Dr. Gyula Bene Department for Theoretical Physics, Loránd Eötvös University Pázmány Péter sétány 1/A, 1117 Budapest 10. week
Weak gravitational fields
If all nearby masses are small, one can find a coordinate system in which
metric does not differ much from Minkowski metric. The choice of such a
coordinate system is not unique, since a coordinate transformation near the
identity transforms to another such coordinate system.
Thus metric corresponding to weak gravitational fields may be written as
\[\begin{align}
g_{ik}=\eta_{ik}+h_{ik}\;,
\end{align}\]
where \(\eta_{ik}\) stands for Minkowski metric and components of \(h_{ik}\) are
small compared unity. Therefore, quadratic and higher order terms may be neglected.
The determinant of the metric tensor is approximately (to first order in \(h_{ik}\))
\[\begin{align}
g=-1-\eta^{ik}h_{ik}\;,
\end{align}\]
while the contravariant metric tensor may be expressed as
\[\begin{align}
g^{ik}=\eta^{ik}-\eta^{ij}\eta^{kn}h_{jn}\;.
\end{align}\]
Rising and lowering of indices will be done with the dominant Minkowski
metric, hence by definition
\[\begin{align}
h^{ik}=\eta^{ij}\eta^{kn}h_{jn}\;.
\end{align}\]
Riemannian to linear order reads
\[\begin{align}
{\mathcal R}_{iklm}=
\frac{1}{2}\left(\frac{\partial^2 h_{im}}{\partial x^k \partial x^l}
+\frac{\partial^2 h_{kl}}{\partial x^i \partial x^m}
-\frac{\partial^2 h_{il}}{\partial x^k \partial x^m}
-\frac{\partial^2 h_{km}}{\partial x^i \partial x^l}
\right)\;,
\end{align}\]
which yields the Ricci tensor via contracting indices with the Minkowski metric:
\[\begin{align}
{\mathcal R}_{km}&=
\frac{1}{2}\eta^{il}\left(\frac{\partial^2 h_{im}}{\partial x^k \partial x^l}
+\frac{\partial^2 h_{kl}}{\partial x^i \partial x^m}
-\frac{\partial^2 h_{il}}{\partial x^k \partial x^m}
-\frac{\partial^2 h_{km}}{\partial x^i \partial x^l}
\right)\\
&=\frac{1}{2}\left(-\eta^{il}\frac{\partial^2 h_{km}}{\partial x^i \partial
x^l}
+\frac{\partial^2 h_{m}^l}{\partial x^l \partial x^k}
+\frac{\partial^2 h_{k}^l}{\partial x^l \partial x^m}
-\frac{\partial^2 h_{l}^l}{\partial x^k \partial x^m}\right)
\;.\phantom{gyeng1b}
\end{align}\]
Performing a near-identity transformation \(x'^i=x^i+\xi^i(\{x^k\})\) where functions \(\xi^i\)
are small, then
\[\begin{align}
h'_{ik}=h_{ik}-\frac{\partial \xi_i}{\partial x^k}-\frac{\partial \xi_k}{\partial x^i}\phantom{gyeng_tr}
\end{align}\]
will be the correction of the metric in the new coordinates, still small.
The possibility of free choice of the four functions \(\xi^i\) may be utilized
to simplify Einstein's equations. Let
\[\begin{align}
\psi^k_i=h^k_i-\frac{1}{2}\delta^k_ih^j_j\;.\phantom{gyeng_psi}
\end{align}\]
that are subject to the four side conditions
\[\begin{align}
\frac{\partial \psi^k_i}{\partial x^k}=0\;.\phantom{gyeng_mellek}
\end{align}\]
Owing to these the last three terms in the expression of the Ricci tensor
cancel each other, thus remains
\[\begin{align}
{\mathcal R}_{km}=\frac{1}{2}\square h_{km}\equiv -\frac{1}{2}\eta^{il}\frac{\partial^2 h_{km}}{\partial x^i \partial
x^l}
\end{align}
\]
Here \(\square=\triangle-1/c^2\partial^2/\partial t^2\) stands for the wave
operator (d'Alambert's operator). Einstein's equations then read as
\[\begin{align}
\frac{1}{2}\square \psi^k_i=\frac{8\pi k}{c^4}T^k_{i}\;.\phantom{gyeng2}
\end{align}\]
Side conditions are compatible with this, as one may prove by taking the
divergence of the equation.
Static gravitational field
In case of a static mass distribution all the other components of the energy
momentum tensor are small compared to \(T^0_0\). Then the linearized Einstein
equations simplify to
\[\begin{align}
\triangle \psi^0_0=\frac{16\pi k}{c^2}\rho\;,\phantom{gyeng3}
\end{align}\]
where \(\rho\) stands for mass density. Other components of Einstein's equations
have vanishing right hand sides, implying that the other components of
\(\psi^k_i\) are zero. The solution for \(\psi^0_0\):
\[\begin{align}
\psi^0_0({\bf r})=-\frac{4 k}{c^2}\int \frac{\rho({\bf r}')d^3{\bf r}'}{|{\bf r}-{\bf r}'|}\;,\phantom{gyeng4}
\end{align}\]
which implies
\[\begin{align}
h^0_0({\bf r})=-h^\alpha_\alpha({\bf r})=-\frac{2 k}{c^2}\int \frac{\rho({\bf
r}')d^3{\bf r}'}{|{\bf r}-{\bf r}'|}\equiv \frac{2 \Phi({\bf r})}{c^2}\;.\phantom{gyeng4a}
\end{align}\]
Here no summation over the spacelike index \(\alpha\) is understood, and
\(\Phi({\bf r})\)
stands for the Newtonian gravitational potential. Thus the interval in a weak
static gravitational field is given by
\[\begin{align}
ds^2 = \left( 1+\frac{2\Phi ({\bf r}) }{c^2} \right) c^2dt^2-\left(
1-\frac{2\Phi({\bf r})}{c^2} \right) \left( dx^2+dy^2+dz^2 \right)
\;.
\end{align}\]
Stationary gravitational field
In this case the components of the energy momentum tensor and those of the
metric are still independent of time (in a suitably chosen coordinate system),
but beside
\(T^0_0\) components \(T^\alpha_0\) and \(T^0_\alpha\) are significant, too.
Thus equations
\[\begin{align}
\triangle \psi^\alpha_0=\frac{16\pi k}{c^3}\rho v^\alpha\phantom{gyeng6}
\end{align}\]
are to be solved, too. The solution reads:
\[\begin{align}
h^\alpha_0({\bf r})=-\frac{4 k}{c^3}\int \frac{\rho({\bf
r}')v^\alpha({\bf
r}')d^3{\bf r}'}{|{\bf r}-{\bf r}'|}\;.\phantom{gyeng7}
\end{align}\]
In case of a rotating sphere of angular momentum \({\bf J}\) this gives
\[\begin{align}
h_{\alpha 0}({\bf r})=\frac{2k}{c^3}\frac{\left({\bf r}\times{\bf J}\right)_\alpha}{r^3}\;.\phantom{gyeng7a}
\end{align}\]
Gravitational waves
Propagation of gravitational waves in Minkowski spacetime
Away from sources Einstein's equations read as
\[\begin{align}
\square \psi^k_i=0\;\phantom{gyeng8}
\end{align}\]
i.e., they are wave equations, which have nontrivial solutions: waves
travelling with the speed of light. If metric is transformed with the previous
near-identity coordinate transformation, transformed metric also
satisfies the side conditions provided that
\[\begin{align}
\square \xi^i=0\;.\phantom{gyeng_m1}
\end{align}\]
This freedom will be used to simplify the metric.
A plane wave solution may be written as
\[\begin{align}
\psi^n_m=A^n_m\exp\left(ik_jx^j\right)\;,\phantom{gyeng9}
\end{align}\]
where the square of the four wave vector is zero:
\[\begin{align}
k_jk^j=0\;,\phantom{gyeng10}
\end{align}\]
and side conditions imply for the constant amplitudes \(A^n_i\)
\[\begin{align}
A^n_mk_n=0\;.\phantom{gyeng11}
\end{align}\]
For the metric we have
\[\begin{align}
h_{nm}=\left(A_{nm}-\frac{1}{2}\eta_{nm}A^j_j\right)\exp\left(ik_jx^j\right)\;.\phantom{gyeng12}
\end{align}\]
Performing now a near-identity coordinate transformation
\(x'^n=x^n+\xi^n(x)\) with
\[\begin{align}
\xi^n=\zeta^n\exp\left(ik_jx^j\right)\;,\phantom{gyeng13}
\end{align}\]
where \(\zeta^n\) is a constant vector, we have
\[\begin{align}
A'_{nm}=A_{nm}+i\eta_{nm}k^j\zeta_j-ik_n\zeta_m-ik_m\zeta_n\phantom{gyeng14}
\end{align}\]
Let us choose our coordinate axes such that axis \(x\) points along the
spacelike part of the wave vector. Then \(k^0=k^1=k\), \(k_2=k_3=0\).
Then we have
\[\begin{align}
A'_{10}&=-A'_{00}=-A'_{11}\\
A'_{20}&=-A'_{21}\\
A'_{30}&=-A'_{31}
\end{align}\]
The four components \(\zeta^n\) may be chosen to set \(A'_{10}\),
\(A'_{20}\), \(A'_{30}\) and \(A'_{22}+A'_{33}\) to zero. Then the trace \(A'^n_n\)
also vanishes. Explicitly, these conditions mean
\[\begin{align}
&A_{10}+ik\zeta_0-ik\zeta_1=0\\
&A_{20}-ik\zeta_2=0\\
&A_{30}-ik\zeta_3=0\\
&A_{22}+A_{33}-2ik(\zeta_0+\zeta_1)=0\;,
\end{align}\]
that may always be solved for \(\zeta_j\). We also have
\[\begin{align}
A'_{23}&=A_{23}\\
A'_{22}-A'_{33}&=A_{22}-A_{33}\;,
\end{align}\]
hence these combinations are not influenced by the coordinate transformation.
Finally, we get for \(h_{nm}\)
\[\begin{align}
\left(
\begin{array}{cccc}
0&0&0&0\\
0&0&0&0\\
0&0&A_{22}&A_{23}\\
0&0&A_{23}&-A_{22}\\
\end{array}
\right)\;,
\end{align}\]
determined by the two independent quantities \(A_{22}\) and \(A_{23}\).
The matrix of amplitudes is orthogonal to the wave vector:
\[A_{n\alpha}k^\alpha=0\]
Thus we have to do with a transversal wave. The two independent quantities
correspond to two possible polarizations (one: \(A_{22}=0\) and \(A_{23}=1\), two: \(A_{23}=0\) and
\(A_{22}=1\)). They transform into each other by a \(45^\circ\) rotation around
the direction of propagation.
Energy carried by a gravitational planar wave
For planar gravitational waves the energy momentum pseudotensor of the
gravitational field simplifies as
\[\begin{align}
t^{ik}=\frac{c^4}{32\pi k}h^{n,i}_mh^{m,k}_n\;.
\end{align}\]
In case of a planar wave propagating along the \(x\) axis we get
\[\begin{align}
ct^{01}=\frac{c^3}{16\pi k}\left[\left(\dot{h}_{23}\right)^2+\frac{1}{4}\left(\dot{h}_{22}-\dot{h}_{33}\right)^2\right]\;.\phantom{gyeng_en}
\end{align}\]
Radiation of gravitational waves
In the presence of sources the general weak gravitational field equations are
to be solved. The are of the same form as that of four potential of
electromagnetic radiation, thus one may write directly
\[\begin{align}
\psi^j_n({\bf r},t)=-\frac{4 k}{c^4}\int \frac{T^j_n({\bf
r}',t')d^3{\bf r}'}{|{\bf r}-{\bf r}'|}\;,\phantom{gyeng_sug1}
\end{align}\]
where
\[\begin{align}
t'=t-\frac{|{\bf r}-{\bf r}'|}{c}\;.\phantom{gyeng_sug2}
\end{align}\]
Provided that the extension of the source is much smaller than the wavelength,
while the distance of the observation point is much larger than the
wavelength (radiation zone), than it simplifies to
\[\begin{align}
\psi^j_n({\bf r},t)=-\frac{4 k}{c^4R_0}\int T^j_n({\bf
r}',t')d^3{\bf r}'\;,\phantom{gyeng_sug3}
\end{align}\]
where
\[\begin{align}
t'=t-\frac{R_0}{c}\;,\phantom{gyeng_sug4}
\end{align}\]
and
\[\begin{align}
R_0=|{\bf r}-{\bf r}_0|\;.\phantom{gyeng_sug4a}
\end{align}\]
Here \({\bf r}_0\) is an arbitrary fixed point inside the source.
It is sufficient to evaluate this expression for spacelike indices only, i.e.,
\[\begin{align}
h_{\alpha\beta}({\bf r},t)=-\frac{4 k}{c^4R_0}\int T_{\alpha\beta}({\bf
r}',t')d^3{\bf r}'\;.\phantom{gyeng_sug5}
\end{align}\]
Continuity equation now reads simply
\[\begin{align}
\frac{\partial T_{00}}{\partial x^0}-\frac{\partial T_{0\mu}}{\partial x^\mu}=0
\end{align}\]
and
\[\begin{align}
\frac{\partial T_{0\mu}}{\partial x^0}-\frac{\partial T_{\mu\nu}}{\partial x^\nu}=0\;,
\end{align}\]
which imply
\[\begin{align}
\frac{\partial^2 T_{00}}{\partial x^{02}}=\frac{\partial^2 T_{0\mu}}{\partial x^\mu\partial x^{0}}=\frac{\partial^2 T_{\mu\nu}}{\partial x^\mu\partial x^\nu}\;.
\end{align}\]
Multiplying both sides with \(x^\alpha x^\beta\), we have
\[\begin{align}
\frac{\partial^2 T_{00}x^\alpha x^\beta}{\partial x^{02}}=x^\alpha x^\beta\frac{\partial^2 T_{\mu\nu}}{\partial x^\mu\partial x^\nu}=\frac{\partial^2 T_{\mu\nu}x^\alpha x^\beta}{\partial x^\mu\partial x^\nu}-2\frac{\partial }{\partial x^\mu}\left(T_{\mu\alpha}x^\beta+T_{\mu\beta}x^\alpha\right)+2T_{\alpha\beta}
\end{align}\]
Integrating over three dimensional space and applying Gauss's theorem we get
\[\begin{align}
\int T_{\alpha\beta}d^3{\bf r}=\frac{1}{2}\left(\frac{\partial }{\partial x^{0}}\right)^2\int x^\alpha x^\beta T_{00}d^3{\bf r}\;.
\end{align}\]
Hence
\[\begin{align}
h_{\alpha\beta}({\bf r},t)=-\frac{2 k}{c^4R_0}\frac{\partial^2 }{\partial
t^{2}}\int \rho({\bf
r}',t')x'^\alpha x'^\beta d^3{\bf r}'\;,\phantom{gyeng_sug6}
\end{align}\]
where \(\rho=T_{00}/c^2\) stands for mass density.
Let us introduce quadrupole momentum 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
\[\begin{align}
ct^{01}=\frac{k}{36\pi c^5 R_0^2}\left[{\dddot{Q}}^2_{23}+\frac{1}{4}\left({\dddot{Q}}_{22}-{\dddot{Q}}_{33}\right)^2\right]\;.
\phantom{gyeng_sug8}
\end{align}\]
The energy carried by gravitational radiation is obtained by integrating over
all directions:
\[\begin{align}
-\frac{dE}{dt}=\frac{k}{45c^5}{\dddot{Q}}^2_{\alpha\beta}\;,
\phantom{gyeng_sug9}
\end{align}\]
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:
\[\begin{align}
-\frac{dE}{dt}=\frac{32k^4m_1^2m_2^2(m_1+m_2)}{5c^5r^5}\;,
\phantom{gyeng_sug10}
\end{align}\]
where \(m_1\) and \(m_2\) are the masses of the objects, \(r\) stand for their
distance, assuming that it changes slowly because the energy loss due to
gravitational radiation is much less even in significant time than the
mechanical energy of the stellar system.
bene@arpad.elte.hu
