , … , N
(xi) The two pulsars with correlated pulse arrival times or redshifts are labeled “1” and “2” and are at distances L 1 L_{1} L 1 L 2 L_{2} L 2 p 1 \mathbf{p}_{1} p 1 p 2 \mathbf{p}_{2} p 2 γ ∈ [ 0 , π ] \gamma\in[0,\pi] γ ∈ [ 0 , π ] cos γ = p 1 ⋅ p 2 \cos\gamma=\mathbf{p}_{1}\cdot\mathbf{p}_{2} cos γ = p 1 ⋅ p 2
(xii) Some calculations incorporate a variable χ \chi χ
(xiii) Unless indicated, we use units in which the speed of light c = 1 c=1 c = 1 10 15 10^{15} 1 0 15 [26 , 27 , 28 ] .
Mean and variance of the Hellings-Downs correlation for one unpolarized GW point source
Here, we briefly describe the mean and variance of the Hellings-Downs correlation for a single unpolarized point source, which are computed in Appendix D and used throughout the paper. In this section, we neglect pulsar terms.
Hellings and Downs calculate the effect (on pulse redshift or timing residual) from a plane GW with unit amplitude traveling in direction Ω \mathbf{\Omega} Ω A = + A=+ A = + A = × A=\times A = ×
F A ( Ω ) = 1 2 p a p b 1 + Ω ⋅ p e a b A ( Ω ) , (2) F^{A}(\mathbf{\Omega})=\frac{1}{2}\frac{p^{a}p^{b}}{1+\mathbf{\Omega}\cdot\mathbf{p}}e^{A}_{ab}(\mathbf{\Omega}), \tag*{(2)} F A ( Ω ) = 2 1 1 + Ω ⋅ p p a p b e ab A ( Ω ) , (2) where p \mathbf{p} p e a b A ( Ω ) e^{A}_{ab}(\mathbf{\Omega}) e ab A ( Ω ) (180) . Clear derivations of these equations starting from first principles may be found in Appendices A and B of Ref. [18 ] , noting the redshift sign correction in [29 ] .
The correlation between two pulsars (subscripts 1 and 2) summed uniformly over both GW polarizations is
ρ = F 1 + ( Ω ) F 2 + ( Ω ) + F 1 × ( Ω ) F 2 × ( Ω ) . (3) \rho=F_{1}^{+}(\mathbf{\Omega})F_{2}^{+}(\mathbf{\Omega})+F_{1}^{\times}(\mathbf{\Omega})F_{2}^{\times}(\mathbf{\Omega}). \tag*{(3)} ρ = F 1 + ( Ω ) F 2 + ( Ω ) + F 1 × ( Ω ) F 2 × ( Ω ) . (3) The Hellings-Downs curve [11 ] , Eq. 5 is obtained by averaging this correlation over the directions of the GWs. We use angle brackets to denote this uniform average over directions: if Q ( Ω ) Q(\mathbf{\Omega}) Q ( Ω )
⟨ Q ⟩ = 1 4 π ∫ d Ω Q ( Ω ) = 1 4 π ∫ 0 π sin θ d θ ∫ 0 2 π d ϕ Q ( θ , ϕ ) . (4) \begin{aligned}
\langle Q\rangle=\frac{1}{4\pi}\int d\mathbf{\Omega}\,Q(\mathbf{\Omega}) \\
&=\frac{1}{4\pi}\int_{0}^{\pi}\sin\theta\,d\theta\int_{0}^{2\pi}d\phi Q(\theta,\phi). \tag*{(4)}
\end{aligned} ⟨ Q ⟩ = 4 π 1 ∫ d Ω Q ( Ω ) = 4 π 1 ∫ 0 π sin θ d θ ∫ 0 2 π d ϕQ ( θ , ϕ ) . (4) If we average the correlation ρ \rho ρ (3) in this way, we obtain the Hellings-Downs curve
μ u ( γ ) = ⟨ ρ ⟩ = 1 4 + 1 12 cos γ + 1 2 ( 1 − cos γ ) log ( 1 − cos γ 2 ) , (5) \mu_u(\gamma)=\langle\rho\rangle
=\frac{1}{4}+\frac{1}{12}\cos\gamma+\frac{1}{2}(1-\cos\gamma)\log\left(\frac{1-\cos\gamma}{2}\right), \tag*{(5)} μ u ( γ ) = ⟨ ρ ⟩ = 4 1 + 12 1 cos γ + 2 1 ( 1 − cos γ ) log ( 2 1 − cos γ ) , (5) where γ \gamma γ cos γ = p 1 ⋅ p 2 \cos\gamma=\mathbf{p}_1\cdot\mathbf{p}_2 cos γ = p 1 ⋅ p 2 § 13 ; this result is obtained in Eq. (203) and is plotted in Figure 1 .
Note that we follow the Hellings-Downs normalization convention μ u ( 0 ) = 1 / 3 \mu_u(0)=1/3 μ u ( 0 ) = 1/3 1 / 2 1/2 1/2 γ = 0 \gamma=0 γ = 0 (136) . This additional delta-function comes from the “pulsar terms,” and doubles the correlation for time-stationary GW sources. It arises from Eq. (48) for individual sources and from Eq. (145) for a Gaussian ensemble of sources.
Our interpretation of the function μ u ( γ ) \mu_u(\gamma) μ u ( γ ) [25 ] . Place an unpolarized point source of GWs at a fixed point on the sky. Then consider the correlation Eq. (3) as a function of the three variables that define the positions of the two pulsars on the sky, with fixed angle γ \gamma γ ρ \rho ρ μ u ( γ ) = ⟨ ρ ⟩ \mu_u(\gamma)=\langle\rho\rangle μ u ( γ ) = ⟨ ρ ⟩ ρ \rho ρ Δ ρ \Delta\rho Δ ρ γ \gamma γ
σ u 2 ( γ ) = ⟨ Δ ρ 2 ⟩ = ⟨ ( ρ − ⟨ ρ ⟩ ) 2 ⟩ = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 , (6) \sigma_u^2(\gamma)=\langle\Delta\rho^2\rangle=\langle(\rho-\langle\rho\rangle)^2\rangle=\langle\rho^2\rangle-\langle\rho\rangle^2, \tag*{(6)} σ u 2 ( γ ) = ⟨ Δ ρ 2 ⟩ = ⟨( ρ − ⟨ ρ ⟩ ) 2 ⟩ = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 , (6) where the subscript “u” means “unpolarized.” This variance cannot be determined from the Hellings-Downs curve itself, which is the first moment ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩ ⟨ ρ 2 ⟩ \langle\rho^2\rangle ⟨ ρ 2 ⟩
There are two equivalent ways to calculate that second moment, which give exactly the same result. One can (a) fix the source position and average over the three variables that define the directions to the pulsars at fixed relative angle γ \gamma γ § 2.1 ) or one can (b) fix the pulsar locations, with angle γ \gamma γ § 13 ). Both approaches give the second moment of Eq. (210) . From that second moment, one finds the variance from Eqs. (5) and (6) , obtaining
σ u 2 ( γ ) = 97 80 + 1 24 cos γ − 839 720 cos 2 γ + 1 12 ( 1 − cos γ ) log ( 1 − cos γ 2 ) × ( 18 − 10 cos γ − 3 ( 1 − cos γ ) log ( 1 − cos γ 2 ) ) . (7) \sigma_u^2(\gamma)=\frac{97}{80}+\frac{1}{24}\cos\gamma-\frac{839}{720}\cos^2\gamma
+\frac{1}{12}(1-\cos\gamma)\log\left(\frac{1-\cos\gamma}{2}\right)
\times\left(18-10\cos\gamma-3(1-\cos\gamma)\log\left(\frac{1-\cos\gamma}{2}\right)\right). \tag*{(7)} σ u 2 ( γ ) = 80 97 + 24 1 cos γ − 720 839 cos 2 γ + 12 1 ( 1 − cos γ ) log ( 2 1 − cos γ ) × ( 18 − 10 cos γ − 3 ( 1 − cos γ ) log ( 2 1 − cos γ ) ) . (7) A detailed calculation is given in Appendix § 13 ; this result is obtained in Eq. (211) .
The square root of the variance is the standard deviation Δ ρ = σ u \Delta\rho=\sigma_u Δ ρ = σ u Figure 1 . The standard deviation indicates the approximate range in which, with the Hellings and Downs assumptions, the correlation for any particular pulsar pair should fall, if the Universe contains a single distant pointlike unpolarized GW source, and the pulsar terms can be neglected. To indicate this graphically, the dashed lines in Figure 1 show the ± 1 Δ ρ \pm1\Delta\rho ± 1Δ ρ ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩ Figure 3 shows a similar plot for a constant amplitude GW source, to which the pulsar terms necessarily contribute.)
Figure 3. The Hellings-Downs correlation mean and standard deviation for a single unpolarized point source, including the pulsar terms. These are given by Eqs. (63) and (67) with χ = 1 \chi=1 χ = 1 H 2 = 2 \mathcal{H}_2=2 H 2 = 2 H 4 = 4 \mathcal{H}_4=4 H 4 = 4 μ = μ u ( γ ) \mu=\mu_u(\gamma) μ = μ u ( γ ) σ 2 = σ u 2 ( γ ) + 3 σ c 2 ( γ ) / 2 \sigma^2=\sigma_u^2(\gamma)+3\sigma_c^2(\gamma)/2 σ 2 = σ u 2 ( γ ) + 3 σ c 2 ( γ ) /2 ± 1 σ \pm1\sigma ± 1 σ σ \sigma σ Figure 1 shows that including the pulsar term increases the variance by about a factor of four.
In Appendix § 2.1 we show that the mean correlation produced by a single (possibly polarized) point source is always described by the Hellings-Downs curve μ u ( γ ) \mu_u(\gamma) μ u ( γ )
Variance of the Hellings-Downs Correlation with Many Sources
The mean and variance of the correlation for a single point source are not directly applicable. We expect that our Universe contains large numbers of PTA point sources, probably in the form of supermassive black-hole binaries [30 , 31 ] with the closest at 50–100 Mpc from Earth, and the most distant at the Hubble radius ≈ 4 \approx4 ≈ 4
We begin by constructing a model Universe containing N N N − Ω j -\mathbf{\Omega}_j − Ω j j = 1 , 2 , … , N j=1,2,\ldots,N j = 1 , 2 , … , N Ω j \mathbf{\Omega}_j Ω j Ω j \mathbf{\Omega}_j Ω j h j + ( t ) h_j^+(t) h j + ( t ) h j × ( t ) h_j^\times(t) h j × ( t ) [18 ] for further discussion.
For this fixed set of GW sources, the pulse redshift of pulsar n = 1 , 2 n = 1, 2 n = 1 , 2 t t t
Z n ( t ) = ∑ j Δ h j + ( t , L n p n ) F n + ( Ω j ) + Δ h j × ( t , L n p n ) F n × ( Ω j ) , (8) Z_n(t) = \sum_j \Delta h_j^+(t,L_n\mathbf{p}_n)F_n^+(\boldsymbol{\Omega}_j) + \Delta h_j^\times(t,L_n\mathbf{p}_n)F_n^\times(\boldsymbol{\Omega}_j), \tag*{(8)} Z n ( t ) = j ∑ Δ h j + ( t , L n p n ) F n + ( Ω j ) + Δ h j × ( t , L n p n ) F n × ( Ω j ) , (8) where L n p n L_n\mathbf{p}_n L n p n n n n p n \mathbf{p}_n p n [18 ] .) The antenna pattern functions F n + , × F^{+,\times}_n F n + , × (2) , or more explicitly as functions of the pulsar positions in Eq. (94) , and the strain differences are
Δ h j A ( t , L p ) = h j A ( t ) − χ h j A ( t − L ( 1 + p ⋅ Ω j ) ) , (9) \Delta h_j^A(t,L\mathbf{p}) = h_j^A(t) - \chi h_j^A\left(t-L(1+\mathbf{p}\cdot\boldsymbol{\Omega}_j)\right), \tag*{(9)} Δ h j A ( t , L p ) = h j A ( t ) − χ h j A ( t − L ( 1 + p ⋅ Ω j ) ) , (9) where the first term is called the “Earth term” and the second is called the “pulsar term.” Here, A = + , × A = +,\times A = + , × (9) is a function of time t t t L p L\mathbf{p} L p L = ( L p ⋅ L p ) 1 / 2 L = (\mathbf{L}\mathbf{p}\cdot\mathbf{L}\mathbf{p})^{1/2} L = ( Lp ⋅ Lp ) 1/2
The constant χ \chi χ (9) allows us to control the pulsar term. Setting χ = 1 \chi= 1 χ = 1 χ = 0 \chi= 0 χ = 0 χ \chi χ χ \chi χ [32 ] .
Note that the ensembles of GW sources which we employ in Secs. III A and III B have frequencies and amplitudes that do not change with time. For such ensembles, or for any time-stationary ensemble such as the Gaussian ensemble of Appendix § 8 , the pulsar terms cannot be neglected when computing the variance, since they make contributions comparable to the Earth terms. Hence, for such ensembles, one must set χ = 1 \chi= 1 χ = 1 χ = 0 \chi= 0 χ = 0
To calculate the correlations between the pulse redshifts, it is sometimes convenient to work in a circular polarization basis rather than in a linear polarization basis. For this, define the complex waveforms
h j ( t ) = h j + ( t ) + i h j × ( t ) (10) h_j(t) = h_j^+(t) + i h_j^\times(t) \tag*{(10)} h j ( t ) = h j + ( t ) + i h j × ( t ) (10) and the corresponding complex antenna pattern functions
F n ( Ω ) = F n + ( Ω ) − i F n × ( Ω ) , (11) F_n(\boldsymbol{\Omega}) = F_n^+(\boldsymbol{\Omega}) - iF_n^\times(\boldsymbol{\Omega}), \tag*{(11)} F n ( Ω ) = F n + ( Ω ) − i F n × ( Ω ) , (11) where n = 1 , 2 n = 1, 2 n = 1 , 2 (8) of pulsar n n n
Z n ( t ) = ℜ ∑ j Δ h j ( t , L n p n ) F n ( Ω j ) , (12) Z_n(t) = \Re\sum_j \Delta h_j(t,L_n\mathbf{p}_n)F_n(\boldsymbol{\Omega}_j), \tag*{(12)} Z n ( t ) = ℜ j ∑ Δ h j ( t , L n p n ) F n ( Ω j ) , (12) where the complex strain differences are
Δ h j ( t , L p ) = Δ h j + ( t , L p ) + i Δ h j × ( t , L p ) = h j ( t ) − χ h j ( t − L ( 1 + p ⋅ Ω ) ) . (13) \Delta h_j(t,L\mathbf{p}) = \Delta h_j^+(t,L\mathbf{p}) + i\Delta h_j^\times(t,L\mathbf{p})
= h_j(t) - \chi h_j\left(t-L(1+\mathbf{p}\cdot\boldsymbol{\Omega})\right). \tag*{(13)} Δ h j ( t , L p ) = Δ h j + ( t , L p ) + i Δ h j × ( t , L p ) = h j ( t ) − χ h j ( t − L ( 1 + p ⋅ Ω ) ) . (13) This can be verified by using Eqs. (10) , (11) , and (13) to obtain Eq. (8) .
We now compute the correlation between two pulsars. For each, take the real part of Eq. (12) by summing half of the rhs and its complex conjugate, then multiply and time average. The resulting correlation is the sum of N 2 N^2 N 2
Z 1 Z 2 ‾ = 1 4 ∑ j = 1 N ∑ k = 1 N [ Δ h j ( t , L 1 p 1 ) Δ h k ( t , L 2 p 2 ) ‾ F 1 ( Ω j ) F 2 ( Ω k ) + Δ h j ∗ ( t , L 1 p 1 ) Δ h k ( t , L 2 p 2 ) ‾ F 1 ∗ ( Ω j ) F 2 ( Ω k ) + Δ h j ( t , L 1 p 1 ) Δ h k ∗ ( t , L 2 p 2 ) ‾ F 1 ( Ω j ) F 2 ∗ ( Ω k ) + Δ h j ∗ ( t , L 1 p 1 ) Δ h k ∗ ( t , L 2 p 2 ) ‾ F 1 ∗ ( Ω j ) F 2 ∗ ( Ω k ) ] , (14) \overline{Z_1Z_2} = \frac{1}{4}\sum_{j=1}^{N}\sum_{k=1}^{N}\left[
\overline{\Delta h_j(t,L_1\mathbf{p}_1)\Delta h_k(t,L_2\mathbf{p}_2)}F_1(\boldsymbol{\Omega}_j)F_2(\boldsymbol{\Omega}_k)
+ \overline{\Delta h_j^*(t,L_1\mathbf{p}_1)\Delta h_k(t,L_2\mathbf{p}_2)}F_1^*(\boldsymbol{\Omega}_j)F_2(\boldsymbol{\Omega}_k)
+ \overline{\Delta h_j(t,L_1\mathbf{p}_1)\Delta h_k^*(t,L_2\mathbf{p}_2)}F_1(\boldsymbol{\Omega}_j)F_2^*(\boldsymbol{\Omega}_k)
+ \overline{\Delta h_j^*(t,L_1\mathbf{p}_1)\Delta h_k^*(t,L_2\mathbf{p}_2)}F_1^*(\boldsymbol{\Omega}_j)F_2^*(\boldsymbol{\Omega}_k)
\right], \tag*{(14)} Z 1 Z 2 = 4 1 j = 1 ∑ N k = 1 ∑ N [ Δ h j ( t , L 1 p 1 ) Δ h k ( t , L 2 p 2 ) F 1 ( Ω j ) F 2 ( Ω k ) + Δ h j ∗ ( t , L 1 p 1 ) Δ h k ( t , L 2 p 2 ) F 1 ∗ ( Ω j ) F 2 ( Ω k ) + Δ h j ( t , L 1 p 1 ) Δ h k ∗ ( t , L 2 p 2 ) F 1 ( Ω j ) F 2 ∗ ( Ω k ) + Δ h j ∗ ( t , L 1 p 1 ) Δ h k ∗ ( t , L 2 p 2 ) F 1 ∗ ( Ω j ) F 2 ∗ ( Ω k ) ] , (14) where the overbars denote the time averages of the quantities beneath them. Here, we explicitly indicate that there are N 2 N^2 N 2 j , k j,k j , k (14) to compute the redshift correlation for any pair of pulsars.
Since we do not know the locations or waveforms of the relevant PTA sources, to estimate the correlation and its variance, we need to adopt some statistical model. This creates an ensemble of universes, which we can then employ for averaging. We will pick the minimum set of assumptions that will permit us to derive an estimate.
Confusion-noise case
PTAs measure timing residuals with a cadence of about a week. Since the observations span several decades, this means that the fluctuations are spread among O ( 10 3 ) O(10^3) O ( 1 0 3 ) O ( 10 6 ) O(10^6) O ( 1 0 6 ) O ( 1000 ) O(1000) O ( 1000 )
For our model, we give all sources an identical fixed GW angular frequency ω \omega ω ω \omega ω 2 π / T 2\pi/T 2 π / T T T T [33 ] . Assume that the sources are unpolarized, with the same intrinsic GW amplitude (at some fiducial distance) in both polarizations. Thus,
h j + ( t ) = A j cos ( ω t + ϕ j ) , and h_j^{+}(t)=\mathcal{A}_j\cos(\omega t+\phi_j),\qquad\text{and} h j + ( t ) = A j cos ( ω t + ϕ j ) , and h j × ( t ) = A j sin ( ω t + ϕ j ) . (15) h_j^{\times}(t)=\mathcal{A}_j\sin(\omega t+\phi_j). \tag*{(15)} h j × ( t ) = A j sin ( ω t + ϕ j ) . (15) (Note that, consistent with Eq. (8) , these equations define plus and cross polarization amplitudes with respect to a polarization basis that depends upon the direction Ω \mathbf{\Omega} Ω (180) : see the first line of Eq. (18) below.) Here, the A j \mathcal{A}_j A j ϕ j \phi_j ϕ j [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ]
We locate the sources uniformly in 3-dimensional space, labeling them with increasing distance from Earth, so j = 1 j=1 j = 1 j = 2 j=2 j = 2
A 1 = A , A 2 = 2 − 1 / 3 A , \mathcal{A}_1=\mathcal{A},\qquad\mathcal{A}_2=2^{-1/3}\mathcal{A}, A 1 = A , A 2 = 2 − 1/3 A , A 3 = 3 − 1 / 3 A , … , A N = N − 1 / 3 A , (16) \mathcal{A}_3=3^{-1/3}\mathcal{A},\ldots,\qquad\mathcal{A}_N=N^{-1/3}\mathcal{A}, \tag*{(16)} A 3 = 3 − 1/3 A , … , A N = N − 1/3 A , (16) where the dimensionless real positive quantity A = A 1 \mathcal{A}=\mathcal{A}_1 A = A 1 [34 ] , the number of sources N N N
Finally, to finish specifying our statistical ensemble of model universes, we need to specify the directions Ω j \mathbf{\Omega}_j Ω j ϕ j \phi_j ϕ j
We begin by computing the time average s = h a b h a b ‾ s=\overline{h_{ab}h^{ab}} s = h ab h ab s s s s s s s s s c 2 ω 2 s / 32 π G c^2\omega^2s/32\pi G c 2 ω 2 s /32 π G G G G c c c
The value of s s s ϕ j \phi_j ϕ j Ω j \mathbf{\Omega}_j Ω j s s s (10) and (15) as
h j ( t ) = h j + ( t ) + i h j × ( t ) = A j e i ( ω t + ϕ j ) . (17) h_j(t)=h_j^{+}(t)+ih_j^{\times}(t)=\mathcal{A}_j e^{i(\omega t+\phi_j)}. \tag*{(17)} h j ( t ) = h j + ( t ) + i h j × ( t ) = A j e i ( ω t + ϕ j ) . (17) The total GW strain at Earth is obtained by summing this over all sources
h a b ( t ) = ∑ j [ h j + ( t ) e a b + ( Ω j ) + h j × ( t ) e a b × ( Ω j ) ] = ℜ ∑ j h j ( t ) e a b ( Ω j ) = 1 2 ∑ j A j [ e i ( ω t + ϕ j ) e a b ( Ω j ) + e − i ( ω t + ϕ j ) e a b ∗ ( Ω j ) ] , (18) \begin{aligned}
h_{ab}(t)=\sum_j\left[h_j^{+}(t)e_{ab}^{+}(\mathbf{\Omega}_j)+h_j^{\times}(t)e_{ab}^{\times}(\mathbf{\Omega}_j)\right] \\
&=\Re\sum_j h_j(t)e_{ab}(\mathbf{\Omega}_j) \\
&=\frac{1}{2}\sum_j\mathcal{A}_j\left[e^{i(\omega t+\phi_j)}e_{ab}(\mathbf{\Omega}_j)+e^{-i(\omega t+\phi_j)}e_{ab}^{*}(\mathbf{\Omega}_j)\right], \tag*{(18)}
\end{aligned} h ab ( t ) = j ∑ [ h j + ( t ) e ab + ( Ω j ) + h j × ( t ) e ab × ( Ω j ) ] = ℜ j ∑ h j ( t ) e ab ( Ω j ) = 2 1 j ∑ A j [ e i ( ω t + ϕ j ) e ab ( Ω j ) + e − i ( ω t + ϕ j ) e ab ∗ ( Ω j ) ] , (18) where e a b = e a b + − i e a b × e_{ab}=e_{ab}^{+}-ie_{ab}^{\times} e ab = e ab + − i e ab ×
s = h a b h a b ‾ = 1 4 ∑ j , k A j A k [ e i ( ϕ j − ϕ k ) e a b ( Ω j ) e a b ∗ ( Ω k ) + e − i ( ϕ j − ϕ k ) e a b ∗ ( Ω j ) e a b ( Ω k ) ] = 2 ∑ j A j 2 + 1 4 ∑ j ≠ k A j A k [ e i ( ϕ j − ϕ k ) e a b ( Ω j ) e a b ∗ ( Ω k ) + e − i ( ϕ j − ϕ k ) e a b ∗ ( Ω j ) e a b ( Ω k ) ] . (19) \begin{aligned}
s=\overline{h_{ab}h^{ab}}=\frac{1}{4}\sum_{j,k}\mathcal{A}_j\mathcal{A}_k\left[e^{i(\phi_j-\phi_k)}e^{ab}(\mathbf{\Omega}_j)e_{ab}^{*}(\mathbf{\Omega}_k)+e^{-i(\phi_j-\phi_k)}e_{ab}^{*}(\mathbf{\Omega}_j)e^{ab}(\mathbf{\Omega}_k)\right] \\
&=2\sum_j\mathcal{A}_j^2+\frac{1}{4}\sum_{j\ne k}\mathcal{A}_j\mathcal{A}_k\left[e^{i(\phi_j-\phi_k)}e^{ab}(\mathbf{\Omega}_j)e_{ab}^{*}(\mathbf{\Omega}_k)+e^{-i(\phi_j-\phi_k)}e_{ab}^{*}(\mathbf{\Omega}_j)e^{ab}(\mathbf{\Omega}_k)\right]. \tag*{(19)}
\end{aligned} s = h ab h ab = 4 1 j , k ∑ A j A k [ e i ( ϕ j − ϕ k ) e ab ( Ω j ) e ab ∗ ( Ω k ) + e − i ( ϕ j − ϕ k ) e ab ∗ ( Ω j ) e ab ( Ω k ) ] = 2 j ∑ A j 2 + 4 1 j = k ∑ A j A k [ e i ( ϕ j − ϕ k ) e ab ( Ω j ) e ab ∗ ( Ω k ) + e − i ( ϕ j − ϕ k ) e ab ∗ ( Ω j ) e ab ( Ω k ) ] . (19) To obtain the second equality of Eq. (19) , square the final line of Eq. (18) , contract polarization tensor indices and average over time: two terms vanish because the frequency is an integer multiple of the inverse observation time. For the final equality, break the sum into diagonal terms (j = k j=k j = k j ≠ k j\ne k j = k e a b ( Ω ) e a b ∗ ( Ω ) = 4 e^{ab}(\mathbf{\Omega})e_{ab}^{*}(\mathbf{\Omega})=4 e ab ( Ω ) e ab ∗ ( Ω ) = 4
While s s s ϕ j \phi_j ϕ j Ω j \mathbf{\Omega}_j Ω j s s s ϕ j \phi_j ϕ j
Q ( ϕ 1 , ϕ 2 , … , ϕ N ) Q(\phi_1,\phi_2,\ldots,\phi_N) Q ( ϕ 1 , ϕ 2 , … , ϕ N )
⟨ Q ⟩ ϕ = ∫ 0 2 π d ϕ 1 2 π ⋯ ∫ 0 2 π d ϕ N 2 π Q ( ϕ 1 , … , ϕ N ) . (20) \langle Q\rangle_{\phi}=\int_0^{2\pi}\frac{d\phi_1}{2\pi}\cdots\int_0^{2\pi}\frac{d\phi_N}{2\pi}Q(\phi_1,\ldots,\phi_N). \tag*{(20)} ⟨ Q ⟩ ϕ = ∫ 0 2 π 2 π d ϕ 1 ⋯ ∫ 0 2 π 2 π d ϕ N Q ( ϕ 1 , … , ϕ N ) . (20) where the subscript on the angle brackets indicates an average over random phases. For example, if we pick Q = e i ( ϕ j − ϕ k ) Q=e^{i(\phi_j-\phi_k)} Q = e i ( ϕ j − ϕ k ) (20) gives an ensemble average
⟨ e i ( ϕ j − ϕ k ) ⟩ ϕ = δ j k , (21) \left\langle e^{i(\phi_j-\phi_k)}\right\rangle_{\phi}=\delta_{jk}, \tag*{(21)} ⟨ e i ( ϕ j − ϕ k ) ⟩ ϕ = δ j k , (21) where δ j k \delta_{jk} δ j k s s s (19) . Evaluating this with Eq. (21) eliminates the off-diagonal terms j ≠ k j \ne k j = k
⟨ s ⟩ ϕ = 2 ∑ j A j 2 = 2 H 2 . (22) \langle s\rangle_{\phi}=2\sum_j A_j^2=2\mathcal{H}_2. \tag*{(22)} ⟨ s ⟩ ϕ = 2 j ∑ A j 2 = 2 H 2 . (22) Since ⟨ s ⟩ ϕ \langle s\rangle_{\phi} ⟨ s ⟩ ϕ Ω j \mathbf{\Omega}_j Ω j ⟨ s ⟩ = ⟨ s ⟩ ϕ \langle s\rangle=\langle s\rangle_{\phi} ⟨ s ⟩ = ⟨ s ⟩ ϕ
H 2 = ∑ j = 1 N A j 2 = A 2 ∑ j = 1 N j − 2 / 3 = A 2 N s . (23) \mathcal{H}_2=\sum_{j=1}^{N}A_j^2=A^2\sum_{j=1}^{N}j^{-2/3}=A^2N_s. \tag*{(23)} H 2 = j = 1 ∑ N A j 2 = A 2 j = 1 ∑ N j − 2/3 = A 2 N s . (23) For any fixed + + + × \times × h a b h a b = 2 ( h + ) 2 + 2 ( h × ) 2 h_{ab}h^{ab}=2(h^+)^2+2(h^\times)^2 h ab h ab = 2 ( h + ) 2 + 2 ( h × ) 2 (22) implies that the average squared strain in either linear polarization is H 2 / 2 \mathcal{H}_2/2 H 2 /2 c 2 ω 2 H 2 / 16 π G c^2\omega^2\mathcal{H}_2/16\pi G c 2 ω 2 H 2 /16 π G (22) .
In Eq. (23) we define the number of shells N s N_s N s
N s = ∑ n = 1 N n − 2 / 3 ≈ ∫ 0 N n − 2 / 3 d n ≈ 3 N 1 / 3 , (24) N_s=\sum_{n=1}^{N}n^{-2/3}\approx\int_0^N n^{-2/3}dn\approx3N^{1/3}, \tag*{(24)} N s = n = 1 ∑ N n − 2/3 ≈ ∫ 0 N n − 2/3 d n ≈ 3 N 1/3 , (24) where the approximation is valid for larger numbers of sources. N s N_s N s
Thus, N s N_s N s N s ≈ 4000 M p c / 50 M p c ≈ 80 N_s\approx4000\ \mathrm{Mpc}/50\ \mathrm{Mpc}\approx80 N s ≈ 4000 Mpc /50 Mpc ≈ 80 N s N_s N s N N N
For later use, it is also helpful to define another measure of strain
H 4 = ∑ j = 1 N A j 4 = A 4 ∑ j = 1 N j − 4 / 3 , (25) \mathcal{H}_4=\sum_{j=1}^{N}A_j^4=A^4\sum_{j=1}^{N}j^{-4/3}, \tag*{(25)} H 4 = j = 1 ∑ N A j 4 = A 4 j = 1 ∑ N j − 4/3 , (25) noting that if there is a single source, then H 4 = H 2 2 = A 4 \mathcal{H}_4=\mathcal{H}_2^2=A^4 H 4 = H 2 2 = A 4
H 4 ≈ A 4 ζ ( 4 / 3 ) , (26) \mathcal{H}_4\approx A^4\zeta(4/3), \tag*{(26)} H 4 ≈ A 4 ζ ( 4/3 ) , (26) where ζ ( 4 / 3 ) ≈ 3.601 \zeta(4/3)\approx3.601 ζ ( 4/3 ) ≈ 3.601 H 2 \mathcal{H}_2 H 2 H 4 \mathcal{H}_4 H 4
Having found the first moment of s = h a b h a b s=h_{ab}h^{ab} s = h ab h ab (22) , we now compute the second moment. For this, we square s s s (19) and then take its ensemble average over the source phases ϕ j \phi_j ϕ j (19) consists of three terms: the square of the first term, twice the cross-term, and the square of the second term. The cross-term has j ≠ k j\ne k j = k (21) . The remaining two terms are
⟨ s 2 ⟩ ϕ = ( 2 ∑ j A j 2 ) 2 + 1 16 ∑ j ≠ k ∑ ℓ ≠ m A j A k A ℓ A m [ e a b ( Ω j ) e a b ∗ ( Ω k ) e c d ( Ω ℓ ) e c d ∗ ( Ω m ) ⟨ e i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ + e a b ∗ ( Ω j ) e a b ( Ω k ) e c d ∗ ( Ω ℓ ) e c d ( Ω m ) ⟨ e − i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ + e a b ( Ω j ) e a b ∗ ( Ω k ) e c d ∗ ( Ω ℓ ) e c d ( Ω m ) ⟨ e i ( ϕ j − ϕ k − ϕ ℓ + ϕ m ) ⟩ ϕ + e a b ∗ ( Ω j ) e a b ( Ω k ) e c d ( Ω ℓ ) e c d ∗ ( Ω m ) ⟨ e − i ( ϕ j − ϕ k − ϕ ℓ + ϕ m ) ⟩ ϕ ] . (27) \begin{aligned}
\langle s^2\rangle_\phi={}&\left(2\sum_j A_j^2\right)^2+\frac{1}{16}\sum_{j\ne k}\sum_{\ell\ne m}A_jA_kA_\ell A_m\Big[ e^{ab}(\mathbf{\Omega}_j)e^*_{ab}(\mathbf{\Omega}_k)e^{cd}(\mathbf{\Omega}_\ell)e^*_{cd}(\mathbf{\Omega}_m)\left\langle e^{i(\phi_j-\phi_k+\phi_\ell-\phi_m)}\right\rangle_\phi\\
&+e^*_{ab}(\mathbf{\Omega}_j)e^{ab}(\mathbf{\Omega}_k)e^*_{cd}(\mathbf{\Omega}_\ell)e^{cd}(\mathbf{\Omega}_m)\left\langle e^{-i(\phi_j-\phi_k+\phi_\ell-\phi_m)}\right\rangle_\phi\\
&+e^{ab}(\mathbf{\Omega}_j)e^*_{ab}(\mathbf{\Omega}_k)e^*_{cd}(\mathbf{\Omega}_\ell)e^{cd}(\mathbf{\Omega}_m)\left\langle e^{i(\phi_j-\phi_k-\phi_\ell+\phi_m)}\right\rangle_\phi\\
&+e^*_{ab}(\mathbf{\Omega}_j)e^{ab}(\mathbf{\Omega}_k)e^{cd}(\mathbf{\Omega}_\ell)e^*_{cd}(\mathbf{\Omega}_m)\left\langle e^{-i(\phi_j-\phi_k-\phi_\ell+\phi_m)}\right\rangle_\phi\Big]. \tag*{(27)}
\end{aligned} ⟨ s 2 ⟩ ϕ = ( 2 j ∑ A j 2 ) 2 + 16 1 j = k ∑ ℓ = m ∑ A j A k A ℓ A m [ e ab ( Ω j ) e ab ∗ ( Ω k ) e c d ( Ω ℓ ) e c d ∗ ( Ω m ) ⟨ e i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ + e ab ∗ ( Ω j ) e ab ( Ω k ) e c d ∗ ( Ω ℓ ) e c d ( Ω m ) ⟨ e − i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ + e ab ( Ω j ) e ab ∗ ( Ω k ) e c d ∗ ( Ω ℓ ) e c d ( Ω m ) ⟨ e i ( ϕ j − ϕ k − ϕ ℓ + ϕ m ) ⟩ ϕ + e ab ∗ ( Ω j ) e ab ( Ω k ) e c d ( Ω ℓ ) e c d ∗ ( Ω m ) ⟨ e − i ( ϕ j − ϕ k − ϕ ℓ + ϕ m ) ⟩ ϕ ] . (27) The four ensemble averages which appear in Eq. (27) are evaluated using Eq. (20) . Since j ≠ k j\ne k j = k ℓ ≠ m \ell\ne m ℓ = m ⟨ e i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ = δ j m δ k ℓ \langle e^{i(\phi_j-\phi_k+\phi_\ell-\phi_m)}\rangle_\phi=\delta_{jm}\delta_{k\ell} ⟨ e i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ = δ j m δ k ℓ δ j ℓ δ k m \delta_{j\ell}\delta_{km} δ j ℓ δ k m (27) simplifies to a double sum. The four products of polarization tensors are equal (just relabel the indices a a a b b b c c c d d d
⟨ s 2 ⟩ ϕ = ( 2 H 2 ) 2 + 1 4 ∑ j ≠ k A j 2 A k 2 e a b ( Ω j ) e c d ∗ ( Ω j ) e c d ( Ω k ) e a b ∗ ( Ω k ) . (28) \begin{aligned}
\langle s^2\rangle_\phi={}&(2\mathcal{H}_2)^2\\
&+\frac{1}{4}\sum_{j\ne k}A_j^2A_k^2e^{ab}(\mathbf{\Omega}_j)e^*_{cd}(\mathbf{\Omega}_j)e^{cd}(\mathbf{\Omega}_k)e^*_{ab}(\mathbf{\Omega}_k). \tag*{(28)}
\end{aligned} ⟨ s 2 ⟩ ϕ = ( 2 H 2 ) 2 + 4 1 j = k ∑ A j 2 A k 2 e ab ( Ω j ) e c d ∗ ( Ω j ) e c d ( Ω k ) e ab ∗ ( Ω k ) . (28) To complete our calculation of the second moment of s s s ⟨ s 2 ⟩ ϕ \langle s^2\rangle_\phi ⟨ s 2 ⟩ ϕ Ω j \mathbf{\Omega}_j Ω j (28) only contains terms with j ≠ k j\ne k j = k (154) from the Appendix. Each average yields a term η a b c d \eta_{abcd} η ab c d (155) with α = − 4 / 15 \alpha=-4/15 α = − 4/15 β = 2 / 5 \beta=2/5 β = 2/5
⟨ s 2 ⟩ = ⟨ s 2 ⟩ ϕ , Ω = ( 2 H 2 ) 2 + 1 4 ∑ j ≠ k A j 2 A k 2 η c d a b η a b c d = 4 H 2 2 + 4 5 ∑ j ≠ k A j 2 A k 2 . (29) \begin{aligned}
\langle s^2\rangle=\langle s^2\rangle_{\phi,\Omega}
&=(2\mathcal{H}_2)^2+\frac{1}{4}\sum_{j\ne k}A_j^2A_k^2\eta^{ab}_{\ \ cd}\eta^{cd}_{\ \ ab}\\
&=4\mathcal{H}_2^2+\frac{4}{5}\sum_{j\ne k}A_j^2A_k^2. \tag*{(29)}
\end{aligned} ⟨ s 2 ⟩ = ⟨ s 2 ⟩ ϕ , Ω = ( 2 H 2 ) 2 + 4 1 j = k ∑ A j 2 A k 2 η c d ab η ab c d = 4 H 2 2 + 5 4 j = k ∑ A j 2 A k 2 . (29) where we have used η c d a b η a b c d = η a b c d η a b c d = 16 / 5 \eta^{ab}_{\ \ cd}\eta^{cd}_{\ \ ab}=\eta_{abcd}\eta^{abcd}=16/5 η c d ab η ab c d = η ab c d η ab c d = 16/5 (159) .
The variance of the average squared strain s = h a b h a b ‾ s=\overline{h_{ab}h^{ab}} s = h ab h ab (22) and (29) , and is
σ s 2 = ⟨ s 2 ⟩ − ⟨ s ⟩ 2 = 4 5 ∑ j ≠ k A j 2 A k 2 = 4 5 ( H 2 2 − H 4 ) , (30) \sigma_s^2=\langle s^2\rangle-\langle s\rangle^2=\frac{4}{5}\sum_{j\ne k}A_j^2A_k^2=\frac{4}{5}\left(\mathcal{H}_2^2-\mathcal{H}_4\right), \tag*{(30)} σ s 2 = ⟨ s 2 ⟩ − ⟨ s ⟩ 2 = 5 4 j = k ∑ A j 2 A k 2 = 5 4 ( H 2 2 − H 4 ) , (30) where we used Eq. (54) to evaluate the double sum. Thus, the fractional (cosmic) variance in the squared strain is
σ s 2 ⟨ s ⟩ 2 = 1 5 ( 1 − H 4 H 2 2 ) . (31) \frac{\sigma_s^2}{\langle s\rangle^2}=\frac{1}{5}\left(1-\frac{\mathcal{H}_4}{\mathcal{H}_2^2}\right). \tag*{(31)} ⟨ s ⟩ 2 σ s 2 = 5 1 ( 1 − H 2 2 H 4 ) . (31) The fractional fluctuation σ s / ⟨ s ⟩ \sigma_s/\langle s\rangle σ s / ⟨ s ⟩ ≈ 21 % \approx21\% ≈ 21% N = 2 N=2 N = 2 ≈ 32 % \approx32\% ≈ 32% N = 4 N=4 N = 4 ≈ 40 % \approx40\% ≈ 40% N = 14 N=14 N = 14 ≈ 45 % \approx45\% ≈ 45% § 8 4 and given by Eq. (161) . The confusion-noise model of this section corresponds to H 4 / H 2 2 = 1 / 2 \mathcal{H}_4/\mathcal{H}_2^2=1/2 H 4 / H 2 2 = 1/2 N N N
We now compute the variance in the Hellings-Downs correlation for this confusion noise model. As a first step, we need to compute the correlation Z 1 Z 2 ‾ \overline{Z_1Z_2} Z 1 Z 2 (14) and its square for any one of our model universes. Inserting Eq. (17) into Eq. (13) , the complex strain differences are
Δ h j ( t , L p ) = A j e i ( ω t + ϕ j ) [ 1 − χ e − i Δ j ( L p ) ] , (32) \Delta h_j(t,L\mathbf{p})=A_je^{i(\omega t+\phi_j)}\left[1-\chi e^{-i\Delta_j(L\mathbf{p})}\right], \tag*{(32)} Δ h j ( t , L p ) = A j e i ( ω t + ϕ j ) [ 1 − χ e − i Δ j ( L p ) ] , (32) where the Earth-pulsar phase offset for a given GW source and pulsar is
Δ j ( L p ) = ω L ( 1 + p ⋅ Ω j ) . (33) \Delta_j(L\mathbf{p})=\omega L(1+\mathbf{p}\cdot\mathbf{\Omega}_j). \tag*{(33)} Δ j ( L p ) = ω L ( 1 + p ⋅ Ω j ) . (33) (As discussed earlier, in Eq. (32) the physically correct value is χ = 1 \chi=1 χ = 1 χ = 0 \chi=0 χ = 0 n n n (12) and (32) as
Z n ( t ) = ℜ ∑ j A j e i ( ω t + ϕ j ) [ 1 − χ e − i Δ j ( L n p n ) ] F n ( Ω j ) = 1 2 ∑ j A j [ e i ( ω t + ϕ j ) R n ( Ω j ) + e − i ( ω t + ϕ j ) R n ∗ ( Ω j ) ] , (34) \begin{aligned}
Z_n(t)=\Re\sum_j A_je^{i(\omega t+\phi_j)}\left[1-\chi e^{-i\Delta_j(L_n\mathbf{p}_n)}\right]F_n(\mathbf{\Omega}_j) \\
&=\frac{1}{2}\sum_j A_j\left[e^{i(\omega t+\phi_j)}R_n(\mathbf{\Omega}_j)+e^{-i(\omega t+\phi_j)}R_n^*(\mathbf{\Omega}_j)\right], \tag*{(34)}
\end{aligned} Z n ( t ) = ℜ j ∑ A j e i ( ω t + ϕ j ) [ 1 − χ e − i Δ j ( L n p n ) ] F n ( Ω j ) = 2 1 j ∑ A j [ e i ( ω t + ϕ j ) R n ( Ω j ) + e − i ( ω t + ϕ j ) R n ∗ ( Ω j ) ] , (34) where the modified antenna pattern function is defined from Eq. (34) by
R n ( Ω j ) = [ 1 − χ e − i Δ j ( L n p n ) ] F n ( Ω j ) . (35) R_n(\mathbf{\Omega}_j)=\left[1-\chi e^{-i\Delta_j(L_n\mathbf{p}_n)}\right]F_n(\mathbf{\Omega}_j). \tag*{(35)} R n ( Ω j ) = [ 1 − χ e − i Δ j ( L n p n ) ] F n ( Ω j ) . (35) Thus, setting n = 1 n=1 n = 1 n = 2 n=2 n = 2 (34) , the redshifts of pulsars 1 and 2 are given by
Z 1 ( t ) = ∑ j [ c j e i ( ω t + ϕ j ) + c j ∗ e − i ( ω t + ϕ j ) ] , and Z_1(t)=\sum_j\left[c_je^{i(\omega t+\phi_j)}+c_j^*e^{-i(\omega t+\phi_j)}\right],
\qquad\text{and} Z 1 ( t ) = j ∑ [ c j e i ( ω t + ϕ j ) + c j ∗ e − i ( ω t + ϕ j ) ] , and Z 2 ( t ) = ∑ k [ d k e i ( ω t + ϕ k ) + d k ∗ e − i ( ω t + ϕ k ) ] , (36) Z_2(t)=\sum_k\left[d_ke^{i(\omega t+\phi_k)}+d_k^*e^{-i(\omega t+\phi_k)}\right], \tag*{(36)} Z 2 ( t ) = k ∑ [ d k e i ( ω t + ϕ k ) + d k ∗ e − i ( ω t + ϕ k ) ] , (36) where the coefficients that appear in the sums are
c j = 1 2 A j [ 1 − χ e − i ω L 1 ( 1 + p 1 ⋅ Ω j ) ] ( F 1 + ( Ω j ) − i F 1 × ( Ω j ) ) , c_j=\frac{1}{2}A_j\left[1-\chi e^{-i\omega L_1(1+\mathbf{p}_1\cdot\mathbf{\Omega}_j)}\right]\left(F_1^+(\mathbf{\Omega}_j)-iF_1^\times(\mathbf{\Omega}_j)\right), c j = 2 1 A j [ 1 − χ e − iω L 1 ( 1 + p 1 ⋅ Ω j ) ] ( F 1 + ( Ω j ) − i F 1 × ( Ω j ) ) , d k = 1 2 A k [ 1 − χ e − i ω L 2 ( 1 + p 2 ⋅ Ω k ) ] ( F 2 + ( Ω k ) − i F 2 × ( Ω k ) ) . (37) d_k=\frac{1}{2}A_k\left[1-\chi e^{-i\omega L_2(1+\mathbf{p}_2\cdot\mathbf{\Omega}_k)}\right]\left(F_2^+(\mathbf{\Omega}_k)-iF_2^\times(\mathbf{\Omega}_k)\right). \tag*{(37)} d k = 2 1 A k [ 1 − χ e − iω L 2 ( 1 + p 2 ⋅ Ω k ) ] ( F 2 + ( Ω k ) − i F 2 × ( Ω k ) ) . (37) Note that the c j c_j c j d k d_k d k (36) , the time-averaged product is
ρ = Z 1 ( t ) Z 2 ( t ) ‾ = ∑ j , k [ c j d k ∗ e i ( ϕ j − ϕ k ) + c j ∗ d k e − i ( ϕ j − ϕ k ) ] = ∑ j ( c j d j ∗ + c j ∗ d j ) + ∑ j ≠ k [ c j d k ∗ e i ( ϕ j − ϕ k ) + c j ∗ d k e − i ( ϕ j − ϕ k ) ] , (38) \begin{aligned}
\rho=\overline{Z_1(t)Z_2(t)}=\sum_{j,k}\left[c_jd_k^*e^{i(\phi_j-\phi_k)}+c_j^*d_ke^{-i(\phi_j-\phi_k)}\right] \\
&=\sum_j\left(c_jd_j^*+c_j^*d_j\right)+\sum_{j\ne k}\left[c_jd_k^*e^{i(\phi_j-\phi_k)}+c_j^*d_ke^{-i(\phi_j-\phi_k)}\right], \tag*{(38)}
\end{aligned} ρ = Z 1 ( t ) Z 2 ( t ) = j , k ∑ [ c j d k ∗ e i ( ϕ j − ϕ k ) + c j ∗ d k e − i ( ϕ j − ϕ k ) ] = j ∑ ( c j d j ∗ + c j ∗ d j ) + j = k ∑ [ c j d k ∗ e i ( ϕ j − ϕ k ) + c j ∗ d k e − i ( ϕ j − ϕ k ) ] , (38) where two terms have been eliminated on the first line of Eq. (38) by assuming that the angular frequency ω \omega ω 2 π 2\pi 2 π (38) we have broken up the double sum over sources into “diagonal” terms for which j = k j=k j = k j ≠ k j\ne k j = k (38) gives the pulsar-pair correlation for any representative member of our ensemble of model universes, which are defined by specific values of the phases ϕ j \phi_j ϕ j Ω j \mathbf{\Omega}_j Ω j
To obtain the mean (first moment) of the Hellings-Downs correlation, we calculate the ensemble average of ρ \rho ρ (38) . Since the Kronecker delta in Eq. (21) vanishes for j ≠ k j\ne k j = k (38) over the random phases eliminates the off-diagonal terms, leaving
⟨ ρ ⟩ ϕ = ∑ j ( c j d j ∗ + c j ∗ d j ) . (39) \langle\rho\rangle_\phi=\sum_j\left(c_jd_j^*+c_j^*d_j\right). \tag*{(39)} ⟨ ρ ⟩ ϕ = j ∑ ( c j d j ∗ + c j ∗ d j ) . (39) To complete the ensemble average, we average this over random source directions Ω j \mathbf{\Omega}_j Ω j
⟨ ρ ⟩ = ⟨ ρ ⟩ ϕ , Ω = ∑ j ( ⟨ c j d j ∗ ⟩ Ω + ⟨ c j ∗ d j ⟩ Ω ) . (40) \langle\rho\rangle=\langle\rho\rangle_{\phi,\Omega}=\sum_j\left(\langle c_jd_j^*\rangle_\Omega+\langle c_j^*d_j\rangle_\Omega\right). \tag*{(40)} ⟨ ρ ⟩ = ⟨ ρ ⟩ ϕ , Ω = j ∑ ( ⟨ c j d j ∗ ⟩ Ω + ⟨ c j ∗ d j ⟩ Ω ) . (40) Here, the angle bracket with subscript Ω \Omega Ω Ω j \mathbf{\Omega}_j Ω j Ω j \mathbf{\Omega}_j Ω j 4 π 4\pi 4 π
To evaluate the source-direction average of the product c j d j ∗ c_jd_j^* c j d j ∗ (40) , first examine the definitions of c c c d d d (37) . If the interpulsar separation L 1 p 1 − L 2 p 2 L_1\mathbf{p}_1-L_2\mathbf{p}_2 L 1 p 1 − L 2 p 2 2 π / ω 2\pi/\omega 2 π / ω (37) can be replaced by unity
[ 1 − χ e − i ω L 1 ( 1 + p 1 ⋅ Ω j ) ] [ 1 − χ e i ω L 2 ( 1 + p 2 ⋅ Ω j ) ] ≈ 1. (41) \left[1-\chi e^{-i\omega L_1(1+\mathbf{p}_1\cdot\mathbf{\Omega}_j)}\right]\left[1-\chi e^{i\omega L_2(1+\mathbf{p}_2\cdot\mathbf{\Omega}_j)}\right]\approx1. \tag*{(41)} [ 1 − χ e − iω L 1 ( 1 + p 1 ⋅ Ω j ) ] [ 1 − χ e iω L 2 ( 1 + p 2 ⋅ Ω j ) ] ≈ 1. (41) This is because the product of the terms in square brackets yields four terms. Three are rapidly oscillating (complex exponential) functions of the source direction Ω j \mathbf{\Omega}_j Ω j [35 ] . Only the unity term remains, so the average over source directions gives
⟨ c j d j ∗ ⟩ Ω = 1 4 A j 2 ⟨ ( F 1 + ( Ω ) − i F 1 × ( Ω ) ) ( F 2 + ( Ω ) + i F 2 × ( Ω ) ) ⟩ Ω = 1 4 A j 2 [ ⟨ F 1 + ( Ω ) F 2 + ( Ω ) + F 1 × ( Ω ) F 2 × ( Ω ) ⟩ Ω + i ⟨ F 1 + ( Ω ) F 2 × ( Ω ) − F 1 × ( Ω ) F 2 + ( Ω ) ⟩ Ω ] = 1 4 A j 2 μ u ( γ ) . (42) \begin{aligned}
\left\langle c_jd_j^*\right\rangle_{\Omega}
&=\frac{1}{4}\mathcal{A}_j^2\left\langle\left(F_1^+(\mathbf{\Omega})-iF_1^\times(\mathbf{\Omega})\right)\left(F_2^+(\mathbf{\Omega})+iF_2^\times(\mathbf{\Omega})\right)\right\rangle_{\Omega}\\
&=\frac{1}{4}\mathcal{A}_j^2\left[\left\langle F_1^+(\mathbf{\Omega})F_2^+(\mathbf{\Omega})+F_1^\times(\mathbf{\Omega})F_2^\times(\mathbf{\Omega})\right\rangle_{\Omega}\right.\\
&\qquad\left.+i\left\langle F_1^+(\mathbf{\Omega})F_2^\times(\mathbf{\Omega})-F_1^\times(\mathbf{\Omega})F_2^+(\mathbf{\Omega})\right\rangle_{\Omega}\right]\\
&=\frac{1}{4}\mathcal{A}_j^2\mu_u(\gamma). \tag*{(42)}
\end{aligned} ⟨ c j d j ∗ ⟩ Ω = 4 1 A j 2 ⟨ ( F 1 + ( Ω ) − i F 1 × ( Ω ) ) ( F 2 + ( Ω ) + i F 2 × ( Ω ) ) ⟩ Ω = 4 1 A j 2 [ ⟨ F 1 + ( Ω ) F 2 + ( Ω ) + F 1 × ( Ω ) F 2 × ( Ω ) ⟩ Ω + i ⟨ F 1 + ( Ω ) F 2 × ( Ω ) − F 1 × ( Ω ) F 2 + ( Ω ) ⟩ Ω ] = 4 1 A j 2 μ u ( γ ) . (42) where γ \gamma γ (42) has two source-direction averages, which are evaluated in Appendixes D and E respectively. The first average is the mean value of the standard Hellings-Downs “unpolarized” correlation. This is given by Eq. (177) , and evaluates to μ u \mu_u μ u (203) . The second average is the mean value of the “polarized” integrand Eq. (212) . That mean value evaluates to zero, as explained after Eq. (214) , which means that the imaginary part of ⟨ c j d j ∗ ⟩ Ω \left\langle c_jd_j^*\right\rangle_{\Omega} ⟨ c j d j ∗ ⟩ Ω
The mean or first moment of the Hellings-Downs correlation is found by substituting Eq. (42) and its complex conjugate into Eq. (40) and summing over the sources:
μ ( γ ) = ⟨ ρ ⟩ = 1 2 H 2 μ u ( γ ) . (43) \mu(\gamma)=\langle\rho\rangle=\frac{1}{2}\mathcal{H}_2\mu_u(\gamma). \tag*{(43)} μ ( γ ) = ⟨ ρ ⟩ = 2 1 H 2 μ u ( γ ) . (43) The result is expressed in terms of the squared strain amplitude at Earth, defined by Eq. (23) .
Next, we compute the second moment of ρ \rho ρ ρ \rho ρ (38) and average it over the random phases. The diagonal term does not depend upon the random phases, so its square appears “as is.” The product of the diagonal and off-diagonal terms contains ⟨ e i ( ϕ j − ϕ k ) ⟩ ϕ \left\langle e^{i(\phi_j-\phi_k)}\right\rangle_\phi ⟨ e i ( ϕ j − ϕ k ) ⟩ ϕ j ≠ k j\ne k j = k (21) . Thus, only the squares of the diagonal and off-diagonal terms remain, giving
⟨ ρ 2 ⟩ ϕ = ( ∑ j ( c j d j ∗ + c j ∗ d j ) ) 2 + ∑ j ≠ k ∑ ℓ ≠ m [ c j d k ∗ c ℓ d m ∗ ⟨ e i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ + c j d k ∗ c ℓ ∗ d m ⟨ e i ( ϕ j − ϕ k − ϕ ℓ + ϕ m ) ⟩ ϕ + c j ∗ d k c ℓ ∗ d m ⟨ e − i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ + c j ∗ d k c ℓ d m ∗ ⟨ e − i ( ϕ j − ϕ k − ϕ ℓ + ϕ m ) ⟩ ϕ ] . (44) \begin{aligned}
\left\langle\rho^2\right\rangle_\phi
={}&\left(\sum_j\left(c_jd_j^*+c_j^*d_j\right)\right)^2\\
&+\sum_{j\ne k}\sum_{\ell\ne m}\left[c_jd_k^*c_\ell d_m^*\left\langle e^{i(\phi_j-\phi_k+\phi_\ell-\phi_m)}\right\rangle_\phi+c_jd_k^*c_\ell^*d_m\left\langle e^{i(\phi_j-\phi_k-\phi_\ell+\phi_m)}\right\rangle_\phi\right.\\
&\qquad\left.+c_j^*d_kc_\ell^*d_m\left\langle e^{-i(\phi_j-\phi_k+\phi_\ell-\phi_m)}\right\rangle_\phi+c_j^*d_kc_\ell d_m^*\left\langle e^{-i(\phi_j-\phi_k-\phi_\ell+\phi_m)}\right\rangle_\phi\right]. \tag*{(44)}
\end{aligned} ⟨ ρ 2 ⟩ ϕ = ( j ∑ ( c j d j ∗ + c j ∗ d j ) ) 2 + j = k ∑ ℓ = m ∑ [ c j d k ∗ c ℓ d m ∗ ⟨ e i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ + c j d k ∗ c ℓ ∗ d m ⟨ e i ( ϕ j − ϕ k − ϕ ℓ + ϕ m ) ⟩ ϕ + c j ∗ d k c ℓ ∗ d m ⟨ e − i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ + c j ∗ d k c ℓ d m ∗ ⟨ e − i ( ϕ j − ϕ k − ϕ ℓ + ϕ m ) ⟩ ϕ ] . (44) The four ensemble averages which appear in Eq. (44) may be easily evaluated using Eq. (20) . Since j ≠ k j\ne k j = k ℓ ≠ m \ell\ne m ℓ = m ⟨ e i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ = δ j m δ k ℓ \left\langle e^{i(\phi_j-\phi_k+\phi_\ell-\phi_m)}\right\rangle_\phi=\delta_{jm}\delta_{k\ell} ⟨ e i ( ϕ j − ϕ k + ϕ ℓ − ϕ m ) ⟩ ϕ = δ j m δ k ℓ δ j ℓ δ k m \delta_{j\ell}\delta_{km} δ j ℓ δ k m δ j m δ k ℓ \delta_{jm}\delta_{k\ell} δ j m δ k ℓ δ j ℓ δ k m \delta_{j\ell}\delta_{km} δ j ℓ δ k m (44) simplifies to the double sum
⟨ ρ 2 ⟩ ϕ = ( ∑ j ( c j d j ∗ + c j ∗ d j ) ) 2 + ∑ j ≠ k [ c j d k ∗ c k d j ∗ + c j d k ∗ c k ∗ d j + c j ∗ d k c k ∗ d j + c j ∗ d k c k d j ∗ ] . (45) \left\langle\rho^2\right\rangle_\phi=\left(\sum_j\left(c_jd_j^*+c_j^*d_j\right)\right)^2+\sum_{j\ne k}\left[c_jd_k^*c_kd_j^*+c_jd_k^*c_k^*d_j+c_j^*d_kc_k^*d_j+c_j^*d_kc_kd_j^*\right]. \tag*{(45)} ⟨ ρ 2 ⟩ ϕ = ( j ∑ ( c j d j ∗ + c j ∗ d j ) ) 2 + j = k ∑ [ c j d k ∗ c k d j ∗ + c j d k ∗ c k ∗ d j + c j ∗ d k c k ∗ d j + c j ∗ d k c k d j ∗ ] . (45) To evaluate the ensemble average over source directions, it is helpful to rewrite the first term (a perfect square) as a double sum, also breaking it into “diagonal” and “off-diagonal” terms. Noting also that the second and fourth term of the double sum in Eq. (45) are each equal to ∣ c j ∣ 2 ∣ d k ∣ 2 \lvert c_j\rvert^2\lvert d_k\rvert^2 ∣ c j ∣ 2 ∣ d k ∣ 2
⟨ ρ 2 ⟩ ϕ = ∑ j ( c j d j ∗ + c j ∗ d j ) 2 + ∑ j ≠ k [ ( c j d j ∗ + c j ∗ d j ) ( c k d k ∗ + c k ∗ d k ) + c j d k ∗ c k d j ∗ + c j ∗ d k c k ∗ d j + 2 ∣ c j ∣ 2 ∣ d k ∣ 2 ] = ∑ j ( c j d j ∗ + c j ∗ d j ) 2 + 2 ∑ j ≠ k [ c j d j ∗ c k d k ∗ + c j ∗ d j c k d k ∗ + c j ∗ d j c k ∗ d k + ∣ c j ∣ 2 ∣ d k ∣ 2 ] . (46) \begin{aligned}
\left\langle\rho^2\right\rangle_\phi
={}&\sum_j\left(c_jd_j^*+c_j^*d_j\right)^2+\sum_{j\ne k}\left[\left(c_jd_j^*+c_j^*d_j\right)\left(c_kd_k^*+c_k^*d_k\right)+c_jd_k^*c_kd_j^*+c_j^*d_kc_k^*d_j+2\lvert c_j\rvert^2\lvert d_k\rvert^2\right]\\
={}&\sum_j\left(c_jd_j^*+c_j^*d_j\right)^2+2\sum_{j\ne k}\left[c_jd_j^*c_kd_k^*+c_j^*d_jc_kd_k^*+c_j^*d_jc_k^*d_k+\lvert c_j\rvert^2\lvert d_k\rvert^2\right]. \tag*{(46)}
\end{aligned} ⟨ ρ 2 ⟩ ϕ = = j ∑ ( c j d j ∗ + c j ∗ d j ) 2 + j = k ∑ [ ( c j d j ∗ + c j ∗ d j ) ( c k d k ∗ + c k ∗ d k ) + c j d k ∗ c k d j ∗ + c j ∗ d k c k ∗ d j + 2 ∣ c j ∣ 2 ∣ d k ∣ 2 ] j ∑ ( c j d j ∗ + c j ∗ d j ) 2 + 2 j = k ∑ [ c j d j ∗ c k d k ∗ + c j ∗ d j c k d k ∗ + c j ∗ d j c k ∗ d k + ∣ c j ∣ 2 ∣ d k ∣ 2 ] . (46) where we have used symmetry in the sum over j j j k k k j ≠ k j\ne k j = k
The source-direction averages that appear in the double-sum terms of Eq. (46) are evaluated by first factoring them (since j ≠ k j \ne k j = k (37) and proceeding exactly as for Eq. (42) . This gives
⟨ ∣ c j ∣ 2 ∣ d k ∣ 2 ⟩ Ω = ⟨ ∣ c j ∣ 2 ⟩ Ω ⟨ ∣ d k ∣ 2 ⟩ Ω = 1 16 A j 2 A k 2 ( 1 + χ 2 ) 2 μ u 2 ( 0 ) , \left\langle|c_j|^2 |d_k|^2 \right\rangle_{\Omega} = \left\langle|c_j|^2 \right\rangle_{\Omega}\left\langle|d_k|^2 \right\rangle_{\Omega} = \frac{1}{16} A_j^2 A_k^2 (1+\chi^2)^2 \mu_u^2(0), ⟨ ∣ c j ∣ 2 ∣ d k ∣ 2 ⟩ Ω = ⟨ ∣ c j ∣ 2 ⟩ Ω ⟨ ∣ d k ∣ 2 ⟩ Ω = 16 1 A j 2 A k 2 ( 1 + χ 2 ) 2 μ u 2 ( 0 ) , ⟨ c j d j ∗ c k d k ∗ ⟩ Ω = ⟨ c j d j ∗ ⟩ Ω ⟨ c k d k ∗ ⟩ Ω = 1 16 A j 2 A k 2 μ u 2 ( γ ) , \left\langle c_j d_j^* c_k d_k^* \right\rangle_{\Omega} = \left\langle c_j d_j^* \right\rangle_{\Omega}\left\langle c_k d_k^* \right\rangle_{\Omega} = \frac{1}{16} A_j^2 A_k^2 \mu_u^2(\gamma), ⟨ c j d j ∗ c k d k ∗ ⟩ Ω = ⟨ c j d j ∗ ⟩ Ω ⟨ c k d k ∗ ⟩ Ω = 16 1 A j 2 A k 2 μ u 2 ( γ ) , ⟨ c j ∗ d j c k d k ∗ ⟩ Ω = ⟨ c j ∗ d j ⟩ Ω ⟨ c k d k ∗ ⟩ Ω = 1 16 A j 2 A k 2 μ u 2 ( γ ) , \left\langle c_j^* d_j c_k d_k^* \right\rangle_{\Omega} = \left\langle c_j^* d_j \right\rangle_{\Omega}\left\langle c_k d_k^* \right\rangle_{\Omega} = \frac{1}{16} A_j^2 A_k^2 \mu_u^2(\gamma), ⟨ c j ∗ d j c k d k ∗ ⟩ Ω = ⟨ c j ∗ d j ⟩ Ω ⟨ c k d k ∗ ⟩ Ω = 16 1 A j 2 A k 2 μ u 2 ( γ ) , ⟨ c j ∗ d j c k ∗ d k ⟩ Ω = ⟨ c j ∗ d j ⟩ Ω ⟨ c k ∗ d k ⟩ Ω = 1 16 A j 2 A k 2 μ u 2 ( γ ) . (47) \left\langle c_j^* d_j c_k^* d_k \right\rangle_{\Omega} = \left\langle c_j^* d_j \right\rangle_{\Omega}\left\langle c_k^* d_k \right\rangle_{\Omega} = \frac{1}{16} A_j^2 A_k^2 \mu_u^2(\gamma). \tag*{(47)} ⟨ c j ∗ d j c k ∗ d k ⟩ Ω = ⟨ c j ∗ d j ⟩ Ω ⟨ c k ∗ d k ⟩ Ω = 16 1 A j 2 A k 2 μ u 2 ( γ ) . (47) In the first line above, the factors of 1 + χ 2 1+\chi^2 1 + χ 2 ∣ c j ∣ 2 |c_j|^2 ∣ c j ∣ 2 (37) give
∣ 1 − χ e − i ω L 1 ( 1 + p 1 ⋅ Ω j ) ∣ 2 = 1 + χ 2 − 2 χ cos ( ω L 1 ( 1 + p 1 ⋅ Ω j ) ) ≈ 1 + χ 2 , (48) \left|1-\chi e^{-i\omega L_1(1+\mathbf{p}_1\cdot\mathbf{\Omega}_j)}\right|^2
=1+\chi^2-2\chi\cos\left(\omega L_1(1+\mathbf{p}_1\cdot\mathbf{\Omega}_j)\right)
\approx1+\chi^2, \tag*{(48)} 1 − χ e − iω L 1 ( 1 + p 1 ⋅ Ω j ) 2 = 1 + χ 2 − 2 χ cos ( ω L 1 ( 1 + p 1 ⋅ Ω j ) ) ≈ 1 + χ 2 , (48) and a similar factor arises within ∣ d k ∣ 2 |d_k|^2 ∣ d k ∣ 2 (41) and (42) , but here two of the four terms contribute, rather than just one.
The source-direction average of the single-sum term in Eq. (46) may be evaluated by first expanding it as
⟨ ( c j d j ∗ + c j ∗ d j ) 2 ⟩ Ω = ⟨ ( c j d j ∗ ) 2 + ( c j ∗ d j ) 2 + 2 ∣ c j ∣ 2 ∣ d j ∣ 2 ⟩ Ω . (49) \left\langle\left(c_j d_j^*+c_j^*d_j\right)^2\right\rangle_{\Omega}
=\left\langle(c_jd_j^*)^2+(c_j^*d_j)^2+2|c_j|^2|d_j|^2\right\rangle_{\Omega}. \tag*{(49)} ⟨ ( c j d j ∗ + c j ∗ d j ) 2 ⟩ Ω = ⟨ ( c j d j ∗ ) 2 + ( c j ∗ d j ) 2 + 2∣ c j ∣ 2 ∣ d j ∣ 2 ⟩ Ω . (49) From the definitions of Eq. (37) , the first two terms on the rhs of Eq. (49) give
⟨ ( c j d j ∗ ) 2 + ( c j ∗ d j ) 2 ⟩ Ω = 1 16 A j 4 ⟨ 2 ( F 1 + ( Ω ) F 2 + ( Ω ) + F 1 × ( Ω ) F 2 × ( Ω ) ) 2 − 2 ( F 1 + ( Ω ) F 2 × ( Ω ) − F 1 × ( Ω ) F 2 + ( Ω ) ) 2 ⟩ Ω = 1 16 A j 4 ( 2 ( μ u 2 ( γ ) + σ u 2 ( γ ) ) − 2 σ p 2 ( γ ) ) = 1 8 A j 4 ( 2 μ u 2 ( γ ) + 2 σ u 2 ( γ ) − σ c 2 ( γ ) ) . (50) \begin{aligned}
\left\langle(c_jd_j^*)^2+(c_j^*d_j)^2\right\rangle_{\Omega}
&=\frac{1}{16}A_j^4\left\langle2\left(F_1^+(\mathbf{\Omega})F_2^+(\mathbf{\Omega})+F_1^\times(\mathbf{\Omega})F_2^\times(\mathbf{\Omega})\right)^2 \right.\\
&\qquad\left.-2\left(F_1^+(\mathbf{\Omega})F_2^\times(\mathbf{\Omega})-F_1^\times(\mathbf{\Omega})F_2^+(\mathbf{\Omega})\right)^2\right\rangle_{\Omega}\\
&=\frac{1}{16}A_j^4\left(2\left(\mu_u^2(\gamma)+\sigma_u^2(\gamma)\right)-2\sigma_p^2(\gamma)\right)\\
&=\frac{1}{8}A_j^4\left(2\mu_u^2(\gamma)+2\sigma_u^2(\gamma)-\sigma_c^2(\gamma)\right). \tag*{(50)}
\end{aligned} ⟨ ( c j d j ∗ ) 2 + ( c j ∗ d j ) 2 ⟩ Ω = 16 1 A j 4 ⟨ 2 ( F 1 + ( Ω ) F 2 + ( Ω ) + F 1 × ( Ω ) F 2 × ( Ω ) ) 2 − 2 ( F 1 + ( Ω ) F 2 × ( Ω ) − F 1 × ( Ω ) F 2 + ( Ω ) ) 2 ⟩ Ω = 16 1 A j 4 ( 2 ( μ u 2 ( γ ) + σ u 2 ( γ ) ) − 2 σ p 2 ( γ ) ) = 8 1 A j 4 ( 2 μ u 2 ( γ ) + 2 σ u 2 ( γ ) − σ c 2 ( γ ) ) . (50) The first term on the rhs of the first line of Eq. (50) is the second moment of the standard unpolarized Hellings-Downs correlation Eq. (179) . This is given in Eq. (210) of Appendix § 13 , and we write it as ⟨ ρ 2 ⟩ = μ u 2 + σ u 2 \langle\rho^2\rangle=\mu_u^2+\sigma_u^2 ⟨ ρ 2 ⟩ = μ u 2 + σ u 2 (50) is the second moment of the “polarized” Hellings-Downs correlation Eq. (212) , evaluated in Appendix § 14 . This is given by the variance σ p 2 \sigma_p^2 σ p 2 (219) , since the first moment vanishes. For the final line of Eq. (50) we have used the identity σ c 2 = σ u 2 + σ p 2 + μ u 2 \sigma_c^2=\sigma_u^2+\sigma_p^2+\mu_u^2 σ c 2 = σ u 2 + σ p 2 + μ u 2 (223) to eliminate σ p 2 \sigma_p^2 σ p 2
The final term of Eq. (49) is evaluated using the definitions of Eqs. (37) and (48) to obtain
⟨ 2 ∣ c j ∣ 2 ∣ d j ∣ 2 ⟩ Ω = 2 16 A j 4 ( 1 + χ 2 ) 2 ⟨ ( F 1 + ( Ω ) F 1 + ( Ω ) + F 1 × ( Ω ) F 1 × ( Ω ) ) ( F 2 + ( Ω ) F 2 + ( Ω ) + F 2 × ( Ω ) F 2 × ( Ω ) ) ⟩ Ω = 1 8 A j 4 ( 1 + χ 2 ) 2 σ c 2 ( γ ) . (51) \begin{aligned}
\left\langle2|c_j|^2|d_j|^2\right\rangle_{\Omega}
&=\frac{2}{16}A_j^4(1+\chi^2)^2\left\langle\left(F_1^+(\mathbf{\Omega})F_1^+(\mathbf{\Omega})+F_1^\times(\mathbf{\Omega})F_1^\times(\mathbf{\Omega})\right)\right.\\
&\qquad\left.\left(F_2^+(\mathbf{\Omega})F_2^+(\mathbf{\Omega})+F_2^\times(\mathbf{\Omega})F_2^\times(\mathbf{\Omega})\right)\right\rangle_{\Omega}\\
&=\frac{1}{8}A_j^4(1+\chi^2)^2\sigma_c^2(\gamma). \tag*{(51)}
\end{aligned} ⟨ 2∣ c j ∣ 2 ∣ d j ∣ 2 ⟩ Ω = 16 2 A j 4 ( 1 + χ 2 ) 2 ⟨ ( F 1 + ( Ω ) F 1 + ( Ω ) + F 1 × ( Ω ) F 1 × ( Ω ) ) ( F 2 + ( Ω ) F 2 + ( Ω ) + F 2 × ( Ω ) F 2 × ( Ω ) ) ⟩ Ω = 8 1 A j 4 ( 1 + χ 2 ) 2 σ c 2 ( γ ) . (51) This is evaluated by noting that the product of antenna pattern functions on the first line is the same as Eq. (220) , whose source-direction average is evaluated in Appendix § 15 , giving Eq. (221) .
Combining the results of Eqs. (49) –(51) , we obtain the source-direction average for the single-sum term of Eq. (46) as
⟨ ( c j d j ∗ + c j ∗ d j ) 2 ⟩ Ω = 1 8 A j 4 ( μ u 2 ( γ ) + σ u 2 ( γ ) − σ p 2 ( γ ) + ( 1 + χ 2 ) 2 σ c 2 ( γ ) ) = 1 8 A j 4 ( 2 μ u 2 ( γ ) + 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) . (52) \begin{aligned}
\left\langle\left(c_jd_j^*+c_j^*d_j\right)^2\right\rangle_{\Omega}
&=\frac{1}{8}A_j^4\left(\mu_u^2(\gamma)+\sigma_u^2(\gamma)-\sigma_p^2(\gamma)+(1+\chi^2)^2\sigma_c^2(\gamma)\right)\\
&=\frac{1}{8}A_j^4\left(2\mu_u^2(\gamma)+2\sigma_u^2(\gamma)+\left((1+\chi^2)^2-1\right)\sigma_c^2(\gamma)\right). \tag*{(52)}
\end{aligned} ⟨ ( c j d j ∗ + c j ∗ d j ) 2 ⟩ Ω = 8 1 A j 4 ( μ u 2 ( γ ) + σ u 2 ( γ ) − σ p 2 ( γ ) + ( 1 + χ 2 ) 2 σ c 2 ( γ ) ) = 8 1 A j 4 ( 2 μ u 2 ( γ ) + 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) . (52) This completes the evaluation of the source-direction averages for the different terms appearing in the Eq. (46) expression of ⟨ ρ 2 ⟩ ϕ \langle\rho^2\rangle_\phi ⟨ ρ 2 ⟩ ϕ
We now compute the total variance for this simple confusion-noise model. The ensemble-averaged second moment ⟨ ρ 2 ⟩ = ⟨ ρ 2 ⟩ ϕ , Ω \langle\rho^2\rangle=\langle\rho^2\rangle_{\phi,\Omega} ⟨ ρ 2 ⟩ = ⟨ ρ 2 ⟩ ϕ , Ω ⟨ ρ 2 ⟩ ϕ \langle\rho^2\rangle_\phi ⟨ ρ 2 ⟩ ϕ (46) , using Eqs. (47) and (52) to average over source directions, and summing over the sources. This gives
⟨ ρ 2 ⟩ = 1 8 H 4 ( 2 μ u 2 ( γ ) + 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) + 1 8 ( H 2 2 − H 4 ) ( 3 μ u 2 ( γ ) + ( 1 + χ 2 ) 2 μ u 2 ( 0 ) ) (53) \langle\rho^2 \rangle= \frac{1}{8}\mathcal{H}_4\left(2\mu_u^2(\gamma)+2\sigma_u^2(\gamma)+\left((1+\chi^2)^2-1\right)\sigma_c^2(\gamma)\right)+\frac{1}{8}\left(\mathcal{H}_2^2-\mathcal{H}_4\right)\left(3\mu_u^2(\gamma)+(1+\chi^2)^2\mu_u^2(0)\right) \tag*{(53)} ⟨ ρ 2 ⟩ = 8 1 H 4 ( 2 μ u 2 ( γ ) + 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) + 8 1 ( H 2 2 − H 4 ) ( 3 μ u 2 ( γ ) + ( 1 + χ 2 ) 2 μ u 2 ( 0 ) ) (53) where we have used Eqs. (23) and (25) to carry out the sums over sources, noting that
∑ j ≠ k A j 2 A k 2 = ∑ j , k A j 2 A k 2 − ∑ j A j 4 = ( ∑ j A j 2 ) 2 − ∑ j A j 4 = H 2 2 − H 4 . (54) \sum_{j\ne k} A_j^2 A_k^2=\sum_{j,k}A_j^2A_k^2-\sum_j A_j^4=\left(\sum_j A_j^2\right)^2-\sum_j A_j^4=\mathcal{H}_2^2-\mathcal{H}_4. \tag*{(54)} j = k ∑ A j 2 A k 2 = j , k ∑ A j 2 A k 2 − j ∑ A j 4 = ( j ∑ A j 2 ) 2 − j ∑ A j 4 = H 2 2 − H 4 . (54) To obtain the variance σ 2 = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 \sigma^2=\langle\rho^2\rangle-\langle\rho\rangle^2 σ 2 = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 (43) from Eq. (53) . This eliminates the first H 4 \mathcal{H}_4 H 4 2 / 3 2/3 2/3 H 2 2 − H 4 \mathcal{H}_2^2-\mathcal{H}_4 H 2 2 − H 4 (53) , giving
σ 2 = 1 8 H 4 ( 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) + 1 8 ( H 2 2 − H 4 ) ( μ u 2 ( γ ) + ( 1 + χ 2 ) 2 μ u 2 ( 0 ) ) . (55) \sigma^2=\frac{1}{8}\mathcal{H}_4\left(2\sigma_u^2(\gamma)+\left((1+\chi^2)^2-1\right)\sigma_c^2(\gamma)\right)+\frac{1}{8}\left(\mathcal{H}_2^2-\mathcal{H}_4\right)\left(\mu_u^2(\gamma)+(1+\chi^2)^2\mu_u^2(0)\right). \tag*{(55)} σ 2 = 8 1 H 4 ( 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) + 8 1 ( H 2 2 − H 4 ) ( μ u 2 ( γ ) + ( 1 + χ 2 ) 2 μ u 2 ( 0 ) ) . (55) This is one of the main results of the paper: the expected variance in the Hellings-Downs correlation for a confusion-noise model containing N N N χ = 1 \chi=1 χ = 1 χ = 0 \chi=0 χ = 0 § 3 , we calculate the cosmic variance, which can be subtracted from Eq. (55) to obtain the pulsar variance.
As a consistency check, consider the case where there is only a single source, so that H 4 = H 2 2 = A 4 \mathcal{H}_4=\mathcal{H}_2^2=A^4 H 4 = H 2 2 = A 4 (55) reduces to
σ 2 = A 4 8 ( 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) . (56) \sigma^2=\frac{A^4}{8}\left(2\sigma_u^2(\gamma)+\left((1+\chi^2)^2-1\right)\sigma_c^2(\gamma)\right). \tag*{(56)} σ 2 = 8 A 4 ( 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) . (56) If the pulsar term is neglected (set χ = 0 \chi=0 χ = 0 σ 2 = A 4 σ u 2 ( γ ) / 4 \sigma^2=A^4\sigma_u^2(\gamma)/4 σ 2 = A 4 σ u 2 ( γ ) /4 (108) : set c 1 = c 2 = A 2 / 2 c_1=c_2=A^2/2 c 1 = c 2 = A 2 /2 c 3 = c 4 = 0 c_3=c_4=0 c 3 = c 4 = 0 (55) includes both the pulsar variance and cosmic variance, because in the case N = 1 N=1 N = 1 γ \gamma γ χ = 1 \chi=1 χ = 1 σ 2 = A 4 ( 2 σ u 2 ( γ ) + 3 σ c 2 ( γ ) ) / 8 \sigma^2=A^4(2\sigma_u^2(\gamma)+3\sigma_c^2(\gamma))/8 σ 2 = A 4 ( 2 σ u 2 ( γ ) + 3 σ c 2 ( γ )) /8 (120) after setting A c = A s = A A_c=A_s=A A c = A s = A
Another important limit is that of many sources. For current PTAs, with at least N = O ( 10 3 ) N=O(10^3) N = O ( 1 0 3 ) (23) and (25) that H 2 2 ≫ H 4 \mathcal{H}_2^2\gg\mathcal{H}_4 H 2 2 ≫ H 4 H 2 \mathcal{H}_2 H 2 N s ∝ N 1 / 3 N_s\propto N^{1/3} N s ∝ N 1/3 H 4 → A 4 ζ ( 4 / 3 ) ≈ 3.6 A 4 \mathcal{H}_4\to A^4\zeta(4/3)\approx3.6A^4 H 4 → A 4 ζ ( 4/3 ) ≈ 3.6 A 4 N N N (55) dominates, and (setting χ = 1 \chi=1 χ = 1
σ 2 ≈ 1 8 H 2 2 ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (57) \sigma^2\approx\frac{1}{8}\mathcal{H}_2^2\left(\mu_u^2(\gamma)+4\mu_u^2(0)\right). \tag*{(57)} σ 2 ≈ 8 1 H 2 2 ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (57) This should be compared with the mean taken from Eq. (43) :
μ ( γ ) = 1 2 H 2 μ u ( γ ) (58) \mu(\gamma)=\frac{1}{2}\mathcal{H}_2\mu_u(\gamma) \tag*{(58)} μ ( γ ) = 2 1 H 2 μ u ( γ ) (58) Notice that for large numbers of sources the standard deviation σ \sigma σ μ \mu μ (57) to the mean value from Eq. (58) , which are shown in Figure 2 . When there are many confusion-noise sources,
Figure 2. The Hellings-Downs correlation mean μ \mu μ (58) ] and standard deviation σ \sigma σ (57) ] for an unpolarized confusion-limited stochastic background from many point sources, where all sources have the same frequency. The dashed lines show a ± 1 σ \pm1\sigma ± 1 σ § 3 , we show that this variance includes both pulsar and cosmic variance. In the plot, μ \mu μ σ \sigma σ H 2 / 2 = ⟨ ( h + ) 2 ⟩ = ⟨ ( h × ) 2 ⟩ \mathcal{H}_2/2=\langle(h^+)^2\rangle=\langle(h^\times)^2\rangle H 2 /2 = ⟨( h + ) 2 ⟩ = ⟨( h × ) 2 ⟩
σ μ ≈ 1 2 [ 1 + 4 μ u 2 ( 0 ) μ u 2 ( γ ) ] 1 / 2 (59) \frac{\sigma}{\mu}\approx\sqrt{\frac{1}{2}\left[1+4\frac{\mu_u^2(0)}{\mu_u^2(\gamma)}\right]}^{1/2} \tag*{(59)} μ σ ≈ 2 1 [ 1 + 4 μ u 2 ( γ ) μ u 2 ( 0 ) ] 1/2 (59) for the sources in one frequency bin. (Note: if the source frequency is not commensurate with the observation time, then the factor 1 / 2 \sqrt{1/2} 1/2 (59) can increase to be as large as unity; see comments and footnote before Eq. (15) and the discussion following Eq. (151) .)
The effective number of frequency bins may be quite small (≲ 4 \lesssim4 ≲ 4 T T T 1 / T 1/T 1/ T [36 ] . For further details, see the “transmission functions” illustrated in Figure 2 of Fig. 2 of [36 ] .
If there are M M M i = 1 , … , M i=1,\ldots,M i = 1 , … , M
μ = μ u ( γ ) ∑ i = 1 M h i 2 , \mu=\mu_u(\gamma)\sum_{i=1}^{M}h_i^2, μ = μ u ( γ ) i = 1 ∑ M h i 2 , σ 2 = 1 2 ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) ∑ i = 1 M h i 4 (60) \sigma^2=\frac{1}{2}\left(\mu_u^2(\gamma)+4\mu_u^2(0)\right)\sum_{i=1}^{M}h_i^4 \tag*{(60)} σ 2 = 2 1 ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) i = 1 ∑ M h i 4 (60) where h i 2 = ( H 2 / 2 ) i h_i^2=(\mathcal{H}^2/2)_i h i 2 = ( H 2 /2 ) i i i i § 8 § 11 : compare Eqs. (C30) and (C32).) If the squared strain is distributed uniformly over frequency, then the fractional fluctuations σ / μ \sigma/\mu σ / μ M − 1 / 2 M^{-1/2} M − 1/2 f i − 2 f_i^{-2} f i − 2 h i 2 h_i^2 h i 2
In Appendix § 8 § 9.2 we calculate the variance for a Gaussian ensemble of sources as employed in [17 , 20 , 21 ] . The Gaussian ensemble variance has the term proportional to μ 2 ( γ ) + 4 μ 2 ( 0 ) \mu^2(\gamma)+4\mu^2(0) μ 2 ( γ ) + 4 μ 2 ( 0 ) (55) . This is because the Gaussian ensemble corresponds to the limit as the number of sources N N N A 2 N 1 / 3 A^2N^{1/3} A 2 N 1/3 H 4 \mathcal{H}^4 H 4
Why does the standard deviation σ \sigma σ μ \mu μ (38) . In our simple model, the diagonal terms have nonzero mean but zero variance. The off-diagonal terms have zero mean but nonzero variance. Moreover, (if we turn off the pulsar terms) the variance of the second quantity is the squared mean of the first quantity. So, on average, the square of the second term contributes as much as the square of the first term. The situation is identical to that which describes the errors in periodogram estimates of power spectra [37 ] see discussion on the final page of Sec. 13.4].
Noninterfering source case
The previous subsection considers a set of sources radiating at the same GW frequency, which generate confusion noise from interference between the sources. Here, we consider the opposite extreme, in which each source is radiating at a different GW frequency, or following a different track in time-frequency space, so that their waveforms are uncorrelated in time.
The calculation is very similar to that of the previous Sec. III A for the confusion-noise case. The fundamental difference is that in the place of the GW waveforms of Eq. (15) , we use waveforms
h j + ( t ) = A j cos ( ω j t + ϕ j ) , and h_j^+(t)=\mathcal{A}_j\cos(\omega_jt+\phi_j),\qquad\text{and} h j + ( t ) = A j cos ( ω j t + ϕ j ) , and h j × ( t ) = A j sin ( ω j t + ϕ j ) , (61) h_j^\times(t)=\mathcal{A}_j\sin(\omega_jt+\phi_j), \tag*{(61)} h j × ( t ) = A j sin ( ω j t + ϕ j ) , (61) where we assume that the GW angular frequencies ω j \omega_j ω j j = 1 , … , N j=1,\ldots,N j = 1 , … , N ω j \omega_j ω j 2 π / T 2\pi/T 2 π / T T T T A j \mathcal{A}_j A j (16) , corresponding to a uniform density of sources, as well as the same set of random phases ϕ j \phi_j ϕ j Ω j \Omega_j Ω j
In this model, it is easy to see that the variance of the time-averaged squared strain s = h ˉ a b h ˉ a b s=\bar{h}_{ab}\bar{h}^{ab} s = h ˉ ab h ˉ ab (19) , all of the j ≠ k j\ne k j = k ⟨ s ⟩ = 2 H 2 \langle s\rangle=2\mathcal{H}_2 ⟨ s ⟩ = 2 H 2 ⟨ s 2 ⟩ = 4 H 2 2 \langle s^2\rangle=4\mathcal{H}_2^2 ⟨ s 2 ⟩ = 4 H 2 2 σ s 2 = 0 \sigma_s^2=0 σ s 2 = 0 s s s
The effect on the Hellings and Downs correlations is more subtle: while the one-pulsar-pair (total) variance does not vanish, it is reduced. Again, the critical difference between the independent source model of this subsection and the confusion-noise model of the previous subsection lies in the form taken by the time-averaged correlation ρ \rho ρ (38) . With the “independent source” assumptions we have made here, because every source is emitting GW at a different frequency, ρ \rho ρ
ρ = Z 1 ( t ) Z 2 ( t ) ‾ = ∑ j ( c j d j ∗ + c j ∗ d j ) . (62) \rho=\overline{Z_1(t)Z_2(t)}=\sum_j\left(c_jd_j^*+c_j^*d_j\right). \tag*{(62)} ρ = Z 1 ( t ) Z 2 ( t ) = j ∑ ( c j d j ∗ + c j ∗ d j ) . (62) The off-diagonal j ≠ k j\ne k j = k (38) are absent. In contrast to the confusion-noise case of Eq. (38) , ρ \rho ρ ϕ j \phi_j ϕ j c j c_j c j d k d_k d k (37) , the quantity ω \omega ω ω j \omega_j ω j ω k \omega_k ω k
For the independent sources which we consider here, the ensemble average of the correlation ρ \rho ρ (62) , since ρ \rho ρ ρ \rho ρ (39) . Thus, after completing the average over source directions, the independent-source case has the same first moment as the confusion-noise case
μ ( γ ) = ⟨ ρ ⟩ = 1 2 H 2 μ u ( γ ) , (63) \mu(\gamma)=\langle\rho\rangle=\frac{1}{2}\mathcal{H}_2\mu_u(\gamma), \tag*{(63)} μ ( γ ) = ⟨ ρ ⟩ = 2 1 H 2 μ u ( γ ) , (63) which is identical to Eq. (43) .
Because ρ \rho ρ (44) , for the case of independent sources, all of the double-sum terms are absent. So, in contrast with Eq. (45) , one has the simpler expression
⟨ ρ 2 ⟩ ϕ = ( ∑ j ( c j d j ∗ + c j ∗ d j ) ) 2 . (64) \langle\rho^2\rangle_\phi=\left(\sum_j(c_jd_j^*+c_j^*d_j)\right)^2. \tag*{(64)} ⟨ ρ 2 ⟩ ϕ = ( j ∑ ( c j d j ∗ + c j ∗ d j ) ) 2 . (64) The notation here correctly indicates an average over the random phases, but note that ρ \rho ρ ρ 2 \rho^2 ρ 2 (64) follows directly from Eq. (62) . Hence, the equivalent of Eq. (46) is
⟨ ρ 2 ⟩ ϕ = ∑ j ( c j d j ∗ + c j ∗ d j ) 2 + ∑ j ≠ k [ c j d j ∗ c k d k ∗ + 2 c j ∗ d j c k d k ∗ + c j ∗ d j c k ∗ d k ] . (65) \begin{aligned}
\langle\rho^2\rangle_\phi={}&\sum_j(c_jd_j^*+c_j^*d_j)^2\\
&+\sum_{j\ne k}\left[c_jd_j^*c_kd_k^*+2c_j^*d_jc_kd_k^*+c_j^*d_jc_k^*d_k\right]. \tag*{(65)}
\end{aligned} ⟨ ρ 2 ⟩ ϕ = j ∑ ( c j d j ∗ + c j ∗ d j ) 2 + j = k ∑ [ c j d j ∗ c k d k ∗ + 2 c j ∗ d j c k d k ∗ + c j ∗ d j c k ∗ d k ] . (65) The average of Eq. (65) over source directions is obtained from Eqs. (47) and (52) . After summing over j j j k k k
⟨ ρ 2 ⟩ = 1 8 H 4 ( 2 μ u 2 ( γ ) + 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) + 1 8 ( H 2 2 − H 4 ) ( 2 μ u 2 ( γ ) ) . (66) \begin{aligned}
\langle\rho^2\rangle={}&\frac{1}{8}\mathcal{H}_4\left(2\mu_u^2(\gamma)+2\sigma_u^2(\gamma)+\left((1+\chi^2)^2-1\right)\sigma_c^2(\gamma)\right)\\
&+\frac{1}{8}\left(\mathcal{H}_2^2-\mathcal{H}_4\right)\left(2\mu_u^2(\gamma)\right). \tag*{(66)}
\end{aligned} ⟨ ρ 2 ⟩ = 8 1 H 4 ( 2 μ u 2 ( γ ) + 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) + 8 1 ( H 2 2 − H 4 ) ( 2 μ u 2 ( γ ) ) . (66) This may be compared with the confusion-noise second moment given in Eq. (53) .
The variance in the Hellings-Downs correlation is obtained by forming σ 2 = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 \sigma^2=\langle\rho^2\rangle-\langle\rho\rangle^2 σ 2 = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 (63) and (66) , which cancels the H 2 2 \mathcal{H}_2^2 H 2 2
σ 2 ( γ ) = 1 8 H 4 ( 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) . (67) \sigma^2(\gamma)=\frac{1}{8}\mathcal{H}_4\left(2\sigma_u^2(\gamma)+\left((1+\chi^2)^2-1\right)\sigma_c^2(\gamma)\right). \tag*{(67)} σ 2 ( γ ) = 8 1 H 4 ( 2 σ u 2 ( γ ) + ( ( 1 + χ 2 ) 2 − 1 ) σ c 2 ( γ ) ) . (67) For any number of sources, this is exactly proportional to the single-point-source variance (without the pulsar term, if χ = 0 \chi=0 χ = 0 χ = 1 \chi=1 χ = 1 (56) . This is shown in Figure 3 , including the pulsar term.
The variance Eq. (67) lacks the term proportional to H 2 2 \mathcal{H}_2^2 H 2 2 (23) and (26) in Eq. (67) , and setting χ = 1 \chi=1 χ = 1
σ μ ≈ ζ ( 4 / 3 ) N s σ u 2 ( γ ) + 3 σ c 2 ( γ ) / 2 μ u 2 ( γ ) , (68) \frac{\sigma}{\mu}\approx\frac{\sqrt{\zeta(4/3)}}{\sqrt{N_s}}\sqrt{\frac{\sigma_u^2(\gamma)+3\sigma_c^2(\gamma)/2}{\mu_u^2(\gamma)}}, \tag*{(68)} μ σ ≈ N s ζ ( 4/3 ) μ u 2 ( γ ) σ u 2 ( γ ) + 3 σ c 2 ( γ ) /2 , (68) where ζ ( 4 / 3 ) ≈ 3.601 \zeta(4/3)\approx3.601 ζ ( 4/3 ) ≈ 3.601 N s N_s N s
For independent noninterfering sources with constant amplitude and frequency, in contrast with the confusion-noise case given in Eq. (59) , the variance is dominated by the closest handful of sources in the numerator, whereas the mean in the denominator is proportional to the total number of radial shells of sources. So, for large numbers of sources, the standard deviation σ \sigma σ μ \mu μ Figure 3 for a single unpolarized point source, but the fractional fluctuations are reduced in proportion to the number of source shells N s N_{\mathrm{s}} N s
Cosmic Variance
The variance quantifies the uncertainty in the correlation between pairs of pulsars. Part of this uncertainty arises because, for a given source, different pairs of pulsars, separated by the same angle on the sky, have different correlations. This is called “pulsar variance.” The remainder of the uncertainty arises because of interference between GW sources radiating at the same frequency, which creates interference patterns on the sky. These do not average to give the Hellings-Downs curve μ u ( γ ) \mu_{\mathrm{u}}(\gamma) μ u ( γ )
How do we eliminate the pulsar variance? The variance calculated in Secs. III A and III B is the sum of both cosmic and pulsar variance. We can remove the pulsar variance from these calculations, by replacing the correlation ρ \rho ρ Γ ( γ ) \Gamma(\gamma) Γ ( γ ) Γ ( γ ) \Gamma(\gamma) Γ ( γ ) ρ \rho ρ γ \gamma γ before determining the ensemble mean and the ensemble second moment. This approach lets us calculate the pure cosmic variance, because it completely removes the pulsar variance.
This also reflects observational reality. In experiments with many pulsars, the correlation is typically “binned” by separation angle and averaged within each bin. In fact, an experiment with access to large numbers of pulsars can produce a correlation curve Γ ( γ ) \Gamma(\gamma) Γ ( γ ) only a function of angle γ \gamma γ , by averaging the measured correlation over all pulsars separated by angle γ \gamma γ [38 ] . It also shows that PTAs are sensitive to GW energy density in a way which differs from “local” measurements of s = h a b h a b s = h_{ab}h^{ab} s = h ab h ab h ˙ a b h ˙ a b \dot{h}_{ab}\dot{h}^{ab} h ˙ ab h ˙ ab
For our Universe, where we believe that the GW sources interfere with one another, this pulsar-averaged correlation curve will probably not agree exactly in shape with the Hellings-Downs curve μ u ( γ ) \mu_{\mathrm{u}}(\gamma) μ u ( γ )
For example, consider the ensemble of independent noninterfering GW point sources used in Sec. III B, and look closely at Eq. (62) , which is the correlation ρ \rho ρ c j c_j c j d k d_k d k (37) . If we average ρ \rho ρ p 1 p_1 p 1 p 2 p_2 p 2 γ \gamma γ [39 ] and we obtain
Γ ( γ ) = ⟨ ρ ⟩ p = ∑ j ⟨ c j d j ∗ + c j ∗ d j ⟩ p = 1 2 ∑ j A j 2 ⟨ F 1 + ( Ω j ) F 2 + ( Ω j ) + F 1 × ( Ω j ) F 2 × ( Ω j ) ⟩ p = 1 2 H 2 μ u ( γ ) . (69) \begin{aligned}
\Gamma(\gamma) &= \langle\rho\rangle_{\mathrm{p}} \\
&= \sum_j \left\langle c_j d_j^* + c_j^* d_j \right\rangle_{\mathrm{p}} \\
&= \frac{1}{2}\sum_j A_j^2 \left\langle F_1^+(\boldsymbol{\Omega}_j)F_2^+(\boldsymbol{\Omega}_j) + F_1^\times(\boldsymbol{\Omega}_j)F_2^\times(\boldsymbol{\Omega}_j) \right\rangle_{\mathrm{p}} \\
&= \frac{1}{2}\mathcal{H}_2\mu_{\mathrm{u}}(\gamma). \tag*{(69)}
\end{aligned} Γ ( γ ) = ⟨ ρ ⟩ p = j ∑ ⟨ c j d j ∗ + c j ∗ d j ⟩ p = 2 1 j ∑ A j 2 ⟨ F 1 + ( Ω j ) F 2 + ( Ω j ) + F 1 × ( Ω j ) F 2 × ( Ω j ) ⟩ p = 2 1 H 2 μ u ( γ ) . (69) Thus, after averaging over pulsar positions, we obtain a correlation curve Γ ( γ ) \Gamma(\gamma) Γ ( γ ) exactly proportional to the Hellings-Downs curve μ u ( γ ) \mu_{\mathrm{u}}(\gamma) μ u ( γ ) any member of the ensemble, regardless of the source locations Ω j \boldsymbol{\Omega}_j Ω j [25 ] and Appendix § 2.1 prove that the pulsar average of the antenna pattern functions in Eq. (69) yields μ u \mu_{\mathrm{u}} μ u Γ ( γ ) \Gamma(\gamma) Γ ( γ ) μ u ( γ ) \mu_{\mathrm{u}}(\gamma) μ u ( γ )
In contrast, consider the confusion-noise-limited ensemble of Sec. III A, where Eq. (38) gives the correlation ρ \rho ρ ρ \rho ρ p 1 p_1 p 1 p 2 p_2 p 2 γ \gamma γ Γ ( γ ) \Gamma(\gamma) Γ ( γ ) not proportional to the Hellings-Downs curve μ u ( γ ) \mu_{\mathrm{u}}(\gamma) μ u ( γ ) [40 ] . However, if we examine many representatives from the ensemble, we’ll find that on the average the curve Γ ( γ ) \Gamma(\gamma) Γ ( γ ) μ u ( γ ) \mu_{\mathrm{u}}(\gamma) μ u ( γ ) Γ ( γ ) \Gamma(\gamma) Γ ( γ )
To calculate the cosmic variance, we must remove the pulsar variance. To do this, begin with the general expression for the correlation between pulsar redshifts given in Eq. (14) , and average it over pulsar positions using the method from [25 ] , which also eliminates the pulsar terms. In Appendix § 2.1 , we show how to compute such averages over the three variables ( θ , ϕ , λ ) (\theta,\phi,\lambda) ( θ , ϕ , λ ) γ \gamma γ § 16 , we evaluate these averages for the four products of real antenna pattern functions which appear in Eq. (14) . The corresponding complex averages are
⟨ F 1 ( Ω j ) F 2 ∗ ( Ω k ) ⟩ p = μ + + ( γ , β j k ) + μ × × ( γ , β j k ) = μ ( γ , β j k ) , and \left\langle F_1(\Omega_j)F_2^*(\Omega_k)\right\rangle_p=\mu_{++}(\gamma,\beta_{jk})+\mu_{\times\times}(\gamma,\beta_{jk})=\mu(\gamma,\beta_{jk}),\qquad\text{and} ⟨ F 1 ( Ω j ) F 2 ∗ ( Ω k ) ⟩ p = μ ++ ( γ , β j k ) + μ ×× ( γ , β j k ) = μ ( γ , β j k ) , and ⟨ F 1 ( Ω j ) F 2 ( Ω k ) ⟩ p = μ + + ( γ , β j k ) − μ × × ( γ , β j k ) = μ ( π − γ , π − β j k ) . (70) \left\langle F_1(\Omega_j)F_2(\Omega_k)\right\rangle_p=\mu_{++}(\gamma,\beta_{jk})-\mu_{\times\times}(\gamma,\beta_{jk})=\mu(\pi-\gamma,\pi-\beta_{jk}). \tag*{(70)} ⟨ F 1 ( Ω j ) F 2 ( Ω k ) ⟩ p = μ ++ ( γ , β j k ) − μ ×× ( γ , β j k ) = μ ( π − γ , π − β j k ) . (70) The two-point function μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) § 16 , and given by Eq. (228) . The individual polarization functions μ + + \mu_{++} μ ++ μ × × \mu_{\times\times} μ ×× (229) and given in Eq. (232) . The function μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) γ \gamma γ β j k \beta_{jk} β j k j j j k k k cos β j k = Ω j ⋅ Ω k \cos\beta_{jk}=\Omega_j\cdot\Omega_k cos β j k = Ω j ⋅ Ω k β = 0 \beta=0 β = 0 μ ( γ , 0 ) = 2 μ + + ( γ , 0 ) = 2 μ × × ( γ , 0 ) = μ u ( γ ) \mu(\gamma,0)=2\mu_{++}(\gamma,0)=2\mu_{\times\times}(\gamma,0)=\mu_u(\gamma) μ ( γ , 0 ) = 2 μ ++ ( γ , 0 ) = 2 μ ×× ( γ , 0 ) = μ u ( γ ) (70) to compute the pulsar average of Eq. (14) gives
Γ ( γ ) = ⟨ Z 1 Z 2 ⟩ p = 1 4 ∑ j , k [ ( h j ( t ) ‾ h k ∗ ( t ) + h j ∗ ( t ) h k ( t ) ) μ ( γ , β j k ) + ( h j ( t ) ‾ h k ( t ) + h j ∗ ( t ) h k ∗ ( t ) ) μ ( π − γ , π − β j k ) ] = ∑ j , k ( h j + ( t ) ‾ h k + ( t ) μ + + ( γ , β j k ) + h j × ( t ) ‾ h k × ( t ) μ × × ( γ , β j k ) ) . (71) \begin{aligned}
\Gamma(\gamma)=\left\langle Z_1Z_2\right\rangle_p={}&\frac{1}{4}\sum_{j,k}\left[\left(\overline{h_j(t)}h_k^*(t)+h_j^*(t)h_k(t)\right)\mu(\gamma,\beta_{jk})\right.\\
&\left.+\left(\overline{h_j(t)}h_k(t)+h_j^*(t)h_k^*(t)\right)\mu(\pi-\gamma,\pi-\beta_{jk})\right]\\
={}&\sum_{j,k}\left(\overline{h_j^+(t)}h_k^+(t)\mu_{++}(\gamma,\beta_{jk})+\overline{h_j^\times(t)}h_k^\times(t)\mu_{\times\times}(\gamma,\beta_{jk})\right). \tag*{(71)}
\end{aligned} Γ ( γ ) = ⟨ Z 1 Z 2 ⟩ p = = 4 1 j , k ∑ [ ( h j ( t ) h k ∗ ( t ) + h j ∗ ( t ) h k ( t ) ) μ ( γ , β j k ) + ( h j ( t ) h k ( t ) + h j ∗ ( t ) h k ∗ ( t ) ) μ ( π − γ , π − β j k ) ] j , k ∑ ( h j + ( t ) h k + ( t ) μ ++ ( γ , β j k ) + h j × ( t ) h k × ( t ) μ ×× ( γ , β j k ) ) . (71) Here, we have given equivalent expressions for complex and real waveforms. As previously mentioned, because the pulsar averaging eliminates the pulsar terms, the time averages in Eq. (71) only include the Earth terms.
We now evaluate this pulsar-averaged correlation for the confusion-noise case of Sec. III A. Substitute the complex GW waveform Eq. (17) into the first line of Eq. (71) . The time averaging eliminates the second term, giving
Γ ( γ ) = 1 4 ∑ j , k A j A k ( e i ( ϕ j − ϕ k ) + e − i ( ϕ j − ϕ k ) ) μ ( γ , β j k ) = 1 2 H 2 μ u ( γ ) + 1 4 ∑ j ≠ k A j A k ( e i ( ϕ j − ϕ k ) + e − i ( ϕ j − ϕ k ) ) μ ( γ , β j k ) . (72) \begin{aligned}
\Gamma(\gamma)&=\frac{1}{4}\sum_{j,k}A_jA_k\left(e^{i(\phi_j-\phi_k)}+e^{-i(\phi_j-\phi_k)}\right)\mu(\gamma,\beta_{jk})\\
&=\frac{1}{2}\mathcal{H}_2\mu_u(\gamma)\\
&\quad+\frac{1}{4}\sum_{j\ne k}A_jA_k\left(e^{i(\phi_j-\phi_k)}+e^{-i(\phi_j-\phi_k)}\right)\mu(\gamma,\beta_{jk}). \tag*{(72)}
\end{aligned} Γ ( γ ) = 4 1 j , k ∑ A j A k ( e i ( ϕ j − ϕ k ) + e − i ( ϕ j − ϕ k ) ) μ ( γ , β j k ) = 2 1 H 2 μ u ( γ ) + 4 1 j = k ∑ A j A k ( e i ( ϕ j − ϕ k ) + e − i ( ϕ j − ϕ k ) ) μ ( γ , β j k ) . (72) On the second line we have partitioned the sum into diagonal and off-diagonal terms; for the diagonal terms with j = k j=k j = k β j j = 0 \beta_{jj}=0 β j j = 0 (72) that the pulsar-averaged correlation curve Γ ( γ ) \Gamma(\gamma) Γ ( γ ) μ u ( γ ) \mu_u(\gamma) μ u ( γ ) β \beta β Figure 12 . Those cross-sections have a shape that differs from μ u ( γ ) \mu_u(\gamma) μ u ( γ )
Figure 12. The two-point function μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) (228) . It is the mean correlation between two pulsars separated by angle γ \gamma γ β \beta β β = 0 \beta=0 β = 0 μ u ( γ ) \mu_u(\gamma) μ u ( γ )
Figure 12 . Those cross-sections have a shape that differs from μ u ( γ ) \mu_u(\gamma) μ u ( γ )
We stress that in obtaining Eq. (72) from Eq. (14) , we have not done anything to the sources, which have the exact waveforms and sky positions appropriate to that representative of the ensemble. All that we have done is to average over the pulsar locations, as would be done if the average correlation at separation angle γ \gamma γ
To compute the ensemble average mean and variance of Γ \Gamma Γ (72) over the random phases. From Eq. (21) we know that ⟨ e i ( ϕ j − ϕ k ) ⟩ ϕ \left\langle e^{i(\phi_j-\phi_k)}\right\rangle_\phi ⟨ e i ( ϕ j − ϕ k ) ⟩ ϕ j ≠ k j\ne k j = k Γ \Gamma Γ
⟨ Γ ( γ ) ⟩ = ⟨ Γ ( γ ) ⟩ ϕ = 1 2 H 2 μ u ( γ ) . (73) \left\langle\Gamma(\gamma)\right\rangle=\left\langle\Gamma(\gamma)\right\rangle_\phi=\frac{1}{2}\mathcal{H}_2\mu_u(\gamma). \tag*{(73)} ⟨ Γ ( γ ) ⟩ = ⟨ Γ ( γ ) ⟩ ϕ = 2 1 H 2 μ u ( γ ) . (73) Thus, on the average, an observer in this ensemble of model universes would obtain a pulsar-averaged correlation curve Γ ( γ ) \Gamma(\gamma) Γ ( γ ) μ u ( γ ) \mu_u(\gamma) μ u ( γ )
To compute the second moment of Γ \Gamma Γ (72) and evaluate the average over random phases. The square of the first term is independent of random phases and averages to itself. The cross term is zero because ⟨ e i ( ϕ j − ϕ k ) ⟩ ϕ \left\langle e^{i(\phi_j-\phi_k)}\right\rangle_\phi ⟨ e i ( ϕ j − ϕ k ) ⟩ ϕ j ≠ k j\ne k j = k (44) , giving
⟨ Γ 2 ( γ ) ⟩ ϕ = 1 4 H 2 2 μ u 2 ( γ ) + 1 8 ∑ j ≠ k ∑ ℓ ≠ m A j A k A ℓ A m ( δ j m δ k ℓ + δ j ℓ δ k m ) μ ( γ , β j k ) μ ( γ , β ℓ m ) = 1 4 H 2 2 μ u 2 ( γ ) + 1 8 ∑ j ≠ k A j 2 A k 2 ( μ ( γ , β j k ) μ ( γ , β j k ) + μ ( γ , β j k ) μ ( γ , β k j ) ) = 1 4 H 2 2 μ u 2 ( γ ) + 1 4 ∑ j ≠ k A j 2 A k 2 μ u 2 ( γ , β j k ) . (74) \begin{aligned}
\left\langle\Gamma^2(\gamma)\right\rangle_\phi={}&\frac{1}{4}\mathcal{H}_2^2\mu_u^2(\gamma)+\frac{1}{8}\sum_{j\ne k}\sum_{\ell\ne m}A_jA_kA_\ell A_m(\delta_{j m}\delta_{k\ell}+\delta_{j\ell}\delta_{k m})\mu(\gamma,\beta_{jk})\mu(\gamma,\beta_{\ell m})\\
={}&\frac{1}{4}\mathcal{H}_2^2\mu_u^2(\gamma)+\frac{1}{8}\sum_{j\ne k}A_j^2A_k^2\left(\mu(\gamma,\beta_{jk})\mu(\gamma,\beta_{jk})+\mu(\gamma,\beta_{jk})\mu(\gamma,\beta_{kj})\right)\\
={}&\frac{1}{4}\mathcal{H}_2^2\mu_u^2(\gamma)+\frac{1}{4}\sum_{j\ne k}A_j^2A_k^2\mu_u^2(\gamma,\beta_{jk}). \tag*{(74)}
\end{aligned} ⟨ Γ 2 ( γ ) ⟩ ϕ = = = 4 1 H 2 2 μ u 2 ( γ ) + 8 1 j = k ∑ ℓ = m ∑ A j A k A ℓ A m ( δ j m δ k ℓ + δ j ℓ δ k m ) μ ( γ , β j k ) μ ( γ , β ℓ m ) 4 1 H 2 2 μ u 2 ( γ ) + 8 1 j = k ∑ A j 2 A k 2 ( μ ( γ , β j k ) μ ( γ , β j k ) + μ ( γ , β j k ) μ ( γ , β k j ) ) 4 1 H 2 2 μ u 2 ( γ ) + 4 1 j = k ∑ A j 2 A k 2 μ u 2 ( γ , β j k ) . (74) where to obtain the final line we have used β j k = β k j \beta_{jk}=\beta_{kj} β j k = β k j (74) over the random directions Ω j \Omega_j Ω j Ω k \Omega_k Ω k β j k = cos − 1 Ω j ⋅ Ω k \beta_{jk}=\cos^{-1}\boldsymbol{\Omega}_j\cdot\boldsymbol{\Omega}_k β j k = cos − 1 Ω j ⋅ Ω k
⟨ Γ 2 ( γ ) ⟩ = 1 4 H 2 2 μ u 2 ( γ ) + 1 4 ∑ j ≠ k A j 2 A k 2 μ ~ 2 ( γ ) (75) \left\langle\Gamma^2(\gamma)\right\rangle=\frac{1}{4}\mathcal{H}_2^2\mu_u^2(\gamma)+\frac{1}{4}\sum_{j\ne k} A_j^2 A_k^2\widetilde{\mu}^2(\gamma) \tag*{(75)} ⟨ Γ 2 ( γ ) ⟩ = 4 1 H 2 2 μ u 2 ( γ ) + 4 1 j = k ∑ A j 2 A k 2 μ 2 ( γ ) (75) where μ ~ 2 ( γ ) \widetilde{\mu}^2(\gamma) μ 2 ( γ ) μ 2 ( γ , β ) \mu^2(\gamma,\beta) μ 2 ( γ , β ) β \beta β sin β d β \sin\beta\,\mathrm{d}\beta sin β d β § 16 and given in Eq. (234) . Subtracting the square of the first moment Eq. (73) from Eq. (75) gives the cosmic variance
σ c o s m i c 2 ( γ ) = ⟨ Γ 2 ( γ ) ⟩ − ⟨ Γ ( γ ) ⟩ 2 = 1 4 ∑ j ≠ k A j 2 A k 2 μ ~ 2 ( γ ) = 1 4 ( H 2 2 − H 4 ) μ ~ 2 ( γ ) (76) \begin{aligned}
\sigma_{\mathrm{cosmic}}^2(\gamma)=\left\langle\Gamma^2(\gamma)\right\rangle-\left\langle\Gamma(\gamma)\right\rangle^2 \\
&=\frac{1}{4}\sum_{j\ne k}A_j^2A_k^2\widetilde{\mu}^2(\gamma) \\
&=\frac{1}{4}\left(\mathcal{H}_2^2-\mathcal{H}_4\right)\widetilde{\mu}^2(\gamma) \tag*{(76)}
\end{aligned} σ cosmic 2 ( γ ) = ⟨ Γ 2 ( γ ) ⟩ − ⟨ Γ ( γ ) ⟩ 2 = 4 1 j = k ∑ A j 2 A k 2 μ 2 ( γ ) = 4 1 ( H 2 2 − H 4 ) μ 2 ( γ ) (76) for our confusion-noise-limited ensemble. This is another one of the main results of this paper: the cosmic variance for the discrete-source confusion-noise model. It may be compared with the total variance given earlier in Eq. (55) . For large numbers of sources H 2 2 ≫ H 4 \mathcal{H}_2^2\gg\mathcal{H}_4 H 2 2 ≫ H 4 (23) , (57) , (73) , and (76) to write the mean, cosmic variance, and total variance as
⟨ Γ ( γ ) ⟩ = μ ( γ ) = 1 2 H 2 μ u ( γ ) = 1 2 N s A 2 μ u ( γ ) , \left\langle\Gamma(\gamma)\right\rangle=\mu(\gamma)=\frac{1}{2}\mathcal{H}_2\mu_u(\gamma)=\frac{1}{2}N_s\mathcal{A}^2\mu_u(\gamma), ⟨ Γ ( γ ) ⟩ = μ ( γ ) = 2 1 H 2 μ u ( γ ) = 2 1 N s A 2 μ u ( γ ) , σ c o s m i c 2 ( γ ) ≈ 1 4 H 2 2 μ ~ 2 ( γ ) = 1 4 N s 2 A 4 μ ~ 2 ( γ ) , \sigma_{\mathrm{cosmic}}^2(\gamma)\approx\frac{1}{4}\mathcal{H}_2^2\widetilde{\mu}^2(\gamma)=\frac{1}{4}N_s^2\mathcal{A}^4\widetilde{\mu}^2(\gamma), σ cosmic 2 ( γ ) ≈ 4 1 H 2 2 μ 2 ( γ ) = 4 1 N s 2 A 4 μ 2 ( γ ) , σ 2 ( γ ) ≈ 1 4 H 2 2 ( μ ~ 2 ( γ ) + 4 μ u 2 ( 0 ) ) = 1 8 N s 2 A 4 ( μ ~ 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (77) \sigma^2(\gamma)\approx\frac{1}{4}\mathcal{H}_2^2\left(\widetilde{\mu}^2(\gamma)+4\mu_u^2(0)\right)
=\frac{1}{8}N_s^2\mathcal{A}^4\left(\widetilde{\mu}^2(\gamma)+4\mu_u^2(0)\right). \tag*{(77)} σ 2 ( γ ) ≈ 4 1 H 2 2 ( μ 2 ( γ ) + 4 μ u 2 ( 0 ) ) = 8 1 N s 2 A 4 ( μ 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (77) Thus, the predicted scale of fluctuations in the pulsar-averaged correlation for the confusion-noise ensemble is
σ c o s m i c ⟨ Γ ⟩ = σ c o s m i c μ ( γ ) = μ ~ 2 ( γ ) μ u 2 ( γ ) . (78) \frac{\sigma_{\mathrm{cosmic}}}{\langle\Gamma\rangle}=\frac{\sigma_{\mathrm{cosmic}}}{\mu(\gamma)}=\sqrt{\frac{\widetilde{\mu}^2(\gamma)}{\mu_u^2(\gamma)}}. \tag*{(78)} ⟨ Γ ⟩ σ cosmic = μ ( γ ) σ cosmic = μ u 2 ( γ ) μ 2 ( γ ) . (78) This is illustrated in Figure 4 , and can be generalized to multiple frequency bins as shown in Eq. (60) . In Appendix § 8 § 11 we use the Gaussian ensemble to derive the same results in the limit of large numbers of weak sources. If the cosmic variance were zero, then an experiment could (at least in principle) average the correlation over large numbers of low-noise pulsar pairs, and approach the Hellings-Downs curve with arbitrary precision. If the cosmic variance is nonzero, then this is not possible. In that case, once enough pulsar pairs have been employed,
Figure 4. The cosmic variance for an ensemble of confusion-noise universes in the limit of large numbers of sources (with the mean GW amplitude at Earth N s A 2 = H 2 = 2 N_s\mathcal{A}^2=\mathcal{H}_2=2 N s A 2 = H 2 = 2 ⟨ Γ ( γ ) ⟩ = μ u ( γ ) \langle\Gamma(\gamma)\rangle=\mu_u(\gamma) ⟨ Γ ( γ )⟩ = μ u ( γ ) μ u ( γ ) ± σ c o s m i c \mu_u(\gamma)\pm\sigma_{\mathrm{cosmic}} μ u ( γ ) ± σ cosmic 5 σ c o s m i c 5\sigma_{\mathrm{cosmic}} 5 σ cosmic (78) and (G11) (scaled × 5 \times5 × 5
the pulsar variance will be reduced to well below the cosmic variance and a fundamental limit is reached. The shape mismatch between the correlation measured in our Universe and the Hellings-Downs curve can no longer be reduced by incorporating additional low-noise pulsar pairs. This topic is studied in more detail in [38 ] , which determines the number of pulsar pairs required to reach the limits imposed by the cosmic variance.
Conclusion
The goal of this paper is to assess how closely the Hellings-Downs correlation curve would be matched in our Universe, in the absence of any experimental or pulsar noise. We show that the variance in the Hellings-Downs correlation has two parts. Pulsar variance arises from the fluctuations that occur when the correlation is measured with pulsar pairs at different sky locations, but separated by the same angle. In principle, this can be eliminated by using many pulsar pairs and averaging. Cosmic variance arises because the waveforms of the different GW sources interfere with one another. This gives rise to a deviation between the angle-average correlation Γ ( γ ) \Gamma(\gamma) Γ ( γ ) μ u ( γ ) \mu_u(\gamma) μ u ( γ ) (55) with χ = 1 \chi=1 χ = 1 (76) .
We also constructed a similar ensemble of point sources, but with each radiating GWs at a different frequency. For these, the variance is given by Eq. (67) . It is proportional to the variance expected for a single unpolarized point source, because it arises from the closest sources only. There is no interference term, and in this ensemble, the cosmic variance vanishes.
Comparison of these two models leads to an important conclusion. For PTAs, information is carried not just in the mean μ ( γ ) \mu(\gamma) μ ( γ ) σ 2 ( γ ) \sigma^2(\gamma) σ 2 ( γ )
Recent work [38 ] provides an alternative way to understand the cosmic variance for a Gaussian ensemble. Suppose that our goal was to determine the pulsar-averaged correlation Γ \Gamma Γ γ \gamma γ Γ ( γ ) \Gamma(\gamma) Γ ( γ ) Γ ( γ ) \Gamma(\gamma) Γ ( γ ) [38 ] , we demonstrate that the variance of that optimal estimator, when there are many pulsars, is exactly σ c o s m i c 2 \sigma^2_{\mathrm{cosmic}} σ cosmic 2
The work in this paper could be usefully extended. Our expectation is that the variance in the Hellings-Downs correlation is not described by either of the simple models presented here, but rather by some combination of them [20 ] . Our formalism can still be employed, but more realistic statistical ensembles of sources are needed. Specifically, we need better models for the ensemble of amplitudes Eq. (15) which create the correlations between the GW waveforms of different sources. One could construct a statistical model containing a realistic distribution of supermassive black-hole binary systems with different masses and other properties [31 ] . This could then be used to define the correlation ρ \rho ρ [41 ] , which in turn can be used to predict the mean of the Hellings-Downs correlation, and the pulsar and cosmic variance. A promising approach for creating such ensembles is to create synthetic catalogs based on cosmological simulations, which can be used to predict and model the mean and variance of the Hellings-Downs correlation [15 ] .
It may be some years before there is a definitive detection of the gravitational-wave background with pulsar timing arrays. Even when that takes place, the correlations between individual pulsar pairs may be dominated by pulsar timing noise and experimental noise. So it is difficult to predict if the variance can be observed in the near future. But we are hopeful that as telescopes improve and pulsar populations increase, this will eventually become possible.
Acknowledgments
I gratefully acknowledge the members of the International Pulsar Timing Array Detection Committee (DC), and the speakers who made presentations to the DC, for teaching me something about PTAs and for stimulating my interest. I thank Xavier Siemens, Alberto Sesana, Curt Cutler, and Serena Valtolina for useful conversations and Joe Romano and Neil Cornish for pointing me to earlier work on this topic and for identifying a significant mistake in the first version of the manuscript. I also acknowledge the support of the anonymous referee, who identified a number of misstatements and omissions. Very special thanks to Joe Romano for a close and critical reading of many versions of the manuscript, and extended discussions which resulted in many corrections, clarifications and improvements.
Here, we derive the general form of the mean and variance of the Hellings-Downs correlation for a single distant GW point source, without making the assumption that the source is unpolarized or that the pulsar terms can be neglected. Such single point source results can be combined for a collection of sources following the argument given in Sec. III.
In electromagnetism, the polarization of a light or radio source may be described in terms of four Stokes parameters. The same is possible for gravitational waves. For example, see Eq. (31) of [42 ] for the GW Stokes parameters of a binary system in a circular orbit emitting GWs.
In the literature an unpolarized source is often assumed to be one for which the time correlation between the cross and plus polarizations vanishes. For example, in [25 ] , the authors remark following their Eq. (10), “it has been assumed that ⟨ R × R × ⟩ = 0 \langle R_\times R_\times\rangle=0 ⟨ R × R × ⟩ = 0 μ u ( γ ) \mu_u(\gamma) μ u ( γ ) l = 0 ∘ l=0^\circ l = 0 ∘ l = 180 ∘ l=180^\circ l = 18 0 ∘
We now derive the general formula for the mean and variance of the Hellings-Downs correlation. Our starting point is the effect on pulse arrival times or redshifts induced by a gravitational wave propagating in direction Ω \boldsymbol{\Omega} Ω Ω \boldsymbol{\Omega} Ω [18 ] , Appendixes A and B]. A plane gravitational wave propagating in direction Ω \boldsymbol{\Omega} Ω h + h^{+} h + h × h^{\times} h ×
h a b ( t , x ) = h + ( t − x ⋅ Ω ) e a b + + h × ( t − x ⋅ Ω ) e a b × , (79) h_{ab}(t,\boldsymbol{x})=h^{+}(t-\boldsymbol{x}\cdot\boldsymbol{\Omega})e^{+}_{ab}+h^{\times}(t-\boldsymbol{x}\cdot\boldsymbol{\Omega})e^{\times}_{ab}, \tag*{(79)} h ab ( t , x ) = h + ( t − x ⋅ Ω ) e ab + + h × ( t − x ⋅ Ω ) e ab × , (79) where t t t x \boldsymbol{x} x (180) .
The effect of the gravitational wave on the pulse arrival times is most simply described by the redshift Z Z Z Z Z Z p L pL p L p \boldsymbol{p} p L L L t t t
Z ( t ) = 1 2 p a p b 1 + Ω ⋅ p Δ h a b ( t ) , (80) Z(t)=\frac{1}{2}\frac{p^{a}p^{b}}{1+\boldsymbol{\Omega}\cdot\boldsymbol{p}}\Delta h_{ab}(t), \tag*{(80)} Z ( t ) = 2 1 1 + Ω ⋅ p p a p b Δ h ab ( t ) , (80) where the gravitational wave enters through two terms
Δ h a b ( t ) = h a b ( t , 0 ) − h a b ( t − L , p L ) , (81) \Delta h_{ab}(t)=h_{ab}(t,0)-h_{ab}(t-L,\boldsymbol{p}L), \tag*{(81)} Δ h ab ( t ) = h ab ( t , 0 ) − h ab ( t − L , p L ) , (81) which are called the “Earth term” and the “pulsar term” respectively, since they are the amplitude of the gravitational wave at the time and spatial location where the pulse is received on Earth, and at the time and spatial location where it departed the pulsar. These terms may be simplified and written in terms of the two free functions that define the wave, by
Δ h a b ( t ) = R + ( t ) e a b + + R × ( t ) e a b × = [ h + ( t ) − h + ( t − L ( 1 + p ⋅ Ω ) ) ] e a b + + [ h × ( t ) − h × ( t − L ( 1 + p ⋅ Ω ) ) ] e a b × . (82) \begin{aligned}
\Delta h_{ab}(t)&=R^{+}(t)e^{+}_{ab}+R^{\times}(t)e^{\times}_{ab}\\
&=\left[h^{+}(t)-h^{+}\left(t-L(1+\boldsymbol{p}\cdot\boldsymbol{\Omega})\right)\right]e^{+}_{ab}\\
&\quad+\left[h^{\times}(t)-h^{\times}\left(t-L(1+\boldsymbol{p}\cdot\boldsymbol{\Omega})\right)\right]e^{\times}_{ab}. \tag*{(82)}
\end{aligned} Δ h ab ( t ) = R + ( t ) e ab + + R × ( t ) e ab × = [ h + ( t ) − h + ( t − L ( 1 + p ⋅ Ω ) ) ] e ab + + [ h × ( t ) − h × ( t − L ( 1 + p ⋅ Ω ) ) ] e ab × . (82) As shown in [18 ] , for a combination of waves propagating in different directions, these effects may be summed. But for our purposes, it is sufficient to consider the waves coming from individual sources, under the assumption that they are uncorrelated over the (decades-long) pulsar observations.
Here, without loss of generality, we will place our source on the − z -z − z Ω = z ^ \boldsymbol{\Omega}=\hat{\boldsymbol{z}} Ω = z ^
e a b + = x ^ a x ^ b − y ^ a y ^ b , and e^{+}_{ab}=\hat{x}_{a}\hat{x}_{b}-\hat{y}_{a}\hat{y}_{b},\qquad\text{and} e ab + = x ^ a x ^ b − y ^ a y ^ b , and e a b × = x ^ a y ^ b + y ^ a x ^ b . (83) e^{\times}_{ab}=\hat{x}_{a}\hat{y}_{b}+\hat{y}_{a}\hat{x}_{b}. \tag*{(83)} e ab × = x ^ a y ^ b + y ^ a x ^ b . (83) Note that GW waveforms are often calculated with respect to one set of polarization axes, which must be rotated (often by an angle denoted ψ \psi ψ x x x y y y
To compute the correlations between pulsars 1 and 2, it is convenient to define “antenna pattern” functions
F 1 + = 1 2 p 1 a p 1 b 1 + Ω ⋅ p 1 e a b + , and F^{+}_{1}=\frac{1}{2}\frac{p_{1}^{a}p_{1}^{b}}{1+\boldsymbol{\Omega}\cdot\boldsymbol{p}_{1}}e^{+}_{ab},\qquad\text{and} F 1 + = 2 1 1 + Ω ⋅ p 1 p 1 a p 1 b e ab + , and F 1 × = 1 2 p 1 a p 1 b 1 + Ω ⋅ p 1 e a b × , (84) F^{\times}_{1}=\frac{1}{2}\frac{p_{1}^{a}p_{1}^{b}}{1+\boldsymbol{\Omega}\cdot\boldsymbol{p}_{1}}e^{\times}_{ab}, \tag*{(84)} F 1 × = 2 1 1 + Ω ⋅ p 1 p 1 a p 1 b e ab × , (84) where p 1 \boldsymbol{p}_{1} p 1 p 2 \boldsymbol{p}_{2} p 2
The correlation ρ = Z 1 Z 2 ‾ \rho=\overline{Z_{1}Z_{2}} ρ = Z 1 Z 2
ρ = c 1 F 1 + F 2 + + c 2 F 1 × F 2 × + c 3 F 1 + F 2 × + c 4 F 1 × F 2 + , (85) \rho=c_{1}F^{+}_{1}F^{+}_{2}+c_{2}F^{\times}_{1}F^{\times}_{2}+c_{3}F^{+}_{1}F^{\times}_{2}+c_{4}F^{\times}_{1}F^{+}_{2}, \tag*{(85)} ρ = c 1 F 1 + F 2 + + c 2 F 1 × F 2 × + c 3 F 1 + F 2 × + c 4 F 1 × F 2 + , (85) where the coefficients are the time averages:
c 1 = R 1 + ( t ) R 2 + ( t ) ‾ , c 2 = R 1 × ( t ) R 2 × ( t ) ‾ , c 3 = R 1 + ( t ) R 2 × ( t ) ‾ , and c 4 = R 1 × ( t ) R 2 + ( t ) ‾ . (86) \begin{aligned}
c_{1}&=\overline{R^{+}_{1}(t)R^{+}_{2}(t)},\\
c_{2}&=\overline{R^{\times}_{1}(t)R^{\times}_{2}(t)},\\
c_{3}&=\overline{R^{+}_{1}(t)R^{\times}_{2}(t)},\qquad\text{and}\\
c_{4}&=\overline{R^{\times}_{1}(t)R^{+}_{2}(t)}. \tag*{(86)}
\end{aligned} c 1 c 2 c 3 c 4 = R 1 + ( t ) R 2 + ( t ) , = R 1 × ( t ) R 2 × ( t ) , = R 1 + ( t ) R 2 × ( t ) , and = R 1 × ( t ) R 2 + ( t ) . (86) Here, the subscripts on R R R (82) . But note that if the pulsar terms can be neglected, then the subscripts on R R R c 3 = c 4 c_{3}=c_{4} c 3 = c 4
Although we are treating the general case, it is helpful to have a specific example in mind. For this, take a pair of orbiting point masses m 1 m_{1} m 1 m 2 m_{2} m 2 r r r ω \omega ω ι \iota ι 0 0 0 180 180 180 x y xy x y x x x ϕ c \phi_c ϕ c
For this system, if we neglect the pulsar terms, we have
R + ( t ) = h + ( t ) = 1 2 A ( 1 + cos 2 ι ) cos 2 ( ω t + ϕ c ) , R × ( t ) = h × ( t ) = A cos ι sin 2 ( ω t + ϕ c ) , (87) R^{+}(t)=h^{+}(t)=\frac{1}{2}\mathcal{A}(1+\cos^{2}\iota)\cos2(\omega t+\phi_c), \qquad
R^{\times}(t)=h^{\times}(t)=\mathcal{A}\cos\iota\sin2(\omega t+\phi_c), \tag*{(87)} R + ( t ) = h + ( t ) = 2 1 A ( 1 + cos 2 ι ) cos 2 ( ω t + ϕ c ) , R × ( t ) = h × ( t ) = A cos ι sin 2 ( ω t + ϕ c ) , (87) where the dimensionless GW strain amplitude is
A = 4 r ( G c 2 ) 5 / 3 ( ω c ) 2 / 3 m 1 m 2 ( m 1 + m 2 ) 1 / 3 . (88) \mathcal{A}=\frac{4}{r}\left(\frac{G}{c^{2}}\right)^{5/3}\left(\frac{\omega}{c}\right)^{2/3}\frac{m_1m_2}{(m_1+m_2)^{1/3}}. \tag*{(88)} A = r 4 ( c 2 G ) 5/3 ( c ω ) 2/3 ( m 1 + m 2 ) 1/3 m 1 m 2 . (88) and we assume that the orbital evolution is slow enough to treat ω \omega ω § 2.2 , we will treat the other case, where the pulsar term is significant. Also, to accompany the gravitational constant G G G c c c A \mathcal{A} A 2 ω 2\omega 2 ω (85) are:
c 1 = 1 8 A 2 ( 1 + cos 2 ι ) 2 , c 2 = 1 2 A 2 cos 2 ι , c 3 = 0 , and c 4 = 0. (89) \begin{aligned}
c_1&=\frac{1}{8}\mathcal{A}^{2}(1+\cos^{2}\iota)^{2},\\
c_2&=\frac{1}{2}\mathcal{A}^{2}\cos^{2}\iota,\\
c_3&=0,\qquad\text{and}\\
c_4&=0. \tag*{(89)}
\end{aligned} c 1 c 2 c 3 c 4 = 8 1 A 2 ( 1 + cos 2 ι ) 2 , = 2 1 A 2 cos 2 ι , = 0 , and = 0. (89) For the most general source, all four coefficients can be nonzero and different from one another.
We want to compute the average correlation and its variance as the two pulsars sweep around the sky with a fixed angle between them. The first pulsar at distance L 1 L_1 L 1 ( θ , λ ) (\theta,\lambda) ( θ , λ )
p 1 = x ^ cos λ sin θ + y ^ sin λ sin θ + z ^ cos θ . (90) \mathbf{p}_1=\hat{\mathbf{x}}\cos\lambda\sin\theta+\hat{\mathbf{y}}\sin\lambda\sin\theta+\hat{\mathbf{z}}\cos\theta. \tag*{(90)} p 1 = x ^ cos λ sin θ + y ^ sin λ sin θ + z ^ cos θ . (90) (Note: it is deliberate that we are using the variable λ \lambda λ ϕ \phi ϕ p 2 \mathbf{p}_2 p 2 θ ′ , ϕ ′ \theta',\phi' θ ′ , ϕ ′
To parametrize the location of the second pulsar, we adopt a nice trick from [25 ] , and use variables γ \gamma γ ϕ \phi ϕ
p 2 = x ^ [ cos λ ( sin θ cos γ − cos θ sin γ cos ϕ ) + sin λ sin γ sin ϕ ] + y ^ [ sin λ ( sin θ cos γ − cos θ sin γ cos ϕ ) − cos λ sin γ sin ϕ ] + z ^ [ cos θ cos γ + sin θ sin γ cos ϕ ] . (91) \mathbf{p}_2=\hat{\mathbf{x}}\left[\cos\lambda(\sin\theta\cos\gamma-\cos\theta\sin\gamma\cos\phi)+\sin\lambda\sin\gamma\sin\phi\right]
+\hat{\mathbf{y}}\left[\sin\lambda(\sin\theta\cos\gamma-\cos\theta\sin\gamma\cos\phi)-\cos\lambda\sin\gamma\sin\phi\right]
+\hat{\mathbf{z}}\left[\cos\theta\cos\gamma+\sin\theta\sin\gamma\cos\phi\right]. \tag*{(91)} p 2 = x ^ [ cos λ ( sin θ cos γ − cos θ sin γ cos ϕ ) + sin λ sin γ sin ϕ ] + y ^ [ sin λ ( sin θ cos γ − cos θ sin γ cos ϕ ) − cos λ sin γ sin ϕ ] + z ^ [ cos θ cos γ + sin θ sin γ cos ϕ ] . (91) The variable γ \gamma γ p 1 \mathbf{p}_1 p 1 p 2 \mathbf{p}_2 p 2 ϕ ∈ [ 0 , 2 π ) \phi\in[0,2\pi) ϕ ∈ [ 0 , 2 π ) Figure 5 , and uniformly parametrizes the position of p 2 \mathbf{p}_2 p 2 p 1 \mathbf{p}_1 p 1 ∂ p 2 / ∂ ϕ \partial\mathbf{p}_2/\partial\phi ∂ p 2 / ∂ ϕ p 1 \mathbf{p}_1 p 1 p 2 \mathbf{p}_2 p 2 p 2 \mathbf{p}_2 p 2 p 1 ⋅ p 2 = cos γ \mathbf{p}_1\cdot\mathbf{p}_2=\cos\gamma p 1 ⋅ p 2 = cos γ
Figure 5. The direction to the first pulsar p 1 \mathbf{p}_1 p 1 ( θ , λ ) (\theta,\lambda) ( θ , λ ) (90) . The direction to the second pulsar p 2 \mathbf{p}_2 p 2 (91) via variables γ \gamma γ ϕ \phi ϕ γ \gamma γ p 1 \mathbf{p}_1 p 1 p 2 \mathbf{p}_2 p 2 ϕ ∈ [ 0 , 2 π ) \phi\in[0,2\pi) ϕ ∈ [ 0 , 2 π ) p 1 \mathbf{p}_1 p 1 sin γ \sin\gamma sin γ
This parametrization of p 2 \mathbf{p}_2 p 2 p 1 \mathbf{p}_1 p 1 γ \gamma γ Q ( θ , ϕ , λ ) Q(\theta,\phi,\lambda) Q ( θ , ϕ , λ ) λ \lambda λ
[ Q ] ( θ , ϕ ) = 1 2 π ∫ 0 2 π d λ Q ( θ , ϕ , λ ) . (92) [Q](\theta,\phi)=\frac{1}{2\pi}\int_{0}^{2\pi}d\lambda\,Q(\theta,\phi,\lambda). \tag*{(92)} [ Q ] ( θ , ϕ ) = 2 π 1 ∫ 0 2 π d λ Q ( θ , ϕ , λ ) . (92) and the second average (over θ \theta θ ϕ \phi ϕ
⟨ [ Q ] ⟩ = 1 4 π ∫ 0 π d θ sin θ ∫ 0 2 π d ϕ [ Q ] ( θ , ϕ ) . (93) \langle[Q]\rangle=\frac{1}{4\pi}\int_{0}^{\pi}d\theta\,\sin\theta\int_{0}^{2\pi}d\phi\,[Q](\theta,\phi). \tag*{(93)} ⟨[ Q ]⟩ = 4 π 1 ∫ 0 π d θ sin θ ∫ 0 2 π d ϕ [ Q ] ( θ , ϕ ) . (93) These averages are properly normalized, [ 1 ] = ⟨ 1 ⟩ = 1 [1]=\langle1\rangle=1 [ 1 ] = ⟨ 1 ⟩ = 1 γ \gamma γ Figure 5 ).
The geometric antenna pattern functions for the first pulsar can now be evaluated by substituting the polarization tensors Eq. (83) and pulsar-position vector Eq. (90) into Eq. (84) . One finds
F 1 + = 1 2 p 1 a p 1 b 1 + Ω ⋅ p 1 e a b + = U cos 2 λ , F 1 × = 1 2 p 1 a p 1 b 1 + Ω ⋅ p 1 e a b × = U sin 2 λ , (94) F_1^{+}=\frac{1}{2}\frac{p_1^{a}p_1^{b}}{1+\boldsymbol{\Omega}\cdot\mathbf{p}_1}e_{ab}^{+}=U\cos2\lambda,\qquad
F_1^{\times}=\frac{1}{2}\frac{p_1^{a}p_1^{b}}{1+\boldsymbol{\Omega}\cdot\mathbf{p}_1}e_{ab}^{\times}=U\sin2\lambda, \tag*{(94)} F 1 + = 2 1 1 + Ω ⋅ p 1 p 1 a p 1 b e ab + = U cos 2 λ , F 1 × = 2 1 1 + Ω ⋅ p 1 p 1 a p 1 b e ab × = U sin 2 λ , (94) where the function U = ( 1 − cos θ ) / 2 U=(1-\cos\theta)/2 U = ( 1 − cos θ ) /2 (91) , and obtain
F 2 + = V cos 2 λ + W sin 2 λ , and F_2^{+}=V\cos2\lambda+W\sin2\lambda,\qquad\text{and} F 2 + = V cos 2 λ + W sin 2 λ , and F 2 × = V sin 2 λ − W cos 2 λ (95) F_2^{\times}=V\sin2\lambda-W\cos2\lambda \tag*{(95)} F 2 × = V sin 2 λ − W cos 2 λ (95) where to simplify notation we have introduced three functions that are independent of λ \lambda λ
U = 1 − cos θ 2 , U=\frac{1-\cos\theta}{2}, U = 2 1 − cos θ , V = 1 2 A 2 − B 2 1 + q , V=\frac{1}{2}\frac{A^2-B^2}{1+q}, V = 2 1 1 + q A 2 − B 2 , W = 1 2 2 A B 1 + q . (96) W=\frac{1}{2}\frac{2AB}{1+q}. \tag*{(96)} W = 2 1 1 + q 2 A B . (96) Here, we have introduced the three functions
A = sin θ cos γ − cos θ sin γ cos ϕ , A=\sin\theta\cos\gamma-\cos\theta\sin\gamma\cos\phi, A = sin θ cos γ − cos θ sin γ cos ϕ , B = sin γ sin ϕ , and B=\sin\gamma\sin\phi,\qquad\text{and} B = sin γ sin ϕ , and q = cos θ cos γ + sin θ sin γ cos ϕ . (97) q=\cos\theta\cos\gamma+\sin\theta\sin\gamma\cos\phi. \tag*{(97)} q = cos θ cos γ + sin θ sin γ cos ϕ . (97) One can use Eq. (97) to verify that
A 2 − B 2 = ( 1 − q ) ( 1 + q ) − 2 sin 2 γ sin 2 ϕ . (98) A^2-B^2=(1-q)(1+q)-2\sin^2\gamma\sin^2\phi. \tag*{(98)} A 2 − B 2 = ( 1 − q ) ( 1 + q ) − 2 sin 2 γ sin 2 ϕ . (98) We are now ready to average over pulsar positions. We first compute the average correlation ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩ ρ \rho ρ (85) , we insert the antenna pattern functions from Eqs. (94) and (95) and integrate over λ \lambda λ
ρ ‾ = 1 2 ( c 1 + c 2 ) U V − 1 2 ( c 3 − c 4 ) U W . (99) \overline{\rho}=\frac{1}{2}(c_1+c_2)UV-\frac{1}{2}(c_3-c_4)UW. \tag*{(99)} ρ = 2 1 ( c 1 + c 2 ) U V − 2 1 ( c 3 − c 4 ) U W . (99) If we now do the remaining average, we obtain
⟨ ρ ‾ ⟩ = 1 2 ( c 1 + c 2 ) ⟨ U V ⟩ , (100) \langle\overline{\rho}\rangle=\frac{1}{2}(c_1+c_2)\langle UV\rangle, \tag*{(100)} ⟨ ρ ⟩ = 2 1 ( c 1 + c 2 ) ⟨ U V ⟩ , (100) because ⟨ U W ⟩ = 0 \langle UW\rangle=0 ⟨ U W ⟩ = 0 ϕ \phi ϕ U W UW U W
q 1 sin ϕ ( 1 + q 2 cos ϕ ) 1 + q 3 cos ϕ , (101) \frac{q_1\sin\phi(1+q_2\cos\phi)}{1+q_3\cos\phi}, \tag*{(101)} 1 + q 3 cos ϕ q 1 sin ϕ ( 1 + q 2 cos ϕ ) , (101) where the q i q_i q i ϕ \phi ϕ U W UW U W π \pi π ϕ → 2 π − ϕ \phi\to2\pi-\phi ϕ → 2 π − ϕ cos ϕ \cos\phi cos ϕ sin ϕ \sin\phi sin ϕ ϕ ∈ [ 0 , 2 π ] \phi\in[0,2\pi] ϕ ∈ [ 0 , 2 π ]
The careful reader will have noticed that the coefficients c 1 , … , c 4 c_1,\ldots,c_4 c 1 , … , c 4 c i c_i c i § 2.2 , we will show how to include these pulsar terms if desired: one simply replaces these constants by their mean values obtained by averaging over the pulsar distances, which eliminates their dependence on the pulsar directions.
It is easy to see that ⟨ U V ⟩ \langle UV\rangle ⟨ U V ⟩ μ u ( γ ) \mu_u(\gamma) μ u ( γ ) (93) to define the average, replace U U U V V V (96) , and use Eq. (98) to simplify the integrand. One immediately obtains the form Eq. (188) computed in Appendix § 13 , where the pulsars are fixed and the average is over wave directions. Thus,
⟨ ρ ‾ ⟩ = 1 2 ( c 1 + c 2 ) μ u ( γ ) . (102) \langle\overline{\rho}\rangle=\frac{1}{2}(c_1+c_2)\mu_u(\gamma). \tag*{(102)} ⟨ ρ ⟩ = 2 1 ( c 1 + c 2 ) μ u ( γ ) . (102) We now compute the second moment of the correlation ρ \rho ρ (85) , replace the antenna pattern functions with Eqs. (94) and (95) , and combine the trig functions to obtain
ρ = 1 2 [ ( c 1 + c 2 ) U V + ( c 4 − c 3 ) U W ] + 1 2 [ ( c 1 − c 2 ) U V − ( c 3 + c 4 ) U W ] cos 4 λ + 1 2 [ ( c 1 − c 2 ) U W + ( c 3 + c 4 ) U V ] sin 4 λ . (103) \begin{aligned}
\rho={}&\frac{1}{2}\left[(c_1+c_2)UV+(c_4-c_3)UW\right]\\
&+\frac{1}{2}\left[(c_1-c_2)UV-(c_3+c_4)UW\right]\cos4\lambda\\
&+\frac{1}{2}\left[(c_1-c_2)UW+(c_3+c_4)UV\right]\sin4\lambda. \tag*{(103)}
\end{aligned} ρ = 2 1 [ ( c 1 + c 2 ) U V + ( c 4 − c 3 ) U W ] + 2 1 [ ( c 1 − c 2 ) U V − ( c 3 + c 4 ) U W ] cos 4 λ + 2 1 [ ( c 1 − c 2 ) U W + ( c 3 + c 4 ) U V ] sin 4 λ . (103) (Note that the average of this over λ \lambda λ (99) . ) Square Eq. (103) and average over λ \lambda λ
⟨ ρ 2 ⟩ = U 2 4 [ ( c 1 + c 2 ) V + ( c 4 − c 3 ) W ] 2 + U 2 8 [ ( c 1 − c 2 ) V − ( c 3 + c 4 ) W ] 2 + U 2 8 [ ( c 1 − c 2 ) W + ( c 3 + c 4 ) V ] 2 . (104) \begin{aligned}
\langle\rho^2\rangle={}&\frac{U^2}{4}\left[(c_1+c_2)V+(c_4-c_3)W\right]^2\\
&+\frac{U^2}{8}\left[(c_1-c_2)V-(c_3+c_4)W\right]^2\\
&+\frac{U^2}{8}\left[(c_1-c_2)W+(c_3+c_4)V\right]^2. \tag*{(104)}
\end{aligned} ⟨ ρ 2 ⟩ = 4 U 2 [ ( c 1 + c 2 ) V + ( c 4 − c 3 ) W ] 2 + 8 U 2 [ ( c 1 − c 2 ) V − ( c 3 + c 4 ) W ] 2 + 8 U 2 [ ( c 1 − c 2 ) W + ( c 3 + c 4 ) V ] 2 . (104) Now average this over θ \theta θ ϕ \phi ϕ (93) . The average ⟨ U 2 V W ⟩ \langle U^2VW\rangle ⟨ U 2 V W ⟩
⟨ ρ 2 ⟩ = 1 8 [ 2 ( c 1 + c 2 ) 2 + ( c 1 − c 2 ) 2 + ( c 3 + c 4 ) 2 ] ⟨ U 2 V 2 ⟩ + 1 8 [ 2 ( c 3 − c 4 ) 2 + ( c 1 − c 2 ) 2 + ( c 3 + c 4 ) 2 ] ⟨ U 2 W 2 ⟩ . (105) \begin{aligned}
\langle\rho^2\rangle={}&\frac{1}{8}\left[2(c_1+c_2)^2+(c_1-c_2)^2+(c_3+c_4)^2\right]\langle U^2V^2\rangle\\
&+\frac{1}{8}\left[2(c_3-c_4)^2+(c_1-c_2)^2+(c_3+c_4)^2\right]\langle U^2W^2\rangle. \tag*{(105)}
\end{aligned} ⟨ ρ 2 ⟩ = 8 1 [ 2 ( c 1 + c 2 ) 2 + ( c 1 − c 2 ) 2 + ( c 3 + c 4 ) 2 ] ⟨ U 2 V 2 ⟩ + 8 1 [ 2 ( c 3 − c 4 ) 2 + ( c 1 − c 2 ) 2 + ( c 3 + c 4 ) 2 ] ⟨ U 2 W 2 ⟩ . (105) In the same way we demonstrated that ⟨ U V ⟩ \langle UV\rangle ⟨ U V ⟩ (188) , one can show that ⟨ U 2 V 2 ⟩ \langle U^2V^2\rangle ⟨ U 2 V 2 ⟩ (205) and (210) , that ⟨ U 2 W 2 ⟩ \langle U^2W^2\rangle ⟨ U 2 W 2 ⟩ (219) , and that ⟨ U 2 ( V 2 + W 2 ) ⟩ \langle U^2(V^2+W^2)\rangle ⟨ U 2 ( V 2 + W 2 )⟩ (221) . This means that the averages above can be replaced by quantities that we calculate in the Appendix:
⟨ U V ⟩ = μ u ( γ ) , \langle UV\rangle= \mu_u(\gamma), ⟨ U V ⟩ = μ u ( γ ) , ⟨ U 2 V 2 ⟩ = μ u 2 ( γ ) + σ u 2 ( γ ) , \langle U^2V^2\rangle= \mu_u^2(\gamma) + \sigma_u^2(\gamma), ⟨ U 2 V 2 ⟩ = μ u 2 ( γ ) + σ u 2 ( γ ) , ⟨ U 2 W 2 ⟩ = σ p 2 ( γ ) , \langle U^2W^2\rangle= \sigma_p^2(\gamma), ⟨ U 2 W 2 ⟩ = σ p 2 ( γ ) , and
⟨ U 2 ( V 2 + W 2 ) ⟩ = σ c 2 ( γ ) = μ u 2 ( γ ) + σ u 2 ( γ ) + σ p 2 ( γ ) . (106) \langle U^2(V^2 + W^2)\rangle= \sigma_c^2(\gamma) = \mu_u^2(\gamma) + \sigma_u^2(\gamma) + \sigma_p^2(\gamma). \tag*{(106)} ⟨ U 2 ( V 2 + W 2 )⟩ = σ c 2 ( γ ) = μ u 2 ( γ ) + σ u 2 ( γ ) + σ p 2 ( γ ) . (106) Because the final line above is the sum of the two previous lines, there are many equivalent ways to write the expressions for the second moment and variance of the correlation ρ \rho ρ (105) and (106) , the second moment may be written
⟨ ρ 2 ⟩ = 1 4 ( c 1 + c 2 ) 2 ( μ u 2 + σ u 2 ) + 1 4 ( c 3 − c 4 ) 2 σ p 2 + 1 8 [ ( c 1 − c 2 ) 2 + ( c 3 + c 4 ) 2 ] σ c 2 . (107) \langle\rho^2\rangle= \frac{1}{4}(c_1+c_2)^2(\mu_u^2+\sigma_u^2) + \frac{1}{4}(c_3-c_4)^2\sigma_p^2 + \frac{1}{8}\left[(c_1-c_2)^2+(c_3+c_4)^2\right]\sigma_c^2. \tag*{(107)} ⟨ ρ 2 ⟩ = 4 1 ( c 1 + c 2 ) 2 ( μ u 2 + σ u 2 ) + 4 1 ( c 3 − c 4 ) 2 σ p 2 + 8 1 [ ( c 1 − c 2 ) 2 + ( c 3 + c 4 ) 2 ] σ c 2 . (107) The variance is calculated by subtracting from the second moment Eq. (107) the square of the mean given in Eq. (102) . We find
μ = 1 2 ( c 1 + c 2 ) μ u ( γ ) , \mu= \frac{1}{2}(c_1+c_2)\mu_u(\gamma), μ = 2 1 ( c 1 + c 2 ) μ u ( γ ) , and
σ 2 = 1 4 ( c 1 + c 2 ) 2 σ u 2 ( γ ) + 1 4 ( c 3 − c 4 ) 2 σ p 2 ( γ ) + 1 8 [ ( c 1 − c 2 ) 2 + ( c 3 + c 4 ) 2 ] σ c 2 ( γ ) , (108) \sigma^2 = \frac{1}{4}(c_1+c_2)^2\sigma_u^2(\gamma) + \frac{1}{4}(c_3-c_4)^2\sigma_p^2(\gamma) + \frac{1}{8}\left[(c_1-c_2)^2+(c_3+c_4)^2\right]\sigma_c^2(\gamma), \tag*{(108)} σ 2 = 4 1 ( c 1 + c 2 ) 2 σ u 2 ( γ ) + 4 1 ( c 3 − c 4 ) 2 σ p 2 ( γ ) + 8 1 [ ( c 1 − c 2 ) 2 + ( c 3 + c 4 ) 2 ] σ c 2 ( γ ) , (108) which are our final expressions for the mean and variance of the Hellings-Downs correlation in the general case. The mean μ u ( γ ) \mu_u(\gamma) μ u ( γ ) (203) , the unpolarized variance σ u 2 \sigma_u^2 σ u 2 (211) , the polarized variance σ p 2 \sigma_p^2 σ p 2 (219) , and the cross variance σ c 2 \sigma_c^2 σ c 2 (221) .
The sums and differences of the c i c_i c i (86) , play the role of Stokes parameters in this formula. If the pulsar terms can be neglected, then c 3 = c 4 c_3=c_4 c 3 = c 4 σ p 2 \sigma_p^2 σ p 2 c 1 = c 2 c_1=c_2 c 1 = c 2 σ u 2 \sigma_u^2 σ u 2
Returning now to the binary inspiral example from Eq. (87) , where pulsar terms are neglected, and inserting the Stokes coefficients from Eq. (89) into Eq. (108) , the mean and variance of the Hellings-Downs correlation for a single source are
μ = ⟨ ρ ⟩ = A 2 16 [ 1 + 6 cos 2 ι + cos 4 ι ] μ u ( γ ) , \mu= \langle\rho\rangle= \frac{\mathcal{A}^2}{16}\left[1+6\cos^2\iota+\cos^4\iota\right]\mu_u(\gamma), μ = ⟨ ρ ⟩ = 16 A 2 [ 1 + 6 cos 2 ι + cos 4 ι ] μ u ( γ ) , and
σ 2 = ⟨ Δ ρ 2 ⟩ = A 4 256 [ 1 + 6 cos 2 ι + cos 4 ι ] 2 σ u 2 ( γ ) + A 4 512 [ sin 8 ι ] σ c 2 ( γ ) . (109) \sigma^2 = \langle\Delta\rho^2\rangle= \frac{\mathcal{A}^4}{256}\left[1+6\cos^2\iota+\cos^4\iota\right]^2\sigma_u^2(\gamma) + \frac{\mathcal{A}^4}{512}\left[\sin^8\iota\right]\sigma_c^2(\gamma). \tag*{(109)} σ 2 = ⟨ Δ ρ 2 ⟩ = 256 A 4 [ 1 + 6 cos 2 ι + cos 4 ι ] 2 σ u 2 ( γ ) + 512 A 4 [ sin 8 ι ] σ c 2 ( γ ) . (109) If ι \iota ι 0 ∘ 0^\circ 0 ∘ 180 ∘ 180^\circ 18 0 ∘ c 1 = c 2 c_1=c_2 c 1 = c 2
To estimate these for a typical circular binary, we average the mean and variance from Eq. (109) over orbital inclinations, assuming that cos ι \cos\iota cos ι [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
μ a v e r a g e = 1 5 A 2 μ u ( γ ) , \mu_{\mathrm{average}} = \frac{1}{5}\mathcal{A}^2\mu_u(\gamma), μ average = 5 1 A 2 μ u ( γ ) , and
σ a v e r a g e 2 = 71 1260 A 4 σ u 2 ( γ ) + 1 1260 A 4 σ c 2 ( γ ) . (110) \sigma_{\mathrm{average}}^2 = \frac{71}{1260}\mathcal{A}^4\sigma_u^2(\gamma) + \frac{1}{1260}\mathcal{A}^4\sigma_c^2(\gamma). \tag*{(110)} σ average 2 = 1260 71 A 4 σ u 2 ( γ ) + 1260 1 A 4 σ c 2 ( γ ) . (110) These are plotted in Figure 6 . In this population average, the sin 8 ι \sin^8\iota sin 8 ι σ c 2 \sigma_c^2 σ c 2 σ c 2 \sigma_c^2 σ c 2 ι = 90 ∘ \iota=90^\circ ι = 9 0 ∘ σ a v e r a g e 2 \sigma_{\mathrm{average}}^2 σ average 2 [43 ] ).
Figure 6. Mean μ a v e r a g e = ⟨ ρ ⟩ \mu_{\mathrm{average}}=\langle\rho\rangle μ average = ⟨ ρ ⟩ σ a v e r a g e = ⟨ Δ ρ 2 ⟩ \sigma_{\mathrm{average}}=\sqrt{\langle\Delta\rho^2\rangle} σ average = ⟨ Δ ρ 2 ⟩ ρ \rho ρ (110) with the amplitude set to A 2 = 15 \mathcal{A}^2=15 A 2 = 15 σ a v e r a g e \sigma_{\mathrm{average}} σ average σ c 2 \sigma_c^2 σ c 2 (110) is set to zero; there has no significant effect on the average. The case where the polarization term has the largest effect (an edge-on binary, ι = 90 ∘ \iota=90^\circ ι = 9 0 ∘ (109) . In this case the amplitude is set to A 2 = 48 \mathcal{A}^2=48 A 2 = 48 μ \mu μ
In the next Appendix, we will consider an example where the pulsar terms have the same magnitude as the Earth terms, so that the polarization term proportional to ( c 3 − c 4 ) 2 (c_3-c_4)^2 ( c 3 − c 4 ) 2
The Hellings-Downs Correlation Mean and Variance Including Pulsar Terms
Here, we calculate the mean and variance of the Hellings-Downs correlation for GWs emitted by a slowly evolving binary system, including the pulsar terms . We will see that the mean is unchanged from the case where only the Earth terms are included, but that the variance gets larger by about a factor of four and has a slightly different functional form.
The GW source is the same binary system as in the previous Appendix [defined around Eq. (A9)] but now we will assume that it is a very long time before the system will merge. This means that the timescale on which the waveform is changing in frequency and amplitude is much longer than the light travel time between the pulsars and Earth. This system violates the original Hellings and Downs assumptions, because (depending upon the exact separations) there can be perfect correlation or anticorrelation (or anything in between) between the Earth and pulsar terms. In this model, the GW waveforms at the Earth and pulsars are identical in frequency and amplitude, and only differ in phase.
The time averages Eq. (86) that define the coefficients c 1 , … , c 4 c_1,\ldots,c_4 c 1 , … , c 4
c 1 = 2 sin Δ 1 sin Δ 2 cos ( Δ 2 − Δ 1 ) A c 2 , (111) c_1 = 2\sin\Delta_1\sin\Delta_2\cos(\Delta_2-\Delta_1)A_c^2, \tag*{(111)} c 1 = 2 sin Δ 1 sin Δ 2 cos ( Δ 2 − Δ 1 ) A c 2 , (111) c 2 = 2 sin Δ 1 sin Δ 2 cos ( Δ 2 − Δ 1 ) A s 2 , c_2 = 2\sin\Delta_1\sin\Delta_2\cos(\Delta_2-\Delta_1)A_s^2, c 2 = 2 sin Δ 1 sin Δ 2 cos ( Δ 2 − Δ 1 ) A s 2 , c 3 = 2 sin Δ 1 sin Δ 2 sin ( Δ 2 − Δ 1 ) A c A s , c_3 = 2\sin\Delta_1\sin\Delta_2\sin(\Delta_2-\Delta_1)A_cA_s, c 3 = 2 sin Δ 1 sin Δ 2 sin ( Δ 2 − Δ 1 ) A c A s , c 4 = − c 3 . c_4 = -c_3. c 4 = − c 3 . Here, the phase offsets are determined by the pulsar positions relative to the GW source. If L 1 L_1 L 1 L 2 L_2 L 2
Δ 1 = ω L 1 ( 1 + z ^ ⋅ p 1 ) , \Delta_1 = \omega L_1(1+\hat{z}\cdot\mathbf{p}_1), Δ 1 = ω L 1 ( 1 + z ^ ⋅ p 1 ) , and
Δ 2 = ω L 2 ( 1 + z ^ ⋅ p 2 ) . (112) \Delta_2 = \omega L_2(1+\hat{z}\cdot\mathbf{p}_2). \tag*{(112)} Δ 2 = ω L 2 ( 1 + z ^ ⋅ p 2 ) . (112) The amplitudes of the two polarization components are
A c = A 1 2 ( 1 + cos 2 ι ) , A_c = \mathcal{A}\frac{1}{2}(1+\cos^2\iota), A c = A 2 1 ( 1 + cos 2 ι ) , and
A s = A cos ι . (113) A_s = \mathcal{A}\cos\iota. \tag*{(113)} A s = A cos ι . (113) Now that we are including the pulsar terms, which have the same amplitude and frequency as the Earth terms (but a different phase) one can see that c 3 c_3 c 3 c 4 c_4 c 4
One can work through the calculation of the previous Appendix, and everything carries through up to the point where one averages over pulsar positions on the sky. Then the calculation breaks down, because the c i c_i c i Δ 1 \Delta_1 Δ 1 Δ 2 \Delta_2 Δ 2 c i c_i c i
However, it is easy to work around this. We are interested in the angular correlation function for pulsars at different distances as well as at different sky locations. So, before averaging over the sky positions of the pulsars, we first average over their distances.
average over their distances. This is easy, because the distances L 1 L_1 L 1 L 2 L_2 L 2 Δ 1 \Delta_1 Δ 1 Δ 2 \Delta_2 Δ 2
We average by letting L 1 L_1 L 1 L 2 L_2 L 2 [ L min , L max ] [L_{\min},L_{\max}] [ L m i n , L m a x ] 0 ≪ ω L min ≪ ω L max 0\ll\omega L_{\min}\ll\omega L_{\max} 0 ≪ ω L m i n ≪ ω L m a x 2 π / ω 2\pi/\omega 2 π / ω Δ 1 − Δ 2 \Delta_1-\Delta_2 Δ 1 − Δ 2 Δ 1 + Δ 2 \Delta_1+\Delta_2 Δ 1 + Δ 2 2 π 2\pi 2 π
This averaging removes the pulsar-position dependence of the c i c_i c i
2 sin Δ 1 sin Δ 2 cos ( Δ 2 − Δ 1 ) = [ cos ( Δ 2 − Δ 1 ) − cos ( Δ 1 + Δ 2 ) ] cos ( Δ 2 − Δ 1 ) , (114) \begin{aligned}
2\sin\Delta_1\sin\Delta_2\cos(\Delta_2-\Delta_1) \\
&= [\cos(\Delta_2-\Delta_1)-\cos(\Delta_1+\Delta_2)]\cos(\Delta_2-\Delta_1), \tag*{(114)}
\end{aligned} 2 sin Δ 1 sin Δ 2 cos ( Δ 2 − Δ 1 ) = [ cos ( Δ 2 − Δ 1 ) − cos ( Δ 1 + Δ 2 )] cos ( Δ 2 − Δ 1 ) , (114) we can immediately see that after averaging over pulsar distances in the expressions for ρ \rho ρ ρ 2 \rho^2 ρ 2
⟨ c 1 + c 2 ⟩ = 1 2 ( A c 2 + A s 2 ) , \langle c_1+c_2\rangle= \frac{1}{2}(A_c^2+A_s^2), ⟨ c 1 + c 2 ⟩ = 2 1 ( A c 2 + A s 2 ) , ⟨ c 1 − c 2 ⟩ = 1 2 ( A c 2 − A s 2 ) , \langle c_1-c_2\rangle= \frac{1}{2}(A_c^2-A_s^2), ⟨ c 1 − c 2 ⟩ = 2 1 ( A c 2 − A s 2 ) , ⟨ ( c 1 + c 2 ) 2 ⟩ = 5 8 ( A c 2 + A s 2 ) 2 , \langle(c_1+c_2)^2\rangle= \frac{5}{8}(A_c^2+A_s^2)^2, ⟨( c 1 + c 2 ) 2 ⟩ = 8 5 ( A c 2 + A s 2 ) 2 , and
⟨ ( c 1 − c 2 ) 2 ⟩ = 5 8 ( A c 2 − A s 2 ) 2 . (115) \langle(c_1-c_2)^2\rangle= \frac{5}{8}(A_c^2-A_s^2)^2. \tag*{(115)} ⟨( c 1 − c 2 ) 2 ⟩ = 8 5 ( A c 2 − A s 2 ) 2 . (115) Here and in the next few equations, the angle brackets mean “average over L 1 L_1 L 1 L 2 L_2 L 2 cos 2 \cos^2 cos 2 1 / 2 1/2 1/2 cos 4 \cos^4 cos 4 3 / 8 3/8 3/8
To do the averaging for c 3 c_3 c 3 c 4 c_4 c 4 c 3 + c 4 = 0 c_3+c_4=0 c 3 + c 4 = 0
⟨ c 3 + c 4 ⟩ = ⟨ ( c 3 + c 4 ) 2 ⟩ = 0. (116) \langle c_3+c_4\rangle= \langle(c_3+c_4)^2\rangle= 0. \tag*{(116)} ⟨ c 3 + c 4 ⟩ = ⟨( c 3 + c 4 ) 2 ⟩ = 0. (116) For the other two terms, use
2 sin Δ 1 sin Δ 2 sin ( Δ 2 − Δ 1 ) = [ cos ( Δ 2 − Δ 1 ) − cos ( Δ 1 + Δ 2 ) ] sin ( Δ 2 − Δ 1 ) (117) \begin{aligned}
2\sin\Delta_1\sin\Delta_2\sin(\Delta_2-\Delta_1) \\
&= [\cos(\Delta_2-\Delta_1)-\cos(\Delta_1+\Delta_2)]\sin(\Delta_2-\Delta_1) \tag*{(117)}
\end{aligned} 2 sin Δ 1 sin Δ 2 sin ( Δ 2 − Δ 1 ) = [ cos ( Δ 2 − Δ 1 ) − cos ( Δ 1 + Δ 2 )] sin ( Δ 2 − Δ 1 ) (117) to obtain
⟨ c 3 − c 4 ⟩ = 0 , \langle c_3-c_4\rangle= 0, ⟨ c 3 − c 4 ⟩ = 0 , and
⟨ ( c 3 − c 4 ) 2 ⟩ = 3 2 A c 2 A s 2 . (118) \langle(c_3-c_4)^2\rangle= \frac{3}{2}A_c^2A_s^2. \tag*{(118)} ⟨( c 3 − c 4 ) 2 ⟩ = 2 3 A c 2 A s 2 . (118) The mean of the Hellings-Downs correlation follows immediately from Eq. (A24). Using ⟨ c 1 + c 2 ⟩ \langle c_{1}+c_{2}\rangle ⟨ c 1 + c 2 ⟩
μ = 1 4 ( A c 2 + A s 2 ) μ u , (119) \mu= \frac{1}{4}\left(A_{c}^{2}+A_{s}^{2}\right)\mu_{u}, \tag*{(119)} μ = 4 1 ( A c 2 + A s 2 ) μ u , (119) which agrees with the result Eq. (A31), where only the Earth terms were included. This makes sense, since after averaging over pulsar distances, the pulsar terms have averaged away. (However, their squares will not average away, so the variance will change.)
To compute the variance, it is tempting to substitute the averages from Eqs. (B5) and (B8) into Eq. (A30). However, those expressions were obtained by subtracting the square of the mean from the second moment, and the pulsar distance averaging has ⟨ ( c 1 + c 2 ) 2 ⟩ ≠ ⟨ ( c 1 + c 2 ) ⟩ 2 \langle(c_{1}+c_{2})^{2}\rangle\ne\langle(c_{1}+c_{2})\rangle^{2} ⟨( c 1 + c 2 ) 2 ⟩ = ⟨( c 1 + c 2 ) ⟩ 2
For this binary-inspiral source model, which includes pulsar terms that have the same magnitude and frequency as the Earth terms, the variance of the Hellings-Downs correlation is
σ 2 ( γ ) = 1 16 ( A c 2 + A s 2 ) 2 σ u 2 ( γ ) − 3 32 ( A c 2 − A s 2 ) 2 σ p 2 ( γ ) + [ 3 32 ( A c 2 + A s 2 ) 2 + 5 64 ( A c 2 − A s 2 ) 2 ] σ c 2 ( γ ) . (120) \begin{aligned}
\sigma^{2}(\gamma)={}&\frac{1}{16}\left(A_{c}^{2}+A_{s}^{2}\right)^{2}\sigma_{u}^{2}(\gamma)-\frac{3}{32}\left(A_{c}^{2}-A_{s}^{2}\right)^{2}\sigma_{p}^{2}(\gamma)\\
&+\left[\frac{3}{32}\left(A_{c}^{2}+A_{s}^{2}\right)^{2}+\frac{5}{64}\left(A_{c}^{2}-A_{s}^{2}\right)^{2}\right]\sigma_{c}^{2}(\gamma). \tag*{(120)}
\end{aligned} σ 2 ( γ ) = 16 1 ( A c 2 + A s 2 ) 2 σ u 2 ( γ ) − 32 3 ( A c 2 − A s 2 ) 2 σ p 2 ( γ ) + [ 32 3 ( A c 2 + A s 2 ) 2 + 64 5 ( A c 2 − A s 2 ) 2 ] σ c 2 ( γ ) . (120) This should be compared with the Earth-term only equivalent Eq. (A31). The presence of pulsar terms (which here have the same magnitude as the Earth term) have left the mean unchanged but have modified the variance.
Averaging Eqs. (B9) and (B10) over the inclination angles ι \iota ι
μ a v e r a g e = 1 5 A 2 μ u ( γ ) , and \mu_{\mathrm{average}}=\frac{1}{5}\mathcal{A}^{2}\mu_{u}(\gamma), \qquad\text{and} μ average = 5 1 A 2 μ u ( γ ) , and σ a v e r a g e 2 = 71 1260 A 4 σ u 2 ( γ ) + 109 1260 A 4 σ c 2 ( γ ) − 1 420 A 4 σ p 2 ( γ ) . (121) \sigma_{\mathrm{average}}^{2}=\frac{71}{1260}\mathcal{A}^{4}\sigma_{u}^{2}(\gamma)+\frac{109}{1260}\mathcal{A}^{4}\sigma_{c}^{2}(\gamma)-\frac{1}{420}\mathcal{A}^{4}\sigma_{p}^{2}(\gamma). \tag*{(121)} σ average 2 = 1260 71 A 4 σ u 2 ( γ ) + 1260 109 A 4 σ c 2 ( γ ) − 420 1 A 4 σ p 2 ( γ ) . (121) These are plotted in Figure 7 and should be compared with Eq. (A32), which only include the Earth term. While the means are identical, the variance is larger by a factor of about four, because of the presence of the pulsar terms. While these terms average to zero (hence not affecting the mean) their squares do not average to zero, so they increase the variance (σ a v e r a g e 2 \sigma_{\mathrm{average}}^{2} σ average 2 [43 ] ).
Figure 7. The effect of the pulsar terms on the mean μ \mu μ σ \sigma σ A 2 = 15 \mathcal{A}^{2}=15 A 2 = 15
The Conventional Ensemble-Average Approach
In this Appendix, we discuss and employ the conventional frequency-domain Gaussian ensemble introduced in [16 , 17 ] and used for example in [21 ] . The Gaussian ensemble is introduced in Appendix C 1. Note that for PTAs these frequency-domain methods may be less appropriate than time-domain ones, see [22 , 44 , 45 , 46 ] .
Appendix B demonstrated that including the pulsar terms in the Hellings-Downs correlation does not affect the mean μ ( γ ) \mu(\gamma) μ ( γ ) σ 2 ( γ ) \sigma^{2}(\gamma) σ 2 ( γ )
In Appendix C 3 we compute the variance of the Hellings-Downs correlation, and show that the Gaussian ensemble results correspond exactly to the discrete confusion-noise source model of Sec. III A, in the limit of large numbers of weak sources. Hence, the Gaussian ensemble corresponds to a confusion-noise limit in which the closest (idealized) sources are arbitrarily close to Earth, and not at a fixed distance as in Sec. III. (Note that our calculations are based on a plane-wave radiation-zone description of GWs [47 , 48 ] . For this to apply, a real GW source would have to be at least a few times as distant as the PTA pulsars, say tens of kpc. This is still about three orders of magnitude closer to Earth than the closest relevant supermassive black-hole binaries [30 , 31 ] .)
In Appendix C 4 we use identical techniques to compute the variance of the time-averaged squared strain s = h ˙ a b h ˙ a b ‾ s=\overline{\dot{h}_{ab}\dot{h}^{ab}} s = h ˙ ab h ˙ ab ρ G W \rho_{\mathrm{GW}} ρ GW h ˙ a b h ˙ a b \dot{h}_{ab}\dot{h}^{ab} h ˙ ab h ˙ ab
In Appendix C 5, we use the pulsar-position-averaging technique introduced in Sec. IV to compute the cosmic variance. As for the mean and total variance, the cosmic variance corresponds exactly to the discrete confusion-noise source model of Sec. III A in the limit of large numbers of sources.
Finally, in Appendix § 8 § 12 , we provide an alternative derivation of the cosmic variance for the Gaussian ensemble, using harmonic analysis methods originally developed for characterizing cosmic background radiation temperature fluctuations.
The Gaussian ensemble
We start by writing the general weak gravitational-wave solution as a plane wave expansion [16 , 17 ] . In this, we include two arbitrary functions associated with any propagation direction as a Fourier integral, obtaining
h a b ( t , x ) = ∑ A ∫ d f ∫ d Ω e 2 π i f ( t − Ω ⋅ x ) h A ( f , Ω ) e a b A ( Ω ) . (122) h_{ab}(t,\mathbf{x}) = \sum_A \int df \int d\boldsymbol{\Omega}\, e^{2\pi i f(t-\boldsymbol{\Omega}\cdot\mathbf{x})} h_A(f,\boldsymbol{\Omega}) e^A_{ab}(\boldsymbol{\Omega}). \tag*{(122)} h ab ( t , x ) = A ∑ ∫ df ∫ d Ω e 2 π i f ( t − Ω ⋅ x ) h A ( f , Ω ) e ab A ( Ω ) . (122) Here, A = + , × A=+,\times A = + , × h A ( f , Ω ) h_A(f,\boldsymbol{\Omega}) h A ( f , Ω ) h A ( f , Ω ) = ( h A ( − f , Ω ) ) ∗ h_A(f,\boldsymbol{\Omega})=(h_A(-f,\boldsymbol{\Omega}))^* h A ( f , Ω ) = ( h A ( − f , Ω ) ) ∗ ∗ * ∗ e a b A ( Ω ) e^A_{ab}(\boldsymbol{\Omega}) e ab A ( Ω ) (180) , integration over frequency is f ∈ ( − ∞ , ∞ ) f\in(-\infty,\infty) f ∈ ( − ∞ , ∞ ) (4) .
Now, we imagine that the GW background arises as a central-limit-theorem process, so the functions h A ( f , Ω ) h_A(f,\boldsymbol{\Omega}) h A ( f , Ω ) E \mathcal{E} E Q Q Q h h h h A ( f , Ω ) h_A(f,\boldsymbol{\Omega}) h A ( f , Ω ) Q [ h ] Q[h] Q [ h ] h h h
⟨ Q ⟩ = 1 ∣ E ∣ ∑ h ∈ E Q [ h ] , (123) \langle Q\rangle= \frac{1}{|\mathcal{E}|}\sum_{h\in\mathcal{E}} Q[h], \tag*{(123)} ⟨ Q ⟩ = ∣ E ∣ 1 h ∈ E ∑ Q [ h ] , (123) where ∣ E ∣ |\mathcal{E}| ∣ E ∣
We make two assumptions about our Gaussian ensemble. First, we assume that the mean of h h h
⟨ h A ( f , Ω ) ⟩ = 0. (124) \langle h_A(f,\boldsymbol{\Omega})\rangle= 0. \tag*{(124)} ⟨ h A ( f , Ω )⟩ = 0. (124) (A nonzero mean would be appropriate to describe fluctuations around a specific GW source.) Second, we assume that
⟨ h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) ⟩ = δ A A ′ δ 2 ( Ω , Ω ′ ) δ ( f − f ′ ) H ( f ) , (125) \langle h_A^*(f,\boldsymbol{\Omega})h_{A'}(f',\boldsymbol{\Omega}')\rangle= \delta_{AA'}\delta^2(\boldsymbol{\Omega},\boldsymbol{\Omega}')\delta(f-f')H(f), \tag*{(125)} ⟨ h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ )⟩ = δ A A ′ δ 2 ( Ω , Ω ′ ) δ ( f − f ′ ) H ( f ) , (125) where H ( f ) H(f) H ( f ) H ( f ) = H ( − f ) H(f)=H(-f) H ( f ) = H ( − f ) δ 2 ( Ω , Ω ′ ) \delta^2(\boldsymbol{\Omega},\boldsymbol{\Omega}') δ 2 ( Ω , Ω ′ )
The delta function in polarization indices means that the statistical properties of the two polarization degrees of freedom are identical but uncorrelated: an unpolarized ensemble. The delta function in frequency implies an ensemble that is second-order stationary in time, and the delta function on the sphere implies an ensemble that is second-order stationary in space. Some further discussion of these points may be found in [17 , 22 ] .
The real function H H H f f f Ω g w ( f ) \Omega_{\rm gw}(f) Ω gw ( f ) [17 ] by
H ( f ) = 3 H 0 2 32 π 3 1 ∣ f ∣ 3 Ω g w ( ∣ f ∣ ) , (126) H(f)=\frac{3H_0^2}{32\pi^3}\frac{1}{|f|^3}\Omega_{\rm gw}(|f|), \tag*{(126)} H ( f ) = 32 π 3 3 H 0 2 ∣ f ∣ 3 1 Ω gw ( ∣ f ∣ ) , (126) where H 0 H_0 H 0 h c ( f ) h_c(f) h c ( f ) [30 , 31 ] . These quantities are related via
h c 2 ( f ) = 3 H 0 2 2 π 2 1 ∣ f ∣ 2 Ω g w ( ∣ f ∣ ) (127) h_c^2(f)=\frac{3H_0^2}{2\pi^2}\frac{1}{|f|^2}\Omega_{\rm gw}(|f|) \tag*{(127)} h c 2 ( f ) = 2 π 2 3 H 0 2 ∣ f ∣ 2 1 Ω gw ( ∣ f ∣ ) (127) or equivalently via
H ( f ) = 1 16 π 1 ∣ f ∣ h c 2 ( ∣ f ∣ ) . (128) H(f)=\frac{1}{16\pi}\frac{1}{|f|}h_c^2(|f|). \tag*{(128)} H ( f ) = 16 π 1 ∣ f ∣ 1 h c 2 ( ∣ f ∣ ) . (128) An estimate of the characteristic strain h c ( f ) h_c(f) h c ( f ) Figure 8 . This is reproduced from a semianalytic estimate of the GW stochastic background [[50], red dashed curve of Fig. 3]. This estimate includes both the effects of attenuation from environmental hardening effects as well as the emission of
Figure 8. Semi-analytic prediction for the spectrum of gravitational waves in the band relevant for PTAs, taken from [50 ] , red dashed curve of Figure 3 . The dashed lines show how we cut off the spectrum to compute the autocorrelation function.
GW energy in n > 2 n > 2 n > 2
Neglecting pulsar terms and the GW autocorrelation function
Now we examine the Hellings-Downs correlation for a Gaussian ensemble. The pulse redshift at Earth time t t t (80) via Eq. (122) and making use of Eq. (81) , giving
Z ( t ) = ∑ A ∫ d f ∫ d Ω F A ( Ω ) h A ( f , Ω ) e 2 π i f t [ 1 − e − 2 π i f L ( 1 + Ω ⋅ p ) ] , (129) Z(t) = \sum_A \int df \int d\Omega F^A(\boldsymbol{\Omega})h_A(f,\boldsymbol{\Omega}) e^{2\pi i f t}\left[1-e^{-2\pi i f L(1+\boldsymbol{\Omega}\cdot\boldsymbol{p})}\right], \tag*{(129)} Z ( t ) = A ∑ ∫ df ∫ d Ω F A ( Ω ) h A ( f , Ω ) e 2 π i f t [ 1 − e − 2 π i f L ( 1 + Ω ⋅ p ) ] , (129) where the antenna pattern functions F A ( Ω ) F^A(\boldsymbol{\Omega}) F A ( Ω ) A = + , × A=+,\times A = + , × (84) , with the polarization tensors and Ω \boldsymbol{\Omega} Ω (180) and (181) . Here, the pulsar is at distance L L L L p L\boldsymbol{p} L p F F F p \boldsymbol{p} p
The corresponding formula for the timing residual is obtained by integrating once with respect to time, giving
Δ τ ( t ) = ∑ A ∫ d f ∫ d Ω F A ( Ω ) h A ( f , Ω ) 2 π i f e 2 π i f t [ 1 − e − 2 π i f L ( 1 + Ω ⋅ p ) ] . (130) \Delta\tau(t) = \sum_A \int df \int d\Omega F^A(\boldsymbol{\Omega})\frac{h_A(f,\boldsymbol{\Omega})}{2\pi i f} e^{2\pi i f t}\left[1-e^{-2\pi i f L(1+\boldsymbol{\Omega}\cdot\boldsymbol{p})}\right]. \tag*{(130)} Δ τ ( t ) = A ∑ ∫ df ∫ d Ω F A ( Ω ) 2 π i f h A ( f , Ω ) e 2 π i f t [ 1 − e − 2 π i f L ( 1 + Ω ⋅ p ) ] . (130) Both of these formulas give us exact values for any representative h A ( f , Ω ) h_A(f,\boldsymbol{\Omega}) h A ( f , Ω )
The expected correlation between the arrival time residuals for pulsars 1 and 2 is the expectation value of the quantity
ρ = Δ τ 1 ( t ) Δ τ 2 ( t ) . (131) \rho= \Delta\tau_1(t)\Delta\tau_2(t). \tag*{(131)} ρ = Δ τ 1 ( t ) Δ τ 2 ( t ) . (131) We can evaluate ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩ (130) into Eq. (131) and using Eq. (125) to evaluate the expectation value, allowing us to carry out two of the integrals. We arrive at
⟨ ρ ⟩ = ∫ d f H ( ∣ f ∣ ) 4 π 2 f 2 ∫ d Ω [ 1 − e − 2 π i f L 1 ( 1 + Ω ⋅ p 1 ) ] [ 1 − e 2 π i f L 2 ( 1 + Ω ⋅ p 2 ) ] ∑ A F 1 A ( Ω ) F 2 A ( Ω ) . (132) \langle\rho\rangle= \int df \frac{H(|f|)}{4\pi^2 f^2}\int d\Omega\left[1-e^{-2\pi i f L_1(1+\boldsymbol{\Omega}\cdot\boldsymbol{p}_1)}\right]\left[1-e^{2\pi i f L_2(1+\boldsymbol{\Omega}\cdot\boldsymbol{p}_2)}\right]\sum_A F_1^A(\boldsymbol{\Omega})F_2^A(\boldsymbol{\Omega}). \tag*{(132)} ⟨ ρ ⟩ = ∫ df 4 π 2 f 2 H ( ∣ f ∣ ) ∫ d Ω [ 1 − e − 2 π i f L 1 ( 1 + Ω ⋅ p 1 ) ] [ 1 − e 2 π i f L 2 ( 1 + Ω ⋅ p 2 ) ] A ∑ F 1 A ( Ω ) F 2 A ( Ω ) . (132) A similar formula can be written for the redshift correlation, which removes the 4 π 2 f 2 4\pi^2 f^2 4 π 2 f 2 ρ \rho ρ [36 , 51 , 52 ] . It is also needed to compute the second moment of ρ \rho ρ
Note that the same formula is obtained if we define (as we have done earlier in this paper) the correlation as the time average of Eq. (131) over the interval [ − T / 2 , T / 2 ] [-T/2,T/2] [ − T /2 , T /2 ] sinc ( π ( f − f ′ ) T ) \operatorname{sinc}(\pi(f-f')T) sinc ( π ( f − f ′ ) T ) sinc ( x ) = sin ( x ) / x \operatorname{sinc}(x)=\sin(x)/x sinc ( x ) = sin ( x ) / x [53 ] . However, this disappears after taking the expectation value. [Many authors define the correlation via a time integral rather than a time average, so a factor of the integration time T T T (132) .]
It is instructive to compare Eq. (132) with the mean of the Hellings-Downs correlation calculated in Appendix § 13 and given by Eq. (177) . This is obtained from Eq. (132) if the product of the two quantities in square brackets is set to unity. This corresponds to discarding the non-Earth terms.
We now return to the question of whether this is justified. The product of the quantities in square brackets on the final line of Eq. (132) yields four terms. The product of “1” with “1” is the “Earth” term, the products of “1” with the exponentials are the Earth-pulsar interference terms, and the product of the two exponentials is the pulsar-pulsar interference term. If one averages over pulsars at different distances, then it is clear that all but the Earth term vanish. However, for a given set of pulsars in a particular experiment, such averaging is not justified: the pulsars have some particular positions and these do not change.
To understand when it might make sense to neglect the final three terms, it is helpful to introduce the autocorrelation function
C ( T ) = 1 4 π 2 ∫ d f e 2 π i f T H ( ∣ f ∣ ) f 2 = 3 H 0 2 128 π 5 ∫ d f e 2 π i f T Ω g w ( ∣ f ∣ ) f 5 = 1 64 π 3 ∫ d f e 2 π i f T h c 2 ( ∣ f ∣ ) f 3 , (133) \begin{aligned}
C(T) = \frac{1}{4\pi^2}\int df\,e^{2\pi i f T}\frac{H(|f|)}{f^2} \\
&= \frac{3H_0^2}{128\pi^5}\int df\,e^{2\pi i f T}\frac{\Omega_{\mathrm{gw}}(|f|)}{f^5} \\
&= \frac{1}{64\pi^3}\int df\,e^{2\pi i f T}\frac{h_c^2(|f|)}{f^3}, \tag*{(133)}
\end{aligned} C ( T ) = 4 π 2 1 ∫ df e 2 π i f T f 2 H ( ∣ f ∣ ) = 128 π 5 3 H 0 2 ∫ df e 2 π i f T f 5 Ω gw ( ∣ f ∣ ) = 64 π 3 1 ∫ df e 2 π i f T f 3 h c 2 ( ∣ f ∣ ) , (133) which we have written in three equivalent forms. The autocorrelation function is well known in digital and analog signal processing, where the Wiener-Khinchin theorem states that the Fourier transform of a signal’s power spectrum is its autocorrelation function, and vice versa.
The autocorrelation function C ( T ) C(T) C ( T ) C ( T ) C(T) C ( T ) C ( T ) = C ( − T ) C(T)=C(-T) C ( T ) = C ( − T ) H ( ∣ f ∣ ) / f 2 H(|f|)/f^2 H ( ∣ f ∣ ) / f 2 C ( T ) C(T) C ( T ) T = 0 T=0 T = 0 C ( T ) C(T) C ( T ) T T T T T T C ( T ) / C ( 0 ) C(T)/C(0) C ( T ) / C ( 0 ) 1 / 2 1/2 1/2 1 / e 1/e 1/ e Figure 9 shows the autocorrelation C ( T ) C(T) C ( T ) Figure 8 .
Figure 9. Normalized autocorrelation function C ( T ) \mathcal{C}(T) C ( T ) T T T
The useful information in the autocorrelation function is the value at the origin C ( 0 ) C(0) C ( 0 ) C ( 0 ) C(0) C ( 0 ) C ( T ) = C ( T ) / C ( 0 ) \mathcal{C}(T)=C(T)/C(0) C ( T ) = C ( T ) / C ( 0 )
Using the autocorrelation function, we can carry out the integration over frequency in Eq. (132) , and the timing residual correlations may be written
⟨ ρ ⟩ = C ( 0 ) ∫ d Ω ∑ A F 1 A ( Ω ) F 2 A ( Ω ) [ 1 + C ( L 1 ( 1 + Ω ⋅ p 1 ) ) + C ( L 2 ( 1 + Ω ⋅ p 2 ) ) + C ( L 1 ( 1 + Ω ⋅ p 1 ) − L 2 ( 1 + Ω ⋅ p 2 ) ) ] . (134) \langle\rho\rangle=C(0)\int d\boldsymbol{\Omega}\sum_A F_1^A(\boldsymbol{\Omega})F_2^A(\boldsymbol{\Omega})\left[1+\mathcal{C}\bigl(L_1(1+\boldsymbol{\Omega}\cdot\mathbf{p}_1)\bigr)+\mathcal{C}\bigl(L_2(1+\boldsymbol{\Omega}\cdot\mathbf{p}_2)\bigr)+\mathcal{C}\bigl(L_1(1+\boldsymbol{\Omega}\cdot\mathbf{p}_1)-L_2(1+\boldsymbol{\Omega}\cdot\mathbf{p}_2)\bigr)\right]. \tag*{(134)} ⟨ ρ ⟩ = C ( 0 ) ∫ d Ω A ∑ F 1 A ( Ω ) F 2 A ( Ω ) [ 1 + C ( L 1 ( 1 + Ω ⋅ p 1 ) ) + C ( L 2 ( 1 + Ω ⋅ p 2 ) ) + C ( L 1 ( 1 + Ω ⋅ p 1 ) − L 2 ( 1 + Ω ⋅ p 2 ) ) ] . (134) The three terms where the autocorrelation function appears are the two pulsar-Earth terms, and the pulsar-pulsar term.
If pulsars 1 and 2 are different, then we are justified in neglecting the pulsar terms in Eq. (134) if C ( T ) \mathcal{C}(T) C ( T ) T T T Figure 9 , this is justified if the typical pulsar distances are thousands of years. In this case, the final three terms in Eq. (134) will be small compared to unity, and it is valid to drop them. Only the Earth term in Eq. (134) survives, and the timing residual correlations become
⟨ ρ ⟩ = C ( 0 ) ∫ d Ω ∑ A F 1 A ( Ω ) F 2 A ( Ω ) = 4 π C ( 0 ) μ u ( γ ) , (135) \begin{aligned}
\langle\rho\rangle&=C(0)\int d\boldsymbol{\Omega}\sum_A F_1^A(\boldsymbol{\Omega})F_2^A(\boldsymbol{\Omega})\\
&=4\pi C(0)\mu_u(\gamma), \tag*{(135)}
\end{aligned} ⟨ ρ ⟩ = C ( 0 ) ∫ d Ω A ∑ F 1 A ( Ω ) F 2 A ( Ω ) = 4 π C ( 0 ) μ u ( γ ) , (135) where the spherical average Eq. (4) of the antenna pattern functions Eq. (3) is given by Eq. (5) as μ u ( γ ) \mu_u(\gamma) μ u ( γ ) C ( 0 ) C(0) C ( 0 ) (133) . This is the Gaussian ensemble equivalent of Eq. (43) in the discrete source calculation of Sec. III.
An interesting special case of Eq. (134) is when pulsars 1 and 2 are the same (identical). Since they have the same direction, p 1 = p 2 \mathbf{p}_1=\mathbf{p}_2 p 1 = p 2 γ = 0 \gamma=0 γ = 0 L 1 = L 2 L_1=L_2 L 1 = L 2 (134) becomes C ( 0 ) = 1 C(0)=1 C ( 0 ) = 1 α \alpha α β \beta β
ρ α β = ⟨ ρ ⟩ = 4 π C ( 0 ) ( μ u ( γ α β ) + δ α β μ u ( 0 ) ) , (136) \rho_{\alpha\beta}=\langle\rho\rangle=4\pi C(0)\left(\mu_u(\gamma_{\alpha\beta})+\delta_{\alpha\beta}\mu_u(0)\right), \tag*{(136)} ρ α β = ⟨ ρ ⟩ = 4 π C ( 0 ) ( μ u ( γ α β ) + δ α β μ u ( 0 ) ) , (136) where γ α β \gamma_{\alpha\beta} γ α β ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩ (135) in the case where α ≠ β \alpha\ne\beta α = β α = β \alpha=\beta α = β
Total variance for the Gaussian ensemble
Here, we derive the total variance of the Hellings-Downs correlation for a Gaussian ensemble. A similar result is also given in the text before Eq. (9) of [21 ] , and comes from a calculation corresponding to the last line of Eq. (5.4) in [17 ] . The result corresponds to a limit of the confusion-noise model for discrete point sources Eq. (55) found in Sec. III A.
For this derivation, and for the cosmic variance calculation in Sec. C 5, we exploit an important feature of a zero-mean Gaussian process first proved by Isserlis [54 ] : the expectation value of products of more than two h h h h h h h h h (125) , and also implies that the expected values of odd powers of h h h
Begin with Eq. (129) for the arrival time redshift of pulses at Earth at t t t t ∈ [ − T / 2 , T / 2 ] t\in[-T/2,T/2] t ∈ [ − T /2 , T /2 ]
ρ = Z 1 Z 2 ‾ \rho=\overline{Z_1Z_2} ρ = Z 1 Z 2 = ∑ A ∑ A ′ ∫ d f ∫ d f ′ ∫ d Ω ∫ d Ω ′ R 1 A ∗ ( f , Ω ) R 2 A ′ ( f ′ , Ω ′ ) × h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) sinc ( π ( f − f ′ ) T ) , (137) \begin{aligned}
&=\sum_A\sum_{A'}\int df\int df'\int d\boldsymbol{\Omega}\int d\boldsymbol{\Omega}'\,R_1^{A*}(f,\boldsymbol{\Omega})R_2^{A'}(f',\boldsymbol{\Omega}')\\
&\qquad\times h_A^*(f,\boldsymbol{\Omega})h_{A'}(f',\boldsymbol{\Omega}')\operatorname{sinc}\bigl(\pi(f-f')T\bigr), \tag*{(137)}
\end{aligned} = A ∑ A ′ ∑ ∫ df ∫ d f ′ ∫ d Ω ∫ d Ω ′ R 1 A ∗ ( f , Ω ) R 2 A ′ ( f ′ , Ω ′ ) × h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) sinc ( π ( f − f ′ ) T ) , (137) where the subscripts “1” and “2” label the pulsars, and sinc ( x ) = sin ( x ) / x \operatorname{sinc}(x)=\sin(x)/x sinc ( x ) = sin ( x ) / x [53 ] . The pulsar term is incorporated via a modified antenna pattern function
R n A ( f , Ω ) = [ 1 − e − 2 π i f L n ( 1 + Ω ⋅ p n ) ] F n A ( Ω ) , (138) R_n^A(f,\boldsymbol{\Omega})=\left[1-e^{-2\pi i f L_n(1+\boldsymbol{\Omega}\cdot\mathbf{p}_n)}\right]F_n^A(\boldsymbol{\Omega}), \tag*{(138)} R n A ( f , Ω ) = [ 1 − e − 2 π i f L n ( 1 + Ω ⋅ p n ) ] F n A ( Ω ) , (138) where n = 1 n=1 n = 1 n = 2 n=2 n = 2 L n L_n L n (137) gives the correlation between the two pulsars. To compute the variance, we need the mean and second moment of Z 1 Z 2 ‾ \overline{Z_1Z_2} Z 1 Z 2 (137) by using Eq. (125) . One obtains
⟨ ρ ⟩ = ⟨ Z 1 Z 2 ⟩ = h 2 μ u ( γ ) , (139) \langle\rho\rangle=\langle Z_1Z_2\rangle=h^2\mu_u(\gamma), \tag*{(139)} ⟨ ρ ⟩ = ⟨ Z 1 Z 2 ⟩ = h 2 μ u ( γ ) , (139) where we have assumed that the normalized autocorrelation function C C C (135) , provided that a factor of ( 2 π f ) 2 (2\pi f)^2 ( 2 π f ) 2 (133) , since redshifts rather than timing residuals are being correlated.
For convenience, the overall scale in Eq. (139) is expressed as
h 2 = 4 π ∫ − ∞ ∞ H ( f ) d f = 8 π ∫ 0 ∞ H ( f ) d f . (140) h^2=4\pi\int_{-\infty}^{\infty}H(f)\,df=8\pi\int_0^\infty H(f)\,df. \tag*{(140)} h 2 = 4 π ∫ − ∞ ∞ H ( f ) df = 8 π ∫ 0 ∞ H ( f ) df . (140) This is a measure of the (squared) strain, expressed as an integral of the spectral function H ( f ) H(f) H ( f ) (140) we have also expressed this in terms of one-sided integrals, making use of the symmetry H ( f ) = H ( − f ) H(f)=H(-f) H ( f ) = H ( − f ) (125) of the two-point function for the Gaussian ensemble. For the second moment, we multiply the correlation Eq. (137) for any representative of the ensemble by its complex conjugate [55 ] , obtaining
ρ 2 = ∑ A ∑ A ′ ∑ A ′ ′ ∑ A ′ ′ ′ ∫ d f ∫ d f ′ ∫ d f ′ ′ ∫ d f ′ ′ ′ ∫ d Ω ∫ d Ω ′ ∫ d Ω ′ ′ ∫ d Ω ′ ′ ′ sinc ( π ( f − f ′ ) T ) sinc ( π ( f ′ ′ − f ′ ′ ′ ) T ) × R 1 A ∗ ( f , Ω ) R 2 A ′ ( f ′ , Ω ′ ) R 1 A ′ ′ ( f ′ ′ , Ω ′ ′ ) R 2 A ′ ′ ′ ( f ′ ′ ′ , Ω ′ ′ ′ ) h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′ ′ ∗ ( f ′ ′ , Ω ′ ′ ) h A ′ ′ ′ ( f ′ ′ ′ , Ω ′ ′ ′ ) . (141) \begin{aligned}
\rho^2={}&\sum_A\sum_{A'}\sum_{A''}\sum_{A'''}\int df\int df'\int df''\int df'''\int d\boldsymbol{\Omega}\int d\boldsymbol{\Omega}'\int d\boldsymbol{\Omega}''\int d\boldsymbol{\Omega}''' \operatorname{sinc}\left(\pi(f-f')T\right)\operatorname{sinc}\left(\pi(f''-f''')T\right)\\
&\times R_1^{A*}(f,\boldsymbol{\Omega})R_2^{A'}(f',\boldsymbol{\Omega}')R_1^{A''}(f'',\boldsymbol{\Omega}'')R_2^{A'''}(f''',\boldsymbol{\Omega}''')h_A^*(f,\boldsymbol{\Omega})h_{A'}(f',\boldsymbol{\Omega}')h_{A''}^*(f'',\boldsymbol{\Omega}'')h_{A'''}(f''',\boldsymbol{\Omega}'''). \tag*{(141)}
\end{aligned} ρ 2 = A ∑ A ′ ∑ A ′′ ∑ A ′′′ ∑ ∫ df ∫ d f ′ ∫ d f ′′ ∫ d f ′′′ ∫ d Ω ∫ d Ω ′ ∫ d Ω ′′ ∫ d Ω ′′′ sinc ( π ( f − f ′ ) T ) sinc ( π ( f ′′ − f ′′′ ) T ) × R 1 A ∗ ( f , Ω ) R 2 A ′ ( f ′ , Ω ′ ) R 1 A ′′ ( f ′′ , Ω ′′ ) R 2 A ′′′ ( f ′′′ , Ω ′′′ ) h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′′ ∗ ( f ′′ , Ω ′′ ) h A ′′′ ( f ′′′ , Ω ′′′ ) . (141) To compute the ensemble average of ρ 2 \rho^2 ρ 2 [54 ] implies that for a Gaussian ensemble the four-point function is the sum of three two-point functions:
⟨ h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′ ′ ( f ′ ′ , Ω ′ ′ ) h A ′ ′ ′ ∗ ( f ′ ′ ′ , Ω ′ ′ ′ ) ⟩ = ⟨ h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) ⟩ ⟨ h A ′ ′ ∗ ( − f ′ ′ , Ω ′ ′ ) h A ′ ′ ′ ( − f ′ ′ ′ , Ω ′ ′ ′ ) ⟩ + ⟨ h A ∗ ( f , Ω ) h A ′ ′ ( f ′ ′ , Ω ′ ′ ) ⟩ ⟨ h A ′ ∗ ( − f ′ , Ω ′ ) h A ′ ′ ′ ( − f ′ ′ ′ , Ω ′ ′ ′ ) ⟩ + ⟨ h A ∗ ( f , Ω ) h A ′ ′ ′ ( − f ′ ′ ′ , Ω ′ ′ ′ ) ⟩ ⟨ h A ′ ∗ ( − f ′ , Ω ′ ) h A ′ ′ ( f ′ ′ , Ω ′ ′ ) ⟩ . (142) \begin{aligned}
\langle h_A^*(f,\boldsymbol{\Omega})h_{A'}(f',\boldsymbol{\Omega}')h_{A''}(f'',\boldsymbol{\Omega}'')h_{A'''}^*(f''',\boldsymbol{\Omega}''')\rangle={}&
\langle h_A^*(f,\boldsymbol{\Omega})h_{A'}(f',\boldsymbol{\Omega}')\rangle\langle h_{A''}^*(-f'',\boldsymbol{\Omega}'')h_{A'''}(-f''',\boldsymbol{\Omega}''')\rangle\\
&+\langle h_A^*(f,\boldsymbol{\Omega})h_{A''}(f'',\boldsymbol{\Omega}'')\rangle\langle h_{A'}^*(-f',\boldsymbol{\Omega}')h_{A'''}(-f''',\boldsymbol{\Omega}''')\rangle\\
&+\langle h_A^*(f,\boldsymbol{\Omega})h_{A'''}(-f''',\boldsymbol{\Omega}''')\rangle\langle h_{A'}^*(-f',\boldsymbol{\Omega}')h_{A''}(f'',\boldsymbol{\Omega}'')\rangle. \tag*{(142)}
\end{aligned} ⟨ h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′′ ( f ′′ , Ω ′′ ) h A ′′′ ∗ ( f ′′′ , Ω ′′′ )⟩ = ⟨ h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ )⟩ ⟨ h A ′′ ∗ ( − f ′′ , Ω ′′ ) h A ′′′ ( − f ′′′ , Ω ′′′ )⟩ + ⟨ h A ∗ ( f , Ω ) h A ′′ ( f ′′ , Ω ′′ )⟩ ⟨ h A ′ ∗ ( − f ′ , Ω ′ ) h A ′′′ ( − f ′′′ , Ω ′′′ )⟩ + ⟨ h A ∗ ( f , Ω ) h A ′′′ ( − f ′′′ , Ω ′′′ )⟩ ⟨ h A ′ ∗ ( − f ′ , Ω ′ ) h A ′′ ( f ′′ , Ω ′′ )⟩ . (142) where we have used h A ( f , Ω ) = h A ∗ ( − f , Ω ) h_A(f,\boldsymbol{\Omega})=h_A^*(-f,\boldsymbol{\Omega}) h A ( f , Ω ) = h A ∗ ( − f , Ω ) (125) . Using Eq. (125) to evaluate Eq. (142) gives
⟨ h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′ ′ ( f ′ ′ , Ω ′ ′ ) h A ′ ′ ′ ∗ ( f ′ ′ ′ , Ω ′ ′ ′ ) ⟩ = δ A A ′ δ A ′ ′ A ′ ′ ′ δ 2 ( Ω , Ω ′ ) δ 2 ( Ω ′ ′ , Ω ′ ′ ′ ) δ ( f − f ′ ) δ ( f ′ ′ − f ′ ′ ′ ) H ( f ) H ( f ′ ′ ) + δ A A ′ ′ δ A ′ A ′ ′ ′ δ 2 ( Ω , Ω ′ ′ ) δ 2 ( Ω ′ , Ω ′ ′ ′ ) δ ( f − f ′ ′ ) δ ( f ′ − f ′ ′ ′ ) H ( f ) H ( f ′ ) + δ A A ′ ′ ′ δ A ′ A ′ ′ δ 2 ( Ω , Ω ′ ′ ′ ) δ 2 ( Ω ′ , Ω ′ ′ ) δ ( f + f ′ ′ ′ ) δ ( f ′ + f ′ ′ ) H ( f ) H ( f ′ ) . (143) \begin{aligned}
\langle h_A^*(f,\boldsymbol{\Omega})h_{A'}(f',\boldsymbol{\Omega}')h_{A''}(f'',\boldsymbol{\Omega}'')h_{A'''}^*(f''',\boldsymbol{\Omega}''')\rangle={}&
\delta_{AA'}\delta_{A''A'''}\delta^2(\boldsymbol{\Omega},\boldsymbol{\Omega}')\delta^2(\boldsymbol{\Omega}'',\boldsymbol{\Omega}''')\delta(f-f')\delta(f''-f''')H(f)H(f'')\\
&+\delta_{AA''}\delta_{A'A'''}\delta^2(\boldsymbol{\Omega},\boldsymbol{\Omega}'')\delta^2(\boldsymbol{\Omega}',\boldsymbol{\Omega}''')\delta(f-f'')\delta(f'-f''')H(f)H(f')\\
&+\delta_{AA'''}\delta_{A'A''}\delta^2(\boldsymbol{\Omega},\boldsymbol{\Omega}''')\delta^2(\boldsymbol{\Omega}',\boldsymbol{\Omega}'')\delta(f+f''')\delta(f'+f'')H(f)H(f'). \tag*{(143)}
\end{aligned} ⟨ h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′′ ( f ′′ , Ω ′′ ) h A ′′′ ∗ ( f ′′′ , Ω ′′′ )⟩ = δ A A ′ δ A ′′ A ′′′ δ 2 ( Ω , Ω ′ ) δ 2 ( Ω ′′ , Ω ′′′ ) δ ( f − f ′ ) δ ( f ′′ − f ′′′ ) H ( f ) H ( f ′′ ) + δ A A ′′ δ A ′ A ′′′ δ 2 ( Ω , Ω ′′ ) δ 2 ( Ω ′ , Ω ′′′ ) δ ( f − f ′′ ) δ ( f ′ − f ′′′ ) H ( f ) H ( f ′ ) + δ A A ′′′ δ A ′ A ′′ δ 2 ( Ω , Ω ′′′ ) δ 2 ( Ω ′ , Ω ′′ ) δ ( f + f ′′′ ) δ ( f ′ + f ′′ ) H ( f ) H ( f ′ ) . (143) where we have used δ ( f ) = δ ( − f ) \delta(f)=\delta(-f) δ ( f ) = δ ( − f ) ρ \rho ρ (141) and use Eq. (143) to obtain three terms. In the same order as Eq. (143) , the three terms are
⟨ ρ 2 ⟩ = ∫ d f H ( f ) ∫ d f ′ ′ H ( f ′ ′ ) ∑ A ∫ d Ω R 1 A ∗ ( f , Ω ) R 2 A ( f , Ω ) ∑ A ′ ′ ∫ d Ω ′ ′ R 1 A ′ ′ ( f ′ ′ , Ω ′ ′ ) R 2 A ′ ′ ∗ ( f ′ ′ , Ω ′ ′ ) + ∫ d f H ( f ) ∫ d f ′ H ( f ′ ) sinc 2 ( π ( f − f ′ ) T ) ∑ A ∫ d Ω R 1 A ∗ ( f , Ω ) R 1 A ( f , Ω ) ∑ A ′ ∫ d Ω ′ R 2 A ′ ( f ′ , Ω ′ ) R 2 A ′ ∗ ( f ′ , Ω ′ ) + ∫ d f H ( f ) ∫ d f ′ H ( f ′ ) sinc 2 ( π ( f − f ′ ) T ) ∑ A ∫ d Ω R 1 A ∗ ( f , Ω ) R 2 A ( f , Ω ) ∑ A ′ ∫ d Ω ′ R 1 A ′ ( f ′ , Ω ′ ) R 2 A ′ ∗ ( f ′ , Ω ′ ) . (144) \begin{aligned}
\langle\rho^2\rangle={}&\int dfH(f)\int df''H(f'')\sum_A\int d\boldsymbol{\Omega}R_1^{A*}(f,\boldsymbol{\Omega})R_2^A(f,\boldsymbol{\Omega})\sum_{A''}\int d\boldsymbol{\Omega}''R_1^{A''}(f'',\boldsymbol{\Omega}'')R_2^{A''*}(f'',\boldsymbol{\Omega}'')\\
&+\int dfH(f)\int df'H(f')\operatorname{sinc}^2\left(\pi(f-f')T\right)\sum_A\int d\boldsymbol{\Omega}R_1^{A*}(f,\boldsymbol{\Omega})R_1^A(f,\boldsymbol{\Omega})\sum_{A'}\int d\boldsymbol{\Omega}'R_2^{A'}(f',\boldsymbol{\Omega}')R_2^{A'*}(f',\boldsymbol{\Omega}')\\
&+\int dfH(f)\int df'H(f')\operatorname{sinc}^2\left(\pi(f-f')T\right)\sum_A\int d\boldsymbol{\Omega}R_1^{A*}(f,\boldsymbol{\Omega})R_2^A(f,\boldsymbol{\Omega})\sum_{A'}\int d\boldsymbol{\Omega}'R_1^{A'}(f',\boldsymbol{\Omega}')R_2^{A'*}(f',\boldsymbol{\Omega}'). \tag*{(144)}
\end{aligned} ⟨ ρ 2 ⟩ = ∫ df H ( f ) ∫ d f ′′ H ( f ′′ ) A ∑ ∫ d Ω R 1 A ∗ ( f , Ω ) R 2 A ( f , Ω ) A ′′ ∑ ∫ d Ω ′′ R 1 A ′′ ( f ′′ , Ω ′′ ) R 2 A ′′ ∗ ( f ′′ , Ω ′′ ) + ∫ df H ( f ) ∫ d f ′ H ( f ′ ) sinc 2 ( π ( f − f ′ ) T ) A ∑ ∫ d Ω R 1 A ∗ ( f , Ω ) R 1 A ( f , Ω ) A ′ ∑ ∫ d Ω ′ R 2 A ′ ( f ′ , Ω ′ ) R 2 A ′ ∗ ( f ′ , Ω ′ ) + ∫ df H ( f ) ∫ d f ′ H ( f ′ ) sinc 2 ( π ( f − f ′ ) T ) A ∑ ∫ d Ω R 1 A ∗ ( f , Ω ) R 2 A ( f , Ω ) A ′ ∑ ∫ d Ω ′ R 1 A ′ ( f ′ , Ω ′ ) R 2 A ′ ∗ ( f ′ , Ω ′ ) . (144) This second moment ⟨ ρ 2 ⟩ \langle\rho^2\rangle ⟨ ρ 2 ⟩ C C C
In this case, replace the modified antenna pattern functions R n A R^A_n R n A (144) by their definitions from Eq. (138) . In the first and third line of Eq. (144) , all of the terms which contain exponentials may be neglected, and only the product of the unity terms remains (see fig:16 of Ref. [35 ] ). In the second line of Eq. (144) , one has
R 1 A ( f , Ω ) R 1 A ( f , Ω ) = 1 − e − 2 π i f L 1 ( 1 + Ω ⋅ p 1 ) 2 − 2 cos ( 2 π f L 1 ( 1 + Ω ⋅ p 1 ) ) F 1 A ( Ω ) F 1 A ( Ω ) = [ 2 − 2 cos ( 2 π f L 1 ( 1 + Ω ⋅ p 1 ) ) ] − 1 F 1 A ( Ω ) F 1 A ( Ω ) × F 1 A ( Ω ) F 1 A ( Ω ) \begin{aligned}
R^A_1(f,\Omega)R^A_1(f,\Omega)=\frac{1-e^{-2\pi i fL_1(1+\Omega\cdot p_1)}}{2-2\cos\left(2\pi fL_1(1+\Omega\cdot p_1)\right)}F^A_1(\Omega)F^A_1(\Omega) \\
&= \left[2-2\cos\left(2\pi fL_1(1+\Omega\cdot p_1)\right)\right]^{-1}F^A_1(\Omega)F^A_1(\Omega) \\
&\times F^A_1(\Omega)F^A_1(\Omega)
\end{aligned} R 1 A ( f , Ω ) R 1 A ( f , Ω ) = 2 − 2 cos ( 2 π f L 1 ( 1 + Ω ⋅ p 1 ) ) 1 − e − 2 π i f L 1 ( 1 + Ω ⋅ p 1 ) F 1 A ( Ω ) F 1 A ( Ω ) = [ 2 − 2 cos ( 2 π f L 1 ( 1 + Ω ⋅ p 1 ) ) ] − 1 F 1 A ( Ω ) F 1 A ( Ω ) × F 1 A ( Ω ) F 1 A ( Ω ) ≈ 2 F 1 A ( Ω ) F 1 A ( Ω ) , (145) \approx2F^A_1(\Omega)F^A_1(\Omega), \tag*{(145)} ≈ 2 F 1 A ( Ω ) F 1 A ( Ω ) , (145) and a similar expression for the second pulsar term, since the rapidly varying cosine term, which arises from the two remaining exponential terms, may be neglected. Thus we obtain
⟨ ρ 2 ⟩ = (146) \langle\rho^2\rangle= \tag*{(146)} ⟨ ρ 2 ⟩ = (146) h 4 ∑ A ∫ d Ω 4 π F 1 A ( Ω ) F 2 A ( Ω ) ∑ A ′ ∫ d Ω ′ 4 π F 1 A ′ ( Ω ′ ) F 2 A ′ ( Ω ′ ) + 4 h 4 ∑ A ∫ d Ω 4 π F 1 A ( Ω ) F 1 A ( Ω ) ∑ A ′ ∫ d Ω ′ 4 π F 2 A ′ ( Ω ′ ) F 2 A ′ ( Ω ′ ) + h 4 ∑ A ∫ d Ω 4 π F 1 A ( Ω ) F 2 A ( Ω ) ∑ A ′ ∫ d Ω ′ 4 π F 1 A ′ ( Ω ′ ) F 2 A ′ ( Ω ′ ) . \begin{aligned}
h^4\sum_A\int\frac{d\Omega}{4\pi}F^A_1(\Omega)F^A_2(\Omega)\sum_{A'}\int\frac{d\Omega'}{4\pi}F^{A'}_1(\Omega')F^{A'}_2(\Omega') \\
&+4h^4\sum_A\int\frac{d\Omega}{4\pi}F^A_1(\Omega)F^A_1(\Omega)\sum_{A'}\int\frac{d\Omega'}{4\pi}F^{A'}_2(\Omega')F^{A'}_2(\Omega') \\
&+h^4\sum_A\int\frac{d\Omega}{4\pi}F^A_1(\Omega)F^A_2(\Omega)\sum_{A'}\int\frac{d\Omega'}{4\pi}F^{A'}_1(\Omega')F^{A'}_2(\Omega').
\end{aligned} h 4 A ∑ ∫ 4 π d Ω F 1 A ( Ω ) F 2 A ( Ω ) A ′ ∑ ∫ 4 π d Ω ′ F 1 A ′ ( Ω ′ ) F 2 A ′ ( Ω ′ ) + 4 h 4 A ∑ ∫ 4 π d Ω F 1 A ( Ω ) F 1 A ( Ω ) A ′ ∑ ∫ 4 π d Ω ′ F 2 A ′ ( Ω ′ ) F 2 A ′ ( Ω ′ ) + h 4 A ∑ ∫ 4 π d Ω F 1 A ( Ω ) F 2 A ( Ω ) A ′ ∑ ∫ 4 π d Ω ′ F 1 A ′ ( Ω ′ ) F 2 A ′ ( Ω ′ ) . The factor of four on the third line arises from (C24) because the pulsar terms contribute the same amount as the Earth terms. Fundamentally, this is because the Gaussian ensemble is stationary in time. Hence, when a pulsar is correlated with itself, the time-averaged square of the pulsar term is equal to the time-averaged square of the Earth term.
Two different scale factors appear in the second moment. The first is the squared strain h 2 h^2 h 2 (140) . The second is
h 4 = ( 4 π ) 2 ∫ d f ∫ d f ′ sinc 2 ( π ( f − f ′ ) T ) H ( f ) H ( f ′ ) , (147) h^4=(4\pi)^2\int df\int df'\,\operatorname{sinc}^2\left(\pi(f-f')T\right)H(f)H(f'), \tag*{(147)} h 4 = ( 4 π ) 2 ∫ df ∫ d f ′ sinc 2 ( π ( f − f ′ ) T ) H ( f ) H ( f ′ ) , (147) which is (another) measure of strain, determined by the spectral function H ( f ) H(f) H ( f ) T T T [38 ] , this integral is evaluated for a simple cosmological model with H ( f ) ∝ ∣ f ∣ − 7 / 3 H(f)\propto|f|^{-7/3} H ( f ) ∝ ∣ f ∣ − 7/3
The second moment Eq. (146) can now be expressed in terms of the Hellings-Downs curve μ u ( γ ) \mu_u(\gamma) μ u ( γ ) (146) has the “unpolarized” Hellings-Downs term in Eq. (3) as an integrand, whose average on the sphere is μ u ( γ ) \mu_u(\gamma) μ u ( γ ) (5) . Thus we find
⟨ ρ 2 ⟩ = h 4 μ u 2 ( γ ) + 4 h 4 μ u 2 ( 0 ) + h 4 μ u 2 ( γ ) . (148) \langle\rho^2\rangle=h^4\mu_u^2(\gamma)+4h^4\mu_u^2(0)+h^4\mu_u^2(\gamma). \tag*{(148)} ⟨ ρ 2 ⟩ = h 4 μ u 2 ( γ ) + 4 h 4 μ u 2 ( 0 ) + h 4 μ u 2 ( γ ) . (148) Note that each of the integrals over solid angle on the second line of Eq. (146) has the same form as the integrals on the following line, but with the directions to pulsars being identical, thus giving μ u ( 0 ) \mu_u(0) μ u ( 0 )
The variance σ 2 \sigma^2 σ 2 (139) from the second moment of Eq. (148) . The variance σ 2 = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 \sigma^2=\langle\rho^2\rangle-\langle\rho\rangle^2 σ 2 = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2
σ 2 ( γ ) = h 4 ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (149) \sigma^2(\gamma)=h^4\left(\mu_u^2(\gamma)+4\mu_u^2(0)\right). \tag*{(149)} σ 2 ( γ ) = h 4 ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (149) These results can be compared to the confusion-noise model of Sec. III A in the limit of large numbers of weak sources. The Gaussian ensemble variance Eq. (149) equals the confusion-noise model variance of Eq. (57) if h 4 = H 2 / 2 h^4=\mathcal{H}_2/2 h 4 = H 2 /2 (139) of the Gaussian model equals the mean of the confusion-noise model Eq. (58) if h 2 = H 2 / 2 h^2=\mathcal{H}_2/2 h 2 = H 2 /2 h 4 = h 4 / 2 h^4=h^4/2 h 4 = h 4 /2
The Gaussian ensemble variance has two interesting limits. To obtain these two limits, it is helpful to rewrite the definition of the strain measure h 4 h^4 h 4 (147) in terms of one-sided integrals, analogous to the second form given in Eq. (140) :
h 4 = 1 2 ( 8 π ) 2 ∫ 0 ∞ d f H ( f ) ∫ 0 ∞ d f ′ H ( f ′ ) × [ sinc 2 ( π ( f − f ′ ) T ) + sinc 2 ( π ( f + f ′ ) T ) ] . (150) \begin{aligned}
h^4=\frac{1}{2}(8\pi)^2\int_0^\infty dfH(f)\int_0^\infty df'H(f') \\
&\times\left[\operatorname{sinc}^2\left(\pi(f-f')T\right)+\operatorname{sinc}^2\left(\pi(f+f')T\right)\right]. \tag*{(150)}
\end{aligned} h 4 = 2 1 ( 8 π ) 2 ∫ 0 ∞ df H ( f ) ∫ 0 ∞ d f ′ H ( f ′ ) × [ sinc 2 ( π ( f − f ′ ) T ) + sinc 2 ( π ( f + f ′ ) T ) ] . (150) Now consider the “single-sided” function H ( f ) H(f) H ( f ) f ≥ 0 f\ge0 f ≥ 0
In the “narrowband” case, we assume that the support of H ( f ) H(f) H ( f ) f ≥ 0 f\ge0 f ≥ 0 Δ f ≪ 1 / T \Delta f\ll1/T Δ f ≪ 1/ T f 0 f_0 f 0 f 0 T = z f_0T=z f 0 T = z z z z (150) gives sinc ( 0 ) = 1 \operatorname{sinc}(0)=1 sinc ( 0 ) = 1 sinc ( 2 π z ) = 0 \operatorname{sinc}(2\pi z)=0 sinc ( 2 π z ) = 0 [53 ] . Comparison of Eqs. (140) and (150) then implies that in this narrowband case h 4 = h 4 / 2 h^4=h^4/2 h 4 = h 4 /2 (139) and the variance of Eq. (149) become
μ ( γ ) = h 2 μ u ( γ ) , and \mu(\gamma)=h^2\mu_u(\gamma),\qquad\text{and} μ ( γ ) = h 2 μ u ( γ ) , and σ 2 ( γ ) = 1 2 h 4 ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (151) \sigma^2(\gamma)=\frac{1}{2}h^4\left(\mu_u^2(\gamma)+4\mu_u^2(0)\right). \tag*{(151)} σ 2 ( γ ) = 2 1 h 4 ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (151) These apply for a narrowband signal, such as that considered in Sec. III A. In fact, Eq. (151) for the Gaussian ensemble mean and variance can be obtained from Eqs. (43) and (55) of Sec. III A, in the limit where N s N_s N s N s A 2 N_s\mathcal{A}^2 N s A 2 h 2 = H 2 / 2 h^2=\mathcal{H}_2/2 h 2 = H 2 /2 H 4 → 0 \mathcal{H}_4\to0 H 4 → 0
The assumption that the narrowband frequency f 0 f_0 f 0 f 0 T = f_0T= f 0 T = sinc 2 \operatorname{sinc}^{2} sinc 2 (150) can be treated as a constant that lies in the range [0, 1]. Thus, allowing f 0 T f_{0}T f 0 T 1 / 2 1/2 1/2 (151) could instead lie anywhere in the interval [ 1 / 2 , 1 ] [1/2,1] [ 1/2 , 1 ]
The second case of interest is the “broadband signal” case, for which the observation time is long compared with the inverse bandwidth: T ≫ 1 / Δ f T \gg1/\Delta f T ≫ 1/Δ f sinc 2 ( π f T ) \operatorname{sinc}^{2}(\pi fT) sinc 2 ( π f T ) δ ( f ) / T \delta(f)/T δ ( f ) / T
h 4 = ( 4 π ) 2 T ∫ H 2 ( f ) d f = 32 π 2 T ∫ 0 ∞ H 2 ( f ) d f . (152) \mathcal{h}^{4} = \frac{(4\pi)^{2}}{T}\int H^{2}(f)\,\mathrm{d}f = \frac{32\pi^{2}}{T}\int_{0}^{\infty} H^{2}(f)\,\mathrm{d}f. \tag*{(152)} h 4 = T ( 4 π ) 2 ∫ H 2 ( f ) d f = T 32 π 2 ∫ 0 ∞ H 2 ( f ) d f . (152) In this broadband case the (single-pair or total) variance of the Hellings and Downs correlation is
σ 2 ( γ ) = 32 π 2 T ∫ 0 ∞ H 2 ( f ) d f ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (153) \sigma^{2}(\gamma) = \frac{32\pi^{2}}{T}\int_{0}^{\infty} H^{2}(f)\,\mathrm{d}f\left(\mu_{u}^{2}(\gamma)+4\mu_{u}^{2}(0)\right). \tag*{(153)} σ 2 ( γ ) = T 32 π 2 ∫ 0 ∞ H 2 ( f ) d f ( μ u 2 ( γ ) + 4 μ u 2 ( 0 ) ) . (153) The relative size of the variance σ \sigma σ μ \mu μ H ( f ) H(f) H ( f ) T T T σ 2 \sigma^{2} σ 2 H 2 ( f ) H^{2}(f) H 2 ( f ) T T T μ 2 \mu^{2} μ 2 H H H
In contrast to the narrowband case of Eq. (151) , the broadband variance Eq. (153) decreases with time. This is because in the Gaussian ensemble, the distinct frequency bands are independent. So, as the observation time T T T 1 / T 1/T 1/ T [45 ] .
Frequency-domain formulas for σ 2 \sigma^{2} σ 2 (151) and (153) are implicitly given in [20 , 21 ] . Those formulas are also obtained from Gaussian ensemble calculations corresponding to the last line of Eq. (5.4) in [17 ] . For example, the text before Eq. (9) of [21 ] gives the variance as σ a b 2 ( f ) = S a a ( f ) S b b ( f ) + S a b 2 ( f ) \sigma_{ab}^{2}(f)=S_{aa}(f)S_{bb}(f)+S_{ab}^{2}(f) σ ab 2 ( f ) = S aa ( f ) S bb ( f ) + S ab 2 ( f ) a a a b b b S a a = S b b = 2 ( 4 π H ( f ) μ u ( 0 ) ) S_{aa}=S_{bb}=2(4\pi H(f)\mu_{u}(0)) S aa = S bb = 2 ( 4 π H ( f ) μ u ( 0 )) S a b ( f ) = 4 π H ( f ) μ u ( γ a b ) S_{ab}(f)=4\pi H(f)\mu_{u}(\gamma_{ab}) S ab ( f ) = 4 π H ( f ) μ u ( γ ab ) σ a b 2 ( f ) = ( 4 π H ( f ) ) 2 ( μ u 2 ( γ a b ) + 4 μ u 2 ( 0 ) ) \sigma_{ab}^{2}(f)=(4\pi H(f))^{2}(\mu_{u}^{2}(\gamma_{ab})+4\mu_{u}^{2}(0)) σ ab 2 ( f ) = ( 4 π H ( f ) ) 2 ( μ u 2 ( γ ab ) + 4 μ u 2 ( 0 )) γ a b = γ \gamma_{ab}=\gamma γ ab = γ a a a b b b γ \gamma γ (151) and (153) , and integrating it over frequency gives the broadband result Eq. (153) .
Variance of the squared GW strain and GW energy-density for the Gaussian ensemble
Here, we find the variance of the time-averaged squared strain s = h a b h a b s=h_{ab}h^{ab} s = h ab h ab ρ G W = c 2 h ˙ a b h ˙ a b / 32 π G \rho_{\mathrm{GW}}=c^{2}\dot{h}_{ab}\dot{h}^{ab}/32\pi G ρ GW = c 2 h ˙ ab h ˙ ab /32 π G [17 ] , Eq. (2.13)], provided that two factors of π c 2 f 2 / 8 G \pi c^{2}f^{2}/8G π c 2 f 2 /8 G
To carry out the calculation, we will need the spherical average Eq. (4) of the product of two polarization tensors, given by Eq. (180) . This spherical average is defined by
η a b c d = 1 4 π ∫ d Ω e a b ( Ω ) e c d ∗ ( Ω ) = 1 4 π ∫ d Ω ( e a b + ( Ω ) e c d + ( Ω ) + e a b × ( Ω ) e c d × ( Ω ) ) . (154) \begin{aligned}
\eta_{abcd}=\frac{1}{4\pi}\int\mathrm{d}\Omega\,e_{ab}(\Omega)e_{cd}^{*}(\Omega) \\
&=\frac{1}{4\pi}\int\mathrm{d}\Omega\left(e_{ab}^{+}(\Omega)e_{cd}^{+}(\Omega)+e_{ab}^{\times}(\Omega)e_{cd}^{\times}(\Omega)\right). \tag*{(154)}
\end{aligned} η ab c d = 4 π 1 ∫ d Ω e ab ( Ω ) e c d ∗ ( Ω ) = 4 π 1 ∫ d Ω ( e ab + ( Ω ) e c d + ( Ω ) + e ab × ( Ω ) e c d × ( Ω ) ) . (154) For completeness we have expressed this average both in terms of the linear polarization basis A = + , × A=+,\times A = + , × e a b = e a b + − i e a b × e_{ab}=e_{ab}^{+}-ie_{ab}^{\times} e ab = e ab + − i e ab × (18) . While η a b c d \eta_{abcd} η ab c d
From rotational symmetry and index symmetries it follows that
η a b c d = α δ a b δ c d + β ( δ a c δ b d + δ a d δ b c ) , (155) \eta_{abcd}=\alpha\delta_{ab}\delta_{cd}+\beta(\delta_{ac}\delta_{bd}+\delta_{ad}\delta_{bc}), \tag*{(155)} η ab c d = α δ ab δ c d + β ( δ a c δ b d + δ a d δ b c ) , (155) where δ a b \delta_{ab} δ ab α \alpha α β \beta β are easily found. Since the polarization tensors are traceless, {xref}Eq. (C33) implies that $\eta^{a}_{\ acd}=0$, which in {xref}Eq. (C34) implies that $3\alpha+2\beta=0$. Since we have normalized the polarization tensors so that $e^{ab}e_{ab}^{*}=4$, {xref}Eq. (C33) implies that $\eta^{ab}_{\ \ ab}=4$, which from {xref}Eq. (C34) implies that $3\alpha+12\beta=4$. The solution to this pair of linear equations is $\alpha=-4/15$ and $\beta=2/5$. This simple formula for $\eta_{abcd}$ may also be obtained by taking the $\alpha\to 0$ limit of the more general overlap reduction function given in {citep}Ref. [17 ] , {xref}Eq. (3.34)`.)
The first and second moments of s = h a b h a b ‾ s=\overline{h_{ab}h^{ab}} s = h ab h ab ρ \rho ρ s s s s s s (137) , is given by
s = ∑ A ∑ A ′ ∫ d f ∫ d f ′ ∫ d Ω ∫ d Ω ′ sinc ( π ( f − f ′ ) T ) e a b A ( Ω ) e a b A ′ ( Ω ′ ) h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) . (156) s=\sum_{A}\sum_{A'}\int\mathrm{d}f\int\mathrm{d}f'\int\mathrm{d}\Omega\int\mathrm{d}\Omega'\,\operatorname{sinc}\left(\pi(f-f')T\right)e_{ab}^{A}(\Omega)e_{ab}^{A'}(\Omega')h_{A}^{*}(f,\Omega)h_{A'}(f',\Omega'). \tag*{(156)} s = A ∑ A ′ ∑ ∫ d f ∫ d f ′ ∫ d Ω ∫ d Ω ′ sinc ( π ( f − f ′ ) T ) e ab A ( Ω ) e ab A ′ ( Ω ′ ) h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) . (156) The ensemble expectation value of s s s (125) . This gives
⟨ s ⟩ = ∑ A ∫ d f H ( f ) ∫ d Ω e a b A ( Ω ) e A a b ( Ω ) = 4 π η a b a b ∫ H ( f ) d f = 4 h 2 , (157) \langle s\rangle= \sum_A \int df H(f) \int d\Omega e_{ab}^A(\Omega)e_A^{ab}(\Omega) = 4\pi\eta_{ab}^{ab}\int H(f)df = 4h^2, \tag*{(157)} ⟨ s ⟩ = A ∑ ∫ df H ( f ) ∫ d Ω e ab A ( Ω ) e A ab ( Ω ) = 4 π η ab ab ∫ H ( f ) df = 4 h 2 , (157) where h 2 h^2 h 2 (140) . The second moment of s s s (141) , which is
s 2 = ∑ A ∑ A ′ ∑ A ′ ′ ∑ A ′ ′ ′ ∫ d f ∫ d f ′ ∫ d f ′ ′ ∫ d f ′ ′ ′ ∫ d Ω ∫ d Ω ′ ∫ d Ω ′ ′ ∫ d Ω ′ ′ ′ sinc ( π ( f − f ′ ) T ) sinc ( π ( f ′ ′ − f ′ ′ ′ ) T ) s^2 = \sum_A\sum_{A'}\sum_{A''}\sum_{A'''}\int df\int df'\int df''\int df'''\int d\Omega\int d\Omega'\int d\Omega''\int d\Omega''' \operatorname{sinc}\left(\pi(f-f')T\right)\operatorname{sinc}\left(\pi(f''-f''')T\right) s 2 = A ∑ A ′ ∑ A ′′ ∑ A ′′′ ∑ ∫ df ∫ d f ′ ∫ d f ′′ ∫ d f ′′′ ∫ d Ω ∫ d Ω ′ ∫ d Ω ′′ ∫ d Ω ′′′ sinc ( π ( f − f ′ ) T ) sinc ( π ( f ′′ − f ′′′ ) T ) × e a b A ( Ω ) e A ′ a b ( Ω ′ ) e c d A ′ ′ ( Ω ′ ′ ) e A ′ ′ ′ c d ( Ω ′ ′ ′ ) h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′ ′ ( f ′ ′ , Ω ′ ′ ) h A ′ ′ ′ ∗ ( f ′ ′ ′ , Ω ′ ′ ′ ) . (158) {}\times e_{ab}^A(\Omega)e_{A'}^{ab}(\Omega')e_{cd}^{A''}(\Omega'')e_{A'''}^{cd}(\Omega''')h_A^*(f,\Omega)h_{A'}(f',\Omega')h_{A''}(f'',\Omega'')h_{A'''}^*(f''',\Omega'''). \tag*{(158)} × e ab A ( Ω ) e A ′ ab ( Ω ′ ) e c d A ′′ ( Ω ′′ ) e A ′′′ c d ( Ω ′′′ ) h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′′ ( f ′′ , Ω ′′ ) h A ′′′ ∗ ( f ′′′ , Ω ′′′ ) . (158) Taking its expectation value for the Gaussian ensemble of universes gives the analog of Eq. (146) , where the four antenna-pattern functions are replaced by contractions of polarization tensors. The three terms arising from Isserlis’ theorem are
⟨ s 2 ⟩ = h 4 η a b a b η c d c d + ℏ 4 ( η a b c d η a b c d + η a b c d η a b c d ) = 16 h 4 + 32 5 ℏ 4 , (159) \langle s^2\rangle= h^4\eta_{ab}^{ab}\eta_{cd}^{cd}+\hbar^4\left(\eta_{abcd}\eta^{abcd}+\eta^{abcd}\eta_{abcd}\right)=16h^4+\frac{32}{5}\hbar^4, \tag*{(159)} ⟨ s 2 ⟩ = h 4 η ab ab η c d c d + ℏ 4 ( η ab c d η ab c d + η ab c d η ab c d ) = 16 h 4 + 5 32 ℏ 4 , (159) where ℏ 4 \hbar^4 ℏ 4 (147) . The final equality follows by using Eq. (155) to compute η a b a b = 3 α + 12 β = 4 \eta_{ab}^{ab}=3\alpha+12\beta=4 η ab ab = 3 α + 12 β = 4 η a b c d η a b c d = 6 ( α + 4 β ) β = 16 / 5 \eta_{abcd}\eta^{abcd}=6(\alpha+4\beta)\beta=16/5 η ab c d η ab c d = 6 ( α + 4 β ) β = 16/5
Note that the ℏ 4 \hbar^4 ℏ 4 (159) are equal. In contrast, when computing the variance of the Hellings and Downs correlation in Eq. (146) , the effect of the “pulsar term” is to make the second term four times as large as the third. Here, the squared strain is a local quantity: there are no pulsar terms.
The (cosmic) variance in s s s (157) and (159) , giving
σ s 2 = ⟨ s 2 ⟩ − ⟨ s ⟩ 2 = 32 5 ℏ 4 . (160) \sigma_s^2=\langle s^2\rangle-\langle s\rangle^2=\frac{32}{5}\hbar^4. \tag*{(160)} σ s 2 = ⟨ s 2 ⟩ − ⟨ s ⟩ 2 = 5 32 ℏ 4 . (160) This implies that a local measurement of the squared GW strain or GW energy density has a fractional uncertainty
σ s ⟨ s ⟩ = 2 5 ℏ 2 h 2 , (161) \frac{\sigma_s}{\langle s\rangle}=\sqrt{\frac{2}{5}}\frac{\hbar^2}{h^2}, \tag*{(161)} ⟨ s ⟩ σ s = 5 2 h 2 ℏ 2 , (161) where we have used Eqs. (157) and (160) .
Corresponding formulas may be obtained for the mean GW energy density and its variance by inserting a factor of π c 2 f 2 / 8 G \pi c^2 f^2/8G π c 2 f 2 /8 G (140) which defines h 2 h^2 h 2 f f f f ′ f' f ′ (147) which defines ℏ 4 \hbar^4 ℏ 4
For the commensurate-frequency narrowband case with ℏ 4 / h 4 = 1 / 2 \hbar^4/h^4=1/2 ℏ 4 / h 4 = 1/2 ± 1 / 5 ≈ ± 45 % \pm1/\sqrt{5}\approx\pm45\% ± 1/ 5 ≈ ± 45%
In contrast, for a broadband signal Eq. (152) shows that the ratio of the fluctuations to the mean decreases proportional to 1 / T 1/\sqrt{T} 1/ T T T T
Cosmic variance of the Hellings and Downs correlation for the Gaussian ensemble
Here, we derive the cosmic variance of the Gaussian ensemble, using the same technique that we used for the discrete-source confusion-noise model in § 3 .
As explained at the start of § 3 , to compute the cosmic variance, we average the pulsar correlation Eq. (137) over all pulsar-pair directions separated by angle γ \gamma γ Γ ( γ ) \Gamma(\gamma) Γ ( γ ) 1 / f 1/f 1/ f (138) are rapidly oscillating functions of the pulsar directions. In contrast, the functions F n A F_n^A F n A R n A R_n^A R n A F n A F_n^A F n A (229) and (233) from Appendix § 16 , the pulsar average converts the antenna patterns in Eq. (137) into the Hellings-Downs two-point functions μ A A ′ ( γ , β ) \mu_{AA'}(\gamma,\beta) μ A A ′ ( γ , β )
Γ ( γ ) = ⟨ Z 1 Z 2 ⟩ p = ∑ A ∑ A ′ ∫ d f ∫ d f ′ sinc ( π ( f − f ′ ) T ) \Gamma(\gamma)=\langle Z_1Z_2\rangle_p=\sum_A\sum_{A'}\int df\int df'\operatorname{sinc}\left(\pi(f-f')T\right) Γ ( γ ) = ⟨ Z 1 Z 2 ⟩ p = A ∑ A ′ ∑ ∫ df ∫ d f ′ sinc ( π ( f − f ′ ) T ) × ∫ d Ω ∫ d Ω ′ μ A A ′ ( γ , β ( Ω , Ω ′ ) ) h A ∗ ( f , Ω ) {}\times\int d\Omega\int d\Omega'\mu_{AA'}(\gamma,\beta(\Omega,\Omega'))h_A^*(f,\Omega) × ∫ d Ω ∫ d Ω ′ μ A A ′ ( γ , β ( Ω , Ω ′ )) h A ∗ ( f , Ω ) × h A ′ ( f ′ , Ω ′ ) , (162) {}\times h_{A'}(f',\Omega'), \tag*{(162)} × h A ′ ( f ′ , Ω ′ ) , (162) where β ( Ω , Ω ′ ) = cos − 1 ( Ω ⋅ Ω ′ ) \beta(\Omega,\Omega')=\cos^{-1}(\boldsymbol{\Omega}\cdot\boldsymbol{\Omega}') β ( Ω , Ω ′ ) = cos − 1 ( Ω ⋅ Ω ′ ) Ω \boldsymbol{\Omega} Ω Ω ′ \boldsymbol{\Omega}' Ω ′ Γ ( γ ) \Gamma(\gamma) Γ ( γ ) Γ ( γ ) \Gamma(\gamma) Γ ( γ ) ⟨ Γ 2 ⟩ − ⟨ Γ ⟩ 2 \langle\Gamma^2\rangle-\langle\Gamma\rangle^2 ⟨ Γ 2 ⟩ − ⟨ Γ ⟩ 2
The ensemble average of Γ ( γ ) \Gamma(\gamma) Γ ( γ ) (162) by using Eq. (125) . It is
⟨ Γ ( γ ) ⟩ = ∑ A ∫ d f ∫ d Ω μ A A ( γ , 0 ) H ( f ) = 4 π ∫ d f H ( f ) μ ( γ , 0 ) = h 2 μ ( γ , 0 ) = h 2 μ u ( γ ) , (163) \langle\Gamma(\gamma)\rangle= \sum_A \int df \int d\Omega\,\mu_{AA}(\gamma,0)H(f)
= 4\pi\int df\,H(f)\mu(\gamma,0)
= \mathfrak{h}^{2}\mu(\gamma,0)=\mathfrak{h}^{2}\mu_u(\gamma), \tag*{(163)} ⟨ Γ ( γ )⟩ = A ∑ ∫ df ∫ d Ω μ AA ( γ , 0 ) H ( f ) = 4 π ∫ df H ( f ) μ ( γ , 0 ) = h 2 μ ( γ , 0 ) = h 2 μ u ( γ ) , (163) where we have used Eq. (230) to evaluate the polarization sum in the first line. Here, μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) § 16 , μ ( γ , 0 ) = μ u ( γ ) \mu(\gamma,0)=\mu_u(\gamma) μ ( γ , 0 ) = μ u ( γ ) h 2 \mathfrak{h}^{2} h 2 (140) . Thus, ⟨ Γ ( γ ) ⟩ \langle\Gamma(\gamma)\rangle ⟨ Γ ( γ )⟩ ρ \rho ρ (139) .
For the second moment, we start with the square of the pulsar-averaged correlation Eq. (162) for any representative of the ensemble
Γ 2 ( γ ) = ∑ A ∑ A ′ ∑ A ′ ′ ∑ A ′ ′ ′ ∫ d f ∫ d f ′ ∫ d f ′ ′ ∫ d f ′ ′ ′ ∫ d Ω ∫ d Ω ′ ∫ d Ω ′ ′ ∫ d Ω ′ ′ ′ × μ A A ′ ( γ , β ( Ω , Ω ′ ) ) μ A ′ ′ A ′ ′ ′ ( γ , β ( Ω ′ ′ , Ω ′ ′ ′ ) ) × h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′ ′ ( f ′ ′ , Ω ′ ′ ) h A ′ ′ ′ ∗ ( f ′ ′ ′ , Ω ′ ′ ′ ) sinc ( π ( f − f ′ ) T ) sinc ( π ( f ′ ′ − f ′ ′ ′ ) T ) . (164) \begin{aligned}
\Gamma^2(\gamma)={}&\sum_A\sum_{A'}\sum_{A''}\sum_{A'''}\int df\int df'\int df''\int df'''\int d\Omega\int d\Omega'\int d\Omega''\int d\Omega'''\\
&\times\mu_{AA'}\bigl(\gamma,\beta(\Omega,\Omega')\bigr)\mu_{A''A'''}\bigl(\gamma,\beta(\Omega'',\Omega''')\bigr)\\
&\times h_A^*(f,\Omega)h_{A'}(f',\Omega')h_{A''}(f'',\Omega'')h_{A'''}^*(f''',\Omega''')\operatorname{sinc}\bigl(\pi(f-f')T\bigr)\operatorname{sinc}\bigl(\pi(f''-f''')T\bigr). \tag*{(164)}
\end{aligned} Γ 2 ( γ ) = A ∑ A ′ ∑ A ′′ ∑ A ′′′ ∑ ∫ df ∫ d f ′ ∫ d f ′′ ∫ d f ′′′ ∫ d Ω ∫ d Ω ′ ∫ d Ω ′′ ∫ d Ω ′′′ × μ A A ′ ( γ , β ( Ω , Ω ′ ) ) μ A ′′ A ′′′ ( γ , β ( Ω ′′ , Ω ′′′ ) ) × h A ∗ ( f , Ω ) h A ′ ( f ′ , Ω ′ ) h A ′′ ( f ′′ , Ω ′′ ) h A ′′′ ∗ ( f ′′′ , Ω ′′′ ) sinc ( π ( f − f ′ ) T ) sinc ( π ( f ′′ − f ′′′ ) T ) . (164) We evaluate its ensemble average by using the four-point ensemble average Eq. (143) . This gives [compare Eq. (165) with Eqs. (C25) and (C27)]
⟨ Γ ( γ ) 2 ⟩ = ∑ A ∑ A ′ ′ ∫ d f H ( f ) ∫ d f ′ ′ H ( f ′ ′ ) ∫ d Ω ∫ d Ω ′ ′ μ A A ( γ , 0 ) μ A ′ ′ A ′ ′ ( γ , 0 ) + ∑ A ∑ A ′ ∫ d f H ( f ) ∫ d f ′ H ( f ′ ) sinc 2 ( π ( f − f ′ ) T ) ∫ d Ω ∫ d Ω ′ μ A A ′ ( γ , β ( Ω , Ω ′ ) ) μ A A ′ ( γ , β ( Ω , Ω ′ ) ) + ∑ A ∑ A ′ ∫ d f H ( f ) ∫ d f ′ H ( f ′ ) sinc 2 ( π ( f − f ′ ) T ) ∫ d Ω ∫ d Ω ′ μ A A ′ ( γ , β ( Ω , Ω ′ ) ) μ A ′ A ( γ , β ( Ω ′ , Ω ) ) = h 4 μ 2 ( γ , 0 ) + 2 h 4 ∫ d Ω 4 π ∫ d Ω ′ 4 π [ μ + + 2 ( γ , β ( Ω , Ω ′ ) ) + μ × × 2 ( γ , β ( Ω , Ω ′ ) ) ] = h 4 μ 2 ( γ , 0 ) + h 4 ∫ 0 π sin β d β [ μ + + 2 ( γ , β ) + μ × × 2 ( γ , β ) ] = h 4 μ u 2 ( γ ) + 2 h 4 μ ~ 2 ( γ ) , (165) \begin{aligned}
\langle\Gamma(\gamma)^2\rangle={}&\sum_{A}\sum_{A''}\int dfH(f)\int df''H(f'')\int d\Omega\int d\Omega''\mu_{AA}(\gamma,0)\mu_{A''A''}(\gamma,0)\\
&+\sum_A\sum_{A'}\int dfH(f)\int df'H(f')\operatorname{sinc}^2\bigl(\pi(f-f')T\bigr)\int d\Omega\int d\Omega'\mu_{AA'}\bigl(\gamma,\beta(\Omega,\Omega')\bigr)\mu_{AA'}\bigl(\gamma,\beta(\Omega,\Omega')\bigr)\\
&+\sum_A\sum_{A'}\int dfH(f)\int df'H(f')\operatorname{sinc}^2\bigl(\pi(f-f')T\bigr)\int d\Omega\int d\Omega'\mu_{AA'}\bigl(\gamma,\beta(\Omega,\Omega')\bigr)\mu_{A'A}\bigl(\gamma,\beta(\Omega',\Omega)\bigr)\\
={}&\mathfrak{h}^{4}\mu^2(\gamma,0)+2\mathfrak{h}^{4}\int\frac{d\Omega}{4\pi}\int\frac{d\Omega'}{4\pi}\left[\mu_{+ +}^2\bigl(\gamma,\beta(\Omega,\Omega')\bigr)+\mu_{\times\times}^2\bigl(\gamma,\beta(\Omega,\Omega')\bigr)\right]\\
={}&\mathfrak{h}^{4}\mu^2(\gamma,0)+\mathfrak{h}^{4}\int_0^\pi\sin\beta\,d\beta\left[\mu_{++}^2(\gamma,\beta)+\mu_{\times\times}^2(\gamma,\beta)\right]\\
={}&\mathfrak{h}^{4}\mu_u^2(\gamma)+2\mathfrak{h}^{4}\widetilde{\mu}^{2}(\gamma), \tag*{(165)}
\end{aligned} ⟨ Γ ( γ ) 2 ⟩ = = = = A ∑ A ′′ ∑ ∫ df H ( f ) ∫ d f ′′ H ( f ′′ ) ∫ d Ω ∫ d Ω ′′ μ AA ( γ , 0 ) μ A ′′ A ′′ ( γ , 0 ) + A ∑ A ′ ∑ ∫ df H ( f ) ∫ d f ′ H ( f ′ ) sinc 2 ( π ( f − f ′ ) T ) ∫ d Ω ∫ d Ω ′ μ A A ′ ( γ , β ( Ω , Ω ′ ) ) μ A A ′ ( γ , β ( Ω , Ω ′ ) ) + A ∑ A ′ ∑ ∫ df H ( f ) ∫ d f ′ H ( f ′ ) sinc 2 ( π ( f − f ′ ) T ) ∫ d Ω ∫ d Ω ′ μ A A ′ ( γ , β ( Ω , Ω ′ ) ) μ A ′ A ( γ , β ( Ω ′ , Ω ) ) h 4 μ 2 ( γ , 0 ) + 2 h 4 ∫ 4 π d Ω ∫ 4 π d Ω ′ [ μ ++ 2 ( γ , β ( Ω , Ω ′ ) ) + μ ×× 2 ( γ , β ( Ω , Ω ′ ) ) ] h 4 μ 2 ( γ , 0 ) + h 4 ∫ 0 π sin β d β [ μ ++ 2 ( γ , β ) + μ ×× 2 ( γ , β ) ] h 4 μ u 2 ( γ ) + 2 h 4 μ 2 ( γ ) , (165) where h 4 \mathfrak{h}^{4} h 4 (147) , and we have used Eq. (233) to set μ + × \mu_{+\times} μ +× μ × + \mu_{\times+} μ ×+ (165) is the square of the mean from Eq. (163) , and that the last two integrals over solid angle are equal. These are evaluated in Eq. (235) , where μ ~ 2 \widetilde{\mu}^{2} μ 2 § 16 and given in Eq. (234) .
We obtain the variance σ 2 \sigma^2 σ 2 ⟨ Γ 2 ⟩ \langle\Gamma^2\rangle ⟨ Γ 2 ⟩ (165) , giving a cosmic variance
σ c o s m i c 2 ( γ ) = ⟨ Γ ( γ ) 2 ⟩ − ⟨ Γ ( γ ) ⟩ 2 = 2 h 4 μ ~ 2 ( γ ) . (166) \sigma^2_{\rm cosmic}(\gamma)=\langle\Gamma(\gamma)^2\rangle-\langle\Gamma(\gamma)\rangle^2=2\mathfrak{h}^{4}\widetilde{\mu}^{2}(\gamma). \tag*{(166)} σ cosmic 2 ( γ ) = ⟨ Γ ( γ ) 2 ⟩ − ⟨ Γ ( γ ) ⟩ 2 = 2 h 4 μ 2 ( γ ) . (166) This should be compared with the total variance (pulsar plus cosmic) given in Eq. (149) . The cosmic variance of Eq. (166) corresponds exactly to the cosmic variance Eq. (76) for the confusion-noise model of Sec. III A in the limit of large numbers of weak sources, where h 4 = H 2 2 / 8 \mathfrak{h}^{4}=\mathcal{H}_2^2/8 h 4 = H 2 2 /8 (150) , the overall amplitude in Eq. (166) takes different forms for narrowband and broadband signals, as in Eqs. (151) and (153) .
Cosmic variance from harmonic analysis
An alternative approach to obtaining the cosmic variance for the Gaussian ensemble uses techniques which were originally developed for the analysis of cosmic background radiation temperature fluctuation maps [57 , 58 , 59 , 60 ] . In [61 , 62 ] , pulsar timing redshift maps are analyzed using these harmonic analysis methods. The existence of such a map is equivalent to assuming that one has data from a large number of low-noise pulsars, uniformly distributed on the sky.
The pulsar redshift Z Z Z Ω \Omega Ω
Z ( Ω ) = ∑ l m a l m Y l m ( Ω ) . (167) Z(\Omega)=\sum_{lm}a_{lm}Y_{lm}(\Omega). \tag*{(167)} Z ( Ω ) = l m ∑ a l m Y l m ( Ω ) . (167) where Ω \Omega Ω S 2 S^2 S 2 θ , ϕ \theta,\phi θ , ϕ a l m a_{lm} a l m a l m a_{lm} a l m Z Z Z (167) is complex because it corresponds to the Fourier transform of data from a single frequency bin [61 ] .
In [61 ] the complex coefficients a l m a_{lm} a l m z z z a l m a_{lm} a l m
C l = 1 2 l + 1 ∑ m = − l l ∣ a l m ∣ 2 . (168) C_l = \frac{1}{2l+1}\sum_{m=-l}^{l}|a_{lm}|^2. \tag*{(168)} C l = 2 l + 1 1 m = − l ∑ l ∣ a l m ∣ 2 . (168) The redshift Z Z Z h a b h_{ab} h ab (80) and (81) . Thus, if the GW background is described by a Gaussian ensemble, it follows from Eq. (167) that the a l m a_{lm} a l m (168) to (statistically) characterize the a l m a_{lm} a l m
For a given value of l l l 2 l + 1 2l+1 2 l + 1 a l m a_{lm} a l m m = − l , … , l m=-l,\ldots,l m = − l , … , l 4 l + 2 4l+2 4 l + 2 4 l + 2 4l+2 4 l + 2 [61 ] as
⟨ C l ⟩ = { 0 if l < 2 , 4 π ( l + 2 ) ( l + 1 ) l ( l − 1 ) h 2 if l ≥ 2. (169) \langle C_l\rangle=
\begin{cases}
0 & \text{if } l<2,\\
\frac{4\pi}{(l+2)(l+1)l(l-1)}h^2 & \text{if } l\ge2.
\end{cases} \tag*{(169)} ⟨ C l ⟩ = { 0 ( l + 2 ) ( l + 1 ) l ( l − 1 ) 4 π h 2 if l < 2 , if l ≥ 2. (169) The statistical independence of the real and imaginary parts of a l m a_{lm} a l m l l l m m m C l C_l C l χ k 2 \chi_k^2 χ k 2 k = 4 l + 2 k=4l+2 k = 4 l + 2 ⟨ C l ⟩ \langle C_l\rangle ⟨ C l ⟩ k = 4 l + 2 k=4l+2 k = 4 l + 2
Note that since we use the normalization μ u ( 0 ) = 1 / 3 \mu_u(0)=1/3 μ u ( 0 ) = 1/3 [61 ] have 1 / 2 1/2 1/2 (169) we have changed their 6 π 6\pi 6 π 4 π 4\pi 4 π h 2 h^2 h 2
The Hellings and Downs correlation for any member of the ensemble is given by Eq. (19) of [61 ] as
μ ( γ ) = ∑ l = 0 ∞ C l 2 l + 1 4 π P l ( cos γ ) , (170) \mu(\gamma)=\sum_{l=0}^{\infty}C_l\frac{2l+1}{4\pi}P_l(\cos\gamma), \tag*{(170)} μ ( γ ) = l = 0 ∑ ∞ C l 4 π 2 l + 1 P l ( cos γ ) , (170) where P l ( x ) P_l(x) P l ( x ) l l l x x x [63 ] it is shown that the expected value of this sum is precisely the Hellings and Downs curve:
⟨ μ ( γ ) ⟩ = h 2 μ u ( γ ) = h 2 ∑ l = 2 ∞ 2 l + 1 ( l + 2 ) ( l + 1 ) l ( l − 1 ) P l ( cos γ ) . (171) \langle\mu(\gamma)\rangle=h^2\mu_u(\gamma)
=h^2\sum_{l=2}^{\infty}\frac{2l+1}{(l+2)(l+1)l(l-1)}P_l(\cos\gamma). \tag*{(171)} ⟨ μ ( γ )⟩ = h 2 μ u ( γ ) = h 2 l = 2 ∑ ∞ ( l + 2 ) ( l + 1 ) l ( l − 1 ) 2 l + 1 P l ( cos γ ) . (171) That is to say, the Hellings-Downs curve may be expressed as this particular sum of Legendre polynomials of cos γ \cos\gamma cos γ
It is also straightforward to compute the variance of the Hellings and Downs correlation from the results of [61 ] . Start with Eq. (170) , take the ensemble average of its square and subtract the square of its ensemble average to obtain
σ 2 ( γ ) = ⟨ μ 2 ⟩ − ⟨ μ ⟩ 2 = ∑ l , l ′ = 0 ∞ ( ⟨ C l C l ′ ⟩ − ⟨ C l ⟩ ⟨ C l ′ ⟩ ) ( 2 l + 1 ) ( 2 l ′ + 1 ) ( 4 π ) 2 P l ( cos γ ) P l ′ ( cos γ ) = ∑ l = 0 ∞ ( ⟨ C l 2 ⟩ − ⟨ C l ⟩ 2 ) ( 2 l + 1 4 π ) 2 P l 2 ( cos γ ) . (172) \begin{aligned}
\sigma^2(\gamma)=\langle\mu^2\rangle-\langle\mu\rangle^2 \\
&=\sum_{l,l'=0}^{\infty}\left(\langle C_lC_{l'}\rangle-\langle C_l\rangle\langle C_{l'}\rangle\right)\frac{(2l+1)(2l'+1)}{(4\pi)^2}
P_l(\cos\gamma)P_{l'}(\cos\gamma) \\
&=\sum_{l=0}^{\infty}\left(\langle C_l^2\rangle-\langle C_l\rangle^2\right)\left(\frac{2l+1}{4\pi}\right)^2P_l^2(\cos\gamma). \tag*{(172)}
\end{aligned} σ 2 ( γ ) = ⟨ μ 2 ⟩ − ⟨ μ ⟩ 2 = l , l ′ = 0 ∑ ∞ ( ⟨ C l C l ′ ⟩ − ⟨ C l ⟩ ⟨ C l ′ ⟩ ) ( 4 π ) 2 ( 2 l + 1 ) ( 2 l ′ + 1 ) P l ( cos γ ) P l ′ ( cos γ ) = l = 0 ∑ ∞ ( ⟨ C l 2 ⟩ − ⟨ C l ⟩ 2 ) ( 4 π 2 l + 1 ) 2 P l 2 ( cos γ ) . (172) To obtain the final line, we have used ⟨ C l C l ′ ⟩ = ⟨ C l ⟩ ⟨ C l ′ ⟩ \langle C_lC_{l'}\rangle=\langle C_l\rangle\langle C_{l'}\rangle ⟨ C l C l ′ ⟩ = ⟨ C l ⟩ ⟨ C l ′ ⟩ l ≠ l ′ l\ne l' l = l ′ C l C_l C l l l l l = l ′ l=l' l = l ′ χ k 2 \chi_k^2 χ k 2 2 k 2k 2 k
⟨ C l 2 ⟩ − ⟨ C l ⟩ 2 = 2 k [ ⟨ C l ⟩ k ] 2 = ⟨ C l ⟩ 2 2 l + 1 . (173) \langle C_l^2\rangle-\langle C_l\rangle^2
=2k\left[\frac{\langle C_l\rangle}{k}\right]^2
=\frac{\langle C_l\rangle^2}{2l+1}. \tag*{(173)} ⟨ C l 2 ⟩ − ⟨ C l ⟩ 2 = 2 k [ k ⟨ C l ⟩ ] 2 = 2 l + 1 ⟨ C l ⟩ 2 . (173) The rescaling factor ⟨ C l ⟩ / k \langle C_l\rangle/k ⟨ C l ⟩ / k (169) ; our Eq. (173) is consistent with Eq. (22) from [61 ] .
The cosmic variance is obtained by substituting Eq. (173) into Eq. (172) and using Eq. (169) , giving
σ c o s m i c 2 ( γ ) = h 4 ∑ l = 2 ∞ 2 l + 1 ( l + 2 ) 2 ( l + 1 ) 2 l 2 ( l − 1 ) 2 P l 2 ( cos γ ) = h 4 μ ~ 2 ( γ ) . (174) \sigma_{\mathrm{cosmic}}^2(\gamma)
=h^4\sum_{l=2}^{\infty}\frac{2l+1}{(l+2)^2(l+1)^2l^2(l-1)^2}P_l^2(\cos\gamma)
=h^4\tilde{\mu}^{\,2}(\gamma). \tag*{(174)} σ cosmic 2 ( γ ) = h 4 l = 2 ∑ ∞ ( l + 2 ) 2 ( l + 1 ) 2 l 2 ( l − 1 ) 2 2 l + 1 P l 2 ( cos γ ) = h 4 μ ~ 2 ( γ ) . (174) While we have not given a rigorous proof, a simple numerical check (it is sufficient to include l = 2 , 3 , … , 10 l=2,3,\ldots,10 l = 2 , 3 , … , 10 (174) . Comparison of Eq. (171) with Eq. (139) and Eq. (174) with Eq. (166) shows that this is exactly the cosmic variance, with h ^ 4 = h 4 / 2 \hat{h}^4=h^4/2 h ^ 4 = h 4 /2 (151) . This agreement makes sense, since as stressed earlier, the existence of the sky map is predicated on the existence of a large number of low-noise pulsars. The cosmic covariance may be found with a similar calculation, see Ref. [38 ] .
(Note: several weeks after Eq. (174) appeared on the arXiv the same result was derived independently in [64].)
Unpolarized Point Source Mean μ u ( γ ) \mu_u(\gamma) μ u ( γ ) σ u 2 ( γ ) \sigma_u^2(\gamma) σ u 2 ( γ )
Here, we calculate the mean and variance of the standard “unpolarized” term in the pulsar redshift or timing residual correlation between two pulsars. The mean is the pulsar correlation for a single GW source of unit amplitude at a fixed point in the sky, averaged over all possible sky positions of the pulsar pair where their separation angle is fixed at γ \gamma γ
As shown in [25 ] , and as we have also demonstrated in Appendix § 2.1 , this is equivalent to fixing the pulsar pair at definite positions, and calculating the mean and variance as the location of the GW source is shifted uniformly around the sky. So this is how our calculation is structured, which is also how it was conceived by Hellings and Downs. Our calculation closely follows that given in Appendix C of Ref. [18 ] , where the mean (but not the variance) is found. Note the sign typo one line before Eq. (130) of Ref. [18 ] .
Our starting point is the correlation of pulsar timing residuals Eq. (135) for an unpolarized unit amplitude source
ρ ( Ω ) = F 1 + ( Ω ) F 2 + ( Ω ) + F 1 × ( Ω ) F 2 × ( Ω ) , (175) \rho(\boldsymbol{\Omega}) = F_1^+(\boldsymbol{\Omega})F_2^+(\boldsymbol{\Omega}) + F_1^\times(\boldsymbol{\Omega})F_2^\times(\boldsymbol{\Omega}), \tag*{(175)} ρ ( Ω ) = F 1 + ( Ω ) F 2 + ( Ω ) + F 1 × ( Ω ) F 2 × ( Ω ) , (175) where 1 and 2 label the pulsars at sky directions p 1 \boldsymbol{p}_1 p 1 p 2 \boldsymbol{p}_2 p 2 ⟨ ρ ⟩ ( γ = 0 ) = μ u ( γ = 0 ) = 1 / 3 \langle\rho\rangle(\gamma=0)=\mu_u(\gamma=0)=1/3 ⟨ ρ ⟩ ( γ = 0 ) = μ u ( γ = 0 ) = 1/3
Here, we use ⟨ Q ⟩ \langle Q\rangle ⟨ Q ⟩ Ω \boldsymbol{\Omega} Ω
⟨ Q ⟩ = 1 4 π ∫ d Ω Q ( Ω ) , (176) \langle Q\rangle= \frac{1}{4\pi}\int d\boldsymbol{\Omega}\,Q(\boldsymbol{\Omega}), \tag*{(176)} ⟨ Q ⟩ = 4 π 1 ∫ d Ω Q ( Ω ) , (176) where the normalization ensures that ⟨ 1 ⟩ = 1 \langle1\rangle=1 ⟨ 1 ⟩ = 1
μ u ( γ ) = ⟨ ρ ⟩ = 1 4 π ∫ d Ω ∑ A F 1 A ( Ω ) F 2 A ( Ω ) , (177) \mu_u(\gamma)=\langle\rho\rangle=\frac{1}{4\pi}\int d\boldsymbol{\Omega}\sum_A F_1^A(\boldsymbol{\Omega})F_2^A(\boldsymbol{\Omega}), \tag*{(177)} μ u ( γ ) = ⟨ ρ ⟩ = 4 π 1 ∫ d Ω A ∑ F 1 A ( Ω ) F 2 A ( Ω ) , (177) and the variance in the Hellings-Downs function is
σ u 2 ( γ ) = ⟨ ( Δ ρ ) 2 ⟩ = ⟨ ( ρ − ⟨ ρ ⟩ ) 2 ⟩ = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 , (178) \sigma_u^2(\gamma)=\langle(\Delta\rho)^2\rangle=\langle(\rho-\langle\rho\rangle)^2\rangle=\langle\rho^2\rangle-\langle\rho\rangle^2, \tag*{(178)} σ u 2 ( γ ) = ⟨( Δ ρ ) 2 ⟩ = ⟨( ρ − ⟨ ρ ⟩ ) 2 ⟩ = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 , (178) where the subscript “u” means “unpolarized” and
⟨ ρ 2 ⟩ = 1 4 π ∫ d Ω ( ∑ A F 1 A ( Ω ) F 2 A ( Ω ) ) 2 (179) \langle\rho^2\rangle=\frac{1}{4\pi}\int d\boldsymbol{\Omega}\left(\sum_A F_1^A(\boldsymbol{\Omega})F_2^A(\boldsymbol{\Omega})\right)^2 \tag*{(179)} ⟨ ρ 2 ⟩ = 4 π 1 ∫ d Ω ( A ∑ F 1 A ( Ω ) F 2 A ( Ω ) ) 2 (179) is the second moment.
To evaluate these quantities, we follow the approach of [18 ] , Appendix C. We write the transverse traceless symmetric polarization tensors as
e a b + ( Ω ) = m a m b − n a n b , and e_{ab}^+(\boldsymbol{\Omega})=m_a m_b-n_a n_b,\qquad\text{and} e ab + ( Ω ) = m a m b − n a n b , and e a b × ( Ω ) = m a n b + n a m b , (180) e_{ab}^\times(\boldsymbol{\Omega})=m_a n_b+n_a m_b, \tag*{(180)} e ab × ( Ω ) = m a n b + n a m b , (180) where n \boldsymbol{n} n m \boldsymbol{m} m Ω \boldsymbol{\Omega} Ω e a b + e_{ab}^+ e ab + e a b × e_{ab}^\times e ab × Ω \boldsymbol{\Omega} Ω
Closed forms for Ω \boldsymbol{\Omega} Ω n \boldsymbol{n} n m \boldsymbol{m} m [17 ] as:
Ω = x ^ cos ϕ sin θ + y ^ sin ϕ sin θ + z ^ cos θ , \boldsymbol{\Omega}=\hat{\boldsymbol{x}}\cos\phi\sin\theta+\hat{\boldsymbol{y}}\sin\phi\sin\theta+\hat{\boldsymbol{z}}\cos\theta, Ω = x ^ cos ϕ sin θ + y ^ sin ϕ sin θ + z ^ cos θ , m = x ^ sin ϕ − y ^ cos ϕ , \boldsymbol{m}=\hat{\boldsymbol{x}}\sin\phi-\hat{\boldsymbol{y}}\cos\phi, m = x ^ sin ϕ − y ^ cos ϕ , n = x ^ cos ϕ cos θ + y ^ sin ϕ cos θ − z ^ sin θ , (181) \boldsymbol{n}=\hat{\boldsymbol{x}}\cos\phi\cos\theta+\hat{\boldsymbol{y}}\sin\phi\cos\theta-\hat{\boldsymbol{z}}\sin\theta, \tag*{(181)} n = x ^ cos ϕ cos θ + y ^ sin ϕ cos θ − z ^ sin θ , (181) The reader unfamiliar with these forms should quickly verify that the three vectors are unit length and mutually orthogonal. We note in passing that m \boldsymbol{m} m n \boldsymbol{n} n
m ′ = m cos ϵ + n sin ϵ , and \boldsymbol{m}'=\boldsymbol{m}\cos\epsilon+\boldsymbol{n}\sin\epsilon,\qquad\text{and} m ′ = m cos ϵ + n sin ϵ , and n ′ = − m sin ϵ + n cos ϵ , (182) \boldsymbol{n}'=-\boldsymbol{m}\sin\epsilon+\boldsymbol{n}\cos\epsilon, \tag*{(182)} n ′ = − m sin ϵ + n cos ϵ , (182) where ϵ \epsilon ϵ Ω \boldsymbol{\Omega} Ω + + + × \times × ϵ \epsilon ϵ arbitrarily as a function of Ω \boldsymbol{\Omega} Ω any GW polarization basis, provided that the two basis vectors are orthogonal to one another, and orthogonal to the wave propagation direction Ω \boldsymbol{\Omega} Ω
Without loss of generality, let pulsars 1 and 2, separated by angle γ \gamma γ
p 1 = z ^ and p 2 = x ^ sin γ + z ^ cos γ . (183) \boldsymbol{p}_1=\hat{\boldsymbol{z}}\quad\text{and}\quad\boldsymbol{p}_2=\hat{\boldsymbol{x}}\sin\gamma+\hat{\boldsymbol{z}}\cos\gamma. \tag*{(183)} p 1 = z ^ and p 2 = x ^ sin γ + z ^ cos γ . (183) Employ the definitions Eq. (84) for F F F (180) for the polarization tensors, and Eq. (181) for the vectors Ω \boldsymbol{\Omega} Ω m \boldsymbol{m} m n \boldsymbol{n} n
F 1 × ( Ω ) = 0 , and F_1^\times(\boldsymbol{\Omega})=0,\qquad\text{and} F 1 × ( Ω ) = 0 , and F 1 + ( Ω ) = 1 2 p 1 a p 1 b 1 + Ω ⋅ p 1 e a b + ( Ω ) = 1 2 ( p 1 ⋅ m ) 2 − ( p 1 ⋅ n ) 2 1 + Ω ⋅ p 1 = − 1 2 sin 2 θ 1 + cos θ = 1 2 ( cos θ − 1 ) . (184) \begin{aligned}
F_1^+(\boldsymbol{\Omega})=\frac{1}{2}\frac{p_1^a p_1^b}{1+\boldsymbol{\Omega}\cdot\boldsymbol{p}_1}e_{ab}^+(\boldsymbol{\Omega}) \\
&=\frac{1}{2}\frac{(\boldsymbol{p}_1\cdot\boldsymbol{m})^2-(\boldsymbol{p}_1\cdot\boldsymbol{n})^2}{1+\boldsymbol{\Omega}\cdot\boldsymbol{p}_1} \\
&=-\frac{1}{2}\frac{\sin^2\theta}{1+\cos\theta} \\
&=\frac{1}{2}(\cos\theta-1). \tag*{(184)}
\end{aligned} F 1 + ( Ω ) = 2 1 1 + Ω ⋅ p 1 p 1 a p 1 b e ab + ( Ω ) = 2 1 1 + Ω ⋅ p 1 ( p 1 ⋅ m ) 2 − ( p 1 ⋅ n ) 2 = − 2 1 1 + cos θ sin 2 θ = 2 1 ( cos θ − 1 ) . (184) For the second pulsar, the expression for F 2 × ( Ω ) F^{\times}_{2}(\boldsymbol{\Omega}) F 2 × ( Ω ) (184) it would be multiplied by zero. The plus polarization term is
F 2 + ( Ω ) = 1 2 ( p 2 ⋅ m ) 2 − ( p 2 ⋅ n ) 2 1 + Ω ⋅ p 2 = ( sin γ sin ϕ ) 2 − ( sin γ cos ϕ cos θ − cos γ sin θ ) 2 2 ( 1 + sin γ cos ϕ sin θ + cos γ cos θ ) = sin 2 γ sin 2 ϕ − sin 2 γ cos 2 ϕ cos 2 θ − cos 2 γ sin 2 θ + 2 sin γ cos γ sin θ cos θ cos ϕ 2 ( 1 + sin γ cos ϕ sin θ + cos γ cos θ ) . (185) F^{+}_{2}(\boldsymbol{\Omega}) = \frac{1}{2}\frac{(\mathbf{p}_{2}\cdot\mathbf{m})^{2}-(\mathbf{p}_{2}\cdot\mathbf{n})^{2}}{1+\boldsymbol{\Omega}\cdot\mathbf{p}_{2}}
= \frac{(\sin\gamma\sin\phi)^{2}-(\sin\gamma\cos\phi\cos\theta-\cos\gamma\sin\theta)^{2}}{2(1+\sin\gamma\cos\phi\sin\theta+\cos\gamma\cos\theta)}
= \frac{\sin^{2}\gamma\sin^{2}\phi-\sin^{2}\gamma\cos^{2}\phi\cos^{2}\theta-\cos^{2}\gamma\sin^{2}\theta+2\sin\gamma\cos\gamma\sin\theta\cos\theta\cos\phi}{2(1+\sin\gamma\cos\phi\sin\theta+\cos\gamma\cos\theta)}. \tag*{(185)} F 2 + ( Ω ) = 2 1 1 + Ω ⋅ p 2 ( p 2 ⋅ m ) 2 − ( p 2 ⋅ n ) 2 = 2 ( 1 + sin γ cos ϕ sin θ + cos γ cos θ ) ( sin γ sin ϕ ) 2 − ( sin γ cos ϕ cos θ − cos γ sin θ ) 2 = 2 ( 1 + sin γ cos ϕ sin θ + cos γ cos θ ) sin 2 γ sin 2 ϕ − sin 2 γ cos 2 ϕ cos 2 θ − cos 2 γ sin 2 θ + 2 sin γ cos γ sin θ cos θ cos ϕ . (185) Since the product of the × \times × (184) and (185) gives ρ ( Ω ) = F 1 + F 2 + \rho(\boldsymbol{\Omega})=F^{+}_{1}F^{+}_{2} ρ ( Ω ) = F 1 + F 2 + ρ ( Ω ) \rho(\boldsymbol{\Omega}) ρ ( Ω ) q q q F 2 + F^{+}_{2} F 2 + (185) , namely
q = sin γ cos ϕ sin θ + cos γ cos θ . (186) q=\sin\gamma\cos\phi\sin\theta+\cos\gamma\cos\theta. \tag*{(186)} q = sin γ cos ϕ sin θ + cos γ cos θ . (186) The product ( 1 + q ) ( q − 1 ) (1+q)(q-1) ( 1 + q ) ( q − 1 )
( 1 + q ) ( q − 1 ) = − sin 2 γ sin 2 ϕ − sin 2 γ cos 2 ϕ cos 2 θ − cos 2 γ sin 2 θ + 2 sin γ cos γ sin θ cos θ cos ϕ , (187) (1+q)(q-1)=-\sin^{2}\gamma\sin^{2}\phi-\sin^{2}\gamma\cos^{2}\phi\cos^{2}\theta-\cos^{2}\gamma\sin^{2}\theta+2\sin\gamma\cos\gamma\sin\theta\cos\theta\cos\phi, \tag*{(187)} ( 1 + q ) ( q − 1 ) = − sin 2 γ sin 2 ϕ − sin 2 γ cos 2 ϕ cos 2 θ − cos 2 γ sin 2 θ + 2 sin γ cos γ sin θ cos θ cos ϕ , (187) which is identical to the numerator that appears in the final line of Eq. (185) , apart from the sign of the first term, sin 2 γ sin 2 ϕ \sin^{2}\gamma\sin^{2}\phi sin 2 γ sin 2 ϕ
ρ ( Ω ) = F 1 + F 2 + = 1 2 ( cos θ − 1 ) ( 1 + q ) ( q − 1 ) + 2 sin 2 γ sin 2 ϕ 2 ( 1 + q ) = 1 4 ( cos θ − 1 ) [ q − 1 + 2 sin 2 γ sin 2 ϕ 1 + q ] . (188) \begin{aligned}
\rho(\boldsymbol{\Omega})&=F^{+}_{1}F^{+}_{2}\\
&=\frac{1}{2}(\cos\theta-1)\frac{(1+q)(q-1)+2\sin^{2}\gamma\sin^{2}\phi}{2(1+q)}\\
&=\frac{1}{4}(\cos\theta-1)\left[q-1+\frac{2\sin^{2}\gamma\sin^{2}\phi}{1+q}\right]. \tag*{(188)}
\end{aligned} ρ ( Ω ) = F 1 + F 2 + = 2 1 ( cos θ − 1 ) 2 ( 1 + q ) ( 1 + q ) ( q − 1 ) + 2 sin 2 γ sin 2 ϕ = 4 1 ( cos θ − 1 ) [ q − 1 + 1 + q 2 sin 2 γ sin 2 ϕ ] . (188) We now proceed to the integrals.
Our goal is to average ρ ( Ω ) \rho(\boldsymbol{\Omega}) ρ ( Ω ) ϕ \phi ϕ ϕ \phi ϕ
ρ ( Ω ) = 1 4 [ u + v cos ϕ + w sin 2 ϕ r + s cos ϕ ] , (189) \rho(\boldsymbol{\Omega})=\frac{1}{4}\left[u+v\cos\phi+w\frac{\sin^{2}\phi}{r+s\cos\phi}\right], \tag*{(189)} ρ ( Ω ) = 4 1 [ u + v cos ϕ + w r + s cos ϕ sin 2 ϕ ] , (189) where the quantities u u u v v v w w w r r r s s s ϕ \phi ϕ (188) and the definition of q q q
u = ( cos θ − 1 ) ( cos γ cos θ − 1 ) , v = ( cos θ − 1 ) sin γ sin θ , w = ( cos θ − 1 ) 2 sin 2 γ , r = 1 + cos γ cos θ , and s = sin γ sin θ . (190) \begin{aligned}
u&=(\cos\theta-1)(\cos\gamma\cos\theta-1),\\
v&=(\cos\theta-1)\sin\gamma\sin\theta,\\
w&=(\cos\theta-1)2\sin^{2}\gamma,\\
r&=1+\cos\gamma\cos\theta,\qquad\text{and}\\
s&=\sin\gamma\sin\theta. \tag*{(190)}
\end{aligned} u v w r s = ( cos θ − 1 ) ( cos γ cos θ − 1 ) , = ( cos θ − 1 ) sin γ sin θ , = ( cos θ − 1 ) 2 sin 2 γ , = 1 + cos γ cos θ , and = sin γ sin θ . (190) In integrating ρ ( Ω ) \rho(\boldsymbol{\Omega}) ρ ( Ω ) ρ 2 ( Ω ) \rho^{2}(\boldsymbol{\Omega}) ρ 2 ( Ω ) ϕ \phi ϕ cos ϕ \cos\phi cos ϕ cos 2 ϕ \cos^{2}\phi cos 2 ϕ
I 1 ( θ ) = ∫ 0 2 π sin 2 ϕ r + s cos ϕ d ϕ , I 2 ( θ ) = ∫ 0 2 π sin 2 ϕ cos ϕ r + s cos ϕ d ϕ , and I 3 ( θ ) = ∫ 0 2 π sin 4 ϕ ( r + s cos ϕ ) 2 d ϕ . (191) \begin{aligned}
I_{1}(\theta)&=\int_{0}^{2\pi}\frac{\sin^{2}\phi}{r+s\cos\phi}\,d\phi,\\
I_{2}(\theta)&=\int_{0}^{2\pi}\frac{\sin^{2}\phi\cos\phi}{r+s\cos\phi}\,d\phi,\qquad\text{and}\\
I_{3}(\theta)&=\int_{0}^{2\pi}\frac{\sin^{4}\phi}{(r+s\cos\phi)^{2}}\,d\phi. \tag*{(191)}
\end{aligned} I 1 ( θ ) I 2 ( θ ) I 3 ( θ ) = ∫ 0 2 π r + s cos ϕ sin 2 ϕ d ϕ , = ∫ 0 2 π r + s cos ϕ sin 2 ϕ cos ϕ d ϕ , and = ∫ 0 2 π ( r + s cos ϕ ) 2 sin 4 ϕ d ϕ . (191) For reasons that will be clear later, we indicate their dependence on θ \theta θ γ \gamma γ [65 ] .) We first carry out this evaluation and then return to the main calculation.
To evaluate the integrals I 1 , 2 , 3 I_{1,2,3} I 1 , 2 , 3 z = e i ϕ z=e^{i\phi} z = e i ϕ Figure 10 , which loops once around the origin in the counterclockwise direction, at unit distance
Figure 10. The contour of integration for the integrals in Eq. (194) goes around the unit circle once in the counterclockwise direction. The integrands have poles along the negative real axis at z = z 1 z=z_{1} z = z 1 z = z 2 z=z_{2} z = z 2 θ < π − γ \theta<\pi-\gamma θ < π − γ z 1 z_{1} z 1 z 2 z_{2} z 2 θ > π − γ \theta>\pi-\gamma θ > π − γ z 2 z_{2} z 2 z 1 z_{1} z 1 Table I .
from the origin. Since cos ϕ = ( z + 1 / z ) / 2 \cos\phi=(z+1/z)/2 cos ϕ = ( z + 1/ z ) /2
r + s cos ϕ = sin θ sin γ 2 z ( z − z 1 ) ( z − z 2 ) . (192) r+s\cos\phi=\frac{\sin\theta\sin\gamma}{2z}(z-z_1)(z-z_2). \tag*{(192)} r + s cos ϕ = 2 z sin θ sin γ ( z − z 1 ) ( z − z 2 ) . (192) or as the square of this quantity. The two roots are found by substituting r r r s s s cos ϕ \cos\phi cos ϕ z z z
z 1 = − ( 1 − cos γ ) ( 1 − cos θ ) sin γ sin θ , z 2 = − ( 1 + cos γ ) ( 1 + cos θ ) sin γ sin θ . (193) z_1=-\frac{(1-\cos\gamma)(1-\cos\theta)}{\sin\gamma\sin\theta},
\qquad
z_2=-\frac{(1+\cos\gamma)(1+\cos\theta)}{\sin\gamma\sin\theta}. \tag*{(193)} z 1 = − sin γ sin θ ( 1 − cos γ ) ( 1 − cos θ ) , z 2 = − sin γ sin θ ( 1 + cos γ ) ( 1 + cos θ ) . (193) Making use of sin ϕ = ( z − 1 / z ) / 2 i \sin\phi=(z-1/z)/2i sin ϕ = ( z − 1/ z ) /2 i d ϕ = − i d z / z d\phi=-idz/z d ϕ = − i d z / z
I 1 = i 2 sin θ sin γ ∮ d z ( z − 1 / z ) 2 ( z − z 1 ) ( z − z 2 ) , I_1=\frac{i}{2\sin\theta\sin\gamma}\oint dz\,\frac{(z-1/z)^2}{(z-z_1)(z-z_2)}, I 1 = 2 sin θ sin γ i ∮ d z ( z − z 1 ) ( z − z 2 ) ( z − 1/ z ) 2 , I 2 = i 4 sin θ sin γ ∮ d z ( z − 1 / z ) 2 ( z + 1 / z ) ( z − z 1 ) ( z − z 2 ) , I 3 = − i 4 sin 2 θ sin 2 γ ∮ d z z ( z − 1 / z ) 4 ( z − z 1 ) 2 ( z − z 2 ) 2 . (194) I_2=\frac{i}{4\sin\theta\sin\gamma}\oint dz\,\frac{(z-1/z)^2(z+1/z)}{(z-z_1)(z-z_2)},
\qquad
I_3=-\frac{i}{4\sin^2\theta\sin^2\gamma}\oint dz\,\frac{z(z-1/z)^4}{(z-z_1)^2(z-z_2)^2}. \tag*{(194)} I 2 = 4 sin θ sin γ i ∮ d z ( z − z 1 ) ( z − z 2 ) ( z − 1/ z ) 2 ( z + 1/ z ) , I 3 = − 4 sin 2 θ sin 2 γ i ∮ d z ( z − z 1 ) 2 ( z − z 2 ) 2 z ( z − 1/ z ) 4 . (194) The path of integration is shown in Figure 10 . Note that the locations of the poles depend upon θ \theta θ γ \gamma γ θ < π − γ \theta<\pi-\gamma θ < π − γ z 1 z_1 z 1 z 2 z_2 z 2 z 2 z_2 z 2 z 1 z_1 z 1
All three integrands are ratios of polynomials, so are analytic on the entire plane apart from the poles at z = 0 z=0 z = 0 z = z 1 z=z_1 z = z 1 z = z 2 z=z_2 z = z 2 2 π i 2\pi i 2 π i Table I . In effect, these integrals are evaluated by taking derivatives to find the residues.
Table I. Residues of integrals in Eq. (194) , without the factors that appear before the integral symbol. Note that the final row can be obtained from the previous row by swapping z 1 z_1 z 1 z 2 z_2 z 2
The first integral is evaluated as follows. For θ < π − γ \theta<\pi-\gamma θ < π − γ z = 0 z=0 z = 0 z = z 1 z=z_1 z = z 1
I 1 < = i ( 2 π i ) ( R 0 + R 1 ) 2 sin θ sin γ = 2 π ( 1 + cos γ ) ( 1 + cos θ ) . (195) I_{1<}=\frac{i(2\pi i)(R_0+R_1)}{2\sin\theta\sin\gamma}
=\frac{2\pi}{(1+\cos\gamma)(1+\cos\theta)}. \tag*{(195)} I 1 < = 2 sin θ sin γ i ( 2 π i ) ( R 0 + R 1 ) = ( 1 + cos γ ) ( 1 + cos θ ) 2 π . (195) whereas for θ > π − γ \theta>\pi-\gamma θ > π − γ z = 0 z=0 z = 0 z = z 2 z=z_2 z = z 2
I 1 > = i ( 2 π i ) ( R 0 + R 2 ) 2 sin θ sin γ = 2 π ( 1 − cos γ ) ( 1 − cos θ ) . (196) I_{1>}=\frac{i(2\pi i)(R_0+R_2)}{2\sin\theta\sin\gamma}
=\frac{2\pi}{(1-\cos\gamma)(1-\cos\theta)}. \tag*{(196)} I 1 > = 2 sin θ sin γ i ( 2 π i ) ( R 0 + R 2 ) = ( 1 − cos γ ) ( 1 − cos θ ) 2 π . (196) Here, R 0 R_0 R 0 R 1 R_1 R 1 R 2 R_2 R 2 Table I , and we have added “< < < > > > I I I θ \theta θ
The behavior is similar for the remaining two integrals. For θ < π − γ \theta<\pi-\gamma θ < π − γ
I 2 < = i ( 2 π i ) ( R 0 + R 1 ) 4 sin θ sin γ = − π ( 1 − cos γ ) ( 1 − cos θ ) sin γ sin θ ( 1 + cos γ ) ( 1 + cos θ ) . (197) I_{2<}=\frac{i(2\pi i)(R_0+R_1)}{4\sin\theta\sin\gamma}
=-\frac{\pi(1-\cos\gamma)(1-\cos\theta)}
{\sin\gamma\sin\theta(1+\cos\gamma)(1+\cos\theta)}. \tag*{(197)} I 2 < = 4 sin θ sin γ i ( 2 π i ) ( R 0 + R 1 ) = − sin γ sin θ ( 1 + cos γ ) ( 1 + cos θ ) π ( 1 − cos γ ) ( 1 − cos θ ) . (197) and for θ > π − γ \theta>\pi-\gamma θ > π − γ
I 2 > = i ( 2 π i ) ( R 0 + R 2 ) 4 sin θ sin γ = − π ( 1 + cos γ ) ( 1 + cos θ ) sin γ sin θ ( 1 − cos γ ) ( 1 − cos θ ) . (198) I_{2>}=\frac{i(2\pi i)(R_0+R_2)}{4\sin\theta\sin\gamma}
=-\frac{\pi(1+\cos\gamma)(1+\cos\theta)}
{\sin\gamma\sin\theta(1-\cos\gamma)(1-\cos\theta)}. \tag*{(198)} I 2 > = 4 sin θ sin γ i ( 2 π i ) ( R 0 + R 2 ) = − sin γ sin θ ( 1 − cos γ ) ( 1 − cos θ ) π ( 1 + cos γ ) ( 1 + cos θ ) . (198) For θ < π − γ \theta<\pi-\gamma θ < π − γ
I 3 < = − i ( 2 π i ) ( R 0 + R 1 ) 4 sin 2 θ sin 2 γ = 3 π ( 1 + cos γ ) 2 ( 1 + cos θ ) 2 . (199) I_{3<}=-\frac{i(2\pi i)(R_0+R_1)}{4\sin^2\theta\sin^2\gamma}
=\frac{3\pi}{(1+\cos\gamma)^2(1+\cos\theta)^2}. \tag*{(199)} I 3 < = − 4 sin 2 θ sin 2 γ i ( 2 π i ) ( R 0 + R 1 ) = ( 1 + cos γ ) 2 ( 1 + cos θ ) 2 3 π . (199) and for θ > π − γ \theta>\pi-\gamma θ > π − γ
I 3 > = − i ( 2 π i ) ( R 0 + R 2 ) 4 sin 2 θ sin 2 γ = 3 π ( 1 − cos γ ) 2 ( 1 − cos θ ) 2 . (200) I_{3>}=-\frac{i(2\pi i)(R_0+R_2)}{4\sin^2\theta\sin^2\gamma}
=\frac{3\pi}{(1-\cos\gamma)^2(1-\cos\theta)^2}. \tag*{(200)} I 3 > = − 4 sin 2 θ sin 2 γ i ( 2 π i ) ( R 0 + R 2 ) = ( 1 − cos γ ) 2 ( 1 − cos θ ) 2 3 π . (200) With these three integrals in hand, only integration over θ \theta θ
We begin by computing the Hellings-Downs function ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩ (176) and (189) , which give
⟨ ρ ⟩ = 1 16 π ∫ d Ω [ u + v cos ϕ + w sin 2 ϕ r + s cos ϕ ] . (201) \langle\rho\rangle=\frac{1}{16\pi}\int d\Omega\left[u+v\cos\phi+w\frac{\sin^2\phi}{r+s\cos\phi}\right]. \tag*{(201)} ⟨ ρ ⟩ = 16 π 1 ∫ d Ω [ u + v cos ϕ + w r + s cos ϕ sin 2 ϕ ] . (201) The integral over ϕ \phi ϕ
⟨ ρ ⟩ = 1 16 π ∫ 0 π d θ sin θ [ 2 π u + w I 1 ( θ ) ] . (202) \langle\rho\rangle=\frac{1}{16\pi}\int_0^\pi d\theta\sin\theta\left[2\pi u+wI_1(\theta)\right]. \tag*{(202)} ⟨ ρ ⟩ = 16 π 1 ∫ 0 π d θ sin θ [ 2 π u + w I 1 ( θ ) ] . (202) where the v cos ϕ v\cos\phi v cos ϕ I 1 I_1 I 1 (191) and u u u w w w (190) . The integral of u sin θ u\sin\theta u sin θ (195) and (D22), the remaining integral must be split into two ranges, giving
μ u ( γ ) = ⟨ ρ ⟩ = 1 4 [ 1 + cos γ 3 + 1 4 π ( ∫ 0 π − γ d θ sin θ w I 1 < ( θ ) + ∫ π − γ π d θ sin θ w I 1 > ( θ ) ) ] = 1 4 [ 1 + 1 3 cos γ + ( 1 − cos γ ) ( 1 + cos γ + 2 log ( 1 − cos γ 2 ) ) − sin 2 γ ] = 1 4 + 1 12 cos γ + 1 2 ( 1 − cos γ ) log ( 1 − cos γ 2 ) . (203) \begin{aligned}
\mu_u(\gamma)=\langle\rho\rangle&=\frac{1}{4}\left[1+\frac{\cos\gamma}{3}+\frac{1}{4\pi}\left(\int_0^{\pi-\gamma}d\theta\sin\theta\,wI_{1<}(\theta)+\int_{\pi-\gamma}^{\pi}d\theta\sin\theta\,wI_{1>}(\theta)\right)\right]\\
&=\frac{1}{4}\left[1+\frac{1}{3}\cos\gamma+(1-\cos\gamma)\left(1+\cos\gamma+2\log\left(\frac{1-\cos\gamma}{2}\right)\right)-\sin^2\gamma\right]\\
&=\frac{1}{4}+\frac{1}{12}\cos\gamma+\frac{1}{2}(1-\cos\gamma)\log\left(\frac{1-\cos\gamma}{2}\right). \tag*{(203)}
\end{aligned} μ u ( γ ) = ⟨ ρ ⟩ = 4 1 [ 1 + 3 cos γ + 4 π 1 ( ∫ 0 π − γ d θ sin θ w I 1 < ( θ ) + ∫ π − γ π d θ sin θ w I 1 > ( θ ) ) ] = 4 1 [ 1 + 3 1 cos γ + ( 1 − cos γ ) ( 1 + cos γ + 2 log ( 2 1 − cos γ ) ) − sin 2 γ ] = 4 1 + 12 1 cos γ + 2 1 ( 1 − cos γ ) log ( 2 1 − cos γ ) . (203) This is the standard Hellings-Downs curve. As γ → 0 \gamma\to0 γ → 0 1 / 3 1/3 1/3 [11 ] ) but note that many authors normalize it to 1 / 2 1/2 1/2
Now we proceed to the second moment and variance. To simplify the notation in what follows, we define the function
Ψ ( γ ) = ( 1 − cos γ ) log ( 1 − cos γ 2 ) = 4 sin 2 ( γ / 2 ) log ( sin ( γ / 2 ) ) . (204) \begin{aligned}
\Psi(\gamma)&=(1-\cos\gamma)\log\left(\frac{1-\cos\gamma}{2}\right)\\
&=4\sin^2(\gamma/2)\log\left(\sin(\gamma/2)\right). \tag*{(204)}
\end{aligned} Ψ ( γ ) = ( 1 − cos γ ) log ( 2 1 − cos γ ) = 4 sin 2 ( γ /2 ) log ( sin ( γ /2 ) ) . (204) To calculate ⟨ ρ 2 ⟩ \langle\rho^2\rangle ⟨ ρ 2 ⟩ (176) and (D15), squaring the latter to obtain
⟨ ρ 2 ⟩ = 1 64 π ∫ d Ω [ u 2 + 2 u v cos ϕ + v 2 cos 2 ϕ + 2 u w sin 2 ϕ r + s cos ϕ + 2 v w cos ϕ sin 2 ϕ r + s cos ϕ + w 2 sin 4 ϕ ( r + s cos ϕ ) 2 ] = 1 64 π ∫ 0 π d θ sin θ [ 2 π u 2 + π v 2 + 2 u w I 1 ( θ ) + 2 v w I 2 ( θ ) + w 2 I 3 ( θ ) ] . (205) \begin{aligned}
\langle\rho^2\rangle={}&\frac{1}{64\pi}\int d\Omega\left[u^2+2uv\cos\phi+v^2\cos^2\phi+2uw\frac{\sin^2\phi}{r+s\cos\phi}+2vw\frac{\cos\phi\sin^2\phi}{r+s\cos\phi}+w^2\frac{\sin^4\phi}{(r+s\cos\phi)^2}\right]\\
={}&\frac{1}{64\pi}\int_0^\pi d\theta\sin\theta\left[2\pi u^2+\pi v^2+2uwI_1(\theta)+2vwI_2(\theta)+w^2I_3(\theta)\right]. \tag*{(205)}
\end{aligned} ⟨ ρ 2 ⟩ = = 64 π 1 ∫ d Ω [ u 2 + 2 uv cos ϕ + v 2 cos 2 ϕ + 2 u w r + s cos ϕ sin 2 ϕ + 2 v w r + s cos ϕ cos ϕ sin 2 ϕ + w 2 ( r + s cos ϕ ) 2 sin 4 ϕ ] 64 π 1 ∫ 0 π d θ sin θ [ 2 π u 2 + π v 2 + 2 u w I 1 ( θ ) + 2 v w I 2 ( θ ) + w 2 I 3 ( θ ) ] . (205) where the functions u u u v v v w w w (190) . On the second line, we integrated over ϕ \phi ϕ (191) .
The θ \theta θ (205) is straightforward. We can integrate the u 2 u^2 u 2 v 2 v^2 v 2
13 120 + 1 12 cos γ + 1 120 cos 2 γ . (206) \frac{13}{120}+\frac{1}{12}\cos\gamma+\frac{1}{120}\cos^2\gamma. \tag*{(206)} 120 13 + 12 1 cos γ + 120 1 cos 2 γ . (206) The remaining three integrals in Eq. (205) , containing I 1 I_1 I 1 I 2 I_2 I 2 I 3 I_3 I 3 ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩ (203) . Each θ \theta θ 0 0 0 π − γ \pi-\gamma π − γ π − γ \pi-\gamma π − γ π \pi π (195) –(D26). The logarithmic divergences around π − γ \pi-\gamma π − γ
I 1 : 1 24 ( 1 − cos 2 γ ) ( 3 + 6 cos γ + cos 2 γ ) + 1 2 ( 1 + cos γ ) Ψ ( γ ) . (207) I_1:\ \frac{1}{24}(1-\cos^2\gamma)(3+6\cos\gamma+\cos^2\gamma)+\frac{1}{2}(1+\cos\gamma)\Psi(\gamma). \tag*{(207)} I 1 : 24 1 ( 1 − cos 2 γ ) ( 3 + 6 cos γ + cos 2 γ ) + 2 1 ( 1 + cos γ ) Ψ ( γ ) . (207) I 2 : 1 24 ( 1 − cos γ ) 2 ( 7 + 8 cos γ + cos 2 γ ) + 1 2 ( 1 − cos γ ) Ψ ( γ ) , (208) I_2:\ \frac{1}{24}(1-\cos\gamma)^2(7+8\cos\gamma+\cos^2\gamma)+\frac{1}{2}(1-\cos\gamma)\Psi(\gamma), \tag*{(208)} I 2 : 24 1 ( 1 − cos γ ) 2 ( 7 + 8 cos γ + cos 2 γ ) + 2 1 ( 1 − cos γ ) Ψ ( γ ) , (208) and
I 3 : 3 4 ( 1 − cos γ ) ( 1 + cos γ + Ψ ( γ ) ) . (209) I_3:\ \frac{3}{4}(1-\cos\gamma)\left(1+\cos\gamma+\Psi(\gamma)\right). \tag*{(209)} I 3 : 4 3 ( 1 − cos γ ) ( 1 + cos γ + Ψ ( γ ) ) . (209) Summing together Eqs. (206) –(D35) yields
⟨ ρ 2 ⟩ = 51 40 + 1 12 cos γ − 139 120 cos 2 γ + 1 4 ( 7 − 3 cos γ ) Ψ ( γ ) . (210) \langle\rho^2\rangle=\frac{51}{40}+\frac{1}{12}\cos\gamma-\frac{139}{120}\cos^2\gamma+\frac{1}{4}(7-3\cos\gamma)\Psi(\gamma). \tag*{(210)} ⟨ ρ 2 ⟩ = 40 51 + 12 1 cos γ − 120 139 cos 2 γ + 4 1 ( 7 − 3 cos γ ) Ψ ( γ ) . (210) This completes the θ \theta θ (205) . From the first and second moments of ρ \rho ρ (203) and (D36), we can compute the variance in the unpolarized term, which is
σ u 2 ( γ ) = ⟨ Δ ρ 2 ⟩ = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 = 97 80 + 1 24 cos γ − 839 720 cos 2 γ + 1 12 ( 18 − 10 cos γ − 3 Ψ ( γ ) ) Ψ ( γ ) . (211) \begin{aligned}
\sigma_u^2(\gamma)=\langle\Delta\rho^2\rangle&=\langle\rho^2\rangle-\langle\rho\rangle^2\\
&=\frac{97}{80}+\frac{1}{24}\cos\gamma-\frac{839}{720}\cos^2\gamma+\frac{1}{12}(18-10\cos\gamma-3\Psi(\gamma))\Psi(\gamma). \tag*{(211)}
\end{aligned} σ u 2 ( γ ) = ⟨ Δ ρ 2 ⟩ = ⟨ ρ 2 ⟩ − ⟨ ρ ⟩ 2 = 80 97 + 24 1 cos γ − 720 839 cos 2 γ + 12 1 ( 18 − 10 cos γ − 3Ψ ( γ )) Ψ ( γ ) . (211) We note that cubic and higher moments of ρ \rho ρ
Polarized Point Source Mean and Variance σ p 2 ( γ ) \sigma_p^2(\gamma) σ p 2 ( γ )
Here, we calculate the mean and variance of the polarized correlation term in the Hellings-Downs correlation
ρ ( Ω ) = F 1 + ( Ω ) F 2 × ( Ω ) − F 1 × ( Ω ) F 2 + ( Ω ) . (212) \rho(\boldsymbol{\Omega})=F_1^+(\boldsymbol{\Omega})F_2^\times(\boldsymbol{\Omega})-F_1^\times(\boldsymbol{\Omega})F_2^+(\boldsymbol{\Omega}). \tag*{(212)} ρ ( Ω ) = F 1 + ( Ω ) F 2 × ( Ω ) − F 1 × ( Ω ) F 2 + ( Ω ) . (212) The formalism and notation are the same as those of Appendix § 13 , apart from using the subscript “p” for “polarized.” We will see that the mean value μ p ( γ ) = ⟨ ρ ⟩ \mu_p(\gamma)=\langle\rho\rangle μ p ( γ ) = ⟨ ρ ⟩ σ p 2 ( γ ) = ⟨ ρ 2 ⟩ \sigma_p^2(\gamma)=\langle\rho^2\rangle σ p 2 ( γ ) = ⟨ ρ 2 ⟩
Following the same steps as in Appendix § 13 , we arrive at
ρ ( Ω ) = 1 4 [ u ˉ sin ϕ + v ˉ sin ϕ cos ϕ r + s cos ϕ ] , (213) \rho(\boldsymbol{\Omega})=\frac{1}{4}\left[\frac{\bar{u}\sin\phi+\bar{v}\sin\phi\cos\phi}{r+s\cos\phi}\right], \tag*{(213)} ρ ( Ω ) = 4 1 [ r + s cos ϕ u ˉ sin ϕ + v ˉ sin ϕ cos ϕ ] , (213) which should be compared with Eq. (189) . Here, we have defined:
u ˉ = − 2 ( cos θ − 1 ) sin γ cos γ sin θ , \bar{u}=-2(\cos\theta-1)\sin\gamma\cos\gamma\sin\theta, u ˉ = − 2 ( cos θ − 1 ) sin γ cos γ sin θ , v ˉ = 2 ( cos θ − 1 ) sin 2 γ cos θ , \bar{v}=2(\cos\theta-1)\sin^2\gamma\cos\theta, v ˉ = 2 ( cos θ − 1 ) sin 2 γ cos θ , r = 1 + cos γ cos θ , and r=1+\cos\gamma\cos\theta,\qquad\text{and} r = 1 + cos γ cos θ , and s = sin γ sin θ . (214) s=\sin\gamma\sin\theta. \tag*{(214)} s = sin γ sin θ . (214) Note that the quantity in square brackets in Eq. (213) is precisely the quantity U W = ( 1 − cos θ ) 2 A B / ( 1 + q ) UW=(1-\cos\theta)^2AB/(1+q) U W = ( 1 − cos θ ) 2 A B / ( 1 + q ) (96) , whose square appears as the polarization term ⟨ U 2 W 2 ⟩ \langle U^2W^2\rangle ⟨ U 2 W 2 ⟩ (106) .
The integral of Eq. (213) over ϕ ∈ [ 0 , 2 π ] \phi\in[0,2\pi] ϕ ∈ [ 0 , 2 π ] ϕ = π \phi=\pi ϕ = π ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩
To calculate the variance ⟨ ρ 2 ⟩ \langle\rho^2\rangle ⟨ ρ 2 ⟩
I 4 ( θ ) = ∫ 0 2 π sin 2 ϕ ( r + s cos ϕ ) 2 d ϕ , I_4(\theta)=\int_0^{2\pi}\frac{\sin^2\phi}{(r+s\cos\phi)^2}\,\mathrm{d}\phi, I 4 ( θ ) = ∫ 0 2 π ( r + s cos ϕ ) 2 sin 2 ϕ d ϕ , I 5 ( θ ) = ∫ 0 2 π sin 2 ϕ cos ϕ ( r + s cos ϕ ) 2 d ϕ , and I_5(\theta)=\int_0^{2\pi}\frac{\sin^2\phi\cos\phi}{(r+s\cos\phi)^2}\,\mathrm{d}\phi,\qquad\text{and} I 5 ( θ ) = ∫ 0 2 π ( r + s cos ϕ ) 2 sin 2 ϕ cos ϕ d ϕ , and I 6 ( θ ) = ∫ 0 2 π sin 2 ϕ cos 2 ϕ ( r + s cos ϕ ) 2 d ϕ . (215) I_6(\theta)=\int_0^{2\pi}\frac{\sin^2\phi\cos^2\phi}{(r+s\cos\phi)^2}\,\mathrm{d}\phi. \tag*{(215)} I 6 ( θ ) = ∫ 0 2 π ( r + s cos ϕ ) 2 sin 2 ϕ cos 2 ϕ d ϕ . (215) Following exactly the same methods as used in Appendix § 13 starting from Eq. (191) , these evaluate to:
I 4 ( θ ) = ∓ 2 π ( cos γ + cos θ ) ( 1 ∓ cos γ ) ( 1 ∓ cos θ ) , I_4(\theta)=\frac{\mp2\pi}{(\cos\gamma+\cos\theta)(1\mp\cos\gamma)(1\mp\cos\theta)}, I 4 ( θ ) = ( cos γ + cos θ ) ( 1 ∓ cos γ ) ( 1 ∓ cos θ ) ∓ 2 π , I 5 ( θ ) = ± 2 π ( 1 ± cos γ ) ( 1 ± cos θ ) ( cos γ + cos θ ) sin γ sin θ ( 1 ∓ cos γ ) ( 1 ∓ cos θ ) , I_5(\theta)=\frac{\pm2\pi(1\pm\cos\gamma)(1\pm\cos\theta)}{(\cos\gamma+\cos\theta)\sin\gamma\sin\theta(1\mp\cos\gamma)(1\mp\cos\theta)}, I 5 ( θ ) = ( cos γ + cos θ ) sin γ sin θ ( 1 ∓ cos γ ) ( 1 ∓ cos θ ) ± 2 π ( 1 ± cos γ ) ( 1 ± cos θ ) , I 6 ( θ ) = ∓ 2 π ( cos γ + cos θ ) ( 1 ∓ cos γ ) ( 1 ∓ cos θ ) − ∓ 3 π ( 1 ∓ cos γ ) 2 ( 1 ∓ cos θ ) 2 . (216) I_6(\theta)=\frac{\mp2\pi}{(\cos\gamma+\cos\theta)(1\mp\cos\gamma)(1\mp\cos\theta)}
-\frac{\mp3\pi}{(1\mp\cos\gamma)^2(1\mp\cos\theta)^2}. \tag*{(216)} I 6 ( θ ) = ( cos γ + cos θ ) ( 1 ∓ cos γ ) ( 1 ∓ cos θ ) ∓ 2 π − ( 1 ∓ cos γ ) 2 ( 1 ∓ cos θ ) 2 ∓ 3 π . (216) Here, the upper sign is for θ > π − γ \theta>\pi-\gamma θ > π − γ θ < π − γ \theta<\pi-\gamma θ < π − γ
The integrals needed to evaluate the variance are
1 4 π ∫ 0 π − γ d θ sin θ [ u ˉ 2 I 4 ( θ ) + 2 u ˉ v ˉ I 5 ( θ ) + v ˉ 2 I 6 ( θ ) ] = 1 3 ( cos 2 γ − 1 ) ( cos 3 γ + 3 cos 2 γ + 58 ) + 4 ( 3 cos γ − 7 ) ( 1 − cos γ ) log ( 1 − cos γ 2 ) (217) \frac{1}{4\pi}\int_0^{\pi-\gamma}\mathrm{d}\theta\,\sin\theta\left[\bar{u}^2I_4(\theta)+2\bar{u}\bar{v}I_5(\theta)+\bar{v}^2I_6(\theta)\right]
=\frac{1}{3}\left(\cos^2\gamma-1\right)\left(\cos^3\gamma+3\cos^2\gamma+58\right)
+4(3\cos\gamma-7)(1-\cos\gamma)\log\left(\frac{1-\cos\gamma}{2}\right) \tag*{(217)} 4 π 1 ∫ 0 π − γ d θ sin θ [ u ˉ 2 I 4 ( θ ) + 2 u ˉ v ˉ I 5 ( θ ) + v ˉ 2 I 6 ( θ ) ] = 3 1 ( cos 2 γ − 1 ) ( cos 3 γ + 3 cos 2 γ + 58 ) + 4 ( 3 cos γ − 7 ) ( 1 − cos γ ) log ( 2 1 − cos γ ) (217) and
1 4 π ∫ π − γ π d θ sin θ [ u ˉ 2 I 4 ( θ ) + 2 u ˉ v ˉ I 5 ( θ ) + v ˉ 2 I 6 ( θ ) ] = 1 3 ( 1 − cos 2 γ ) ( cos 3 γ + 3 cos 2 γ + 2 ) . (218) \frac{1}{4\pi}\int_{\pi-\gamma}^{\pi}\mathrm{d}\theta\,\sin\theta\left[\bar{u}^2I_4(\theta)+2\bar{u}\bar{v}I_5(\theta)+\bar{v}^2I_6(\theta)\right]
=\frac{1}{3}\left(1-\cos^2\gamma\right)\left(\cos^3\gamma+3\cos^2\gamma+2\right). \tag*{(218)} 4 π 1 ∫ π − γ π d θ sin θ [ u ˉ 2 I 4 ( θ ) + 2 u ˉ v ˉ I 5 ( θ ) + v ˉ 2 I 6 ( θ ) ] = 3 1 ( 1 − cos 2 γ ) ( cos 3 γ + 3 cos 2 γ + 2 ) . (218) Note that in Appendix § 13 , the integrals for I 1 I_1 I 1 I 2 I_2 I 2 I 3 I_3 I 3 I 4 I_4 I 4 I 5 I_5 I 5 I 6 I_6 I 6 θ = π − γ \theta=\pi-\gamma θ = π − γ
Adding together Eqs. (217) and (218) and reinserting the factor of ( 1 / 4 ) 2 (1/4)^2 ( 1/4 ) 2 (213) , we obtain the second moment ⟨ ρ 2 ⟩ \langle\rho^2\rangle ⟨ ρ 2 ⟩ ⟨ ρ ⟩ \langle\rho\rangle ⟨ ρ ⟩
σ p 2 ( γ ) = ⟨ ρ 2 ⟩ = 7 6 ( cos 2 γ − 1 ) + 1 4 ( 3 cos γ − 7 ) ( 1 − cos γ ) log ( 1 − cos γ 2 ) (219) \sigma_p^2(\gamma)=\langle\rho^2\rangle
=\frac{7}{6}\left(\cos^2\gamma-1\right)+\frac{1}{4}(3\cos\gamma-7)(1-\cos\gamma)\log\left(\frac{1-\cos\gamma}{2}\right) \tag*{(219)} σ p 2 ( γ ) = ⟨ ρ 2 ⟩ = 6 7 ( cos 2 γ − 1 ) + 4 1 ( 3 cos γ − 7 ) ( 1 − cos γ ) log ( 2 1 − cos γ ) (219) for the variance of the polarized term in the Hellings-Downs correlation.
The square root of this quantity (the standard deviation) is shown in Figure 11 , where it is compared with that of the unpolarized Hellings-Downs expressions in Eqs. (5) and (7) .
Figure 11. The standard deviation σ p ( γ ) \sigma_p(\gamma) σ p ( γ ) μ ( γ ) \mu(\gamma) μ ( γ )
Cross-Polarized Variance σ c 2 ( γ ) \sigma_c^2(\gamma) σ c 2 ( γ )
The final quantity needed to describe the variance of the Hellings-Downs correlation for a general polarized GW source is the average value of the cross term
( F 1 + F 2 + + F 1 × F 2 × ) 2 + ( F 1 + F 2 × − F 1 × F 2 + ) 2 = ( F 1 + F 1 + + F 1 × F 1 × ) ( F 2 + F 2 + + F 2 × F 2 × ) . (220) \begin{aligned}
\left(F_1^+F_2^+ + F_1^\times F_2^\times\right)^2 + \left(F_1^+F_2^\times- F_1^\times F_2^+\right)^2 \\
&= \left(F_1^+F_1^+ + F_1^\times F_1^\times\right)\left(F_2^+F_2^+ + F_2^\times F_2^\times\right). \tag*{(220)}
\end{aligned} ( F 1 + F 2 + + F 1 × F 2 × ) 2 + ( F 1 + F 2 × − F 1 × F 2 + ) 2 = ( F 1 + F 1 + + F 1 × F 1 × ) ( F 2 + F 2 + + F 2 × F 2 × ) . (220) In the final line, the terms associated with the first pulsar are collected on the left, and those with the second pulsar on the right, which is particularly convenient for computation.
Starting from the final line of Eq. (220) , the cross term in the variance is defined as the sky average
σ c 2 ( γ ) = 1 4 π ∫ d Ω ( F 1 + ( Ω ) F 1 + ( Ω ) + F 1 × ( Ω ) F 1 × ( Ω ) ) ( F 2 + ( Ω ) F 2 + ( Ω ) + F 2 × ( Ω ) F 2 × ( Ω ) ) = 1 4 π ∫ d Ω ( 1 − cos θ 2 ) 2 ( 1 − sin γ cos ϕ sin θ − cos γ cos θ 2 ) 2 = 1 32 ∫ 0 π d θ sin θ ( 1 − cos θ ) 2 ( ( 1 − cos γ cos θ ) 2 + 1 2 sin 2 γ sin 2 θ ) = 13 120 + 1 12 cos γ + 1 120 cos 2 γ . (221) \begin{aligned}
\sigma_c^2(\gamma) &= \frac{1}{4\pi}\int d\Omega\left(F_1^+(\Omega)F_1^+(\Omega)+F_1^\times(\Omega)F_1^\times(\Omega)\right)\left(F_2^+(\Omega)F_2^+(\Omega)+F_2^\times(\Omega)F_2^\times(\Omega)\right) \\
&= \frac{1}{4\pi}\int d\Omega\left(\frac{1-\cos\theta}{2}\right)^2\left(\frac{1-\sin\gamma\cos\phi\sin\theta-\cos\gamma\cos\theta}{2}\right)^2 \\
&= \frac{1}{32}\int_0^\pi d\theta\sin\theta(1-\cos\theta)^2\left((1-\cos\gamma\cos\theta)^2+\frac{1}{2}\sin^2\gamma\sin^2\theta\right) \\
&= \frac{13}{120}+\frac{1}{12}\cos\gamma+\frac{1}{120}\cos^2\gamma. \tag*{(221)}
\end{aligned} σ c 2 ( γ ) = 4 π 1 ∫ d Ω ( F 1 + ( Ω ) F 1 + ( Ω ) + F 1 × ( Ω ) F 1 × ( Ω ) ) ( F 2 + ( Ω ) F 2 + ( Ω ) + F 2 × ( Ω ) F 2 × ( Ω ) ) = 4 π 1 ∫ d Ω ( 2 1 − cos θ ) 2 ( 2 1 − sin γ cos ϕ sin θ − cos γ cos θ ) 2 = 32 1 ∫ 0 π d θ sin θ ( 1 − cos θ ) 2 ( ( 1 − cos γ cos θ ) 2 + 2 1 sin 2 γ sin 2 θ ) = 120 13 + 12 1 cos γ + 120 1 cos 2 γ . (221) Note that the second line follows directly from the equality
( p ⋅ m ) 2 + ( p ⋅ n ) 2 = 1 − ( p ⋅ Ω ) 2 = ( 1 + p ⋅ Ω ) ( 1 − p ⋅ Ω ) , (222) (\boldsymbol{p}\cdot\boldsymbol{m})^2+(\boldsymbol{p}\cdot\boldsymbol{n})^2=1-(\boldsymbol{p}\cdot\boldsymbol{\Omega})^2=(1+\boldsymbol{p}\cdot\boldsymbol{\Omega})(1-\boldsymbol{p}\cdot\boldsymbol{\Omega}), \tag*{(222)} ( p ⋅ m ) 2 + ( p ⋅ n ) 2 = 1 − ( p ⋅ Ω ) 2 = ( 1 + p ⋅ Ω ) ( 1 − p ⋅ Ω ) , (222) which holds for the unit vectors pointing to either pulsar, since m \boldsymbol{m} m n \boldsymbol{n} n Ω \boldsymbol{\Omega} Ω (221) is obtained by integrating ϕ ∈ [ 0 , 2 π ] \phi\in[0,2\pi] ϕ ∈ [ 0 , 2 π ] θ \theta θ
A helpful consistency check follows from the first line of Eq. (220) , which implies that
σ c 2 ( γ ) = μ u 2 ( γ ) + σ u 2 ( γ ) + σ p 2 ( γ ) . (223) \sigma_c^2(\gamma)=\mu_u^2(\gamma)+\sigma_u^2(\gamma)+\sigma_p^2(\gamma). \tag*{(223)} σ c 2 ( γ ) = μ u 2 ( γ ) + σ u 2 ( γ ) + σ p 2 ( γ ) . (223) The first two terms on the rhs of Eq. (223) are given by Eq. (210) . The third term on the rhs of Eq. (223) is given by Eq. (219) . Adding them together correctly reproduces the final line of Eq. (221) .
The Two-Point Hellings-Downs Function μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) μ ~ ( γ ) \tilde{\mu}(\gamma) μ ~ ( γ ) μ ~ 2 ( γ ) \tilde{\mu}^2(\gamma) μ ~ 2 ( γ )
Here, we calculate the two-point Hellings-Downs function μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) β \beta β [66 ] at different locations on the sky. It is defined by
μ ( γ , β ) = ⟨ F 1 + ( Ω A ) F 2 + ( Ω B ) + F 1 × ( Ω A ) F 2 × ( Ω B ) ⟩ , (224) \mu(\gamma,\beta)=\left\langle F_1^+(\Omega_A)F_2^+(\Omega_B)+F_1^\times(\Omega_A)F_2^\times(\Omega_B)\right\rangle, \tag*{(224)} μ ( γ , β ) = ⟨ F 1 + ( Ω A ) F 2 + ( Ω B ) + F 1 × ( Ω A ) F 2 × ( Ω B ) ⟩ , (224) where the average is over all pulsar directions p 1 \boldsymbol{p}_1 p 1 p 2 \boldsymbol{p}_2 p 2 γ \gamma γ cos γ = p 1 ⋅ p 2 \cos\gamma=\boldsymbol{p}_1\cdot\boldsymbol{p}_2 cos γ = p 1 ⋅ p 2 cos β = Ω A ⋅ Ω B \cos\beta=\boldsymbol{\Omega}_A\cdot\boldsymbol{\Omega}_B cos β = Ω A ⋅ Ω B β \beta β A A A B B B
A potentially related function of two variables, denoted by γ ( x , y ) \gamma(x,y) γ ( x , y ) Figure 6 of Ref. [25 ] . (Note that in Ref. [25 ] , the symbol γ \gamma γ γ \gamma γ γ ( x , y ) \gamma(x,y) γ ( x , y ) [25 ] , and we have been unable to establish, based on its definition, if γ ( x , y ) \gamma(x,y) γ ( x , y ) μ ( x , y ) \mu(x,y) μ ( x , y ) γ ( x , y ) = c ( μ ( x , y ) ) 2 \gamma(x,y)=c(\mu(x,y))^2 γ ( x , y ) = c ( μ ( x , y ) ) 2 c c c
Our computation of μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) (90) and (91) , and place the GW sources at Ω A = x ^ sin β + z ^ cos β \boldsymbol{\Omega}_{A}=\hat{x}\sin\beta+\hat{z}\cos\beta Ω A = x ^ sin β + z ^ cos β Ω B = z ^ \boldsymbol{\Omega}_{B}=\hat{z} Ω B = z ^ m \boldsymbol{m} m n \boldsymbol{n} n (181) and polarization tensors from Eq. (180) . The antenna pattern functions are then defined by Eq. (84) and the combination ρ = F 1 + ( Ω A ) F 2 + ( Ω B ) + F 1 × ( Ω A ) F 2 × ( Ω B ) \rho=F_{1}^{+}(\boldsymbol{\Omega}_{A})F_{2}^{+}(\boldsymbol{\Omega}_{B})+F_{1}^{\times}(\boldsymbol{\Omega}_{A})F_{2}^{\times}(\boldsymbol{\Omega}_{B}) ρ = F 1 + ( Ω A ) F 2 + ( Ω B ) + F 1 × ( Ω A ) F 2 × ( Ω B )
We now need to average ρ \rho ρ θ \theta θ ϕ \phi ϕ λ \lambda λ γ \gamma γ (92) followed by Eq. (93) defines the average. The first integral to perform is over λ \lambda λ z = exp ( i λ ) z=\exp(i\lambda) z = exp ( iλ ) ρ \rho ρ z z z ρ \rho ρ z z z (192) and (193) with γ \gamma γ β \beta β Figure 10 , provided that in the caption of that figure, γ \gamma γ β \beta β θ < π − β \theta<\pi-\beta θ < π − β θ > π − β \theta>\pi-\beta θ > π − β θ < π − β \theta<\pi-\beta θ < π − β θ > π − β \theta>\pi-\beta θ > π − β 2 π i 2\pi i 2 π i
The integral over ϕ \phi ϕ z = exp ( i ϕ ) z=\exp(i\phi) z = exp ( i ϕ ) (192) and (193) , and now the description of Figure 10 applies without modification. The integrand has a pole of second order at z = 0 z=0 z = 0 z 1 z_{1} z 1 z 2 z_{2} z 2 θ \theta θ
I ( θ , γ , β ) = { 0 for θ > π − β , I < ( θ , γ , β ) = ( cos θ − 1 ) 2 ( cos θ + cos β ) ( cos θ cos γ + 2 cos γ − 1 ) ( 1 + cos θ ) 2 ( 1 + cos β ) 2 for θ ≤ π − β and θ < π − γ , I > ( θ , γ , β ) = ( cos θ + cos β ) ( cos θ cos γ − 2 cos γ − 1 ) ( 1 + cos β ) 2 for θ ≤ π − β and θ > π − γ . (225) I(\theta,\gamma,\beta)=
\begin{cases}
0 & \text{for }\theta>\pi-\beta,\\
I_{<}(\theta,\gamma,\beta)=\dfrac{(\cos\theta-1)^{2}(\cos\theta+\cos\beta)(\cos\theta\cos\gamma+2\cos\gamma-1)}{(1+\cos\theta)^{2}(1+\cos\beta)^{2}} & \text{for }\theta\le\pi-\beta\text{ and }\theta<\pi-\gamma,\\
I_{>}(\theta,\gamma,\beta)=\dfrac{(\cos\theta+\cos\beta)(\cos\theta\cos\gamma-2\cos\gamma-1)}{(1+\cos\beta)^{2}} & \text{for }\theta\le\pi-\beta\text{ and }\theta>\pi-\gamma.
\end{cases} \tag*{(225)} I ( θ , γ , β ) = ⎩ ⎨ ⎧ 0 I < ( θ , γ , β ) = ( 1 + cos θ ) 2 ( 1 + cos β ) 2 ( cos θ − 1 ) 2 ( cos θ + cos β ) ( cos θ cos γ + 2 cos γ − 1 ) I > ( θ , γ , β ) = ( 1 + cos β ) 2 ( cos θ + cos β ) ( cos θ cos γ − 2 cos γ − 1 ) for θ > π − β , for θ ≤ π − β and θ < π − γ , for θ ≤ π − β and θ > π − γ . (225) Note that (because of the integral over λ \lambda λ θ > π − β \theta>\pi-\beta θ > π − β
Integration of these expressions then yields μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) β < γ \beta<\gamma β < γ
μ ( γ , β ) = 1 4 ∫ 0 π − γ d θ sin θ I < ( θ , γ , β ) + 1 4 ∫ π − γ π − β d θ sin θ I > ( θ , γ , β ) . (226) \mu(\gamma,\beta)=\frac{1}{4}\int_{0}^{\pi-\gamma}d\theta\,\sin\theta\,I_{<}(\theta,\gamma,\beta)+\frac{1}{4}\int_{\pi-\gamma}^{\pi-\beta}d\theta\,\sin\theta\,I_{>}(\theta,\gamma,\beta). \tag*{(226)} μ ( γ , β ) = 4 1 ∫ 0 π − γ d θ sin θ I < ( θ , γ , β ) + 4 1 ∫ π − γ π − β d θ sin θ I > ( θ , γ , β ) . (226) and for β > γ \beta>\gamma β > γ
μ ( γ , β ) = 1 4 ∫ 0 π − β d θ sin θ I < ( θ , γ , β ) . (227) \mu(\gamma,\beta)=\frac{1}{4}\int_{0}^{\pi-\beta}d\theta\,\sin\theta\,I_{<}(\theta,\gamma,\beta). \tag*{(227)} μ ( γ , β ) = 4 1 ∫ 0 π − β d θ sin θ I < ( θ , γ , β ) . (227) These integrals evaluate to
μ ( γ , β ) = { 1 48 ( 33 − 18 cos β − 3 cos 2 β + ( 32 − 21 cos β − 6 cos 2 β − cos 3 β ) cos γ ) sec 4 ( β 2 ) + ( 1 − 1 2 cos β − 1 2 cos γ ) sec 4 ( β 2 ) log ( sin 2 ( γ 2 ) ) for β < γ , and 1 24 ( 33 − 3 cos β − ( 16 + 5 cos β + cos 2 β ) cos γ ) sec 2 ( β 2 ) + ( 1 − 1 2 cos β − 1 2 cos γ ) sec 4 ( β 2 ) log ( sin 2 ( β 2 ) ) for β > γ . (228) \mu(\gamma,\beta)=
\begin{cases}
\frac{1}{48}\left(33-18\cos\beta-3\cos^{2}\beta+(32-21\cos\beta-6\cos^{2}\beta-\cos^{3}\beta)\cos\gamma\right)\sec^{4}\left(\frac{\beta}{2}\right)\\
\qquad+\left(1-\frac{1}{2}\cos\beta-\frac{1}{2}\cos\gamma\right)\sec^{4}\left(\frac{\beta}{2}\right)\log\left(\sin^{2}\left(\frac{\gamma}{2}\right)\right) & \text{for }\beta<\gamma,\text{ and}\\
\frac{1}{24}\left(33-3\cos\beta-(16+5\cos\beta+\cos^{2}\beta)\cos\gamma\right)\sec^{2}\left(\frac{\beta}{2}\right)\\
\qquad+\left(1-\frac{1}{2}\cos\beta-\frac{1}{2}\cos\gamma\right)\sec^{4}\left(\frac{\beta}{2}\right)\log\left(\sin^{2}\left(\frac{\beta}{2}\right)\right) & \text{for }\beta>\gamma.
\end{cases} \tag*{(228)} μ ( γ , β ) = ⎩ ⎨ ⎧ 48 1 ( 33 − 18 cos β − 3 cos 2 β + ( 32 − 21 cos β − 6 cos 2 β − cos 3 β ) cos γ ) sec 4 ( 2 β ) + ( 1 − 2 1 cos β − 2 1 cos γ ) sec 4 ( 2 β ) log ( sin 2 ( 2 γ ) ) 24 1 ( 33 − 3 cos β − ( 16 + 5 cos β + cos 2 β ) cos γ ) sec 2 ( 2 β ) + ( 1 − 2 1 cos β − 2 1 cos γ ) sec 4 ( 2 β ) log ( sin 2 ( 2 β ) ) for β < γ , and for β > γ . (228) which is shown in Figure 12 . The function μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) β → 0 \beta\to0 β → 0
In addition to the two-point function defined by the sum in Eq. (224) we also require the two-point functions for the individual polarizations. These are defined by
μ + + ( γ , β ) = ⟨ F 1 + ( Ω A ) F 2 + ( Ω B ) ⟩ , and \mu_{++}(\gamma,\beta)=\left\langle F_{1}^{+}(\boldsymbol{\Omega}_{A})F_{2}^{+}(\boldsymbol{\Omega}_{B})\right\rangle,\qquad\text{and} μ ++ ( γ , β ) = ⟨ F 1 + ( Ω A ) F 2 + ( Ω B ) ⟩ , and μ × × ( γ , β ) = ⟨ F 1 × ( Ω A ) F 2 × ( Ω B ) ⟩ . (229) \mu_{\times\times}(\gamma,\beta)=\left\langle F_{1}^{\times}(\boldsymbol{\Omega}_{A})F_{2}^{\times}(\boldsymbol{\Omega}_{B})\right\rangle. \tag*{(229)} μ ×× ( γ , β ) = ⟨ F 1 × ( Ω A ) F 2 × ( Ω B ) ⟩ . (229) quadratic function of z z z (192) and (193) with γ \gamma γ β \beta β Figure 10 , provided that in the caption of that figure, γ \gamma γ β \beta β θ < π − β \theta<\pi-\beta θ < π − β θ > π − β \theta>\pi-\beta θ > π − β θ < π − β \theta<\pi-\beta θ < π − β θ > π − β \theta>\pi-\beta θ > π − β 2 π i 2\pi i 2 π i
The integral over ϕ \phi ϕ z = exp ( i ϕ ) z=\exp(i\phi) z = exp ( i ϕ ) (192) and (193) , and now the description of Figure 10 applies without modification. The integrand has a pole of second order at z = 0 z=0 z = 0 z 1 z_{1} z 1 z 2 z_{2} z 2 θ \theta θ
where the averages are over all pulsar pairs whose directions are separated by angle γ \gamma γ β \beta β Ω A \boldsymbol{\Omega}_{A} Ω A Ω B \boldsymbol{\Omega}_{B} Ω B (224) and (229) that
μ ( γ , β ) = μ + + ( γ , β ) + μ × × ( γ , β ) . (230) \mu(\gamma,\beta)=\mu_{++}(\gamma,\beta)+\mu_{\times\times}(\gamma,\beta). \tag*{(230)} μ ( γ , β ) = μ ++ ( γ , β ) + μ ×× ( γ , β ) . (230) With some simple symmetry arguments, we can obtain both μ + + \mu_{++} μ ++ μ × × \mu_{\times\times} μ ×× μ \mu μ Ω A \mathbf{\Omega}_A Ω A γ → π − γ \gamma\to\pi-\gamma γ → π − γ β → π − β \beta\to\pi-\beta β → π − β π / 2 \pi/2 π /2
Under this transformation, the polarization basis vectors corresponding to the source direction Ω A \mathbf{\Omega}_A Ω A m → − m \mathbf{m}\to-\mathbf{m} m → − m n → n \mathbf{n}\to\mathbf{n} n → n Ω \mathbf{\Omega} Ω − Ω -\mathbf{\Omega} − Ω ϕ → ϕ + π \phi\to\phi+\pi ϕ → ϕ + π θ → π − θ \theta\to\pi-\theta θ → π − θ (181) . This means that the polarization basis tensors defined in Eq. (180) transform as e a b + ( Ω A ) → e a b + ( Ω A ) e^+_{ab}(\mathbf{\Omega}_A)\to e^+_{ab}(\mathbf{\Omega}_A) e ab + ( Ω A ) → e ab + ( Ω A ) e a b × ( Ω A ) → − e a b × ( Ω A ) e^\times_{ab}(\mathbf{\Omega}_A)\to-e^\times_{ab}(\mathbf{\Omega}_A) e ab × ( Ω A ) → − e ab × ( Ω A ) p 1 \mathbf{p}_1 p 1 Ω A \mathbf{\Omega}_A Ω A 1 + p 1 ⋅ Ω A 1+\mathbf{p}_1\cdot\mathbf{\Omega}_A 1 + p 1 ⋅ Ω A (84) transform as F 1 + ( Ω A ) → F 1 + ( Ω A ) F^+_1(\mathbf{\Omega}_A)\to F^+_1(\mathbf{\Omega}_A) F 1 + ( Ω A ) → F 1 + ( Ω A ) F 1 × ( Ω A ) → − F 1 × ( Ω A ) F^\times_1(\mathbf{\Omega}_A)\to-F^\times_1(\mathbf{\Omega}_A) F 1 × ( Ω A ) → − F 1 × ( Ω A ) (229) this implies that
μ + + ( γ , β ) = μ + + ( π − γ , π − β ) , and \mu_{++}(\gamma,\beta)=\mu_{++}(\pi-\gamma,\pi-\beta), \qquad\text{and} μ ++ ( γ , β ) = μ ++ ( π − γ , π − β ) , and μ × × ( γ , β ) = − μ × × ( π − γ , π − β ) . (231) \mu_{\times\times}(\gamma,\beta)=-\mu_{\times\times}(\pi-\gamma,\pi-\beta). \tag*{(231)} μ ×× ( γ , β ) = − μ ×× ( π − γ , π − β ) . (231) since the antenna pattern functions at the second pulsar are unaffected by the transformation.
These simple transformation rules make it straightforward to obtain both μ × × \mu_{\times\times} μ ×× μ + + \mu_{++} μ ++ μ \mu μ (230) and (231) that the two-point functions for the individual polarizations may be obtained from Eq. (228) as
μ + + ( γ , β ) = 1 2 ( μ ( γ , β ) + μ ( π − γ , π − β ) ) , and \mu_{++}(\gamma,\beta)=\frac{1}{2}\left(\mu(\gamma,\beta)+\mu(\pi-\gamma,\pi-\beta)\right), \qquad\text{and} μ ++ ( γ , β ) = 2 1 ( μ ( γ , β ) + μ ( π − γ , π − β ) ) , and μ × × ( γ , β ) = 1 2 ( μ ( γ , β ) − μ ( π − γ , π − β ) ) . (232) \mu_{\times\times}(\gamma,\beta)=\frac{1}{2}\left(\mu(\gamma,\beta)-\mu(\pi-\gamma,\pi-\beta)\right). \tag*{(232)} μ ×× ( γ , β ) = 2 1 ( μ ( γ , β ) − μ ( π − γ , π − β ) ) . (232) If desired, these relations may be used to obtain explicit closed forms analogous to Eq. (228) .
Similar symmetry arguments show that if we average uniformly over all pulsar locations at fixed angular separation γ \gamma γ
μ + × ( γ , β ) = ⟨ F 1 + ( Ω A ) F 2 × ( Ω B ) ⟩ = 0 , and \mu_{+\times}(\gamma,\beta)=\left\langle F^+_1(\mathbf{\Omega}_A)F^\times_2(\mathbf{\Omega}_B)\right\rangle=0, \qquad\text{and} μ +× ( γ , β ) = ⟨ F 1 + ( Ω A ) F 2 × ( Ω B ) ⟩ = 0 , and μ × + ( γ , β ) = ⟨ F 1 × ( Ω A ) F 2 + ( Ω B ) ⟩ = 0. (233) \mu_{\times+}(\gamma,\beta)=\left\langle F^\times_1(\mathbf{\Omega}_A)F^+_2(\mathbf{\Omega}_B)\right\rangle=0. \tag*{(233)} μ ×+ ( γ , β ) = ⟨ F 1 × ( Ω A ) F 2 + ( Ω B ) ⟩ = 0. (233) The argument is straightforward: shift the pulsar locations p 1 \mathbf{p}_1 p 1 p 2 \mathbf{p}_2 p 2 − p 1 -\mathbf{p}_1 − p 1 − p 2 -\mathbf{p}_2 − p 2 Ω A \mathbf{\Omega}_A Ω A Ω B \mathbf{\Omega}_B Ω B − Ω A -\mathbf{\Omega}_A − Ω A − Ω B -\mathbf{\Omega}_B − Ω B β \beta β γ \gamma γ F + F^+ F + F × F^\times F ×
We stress that the relationships expressed in Eqs. (232) and (233) have nothing to do with the GW sources, which might all be highly polarized and not symmetrically placed or distributed. They are consequences of the pulsar averaging, which erases these terms for any configuration of sources.
In § 3 , the second moment of μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) β \beta β β j k \beta_{jk} β j k cos β j k \cos\beta_{jk} cos β j k [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] [67 ]
μ 2 ~ ( γ ) = 1 2 ∫ 0 π d β sin β μ 2 ( γ , β ) = − 5 48 + 49 432 cos 2 γ − 1 6 ( cos 2 γ + 3 ) log ( 1 − cos γ 2 ) log ( 1 + cos γ 2 ) \begin{aligned}
\widetilde{\mu^2}(\gamma)=\frac{1}{2}\int_0^\pi d\beta\,\sin\beta\,\mu^2(\gamma,\beta) \\
&= -\frac{5}{48}+\frac{49}{432}\cos^2\gamma-\frac{1}{6}\left(\cos^2\gamma+3\right)\log\left(\frac{1-\cos\gamma}{2}\right)\log\left(\frac{1+\cos\gamma}{2}\right)
\end{aligned} μ 2 ( γ ) = 2 1 ∫ 0 π d β sin β μ 2 ( γ , β ) = − 48 5 + 432 49 cos 2 γ − 6 1 ( cos 2 γ + 3 ) log ( 2 1 − cos γ ) log ( 2 1 + cos γ ) + 1 12 ( cos γ − 1 ) ( cos γ + 3 ) log ( 1 − cos γ 2 ) + 1 12 ( cos γ + 1 ) ( cos γ − 3 ) log ( 1 + cos γ 2 ) . (234) \qquad+\frac{1}{12}(\cos\gamma-1)(\cos\gamma+3)\log\left(\frac{1-\cos\gamma}{2}\right)+\frac{1}{12}(\cos\gamma+1)(\cos\gamma-3)\log\left(\frac{1+\cos\gamma}{2}\right). \tag*{(234)} + 12 1 ( cos γ − 1 ) ( cos γ + 3 ) log ( 2 1 − cos γ ) + 12 1 ( cos γ + 1 ) ( cos γ − 3 ) log ( 2 1 + cos γ ) . (234) which is illustrated in Figure 4 . The second moment μ ~ 2 ( γ ) \tilde{\mu}^{2}(\gamma) μ ~ 2 ( γ ) γ = π / 2 \gamma= \pi/2 γ = π /2 γ → π − γ \gamma\to\pi- \gamma γ → π − γ cos γ → − cos γ \cos\gamma\to-\cos\gamma cos γ → − cos γ
For computing the cosmic variance, we also need the spherical average of μ + + 2 + μ × × 2 \mu_{++}^{2}+\mu_{\times\times}^{2} μ ++ 2 + μ ×× 2
1 2 ∫ 0 π d β sin β [ μ + + 2 ( γ , β ) + μ × × 2 ( γ , β ) ] = 1 4 ∫ 0 π d β sin β [ μ 2 ( γ , β ) + μ 2 ( π − γ , π − β ) ] = 1 4 ∫ 0 π d β sin β [ μ 2 ( γ , β ) + μ 2 ( π − γ , β ) ] = 1 2 [ μ ~ 2 ( γ ) + μ ~ 2 ( π − γ ) ] = μ ~ 2 ( γ ) . (235) \begin{aligned}
\frac{1}{2}\int_{0}^{\pi}d\beta\sin\beta\left[\mu_{++}^{2}(\gamma,\beta)+\mu_{\times\times}^{2}(\gamma,\beta)\right]
&=\frac{1}{4}\int_{0}^{\pi}d\beta\sin\beta\left[\mu^{2}(\gamma,\beta)+\mu^{2}(\pi-\gamma,\pi-\beta)\right]\\
&=\frac{1}{4}\int_{0}^{\pi}d\beta\sin\beta\left[\mu^{2}(\gamma,\beta)+\mu^{2}(\pi-\gamma,\beta)\right]\\
&=\frac{1}{2}\left[\tilde{\mu}^{2}(\gamma)+\tilde{\mu}^{2}(\pi-\gamma)\right]=\tilde{\mu}^{2}(\gamma). \tag*{(235)}
\end{aligned} 2 1 ∫ 0 π d β sin β [ μ ++ 2 ( γ , β ) + μ ×× 2 ( γ , β ) ] = 4 1 ∫ 0 π d β sin β [ μ 2 ( γ , β ) + μ 2 ( π − γ , π − β ) ] = 4 1 ∫ 0 π d β sin β [ μ 2 ( γ , β ) + μ 2 ( π − γ , β ) ] = 2 1 [ μ ~ 2 ( γ ) + μ ~ 2 ( π − γ ) ] = μ ~ 2 ( γ ) . (235) On the second line of Eq. (235) we used Eq. (232) to write μ + + \mu_{++} μ ++ μ × × \mu_{\times\times} μ ×× μ \mu μ μ = π − β \mu=\pi-\beta μ = π − β sin β = sin u \sin\beta=\sin u sin β = sin u d β = − d u d\beta=-du d β = − d u (234) of μ ~ 2 \tilde{\mu}^{2} μ ~ 2 μ ~ 2 ( γ ) \tilde{\mu}^{2}(\gamma) μ ~ 2 ( γ ) γ = π / 2 \gamma=\pi/2 γ = π /2
We conclude this part of the Appendix with some comments about the behavior of the two-point functions under a change of GW polarization basis as given by Eq. (182) . A useful discussion of the physical interpretation (in terms of the spin) may be found in Sec. 3.4 of [68 ] . This change of polarization basis corresponds to a local gauge transformation of the gravitational field: any physical observables such as the cosmic variance must be invariant under such a basis change. The calculations in this paper correspond to a particular choice of GW gauge: in Eqs. (180) and (D7) we have defined a specific basis for the polarization vectors and tensors, as a function of the GW propagation direction Ω \mathbf{\Omega} Ω (182) where the angle ϵ ( Ω ) \epsilon(\mathbf{\Omega}) ϵ ( Ω ) function of the propagation direction.
Under this gauge transformation of the gravitational field, the quantities F 1 + ( Ω ) F 2 + ( Ω ) F_{1}^{+}(\mathbf{\Omega})F_{2}^{+}(\mathbf{\Omega}) F 1 + ( Ω ) F 2 + ( Ω ) F 1 × ( Ω ) F 2 × ( Ω ) F_{1}^{\times}(\mathbf{\Omega})F_{2}^{\times}(\mathbf{\Omega}) F 1 × ( Ω ) F 2 × ( Ω ) F 1 + ( Ω ) F 2 + ( Ω ′ ) + F 1 × ( Ω ) F 2 × ( Ω ′ ) F_{1}^{+}(\mathbf{\Omega})F_{2}^{+}(\mathbf{\Omega}')+F_{1}^{\times}(\mathbf{\Omega})F_{2}^{\times}(\mathbf{\Omega}') F 1 + ( Ω ) F 2 + ( Ω ′ ) + F 1 × ( Ω ) F 2 × ( Ω ′ ) μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) ϵ ( Ω ) \epsilon(\mathbf{\Omega}) ϵ ( Ω ) ϵ ( Ω ′ ) \epsilon(\mathbf{\Omega}') ϵ ( Ω ′ ) (235) are gauge-invariant, as can be easily seen from working in a circular polarization basis.
[32 ] In Appendix C2 we show that neglecting the pulsar terms should not affect the mean correlation if the distance to typical pulsars is thousands of years. However, in Appendix § 2.2 we show that for GW sources that are evolving very slowly with time, the pulsar terms will increase the variance by about a factor of four.
[33 ] For large numbers of sources, if the frequency is incommensurate with the observation time, the effect is to increase the total variance and the cosmic variance by up to a factor of two for a fixed mean correlation μ \mu μ (151) .
[34 ] E. Harrison, Darkness at Night: A Riddle of the Universe (Harvard University Press, 1987), https://books.google.de/books?id=IRKYuWeWftYC .
[35 ] J. D. Romano, Searches for stochastic gravitational-wave backgrounds, Les Houches Summer School 2019 Lectures, arXiv:1909.00269 .
[36 ] J. S. Hazboun, J. D. Romano, and T. L. Smith, Realistic sensitivity curves for pulsar timing arrays, Phys. Rev. D 100, 104028 (2019).
[37 ] W. H. Press, S. A. Teukolsky, W. T. Vetterling, and B. P. Flannery, Numerical Recipes in C , 2nd ed. [37 ] , http://numerical.recipes/ .
[38 ] B. Allen and J. D. Romano, The Hellings and Downs correlation of an arbitrary set of pulsars, arXiv:2208.07230 .
[39 ] One simple way to see this is to also average over the pulsar distances L 1 L_1 L 1 L 2 L_2 L 2
[40 ] This is clearly visible in the average of Eq. (38) over pulsar positions. The average of the diagonal term μ ( γ ) \mu(\gamma) μ ( γ ) μ u ( γ ) \mu_u(\gamma) μ u ( γ ) (69) . But the average of the off-diagonal terms gives x = ∑ j ≠ k A j A k cos ( ϕ j − ϕ k ) μ ( β j k ) / 2 x=\sum_{j\ne k}A_jA_k\cos(\phi_j-\phi_k)\mu(\beta_{jk})/2 x = ∑ j = k A j A k cos ( ϕ j − ϕ k ) μ ( β j k ) /2 β k \boldsymbol{\beta}_k β k cos ( β j k ) = Ω j ⋅ Ω k \cos(\beta_{jk})=\boldsymbol{\Omega}_j\cdot\boldsymbol{\Omega}_k cos ( β j k ) = Ω j ⋅ Ω k μ ( γ , β ) \mu(\gamma,\beta) μ ( γ , β ) § 16 . In the confusion-noise ensemble, the sum of these off-diagonal terms has a probability distribution function p ( x ) d x = exp ( − x / μ − 1 ) d x / μ p(x)dx=\exp(-x/\mu-1)dx/\mu p ( x ) d x = exp ( − x / μ − 1 ) d x / μ x ≥ − μ x\ge-\mu x ≥ − μ p ( x ) = 0 p(x)=0 p ( x ) = 0 x < − μ x<-\mu x < − μ μ \mu μ (63) . This PDF has zero mean and a variance/second moment μ 2 \mu^2 μ 2 Γ ( γ ) \Gamma(\gamma) Γ ( γ )
[41 ] One approach would be to assume that supermassive black-hole binaries are slowly evolving with time, so they follow time-frequency tracks which are approximately linear. The time average of products of those waveforms has a simple analytic form.
[42 ] C. Conneely, A. H. Jaffe, and C. M. F. Mingarelli, On the amplitude and Stokes parameters of a stochastic gravitational-wave background, Mon. Not. R. Astron. Soc. 487, 562 (2019).
[43 ] Here, we have defined σ a v e r a g e 2 \sigma_{\mathrm{average}}^2 σ average 2 σ 2 \sigma^2 σ 2 ⟨ ρ 2 ⟩ \langle\rho^2\rangle ⟨ ρ 2 ⟩ l l l 103 A 4 μ u 2 ( γ ) / 6300 103\mathcal{A}^4\mu_u^2(\gamma)/6300 103 A 4 μ u 2 ( γ ) /6300 σ a v e r a g e 2 \sigma_{\mathrm{average}}^2 σ average 2 (110) and Eq. (121) .
[44 ] S. J. Chamberlin, J. D. E. Creighton, X. Siemens, P. Demorest, J. Ellis, L. R. Price, and J. D. Romano, Time-domain implementation of the optimal cross-correlation statistic for stochastic gravitational-wave background
