Dr. Gyula Bene Department for Theoretical Physics, Loránd Eötvös University Pázmány Péter sétány 1/A, 1117 Budapest 12. week
Experimental evidence of general relativity II. (calculations)
Precession of a free symmetric top: the Gravity Probe B experiment
Gravitational field of a rotating sphere
We start with
\[\begin{align}
h^\alpha_0({\bf r})=-\frac{4 k}{c^3}\int \frac{\rho({\bf
r}')v^\alpha({\bf
r}')d^3{\bf r}'}{|{\bf r}-{\bf r}'|}\;,\phantom{gyeng7}
\end{align}\]
or
\[\begin{align}
h_{\alpha 0}({\bf r})=\frac{4 k}{c^3}\int \frac{\rho({\bf
r}')v^\alpha({\bf
r}')d^3{\bf r}'}{|{\bf r}-{\bf r}'|}\;.\phantom{gyeng7}
\end{align}\]
Suppose that the density distribution is spherically symmetric, i.e.
\[\rho({\bf
r}')=\rho(|{\bf
r}'|)\;.\]
Velocity due to rotation:
\[{\bf v}({\bf r}')={\bf \omega}\times{\bf r}'\;.\]
We make use of the expansion
\[\frac{1}{|{\bf r}-{\bf r}'|}=\sum_{l=0}^\infty\sum_{m=-l}^l\frac{4\pi}{2l+1}\frac{r_<^l}{r_>^{l+1}}Y_l^m(\vartheta,\varphi)Y_l^{m*}(\vartheta',\varphi')\;,
\]
where \(\;r_<\;\) and \(\;r_>\;\) is the smaller and larger, respectively, of
\(\;r=|{\bf r}|\;\) and \(\;r'=|{\bf r}'|\;\). Henceforth we assume that \(\;r>r'\;\). Since \(\;{\bf v}({\bf r}')\;\) is linear
in \(\;{\bf r}'\;\), only the \(\;l=1\;\) terms contribute to the integral. These
terms can also be written as
\[\frac{{\bf r}\cdot{\bf r}'}{r^3}\;,\]
hence
\[\begin{align}
h_{\alpha 0}({\bf r})=\frac{4 k}{c^3
r^3}\epsilon^{\alpha\beta\gamma}\omega^\beta r^\mu \int \rho(|{\bf
r}'|){ r}'^\gamma{ r}'^\mu d^3{\bf r}'\;.\phantom{gyeng7}
\end{align}\]
The integral may be expressed as
\[\int \rho(|{\bf
r}'|){ r}'^\gamma{ r}'^\mu d^3{\bf r}'=\frac{1}{3}\delta_{\gamma\mu}\int \rho(|{\bf
r}'|){ r}'^2 d^3{\bf r}'\;,
\]
thus
\[\begin{align}
h_{\alpha 0}({\bf r})=\frac{4 k}{c^3
r^3}\epsilon^{\alpha\beta\gamma}\omega^\beta r^\gamma \frac{4\pi}{3} \int \rho(|{\bf
r}'|){ r}'^4d{ r}'\;.\phantom{gyeng7}
\end{align}\]
Because the inertial momentum is
\[\Theta=\int \rho(|{\bf
r}'|){ r}'^2 \sin^2\vartheta d^3{\bf r}'=\frac{8\pi}{3}\int \rho(|{\bf
r}'|){ r}'^4 d{ r}'\;,\]
we have
\[\begin{align}
h_{\alpha 0}({\bf r})=\frac{2k}{c^3}\frac{\left({\bf r}\times{\bf J}\right)_\alpha}{r^3}\;.\phantom{gyeng7a}
\end{align}\]
Here \(\;{\bf J}=\Theta {\bf \omega}\;\) stands or the angular momentum of the
rotating sphere.
Precession of a symmetric top
\[\begin{align}
\frac{d n^\alpha}{dx^0}=\left(-\Gamma^\alpha_{\beta 0}+\Gamma^\alpha_{0 0}\frac{g_{0\beta}}{g_{00}}\right)n^\beta\phantom{porg1}
\end{align}\]
In a weak field \(\;\Gamma^\alpha_{0 0}\;\) and \(\;g_{0\beta}\;\) are both of
first
order. Hence only the term
\[\Gamma^\alpha_{\beta 0}=-\frac{1}{2}g_{\alpha \beta,0}-\frac{1}{2}g_{\alpha
0, \beta}+\frac{1}{2}g_{\beta 0,\alpha }=-\frac{1}{2}g_{\alpha
0, \beta}+\frac{1}{2}g_{\beta 0,\alpha }\]
contributes. Note that the metric is stationary. So we have
\[\begin{align}
\frac{dn^\alpha}{dx^0}=\frac{1}{2}\left(g_{0\alpha,\beta}-g_{0\beta,\alpha}\right)n^\beta
=\frac{1}{2}\left(h_{0\alpha,\beta}-h_{0\beta,\alpha}\right)n^\beta
\;.\phantom{probe1}
\end{align}\]
We insert here the metric created by a rotating sphere. We get
\[\begin{eqnarray}
\dot{\bf n}&=&\frac{k}{c^2r^5}\left\{-2r^2\left({\bf n}\times{\bf
J}\right)-3\left[{\bf n}\cdot\left({\bf r}\times{\bf J}\right)\right]{\bf r}
+3\left[{\bf r}\cdot{\bf n}\right]\left({\bf r}\times{\bf J}\right)
\right\}\\
&=&\frac{k}{c^2r^5}\left\{-2r^2\left({\bf n}\times{\bf
J}\right)-3{\bf n}\times\left[{\bf r}\times\left({\bf r}\times{\bf J}\right)\right]
\right\}\\
&=&\frac{k}{c^2r^5}\left\{-2r^2\left({\bf n}\times{\bf
J}\right)-3{\bf n}\times\left[\left({\bf r}\cdot {\bf J}\right){\bf r}-r^2 {\bf J}\right]
\right\}\\
&=&\frac{k}{c^2r^5}\left\{r^2\left({\bf n}\times{\bf
J}\right)-3{\bf n}\times\left[\left({\bf r}\cdot {\bf J}\right){\bf r}\right]
\right\}
\;.\phantom{eqnarray}
\end{eqnarray}\]
Radiation of gravitational waves: the Hulse-Taylor pulsar
Total radiated energy in terms of the quadruple moments
Quadruple moment tensor:
\[\begin{align}
Q_{\alpha\beta}=\int \rho(3 x^\alpha x^\beta -\delta_{\alpha\beta} x_\gamma^2)d^3{\bf r}\;.
\end{align}\]
Then we have in an observation point along the \(\;x\;\) axis
\[\begin{align}
h_{23}=-\frac{2 k}{3c^4R_0}\ddot{Q}_{23}\phantom{gyeng_sug7a}
\end{align}\]
and
\[\begin{align}
h_{22}-h_{33}=-\frac{2 k}{3c^4R_0}\left(\ddot{Q}_{22}-\ddot{Q}_{33}\right)\;.\phantom{gyeng_sug7b}
\end{align}\]
Energy current density then reads
\[\begin{align}
ct^{01}=\frac{k}{36\pi c^5 R_0^2}\left[{\dddot{Q}}^2_{23}+\frac{1}{4}\left({\dddot{Q}}_{22}-{\dddot{Q}}_{33}\right)^2\right]\;.
\phantom{gyeng_sug8}
\end{align}\]
This refers to the case when the wave along the \(\;x\;\) axis is
considered. If a general direction is considered, given by unit vector
\(\;\hat{\bf k}\;\), then it is suitable to introduce a tensor (the
transversal projection of the quadruple momentum tensor)
\[\tilde{Q}_{\alpha\beta}=(\delta_{\alpha\mu}-\hat{\bf k}_\alpha\hat{\bf k}_\mu)Q_{\mu\nu}(\delta_{\nu\beta}-\hat{\bf k}_\nu\hat{\bf k}_\beta)\;.\]
Indeed, if \(\;\hat{\bf k}\;\) points in the \(\;x\;\) direction, we have
\[{\dddot{Q}}^2_{23}+\frac{1}{4}\left({\dddot{Q}}_{22}-{\dddot{Q}}_{33}\right)^2=\frac{1}{2}\dddot{\tilde{Q}}^2_{\alpha\beta}-\frac{1}{4}\left(\dddot{\tilde{Q}}_{\alpha\alpha}\right)^2\;.\]
Thus, in a general direction we get for the energy current density
\[\begin{eqnarray} && \frac{k}{36\pi c^5
R_0^2}\left[\frac{1}{2}\dddot{\tilde{Q}}^2_{\alpha\beta}-\frac{1}{4}\left(\dddot{\tilde{Q}}_{\alpha\alpha}\right)^2\right]\\
= && \frac{k}{36\pi c^5
R_0^2}\left[\frac{1}{2}\dddot{Q}^2_{\alpha\beta}-\dddot{Q}_{\alpha\mu}\dddot{Q}_{\alpha\nu}\hat{\bf
k}_\mu\hat{\bf k}_\nu
+\frac{1}{4}\dddot{Q}_{\alpha\mu}\dddot{Q}_{\beta\nu}\hat{\bf k}_\alpha
\hat{\bf k}_\beta \hat{\bf k}_\mu \hat{\bf k}_\nu \right]\;.\end{eqnarray}\]
The energy carried away by gravitational radiation in unit time is obtained by
integrating this over
all directions, i.e., averaging over \(\;\hat{\bf k}\;\) and multiplying the
result with
\(\;4\pi \;\). Since
\[\overline{\hat{\bf k}_\mu \hat{\bf k}_\nu}=\frac{1}{3}\delta_{\mu\nu}\]
and
\[\overline{\hat{\bf k}_\alpha \hat{\bf k}_\beta\hat{\bf k}_\mu \hat{\bf k}_\nu}=\frac{1}{15}\left(\delta_{\alpha\beta}\delta_{\mu\nu}+\delta_{\alpha\mu}\delta_{\beta\nu}+\delta_{\alpha\nu}\delta_{\mu\beta}\right)\;\]
we get
\[\begin{align}
-\frac{dE}{dt}=\frac{k}{45c^5}{\dddot{Q}}^2_{\alpha\beta}\;,
\phantom{gyeng_sug9}
\end{align}\]
where a summation is understood over \(\alpha\) and \(\beta\).
This formula may be evaluated for the case of two stellar objects revolving
around each other, and gives the energy loss due to gravitational radiation.
To this end we calculate the quadruple momentum tensor in a frame whose origin
coincides with the center of mass of the binary system and its \(\;x\;\) axis
is orthogonal to the plane of motion. Provided that the orbits are circular
and the distance of the stars is \(\;r\;\), one with mass \(\;m_1\;\) is
orbiting on a circle of radius \(\;m_2/(m_1+m_2)r\;\), and the other with mass \(\;m_2\;\)
on a circle of radius \(\;m_1/(m_1+m_2)r\;\), always oppositely to the
former. Hence the \(\;y,z\;\) coordinates of the stars are
\[\begin{eqnarray}
y_1 &=& \frac{m_2}{m_1+m_2}r\cos\omega t\\
z_1 &=& \frac{m_2}{m_1+m_2}r\sin\omega t\\
y_2 &=& -\frac{m_1}{m_1+m_2}r\cos\omega t\\
z_2 &=& -\frac{m_1}{m_1+m_2}r\sin\omega t
\end{eqnarray}\]
Here \(\;\omega\;\) stands for the angular frequency of revolution:
\[\omega=\sqrt{\frac{k(m_1+m_2)}{r^3}}\;.\]
Now we have
\[\int \rho({\bf r}) r^\alpha r^\beta d^3{\bf
r}=\frac{m_1m_2\;r^2}{m_1+m_2}\left(\begin{array}{cc}
0 & 0 & 0 \\ 0 & \cos^2\omega t &
\sin\omega t\cos\omega t\\0 & \sin\omega t\cos\omega t & \sin^2\omega t\end{array}\right)\]
and the quadruple momentum tensor:
\[Q_{\alpha\beta} = \frac{m_1m_2\;r^2}{m_1+m_2}\left(\begin{array}{cc}
-1 & 0 & 0 \\ 0 & \frac{1}{2}+\frac{3}{2}\cos 2\omega t &
\frac{3}{2}\sin 2\omega t\\0 & \frac{3}{2}\sin 2\omega t &
\frac{1}{2}-\frac{3}{2}\cos 2\omega t\end{array}\right)\]
Its third time derivative is
\[\dddot{Q}_{\alpha\beta} = \frac{12\omega^3 m_1m_2\;r^2}{m_1+m_2}\left(\begin{array}{cc}
0 & 0 & 0 \\ 0 & \sin 2\omega t &
-\cos 2\omega t\\0 & -\cos 2\omega t &
-\sin 2\omega t\end{array}\right)\]
From this we get
\[\dddot{Q}^2_{\alpha\beta} =\frac{288\omega^6
m_1^2m_2^2\;r^4}{(m_1+m_2)^2}=\frac{288 k^3 m_1^2m_2^2(m_1+m_2)}{r^5}\;.\]
Inserting this into the previous expression of the total intensity, we get
\[\begin{align}
-\frac{dE}{dt}=\frac{32k^4m_1^2m_2^2(m_1+m_2)}{5c^5r^5}\;.
\phantom{gyeng_sug10}
\end{align}\]
Change of the orbital period:
Since
\[T=\frac{2\pi r^{3/2}}{\sqrt{k(m_1+m_2)}}\]
\[\frac{dT}{dt}=\frac{3\pi r^{1/2}}{\sqrt{k(m_1+m_2)}}\frac{dr}{dt}\]
\[E=-\frac{km_1m_2}{2r}\]
\[\frac{dE}{dt}=\frac{km_1m_2}{2r^2}\frac{dr}{dt}\;,\]
we have
\[\frac{dT}{dt}=\frac{6\pi r^{5/2}}{k^{3/2}m_1 m_2
\sqrt{m_1+m_2}}\frac{dE}{dt}=
-\frac{192\pi k^{5/2} m_1m_2\sqrt{m_1+m_2}}{5c^5r^{5/2}}\;.
\]
Orbital frequency:
\[f=\frac{1}{T}\propto r^{-3/2}\]
and its temporal change:
\[\frac{df}{dt}=-\frac{1}{T^2}\frac{dT}{dt}\propto r^{-11/2}\;.\]
Therefore
\[\frac{df}{dt}f^{-11/3}\propto const\;.\]
Chirp mass:
\[\frac{(m_1 m_2)^{3/5}}{(m_1+m_2)^{1/5}}=\frac{c^3}{k}\left[\frac{9\pi^{-8/3}}{96}\frac{df}{dt}f^{-11/3}\right]^{3/5}\;.\]
Note that a similar calculation for elliptic orbits yields
\[\begin{align}
-\frac{dE}{dt}=\frac{32k^4m_1^2m_2^2(m_1+m_2)}{5c^5a^5}\left(1-e^2\right)^{-7/2}\left(1+\frac{73}{24}e^2+\frac{37}{96}e^4\right)\;,
\phantom{gyeng_sug10}
\end{align}\]
where \(\;a\;\) is half of the major axis and \(\;e\;\) stands for excentricity.
For the loss of angular momentum (due to gravitational radiation) we have
\[\begin{align}
-\frac{dJ}{dt}=\frac{32k^{7/2}m_1^2m_2^2\sqrt{m_1+m_2}}{5c^5a^{7/2}}\left(1-e^2\right)^{-2}\left(1+\frac{7}{4}e^2\right)\;.
\phantom{gyeng_sug10}
\end{align}\]
Since
\[E=-\frac{km_1m_2}{2a}\]
and
\[e=\sqrt{1+\frac{2EJ^2(m_1+m_2)}{k^2m_1^3m_2^3}}\;,\]
the temporal change of the orbit parameters can be calculated:
\[\begin{eqnarray}
\dot{a} &=& -\frac{64k^3m_1m_2(m_1+m_2)}{5c^5
a^3}(1-e^2)^{-7/2}\left(1+\frac{73}{24}e^2+\frac{37}{96}e^4\right)\;,\\
\dot{e} &=& -\frac{44k^3m_1m_2(m_1+m_2)}{3c^5
a^4}(1-e^2)^{-7/2}\left(1+\frac{41}{44}e^2\right)e\;.
\end{eqnarray}\]
This approximation obviously fails if
\[a(1-e)<\frac{k(m_1+m_2)}{c^2}\;.\]
bene@arpad.elte.hu
