The propagation delay in the timing of a pulsar orbiting a supermassive black hole

Abstract

The observation of a pulsar closely orbiting the galactic center supermassive black hole would open the window for an accurate determination of the black hole parameters and for new tests of General Relativity. An important relativistic effect which has to be taken into account in the timing model is the propagation delay of the pulses in the gravitational field of the black hole. Due to the extreme mass ratio of the pulsar and the supermassive back hole we use the test particle limit to derive an exact analytical formula for the propagation delay in a Schwarzschild spacetime. We then compare this result to the propagation delays derived in the usually employed post-Newtonian approximation, in particular to the Shapiro delay up to the second post-Newtonian order. For edge-on orbits we also consider modifications of the Shapiro delay which take the lensing effects into account. Our results are then used to assess the accuracy of the different orders of the post-Newtonian approximation of the propagation delay. This comparison indicates that for (nearly) edge-on orbits the new exact delay formula should be used.

Appendix A: solution to the time integral

The integrals in Eq. (12) are of the form

$$\begin{aligned} T(r,b) := \int _{r_4}^r \frac{r^2 dr}{b \left( 1-\frac{2}{r}\right) \sqrt{R(r)}} \end{aligned}$$

(A1)

with $r=r_e$ or $r=\infty$, and R is defined in Eq. (6). Here we normalised all quantities such that they are dimensionless, i.e. $r={\tilde{r}}/m$, $b={\tilde{b}}/m$ where the twiddled quantities are in geometrised units. This integral can be analytically solved in terms of elliptic integrals. The substitution

$$\begin{aligned} x^2&= \frac{(r-r_4)(r_3-r_1)}{(r-r_3)(r_4-r_1)} \end{aligned}$$

(A2)

with the roots $r_i$ of R chosen as in Eq. (9), casts the integral in the Legendre form

$$\begin{aligned} T(r,b)&= \frac{2}{\sqrt{r_4(r_3-r_1)}} \int _{0}^{x(r)} \frac{f(x) dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\,, \end{aligned}$$

(A3)

where

$$\begin{aligned} k^2&= \frac{r_3(r_4-r_1)}{r_4(r_3-r_1)}\,, \end{aligned}$$

(A4)

$$\begin{aligned} f(x)&= \frac{r_3^3}{r_3-2} + \frac{A_1}{1-c_1x^2} + \frac{A_2}{1-c_2x^2} + \frac{A_3}{(1-c_3x^2)^2} \end{aligned}$$

(A5)

with the constants

$$\begin{aligned} A_1&= 2(r_4-r_3)(r_3+1),\quad&c_1&= \frac{r_4-r_1}{r_3-r_1},\nonumber \\ A_2&= \frac{8(r_3-r_4)}{(r_3-2)(r_4-2)},\quad&c_2&= \frac{(r_4-r_1)(r_3-2)}{(r_3-r_1)(r_4-2)} \\ A_3&= (r_4-r_3)^2,\quad&c_3&= c_1 \nonumber . \end{aligned}$$

(A6)

We note that for $r=\infty$ Eq. (A2) reduces to

$$\begin{aligned} x_{\infty }^2&:= x(r=\infty )^2 = \frac{r_3-r_1}{r_4-r_1} = \frac{1}{c_1}\,. \end{aligned}$$

(A7)

With the Jacobi elliptic integrals

$$\begin{aligned} F(x,k)&= \int _0^x \frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\,, \end{aligned}$$

(A8)

$$\begin{aligned} E(x,k)&= \int _0^x \sqrt{1-k^2x^2} dx\,, \end{aligned}$$

(A9)

$$\begin{aligned} \Pi (x,c,k)&= \int _0^x \frac{dx}{(1-cx^2)\sqrt{(1-x^2)(1-k^2x^2)}} \,, \end{aligned}$$

(A10)

we find

$$\begin{aligned} T(r,b)&= \frac{2}{\sqrt{r_4(r_3-r_1)}} \left[ \frac{r_3^3}{r_3-2} F(x,k) + A_1 \Pi (x,c_1,k) + A_2 \Pi (x,c_2,k) \right. \nonumber \\&\quad + \frac{A_3}{2(c_3-1)} \left( \frac{x c_3^2\sqrt{(1-x^2)(1-k^2x^2)}}{(1-c_3x^2)(c_3-k^2)}\right. \nonumber \\&\quad \left. \left. + F(x,k)-\frac{c_3}{c_3-k^2} E(x,k) + \frac{c_3^2+3k^2-2c_3(1+k^2)}{c_3-k^2} \Pi (x,c_3,k) \right) \right] \nonumber \\&= \frac{2}{\sqrt{r_4(r_3-r_1)}} \left[ \left( \frac{r_3^3}{r_3-2}+\frac{(r_4-r_3)(r_3-r_1)}{2}\right) F(x,k)\right. \nonumber \\&\quad - \frac{1}{2} r_4(r_3-r_1) E(x,k) + 2 (r_4-r_3) \Pi (x,c_1,k)\nonumber \\&\quad \left. - \frac{8(r_4-r_3)}{(r_4-2)(r_3-2)} \Pi (x,c_2,k) \right] + \frac{b\sqrt{R(r)}}{r-r_3}\,, \end{aligned}$$

(A11)

where x is related to r via (A2). Note that the Jacobi elliptic integrals can be evaluated without using a numeric integration, and can therefore be considered as an exact analytical solution to the integral T.

Note that the last term in (A11) diverges linearly for $r \rightarrow \infty$. As well, $\Pi (x,c,k)$ diverges logarithmically for $x^2=1/c$, which happens in our case for $x=x_\infty$ and $c=c_1=c_3$. Therefore, the time for reaching $r=\infty$ diverges as expected. To isolate the diverging terms we apply an identity,

$$\begin{aligned} \Pi (x,c,k)&= F(x,k) - \Pi \left( x,\frac{k^2}{c},k\right) + \frac{\ln (Z)}{2P} \end{aligned}$$

(A12)

where

$$\begin{aligned} Z&= \frac{\sqrt{(1-x^2)(1-k^2x^2)}+Px}{\sqrt{(1-x^2)(1-k^2x^2)}-Px}\,, \end{aligned}$$

(A13)

$$\begin{aligned} P^2&= \frac{(c-1)(c-k^2)}{c} = \frac{(r_4-r_3)^2}{r_4(r_3-r_1)}\,. \end{aligned}$$

(A14)

for $c=c_1$. With this Eq. (A11) becomes

$$\begin{aligned} T(r,b)&= \frac{2}{\sqrt{r_4(r_3-r_1)}} \left[ \left( \frac{r_3^3}{r_3-2}+\frac{1}{2}(r_4-r_3)(r_3-r_1+4)\right) F(x,k)\right. \nonumber \\&\quad - \frac{1}{2} r_4(r_3-r_1) E(x,k) - 2 (r_4-r_3) \Pi \left( x,\frac{k^2}{c_1},k\right) \nonumber \\&\quad \left. - \frac{8(r_4-r_3)}{(r_4-2)(r_3-2)} \Pi (x,c_2,k)\right] + \frac{b\sqrt{R(r)}}{r-r_3} \nonumber \\&\quad + 2\ln \left( \frac{\sqrt{r(r-r_1)}+\sqrt{(r-r_4)(r-r_3)}}{\sqrt{r(r-r_1)}-\sqrt{(r-r_4)(r-r_3)}} \right) \end{aligned}$$

(A15)

where the last two terms diverge for $x=x_{\infty }$. We find the Taylor expansions of these terms as

$$\begin{aligned} \frac{b\sqrt{R(r)}}{r-r_3}&= r+r_3+{\mathcal {O}}\left( \frac{1}{r}\right) \,, \end{aligned}$$

(A16)

$$\begin{aligned} 2\ln (Z)&= 2\ln \left( \frac{2}{r_4+r_3}\right) + 2\ln r + {\mathcal {O}}\left( \frac{1}{r}\right) \,. \end{aligned}$$

(A17)