2023-08-28

title: "Advance of periastron with radiation backreaction"
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Additional scalar dipole radiation brings away energy additionally. Since dipole radiation has slower attenuation than quadruple radiation, the scalar radiation might have much larger observational effect at a large orbit comparing to the gravitational wave.

A. The radiation power of different sources

The power of scalar dipole radiation under $m_\phi=0$ limit and gravitational waves up to $\ell=2$ is given by

$P_{\mathrm{GW}}=\frac{32}{5}G\mu ^2r^4\Omega ^6 \tag{A.1}$
$P_{\mathrm{SR}}^{\left( 1 \right)}=\frac{1}{12\pi}\frac{\left( Q_1M_2-Q_2M_1 \right) ^2}{M^2}r^2\Omega ^4 \tag{A.2}$
$P_{\mathrm{SR}}^{\left( 2 \right)}=\frac{4}{15\pi}\frac{\left( Q_1M_{2}^{2}+Q_2M_{1}^{2} \right) ^2}{M^4}r^4\Omega ^6 \tag{A.3}$

For a system only has one charged body, i.e., the relation of the stellar angular frequency $(\Omega)$ and the orbit radius ( $r$ ) in the center-of-mass frame is given by the Kelper's relation in high accuracy as

$\Omega ^2=\frac{GM}{r^3} \tag{A.4}$

Besides, $Q_1=\sqrt{4\pi G}\gamma M_1$ , $Q_2=0$ . Additionally, in an extreme mass ratio inspiral (EMRI) system, $M_1\approx M$ , $M_2\approx \mu$ , and the mass ratio $q\equiv M_2/M_1\ll 1$ . (A.1) - (A.3) could be simplified as
$P_{\mathrm{GW}}=\frac{2}{5}q\left( \frac{r}{2GM} \right) ^{-4}\mu r^{-1}, \tag{A.5}$

$P_{\mathrm{SR}}^{\left( 1 \right)}=\frac{1}{24}\gamma ^2q\left( \frac{r}{2GM} \right) ^{-3}\mu r^{-1}, \tag{A.6}$

$P_{\mathrm{SR}}^{\left( 2 \right)}=\frac{1}{15}\gamma ^2q^3\left( \frac{r}{2GM} \right) ^{-4}\mu r^{-1}. \tag{A.7}$

Comparison between radiation power from different sources is calculated as
$\frac{P_{\mathrm{SR}}^{\left( 1 \right)}}{P_{\mathrm{GW}}}=\frac{5}{48}\gamma ^2\left( \frac{r}{2GM} \right) \tag{A.8}$
For typical $\bar{r}=10^3$ , $\gamma^2=0.1$ , the power of the scalar radiation is about $\mathcal{O}(10)$ times larger than gravitational wave.

Similarly, ratio between the scalar quadruple radiation and gravitational wave is about
$\frac{P_{\mathrm{SR}}^{\left( 2 \right)}}{P_{\mathrm{GW}}}= \frac{\gamma ^2}{6}q^2 \tag{A.9}$
For EMRI system with mass ratio $q\ll 1$ , the scalar quadruple radiation is negligible.

B. First order post-Newtonian approximation

The total energy is made up with two parts, as
$E=E_0+E_1 \tag{B.1}$
where

$E_0=\frac{1}{2}\mu r^2\Omega ^2-\frac{GM\mu}{r}=-\frac{1}{2}\frac{GM\mu}{r} \tag{B.2}$
Satisfies Virial theorem.

The first-order Post-Newtonian (1PN) correction for a circular orbit is given by (Gtavitational Waves Vol.I, pp.246)
$E_1=\frac{1}{8}\mu r^4\Omega ^4-\frac{1}{2}\frac{GM\mu}{r}\left( 3r^2\Omega ^2-\frac{GM}{r} \right) =-\frac{7}{8}\frac{G^2M^2\mu}{r^2} \tag{B.3}$
For $r=\mathcal{O}(1000)\times 2GM$ , 1PN will bring in a $\mathcal{0.1\%}$ contribution.

C. The backreaction of scalar dipole radiation

Under Newtonian mechanics, the system energy is made up with gravitational potential
$U\left( r \right) =-\frac{GM\mu}{r}, \tag{C.1}$

kinetic energy
$K\left( r \right) =\frac{1}{2}\mu \left( v_{r}^{2}+v_{\varphi}^{2} \right), \tag{C.2}$

where $u_r\equiv \mathrm{d}r/\mathrm{d}t$ , $u_\varphi\equiv r\mathrm{d}\varphi /\mathrm{d}t$ . Besides, there is additional energy loss due to the scalar dipole radiation, as
$S\left( r \right) \equiv -\int{P_{\mathrm{SR}}^{\left( 1 \right)}\left( r \right) \mathrm{d}r}. \tag{C.3}$

The conservation of energy shows
$K(r)+U(r)-S(r)=\mathrm{const.}, \tag{C.4}$

or
$\frac{1}{2}\mu \left[ \left( \frac{\mathrm{d}r}{\mathrm{d}t} \right) ^2+r^2\left( \frac{\mathrm{d}\varphi}{\mathrm{d}t} \right) ^2 \right] -\frac{GM\mu}{r}-\int{P_{\mathrm{SR}}^{\left( 1 \right)}\mathrm{d}r}=A. \tag{C.5}$

Define the system angular momentum as $L\equiv r\mu v_\varphi=\mu r^2\mathrm{d}\varphi /\mathrm{d}t$ , the energy conservation condition then reads
$\left( \frac{\mathrm{d}r}{\mathrm{d}\varphi} \right) ^2+r^2=\frac{2\mu r^4}{L^2}\left( A+\frac{GM\mu}{r}+\int{P_{\mathrm{SR}}^{\left( 1 \right)}\mathrm{d}r} \right). \tag{C.6}$

(Notice that the units $[A]=M, [L]=1, [GM]=M^{-1}, [r]=M^{-1}$ ). Define $u\equiv r^{-1}$ , the orbit equation becomes
$\left( \frac{\mathrm{d}u}{\mathrm{d}\varphi} \right) ^2\frac{1}{u^4}+\frac{1}{u^2}=\frac{2\mu A}{L^2}\frac{1}{u^4}+\frac{2GM\mu ^2}{L^2}\frac{1}{u^3}-\left( \int{P_{\mathrm{SR}}^{\left( 1 \right)}\frac{\mathrm{d}u}{u^2}} \right) \frac{2\mu ^2}{L^2}\frac{1}{u^4}. \tag{C.7}$

Re-arrange as
$\left( \frac{\mathrm{d}u}{\mathrm{d}\varphi} \right) ^2+u^2=\frac{2\mu A}{L^2}+\frac{2GM\mu ^2}{L^2}u-\left( \int{P_{\mathrm{SR}}^{\left( 1 \right)}\frac{\mathrm{d}u}{u^2}} \right) \frac{2\mu ^2}{L^2}. \tag{C.8}$

Calculate the derivative with respect to $\varphi$ on both sides. On the right handed side, the first term $2\mu A/L^2$ vanishes. The orbit equation then becomes
$\frac{\mathrm{d}u}{\mathrm{d}\varphi}\left( \frac{\mathrm{d}^2u}{\mathrm{d}\varphi ^2}+u \right) =\frac{\mathrm{d}u}{\mathrm{d}\varphi}\left( \frac{GM\mu ^2}{L^2}-\frac{\mu}{L^2}\frac{P_{\mathrm{SR}}^{\left( 1 \right)}}{u^2} \right) . \tag{C.9}$

The first solution is $\mathrm{d}u/\mathrm{d}\varphi=0$ , which refers to a circular orbit. When the orbit is not circular, after substituting the radiation power in (A.2), the orbital equation is given as
$\frac{\mathrm{d}^2u}{\mathrm{d}\varphi ^2}+u=\frac{GM\mu ^2}{L^2}-\frac{G^3M^2\mu ^3}{3L^2}\gamma ^2u^{-2}\equiv K_1-K_2u^{-2}. \tag{C.10}$

Take $K_2 u^{-2}$ as small perturbation, the leading order solution reads
$\frac{\mathrm{d}^2u_0}{\mathrm{d}\varphi}+u_0=K_1\,\,\Longrightarrow \,\,u_0=K_1\left( 1+e\cos \varphi \right), \tag{C.11}$

where $e$ is the eccentricity. The leading solution refers to Keplerian elliptic orbits when $0<e<1$ Substitute $u_0$ to the orbital equation, as
$r.h.s.=K_1-K_{1}^{2}K_2\left( 1+2e\cos \varphi +e^2\cos ^2\varphi \right) \approx K_1-2K_{1}^{2}K_2e\cos \varphi . \tag{C.12}$

The orbital equation could then be linearly separated into two parts, as
$\frac{\mathrm{d}^2u_0}{\mathrm{d}\varphi ^2}+u_0=K_1, \tag{C.13}$
$\frac{\mathrm{d}^2u_1}{\mathrm{d}\varphi ^2}+u_1=-2K_{1}^{2}K_2e\cos \varphi. \tag{C.14}$

The latter has a special solution as
$u_1=-K_{1}^{2}K_2e\varphi \sin \varphi . \tag{C.15}$

Then the EMRI orbit with additional scalar radiation energy loss is shown as
$u\left( \varphi \right) =u_0\left( \varphi \right) +u_1\left( \varphi \right) =K_1\left[ \left( 1+e\cos \varphi \right) -K_1K_2e\varphi \sin \varphi \right]. \tag{C.16}$

or
$u\left( \varphi \right) =\frac{GM\mu ^2}{L^2}\left[ \left( 1+e\cos \varphi \right) -\frac{G^4M^3\mu ^5}{3L^4}\gamma ^2e\varphi \sin \varphi \right] . \tag{C.17}$

Comparing with the Schawarzschild periastron, as
$u\left( \varphi \right) =\frac{GM\mu ^2}{L^2}\left[ \left( 1+e\cos \varphi \right) +3\frac{G^2M^2\mu ^2}{L^2}e\varphi \sin \varphi \right]. \tag{C.18}$

The periastron term has the ratio
$\frac{\mathcal{P} _{\mathrm{SR}}}{\mathcal{P} _{\mathrm{Sch}}}=-\frac{G^2M\mu ^3}{9L^2}\gamma ^2. \tag{C.19}$

Notice that
$L=\mu r^2\Omega =\mu r^2\sqrt{\frac{GM}{r^3}}=GM\mu \sqrt{2\left( \frac{r}{2GM} \right)}, \tag{C.20}$

The ratio reads
$\frac{\mathcal{P} _{\mathrm{SR}}}{\mathcal{P} _{\mathrm{Sch}}}=-\frac{\mu}{M}\frac{\gamma ^2}{18}\left( \frac{r}{2GM} \right) ^{-1}. \tag{C.21}$

Not only suppressed by the mass ratio $q\approx \mu/M$ , but suppressed by dimensionless radius $\bar{r}^{-1}$ as well.

D. The periastron angles

Notice that the series expansion
$\cos \left[ \left( 1+K \right) \varphi \right] =\cos \varphi -\varphi K\sin \varphi -\frac{1}{2}\varphi ^2K^2\cos \varphi +\mathcal{O} \left( K^3 \right). \tag{D.1}$

The periastron angle (C.17) is given by
$u\left( \varphi \right) =\frac{GM\mu ^2}{L^2}\left\{ 1+e\cos \left[ \left( 1+\frac{G^4M^3\mu ^5}{3L^4}\gamma ^2 \right) \varphi \right] \right\}. \tag{D.2}$

The periastron point is given by
$\left( 1+\frac{G^4M^3\mu ^5}{3L^4}\gamma ^2 \right) \varphi =2n\pi , n=0,1,2,\cdots \tag{D.3}$

Or
$\varphi =\frac{2n\pi}{1+\frac{G^4M^3\mu ^5}{3L^4}\gamma ^2}= 2n\pi\left(1-\frac{G^4M^3\mu ^5}{3L^4}\gamma ^2\right)+\mathcal{O}(\gamma^4). \tag{D.4}$

The precession angle
$\left|\Delta \varphi _{\mathrm{SR}}\right|=\frac{\pi}{24}\frac{\mu}{M}\gamma ^2\frac{\left( 2GM\mu \right) ^4}{L^4}. \tag{D.5}$

Comparing to such precession from GR prediction
$\left|\Delta \varphi _{\mathrm{Sch}}\right|=\frac{3\pi}{2}\frac{\left( 2GM\mu \right) ^2}{L^2}. \tag{D.6}$

This also results in
$\frac{\Delta \varphi _{\mathrm{SR}}}{\Delta \varphi _{\mathrm{Sch}}}=-q\frac{\gamma ^2}{18}\left( \frac{r}{2GM} \right) ^{-1}, \tag{D.7}$

as same as (C.21).

E. The shrinked radius

Due to the additional energy loss channel, the inspiral radius will gradually shrink.

The total energy of the system with 1PN correction, (B.2), (B.3)
$E_{tot}=-\frac{1}{2}\frac{GM\mu}{r}-\frac{7}{8}\frac{G^2M^2\mu}{r^2} \tag{E.1}$

With $\bar{E}\equiv E/\mu$ , $\bar{r}\equiv r/(2GM)$ and $\bar{t}\equiv t/(2GM)$ , the total (dimensionless) energy
$\bar{E}_{tot}\left( \bar{r} \right) =-\frac{1}{4\bar{r}}\left( 1+\frac{7}{8}\frac{1}{\bar{r}} \right) \tag{E.2}$

Thus
$\frac{\mathrm{d}\bar{E}_{tot}}{\mathrm{d}\bar{t}}=\frac{\mathrm{d}\bar{E}_{tot}}{\mathrm{d}\bar{r}}\frac{\mathrm{d}\bar{r}}{\mathrm{d}\bar{t}}=\frac{4\bar{r}+7}{16\bar{r}^3}\frac{\mathrm{d}\bar{r}}{\mathrm{d}\bar{t}} \tag{E.3}$

On the other hand, the source term (scalar dipole radiation and GW)
$\frac{\mathrm{d}E_{sou}}{\mathrm{d}t}=P_{\mathrm{SR}}^{\left( 1 \right)}+P_{\mathrm{GW}}=\frac{32}{5}G\mu ^2r^4\Omega ^6+\frac{1}{12\pi}\frac{4\pi G\gamma ^2M_{1}^{2}M_{2}^{2}}{M^2}r^2\Omega ^4, \tag{E.4}$

or
$\frac{\mathrm{d}\bar{E}_{sou}}{\mathrm{d}\bar{t}}=\frac{2}{5}\bar{r}^{-5}\left( 1+\frac{5}{48}\gamma ^2\bar{r} \right). \tag{E.5}$

From (E.2) and (E.5), the circular orbit radius evolving with time is given by
$\frac{\mathrm{d}\bar{r}}{\mathrm{d}\bar{t}}=-\frac{2}{15}\frac{5\gamma ^2\bar{r}+48}{\bar{r}^2\left( 4\bar{r}+7 \right)}. \tag{E.6}$

In one complete circle, the decrease of the orbit is approximately given by
$\Delta \bar{r}\approx \frac{\mathrm{d}\bar{r}}{\mathrm{d}\bar{t}}\Delta \bar{t}, \tag{E.7}$

where the time period $\Delta\bar{t}=2\pi/\bar{\Omega}=2\sqrt{2}\pi \bar{r}^{3/2}$ . The change of the radius is given by
$\Delta \bar{r}=-\frac{4\sqrt{2}\pi}{15}\frac{48\bar{r}^{3/2}}{\bar{r}^2\left( 4\bar{r}+7 \right)}\left( 1+\frac{5\gamma ^2}{48}\bar{r} \right). \tag{E.8}$

When $\bar{r}=1000$ , $\gamma^2=1$ , $\left|\Delta r \right|\approx 0.047\times2GM$ .

Consider that the Schwarzschild radius (2GM) of Sgr A* is about $10^{-7}$ pc, the radius should decrease around $10^{-8}$ pc. However, the orbit error is around 1 mas at 8 kpc distance, which corresponds to around $10^{-5}$ pc. This means the decreasing of the radius is hard to detect by current observations.

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