A nontrivial footprint of standard cosmology in the future observations of low-frequency gravitational waves

Abstract

Recent research shows that the cosmological components of the Universe should influence on the propagation of Gravitational Waves (GWs) and even it has been proposed a new way to measure the cosmological constant using Pulsar Timing Arrays (PTAs). However, these results have considered very particular cases (e.g. a de Sitter Universe or a mixing with non-relativistic matter). In this work we propose an extension of these results, using the Hubble constant as the natural parameter that includes all the cosmological information and studying its effect on the propagation of GWs. Using linearized gravity we considered a mixture of perfect fluids permeating the spacetime and studied the propagation of GWs within the context of the $\Lambda \hbox {CDM}$ model. We found from numerical simulations that the timing residual of local pulsars should present a distinguishable peak depending on the local value of the Hubble constant. As a consequence, when assuming the standard $\Lambda \hbox {CDM}$ model, our result predicts that the region of maximum timing residual is determined by the redshift of the source. This framework represents an alternative test for the standard cosmological model, and it can be used to facilitate the measurements of gravitational waves by ongoing PTAs projects.

Appendix A: On the equations of standard cosmology

The components of the stress-energy tensor that appears in (3) can be found by considering a perfect fluid (i.e. a fluid that does not have viscosity and does not conduct heat), with energy density $\rho$ and isotropic pressure p, filling the whole Universe. The components of $T^{\mu \nu }$ take the following form,

$$\begin{aligned} T^{\mu \nu } = (\rho + p) U^\mu U^\nu - pg^{\mu \nu }, \end{aligned}$$

(A1)

where $U^\mu$ are the components of the 4-velocity of the fluid. When this expression is inserted into the Einstein’s Field equations we obtain the two Friedmann equations: The first equation obtained from the 00 component of (1) and the second from the combination between the trace of the field equations and the first Friedmann equation, giving,

$$\begin{aligned} \left( \frac{\dot{a}}{a}\right) ^{2}&= \frac{\kappa }{3} (\rho _{i} + \rho _\Lambda )\equiv H^{2}(T) \end{aligned}$$

(A2)

$$\begin{aligned} \left( \frac{\ddot{a}}{a}\right)&= \kappa \left( \frac{\rho _\Lambda }{3} - \frac{\rho _{i}}{6} - \frac{p_{i}}{2} \right) , \end{aligned}$$

(A3)

where $\rho _{i}$ and $p_{i}$ are the energy density and the isotropic pressure of the i-th fluid respectively and H(T) is the Hubble parameter, which value at the present day, i.e. $H(T_{0}) \equiv H_{0}$, is known as the Hubble constant $H_{0}$. In order to obtain the time evolution of the scale factor an equation of state must be provided. In [36] was used $p=0$, which corresponds to the equation of state of non-relativistic dust. However, in this work we will use $p_{i}=\chi _{i}\rho _{i}$, with $\chi _{i}$ constant, in order to develop a general discussion of the phenomenon. Using the Friedmann equations we can find that

$$\begin{aligned} \frac{\rho _{i}}{\rho _{0}} = \left( \frac{a(T)}{a_{0}} \right) ^{-3(\chi _{i}+1)}, \end{aligned}$$

(A4)

where $\rho _{0} = \rho (T_{0})$ is the current energy density of the i-th fluid and $a_{0} = a(T_{0})$ is the current scale factor (usually taken as 1), and both are integration constants. Replacing this expression into (A2) provides a solution of the scale factor in terms of the comoving time and the equation of state,

$$\begin{aligned} a(T) = {\left\{ \begin{array}{ll} a_{0} \left( \frac{T}{T_{0}}\right) ^{\frac{2}{3(\chi _{i}+1)}} &{} \text {if} \chi _{i} \ne -1\\ a_{0}\exp ({\sqrt{\frac{\Lambda }{3}} (T-T_{0})}) &{} \text {if} \chi _{i} = -1 \end{array}\right. }. \end{aligned}$$

(A5)

In the $\chi _{i}\ne -1$ case, when we combine (A4) with (A5) we can obtain the general form of the energy density of the i-th fluid,

$$\begin{aligned} \rho _{i} = {\left\{ \begin{array}{ll} \frac{4}{3(\chi _{i}+1)^{2} \kappa T^{2}} &{} \text {if} \chi _{i} \ne -1\\ \Lambda /\kappa &{} \text {if} \chi _{i} = -1 \end{array}\right. }. \end{aligned}$$

(A6)

Currently, the standard cosmological model is the $\Lambda \hbox {CDM}$: Includes a positive cosmological constant $\Lambda$ (which represents the so-called Dark Energy) and Cold Dark Matter (which is composed of baryonic matter and non-relativistic dark matter). In the $\Lambda \hbox {CDM}$ model, when a global flat geometry is considered, we can use (A4) to write an effective energy density in terms of the scale factor and the currently evaluated energy densities,

$$\begin{aligned} \rho _{\text {eff}}&= \rho _{\Lambda } + \rho _{d} + \rho _{r} \nonumber \\&= \rho _{\Lambda } + \rho _{d0} \left[ \frac{a_{0}}{a(T)}\right] ^3 + \rho _{r0}\left[ \frac{a_{0}}{a(T)}\right] ^4, \end{aligned}$$

(A7)

where $\rho _{\Lambda } = \Lambda /\kappa$, $\rho _{d0}$ is the current density of non-relativistic matter (i.e. Cold Dark Matter and baryonic matter, $\chi _d = 0$) and $\rho _{r0}$ is the current radiation density ($\chi _r=1/3$). These expressions will be used in order to construct a spherically symmetric metric which reproduces the corresponding geometry of a FLRW metric for a perfect fluid with an arbitrary $\chi _{i}$.

1.1 Appendix B: On the derivation of the $SS\chi$ metric

As we have to impose a spherically symmetric geometry we will have the transformation $r^{2} \, d\Omega ^{2} \rightarrow a(T)^{2} R^{2} \, d\Omega ^{2}$. Using the second rank tensor property of the metric tensor when we perform coordinate transformations,

$$\begin{aligned} g_{\mu ' \nu '} = \frac{\partial X^\mu }{\partial x^{\mu '}}\frac{\partial X^\nu }{\partial x^{\nu '}} g_{\mu \nu }, \end{aligned}$$

(B1)

and the requirement that the new metric must be diagonal, we obtain the relation

$$\begin{aligned} 0&= \frac{\partial T}{\partial t}\frac{\partial T}{\partial r}g_{TT} + \frac{\partial R}{\partial t} \frac{\partial R}{\partial r} g_{RR}. \end{aligned}$$

(B2)

By computing the partial derivatives we obtain the expressions

$$\begin{aligned} \frac{\partial R}{\partial r}&= -\frac{1}{3} \frac{2r \frac{\partial T}{\partial r} -3T (\chi _{i}+1) }{a(T) (\chi _{i}+1) T} \end{aligned}$$

(B3)

$$\begin{aligned} \frac{\partial R}{\partial t}&= - \frac{2}{3} \frac{r \frac{\partial T}{\partial t}}{a(T) (\chi _{i}+1) T}, \end{aligned}$$

(B4)

and from (B2) we find that

$$\begin{aligned} \frac{\partial T}{\partial r} = \frac{a(T)^{2}}{\frac{\partial T}{\partial t}} \frac{\partial R}{\partial t} \frac{\partial R}{\partial r}. \end{aligned}$$

(B5)

Thus, from the last equation, $\frac{\partial T}{\partial r}$ becomes

$$\begin{aligned} \frac{\partial T}{\partial r} = \frac{6rT (\chi _{i}+1)}{4r^{2} - 9 (\chi _{i}+1)^{2} T^{2}}, \end{aligned}$$

(B6)

and using (B1) we can obtain the components of the metric,

$$\begin{aligned} g_{tt}&=- \Big ( \frac{\partial T}{\partial t} \Big )^{2} \left[ \frac{9 (\chi _{i}+1)^{2} T^{2} - 4 r^{2}}{9(\chi _{i}+1)^{2} T^{2}} \right] \end{aligned}$$

(B7)

$$\begin{aligned} g_{rr}&= \frac{9(\chi _{i}+1)^{2} T^{2}}{9(\chi _{i}+1)^{2} T^{2}- 4r^{2}}. \end{aligned}$$

(B8)

From (A4) we can write the SS$\chi _{i}$ metric as

$$\begin{aligned} ds^{2} =&-\frac{(\partial _t \rho _{i})^{2}}{3\kappa \rho _{i}^3(\chi _{i}+1)^{2}} \left[ 1-\frac{\kappa \rho _{i} r^{2}}{3} \right] \, dt^{2} + \frac{dr^{2}}{1-\frac{\kappa \rho _{i} r^{2}}{3}} +r^{2} \, d\Omega ^{2}, \end{aligned}$$

(B9)

but, using (A4) and (B6), we get

$$\begin{aligned} \frac{\partial \rho _{i}}{\partial r} = \frac{(\chi _{i}+1)\kappa \rho _{i}^{2} r}{1-\frac{\kappa \rho _{i}}{3} r^{2}}. \end{aligned}$$

(B10)

If we properly redefine $\tilde{\rho }_{i} \equiv \kappa \rho _{i}$, the last expression becomes

$$\begin{aligned} \frac{\partial \tilde{\rho }_{i}}{\partial r} = \frac{ (\chi _{i}+1)\tilde{\rho }_{i}^{2} r}{1-\frac{\tilde{\rho }_{i}}{3} r^{2}}, \end{aligned}$$

(B11)

but it can be noticed from (B11) that we can form the expression

$$\begin{aligned} \frac{\partial }{\partial r} \left[ \frac{c+r^{2} \tilde{\rho }_{i}}{\tilde{\rho }_{i}^{n}} \right] = 0, \end{aligned}$$

(B12)

where c and n are unknown constants. Unfolding the last expression and using the linear independence of r, we obtain that the constants are

$$\begin{aligned} c = \frac{6}{3\chi _{i}+1} \qquad n = \frac{3\chi _{i}+1}{3(\chi _{i}+1)}. \end{aligned}$$

(B13)

Therefore, we can integrate (B12) and write

$$\begin{aligned} \frac{c+r^{2} \tilde{\rho }_{i}}{\tilde{\rho }_{i}^{n}} = F(t), \end{aligned}$$

(B14)

where F(t) is a function of t. By a dimensional analysis, we note that in natural units $[\tilde{\rho }_{i}] = L^{-2}$ and therefore $[F(t)] = L^{2n}$. As there is no other parameter involved apart from t, and also as $[t] = L$ in natural units, then we set $F(t) = A t^{2n}$, with A as a dimensionless arbitrary constant. For any fluid we can expect that at later stage it will be diluted homogeneously, which implies that for $t\rightarrow \infty$ the metric (B9) is almost flat. Then,

$$\begin{aligned} \lim _{t\rightarrow \infty (\rho _{i}\rightarrow 0)} \frac{(\partial _t \rho _{i})^{2}}{3\kappa \rho _{i}^3 (\chi _{i}+1)^{2}} = 1. \end{aligned}$$

(B15)

On the other hand, (B14) can be written as

$$\begin{aligned} \frac{c+r^{2} \kappa \rho _{i}}{(\kappa \rho _{i})^{n}} = At^{2n}, \end{aligned}$$

(B16)

but when we take the derivative with respect to t and solving for $\partial _t \rho _{i}$, we obtain

$$\begin{aligned} \frac{\partial \rho _{i}}{\partial t} = - \frac{2nA t^{2n-1} (\kappa \rho _{i})^n \rho _{i}}{\kappa \rho _{i} nr^{2} - r^{2} \kappa \rho _{i} + cn }, \end{aligned}$$

(B17)

and if we square, divide by $3\kappa \rho _{i}^3$ and replace the previous results, we can found the following equality

$$\begin{aligned} \frac{(\partial _t \rho _{i})^{2}}{3\kappa \rho _{i}^3(\chi _{i}+1)^{2}} = \frac{4n^{2} A^{1/n} (\kappa r^{2} \rho _{i} + c)^{\frac{2n-1}{n}}}{3(\chi _{i}+1)^{2} [(n-1)\kappa r^{2} \rho _{i} + cn]}. \end{aligned}$$

(B18)

Computing the limit $\rho _{i} \rightarrow 0$ as the fluid dilutes at distant times, we can set A,

$$\begin{aligned} \lim _{t\rightarrow \infty (\rho _{i}\rightarrow 0)} \frac{(\partial _t \rho _{i})^{2}}{3\kappa \rho _{i}^3(\chi _{i}+1)^{2}} = \frac{4n^{2} A^{1/n} c^{\frac{2n-1}{n}}}{3(\chi _{i}+1)^{2} (cn)^{2}}, \end{aligned}$$

(B19)

and using that the metric is asymptotically flat, which implies that the previous limit is equal to one, we get the value of A,

$$\begin{aligned} A = c \left( \frac{3}{4}\right) ^n (\chi _{i}+1)^{2n}. \end{aligned}$$

(B20)

Finally, with the constant A known, we can provide an exact expression for the $\hbox {SS}\chi$ metric, which becomes

$$\begin{aligned} ds^{2}&= -\frac{dt^{2}}{\left( 1-\dfrac{\kappa \rho _{i}r^{2}}{3} \right) \left( 1 + \dfrac{\kappa \rho _{i} r^{2} (3\chi _{i}+1)}{6} \right) ^{\frac{1-3\chi _{i}}{1+3\chi _{i}}}}\nonumber \\&+ \frac{dr^{2}}{1-\dfrac{\kappa \rho _{i}r^{2}}{3}} + r^{2} (d{\theta }^{2} + \sin ^{2}(\theta ) d{\phi }^{2}) , \end{aligned}$$

(B21)

and from (B16) we can express the coordinate transformation between the $\hbox {SS}\chi _{i}$ and the FLRW frames in terms of $\rho _{i}$ y $\rho _{i0}=\rho _{i}(T_{0})$,

$$\begin{aligned} t&= \frac{\left[ c + R^{2} (\kappa \rho _{i0})^{\frac{2}{3(\chi _{i}+1)}}(\kappa \rho _{i})^{\frac{3\chi _{i}+1}{3(\chi _{i}+1)}} \right] ^{\frac{1}{2n}}}{\left( A^{\frac{1}{2n}} \right) \sqrt{\kappa \rho _{i}}} \end{aligned}$$

(B22)

$$\begin{aligned} r&= R\left( \frac{\rho _{i0}}{\rho _{i}} \right) ^{\frac{1}{3(\chi _{i}+1)}}. \end{aligned}$$

(B23)

1.2 Appendix C: On the accuracy in the approximation of $H_{0}$

In order to simplify the computation, we can omit the geometrical prefactor that appears in (31), because it is common to every observation and is $H_{0}$-independent. Therefore, we define a reduced timing residual,

$$\begin{aligned} \tau _{\text {GW}}^{\text {red}}&\equiv \int _{-1}^{0} \frac{1 + H_{0} \left[ T_e + \frac{xL}{c} \right] }{Z + xL\cos \alpha } \sin (\frac{\pi }{4} + \Theta (x,\alpha )) dx \approx R_{1} \nonumber \\&\quad +\,\left( \frac{1+ \frac{H_{0} Z}{c}}{Z}\right) \int _{-1}^{0} \sin (\frac{\pi }{4} + \Theta (x,\alpha )) dx , \end{aligned}$$

(C1)

with $|R_1| \le \frac{L H_{0}}{c Z} \sim 10^{-31}$ s. Then, we take the reduced timing residual from (C1) and note that $R_1$ is given by

$$\begin{aligned} R_{1} = \int _{-1}^{0} dx \sin \left( \Theta ( x, \alpha ) + \frac{\pi }{4} \right) \left[ \frac{1+H_{0} \left[ \frac{Z_{e}}{c} + \frac{L}{c} x \right] }{Z_{e} +x L \cos \alpha } - \frac{1+H_{0} \frac{Z_{e}}{c}}{Z_{e}} \right] . \end{aligned}$$

(C2)

Thus we can bound the value of $R_1$ by

$$\begin{aligned} | R_{1} |&\le \frac{L}{Z_{e}} \int _{-1}^{0} \left| \sin \left( \Theta ( x, \alpha ) + \frac{\pi }{4} \right) \right| \nonumber \\&\times \left| \frac{H_{0} \left[ \frac{1}{c} x \right] -x \frac{L}{Z_{e}^{2}} \cos \alpha -x \frac{L}{Z_{e}} \cos \alpha H_{0} \frac{1}{c}}{\left( 1+x \frac{L}{Z_{e}} \cos \alpha \right) } \right| dx \nonumber \\&\le \frac{LH_{0}}{2Z c} + \mathcal {O}(\frac{L^{2}}{Z^3}) \sim 10^{-31}\,\text {s}. \end{aligned}$$

(C3)

Then we can reasonable neglect $R_1$ in the equation (C1). Now we can express $\tau _{\text {GW}}^{\text {red}}$ in terms of the imaginary part of the complex exponential and write, since $\Theta ( x, \alpha )$ is quadratic in x:

$$\begin{aligned} \tau _{\text {GW}}^{\text {red}}&= \mathfrak {I}{ \int _{-1}^{0} dx e^{i \left( \Theta ( x, \alpha ) + \frac{\pi }{4} \right) } }= \mathfrak {I}{ B(\alpha ) e^{i \left( \Theta ( x^{*} , \alpha ) + \frac{\pi }{4} \right) } } , \end{aligned}$$

(C4)

where $B(\alpha )$ is defined as $B(\alpha ) \equiv \int _{-1}^{0} dx e^{i \lambda ( x-x^{*} )^{2}}$, $x^*$ satisfies ${\partial {\Theta (x,\alpha )}/{\partial x}}_{x=x^*} = 0$, thus

$$\begin{aligned} x^* = \frac{-c+c \cos \alpha + {Z}_{e} {H_{0}}}{\left( \cos \alpha ^{2} -2 \cos \alpha \right) {H_{0}} L}, \end{aligned}$$

(C5)

and $\lambda$ is given by

$$\begin{aligned} \lambda = \frac{1}{2} \frac{\partial ^{2} \Theta (x,\alpha )}{\partial x^{2}} = \frac{1}{2} \frac{\Omega H_{0} L^{2}}{c^{2}} ( \cos \alpha ^{2} -2 \cos \alpha ). \end{aligned}$$

(C6)

The integral $B(\alpha )$ can be written in terms of the error function, giving

$$\begin{aligned} B(\alpha ) = \frac{\sqrt{2 \pi }}{4} ( 1+i ) \frac{1}{\sqrt{\lambda }} \left[ - \text {erf} \left( \frac{\sqrt{2}}{2} ( 1-i ) u^{*} \right) + \text {erf} \left( \frac{\sqrt{2}}{2} ( 1-i ) \left( \sqrt{\lambda } +u^{*} \right) \right) \right] , \end{aligned}$$

(C7)

where $u^* \equiv \sqrt{\lambda } x^*$. Using the asymptotic expansion of the error functions for $u^* \gg 1$, e.g. see [46], we can write

$$\begin{aligned} B(\alpha ) \approx e^{-z_{1}^{2}} \left( 1+ \frac{1}{2z_{1}} \right) \qquad z_1 \equiv \frac{\sqrt{2}}{2} ( 1-i ) u^{*}. \end{aligned}$$

(C8)

Inserting the last expression into (C4), $\tau _{\text {GW}}^{\text {red}}$ becomes

$$\begin{aligned} \tau _{\text {GW}}^{\text {red}} \approx \sin \left( C+ \frac{\pi }{4} \right) + \frac{1}{2 | u^{*} |} \sin C, \end{aligned}$$

(C9)

where $C=H_{0} Z^{2} \Omega /2c^{2}$. From this expression we can see that the maximum of $\tau _{\text {GW}}^{\text {red}}$ clearly happens for $u^*\rightarrow 0$. This condition implies, from (C6), that the angle corresponding to the maximum absolute value of $\tau _{\text {GW}}$ satisfies $x^*=0$, or, rearranging the terms, the approximation formula (33). In order to justify the validity of the asymptotic expansion, we can explore around $u^*=0$, finding that for a variation in the angle $\Delta \alpha$, then $u^* \sim i \sqrt{\frac{Z\Omega }{c}} \Delta \alpha \sim 10^{4} \Delta \alpha$. Thus, the expansion is well defined for $\Delta \alpha \gg 10^{-4}$.

1.3 Appendix D: Table of pulsars of the ATNF catalog

Table 4 List of randomly distributed pulsars averaged for an hypothetical source. The galactic longitude is denoted by $\theta$ and the galactic latitude by $\phi$. More information about the pulsars can be found here