Difference between revisions of "Gravitational waves and matter"

1) The geodesic equation is \[\frac{du^\lambda}{d\tau} =-\Gamma^{\lambda}_{\mu\nu}u^\mu u^\nu.\] In the TT gauge only $h_{\alpha\beta}$ are non-zero, so the only $\Gamma^{\lambda}_{\mu\nu}$ different from zero are those with at least two spatial indexes. But then in the non-relativistic limit $u^\alpha \sim v/c \ll 1$ \begin{align*} \frac{du^0}{d\tau}&\sim \Gamma^{0}_{\alpha\beta}u^{\alpha}u^\beta \sim \partial_0 h_{\alpha\beta} u^\alpha u^\beta \sim h_{\alpha\beta} v^2,\\ \frac{du^\gamma}{d\tau}&\sim \Gamma^{\gamma}_{\alpha\mu}u^{\alpha}u^{\mu} \sim \partial_0 h_{\alpha\gamma} u^\alpha \sim h_{\alpha\gamma} v. \end{align*} For particles at rest $v=0$ and therefore $u^\mu =const$. Thus the TT frame (the frame in which the TT gauge conditions hold) appears to be comoving with the test particles. 2) Now consider the congruence of particles, which are at the initial moment at rest in the TT frame. The Raychaudhuri equation for the geodesic deviation $s^\mu$ is \[\frac{D^2 s^\mu}{D\tau^2}= {R^{\mu}}_{\nu\rho\sigma}u^{\nu}u^{\rho}s^{\sigma} ={R^{\mu}}_{00\sigma}s^{\sigma},\] where $\frac{D}{D\tau}=u^\mu \nabla_\mu$. As initially $s^0 =0$, the non-trivial equations are \[\frac{D^2 s^\alpha}{D\tau^2} =R_{0\alpha 0\beta}s^{\beta} =-\tfrac{1}{2}\partial_0^2 h_{\alpha\beta}s^\beta,\] where we used the expression for the curvature tensor components through the metric in the TT gauge. On the right hand side we have (dot denotes $\partial_0$) \begin{align*} \frac{Ds^\alpha}{D\tau}&=u^\nu \nabla_\nu s^\alpha =\nabla_0 s^\alpha =\partial_0 s^\alpha +{\Gamma^{\alpha}}_{0\mu}s^{\nu} =\dot{s}^\alpha-\Gamma_{\alpha,0\beta}s^\beta =\dot{s}^\alpha -\tfrac12 \dot{h}_{\alpha\beta}s^\beta;\\ \frac{D^2 s^\alpha}{D\tau^2}& =\nabla_0 \Big(\frac{D s^\alpha}{D\tau}\Big) =\partial_0 \Big(\frac{D s^\alpha}{D\tau}\Big) +{\Gamma^\alpha}_{0\beta}\frac{D s^\beta}{D\tau} \approx\partial_0 (\dot{s}^\alpha -\tfrac12 \dot{h}_{\alpha\beta}s^\beta) -\tfrac12 \dot{h}_{\alpha\beta}\dot{s}^\beta\\ &=\ddot{s}^{\alpha}- \dot{h}_{\alpha\beta}\dot{s}^\beta -\tfrac12 \ddot{h}_{\alpha\beta}s^\beta, \end{align*} so the equation is reduced to \[\ddot{s}^\alpha =\dot{h}_{\alpha\beta}\dot{s}^\beta.\] However, if the variation of $s$ is of the first order by $h$, then the right hand side is of the second order, and thus should be also discarded. Therefore in the linear approximation $s^\mu =const$ and the geodesic deviation in the TT gauge vanishes, in accordance with the geodesic equation considered above.

The proper distance between two particles with spatial coordinates $(x,y,z)$ and $(x+l,y,z)$ is \[L=\int \limits_{0}^{l}dx \sqrt{g_{xx}}\approx \int \limits_{0}^{l}dx(1+\tfrac12 h_{xx}) \approx l(1+\tfrac12 h_{xx}),\] so for two particles at arbitrary positions the fractional change $\delta l_{x}/l_{x}\equiv (L-l)/l$ in each direction is given by \[\frac{\delta l_{x}}{l_{x}}=\frac12 h_{xx},\quad \frac{\delta l_{y}}{l_{y}}=\frac12 h_{yy},\quad \frac{\delta l_{z}}{l_{z}}=0.\] Note, that fractional change in length of a solid body produces mechanical strain, which, in principle, can be measured directly. For the plane wave with "$+$" polarization $h_{xx}=-h_{yy}$ passing through a ring of particles $l_{0x}=R\cos\phi$, $l_{0y}=r\sin\phi$, so \[l_{x}\approx r\cos\phi \cdot (1+\tfrac{h}{2}e^{i \omega t}),\quad l_{y}\approx r\sin\phi \cdot (1-\tfrac{h}{2}e^{i \omega t}).\]

The metric is \[ds^{2}=dt^{2}-\left[ 1+f(t-z) \right]dx^{2} -\left[ 1-f(t-z) \right]dy^{2}-dz^{2}\] Making the change of variables to \[x=\frac{X}{1+\frac{1}{2}f}, \quad y=\frac{Y}{1-\frac{1}{2}f},\] we find \[ds^{2}\approx dt^{2} -\left[ 1-f^{2} \right]dX^{2} -\left[ 1-f^{2} \right]dY^{2}-dz^{2}.\] In the first order by $f$ it is Euclidean.

In Lorenz gauge the Einstein equations for $h_{\mu\nu}$ are reduced to wave equation ($c=1$) \[\square \bar{h}_{\mu\nu}=16\pi G\;T_{\mu\nu}.\] The retarded solution of the wave equation is \[\bar{h}_{\mu\nu}(t,\mathbf{x})=4G\int d^3 y\; \frac{T_{\mu\nu}(t-|\mathbf{x}-\mathbf{y}|, \mathbf{y})}{|\mathbf{x}-\mathbf{y}|}.\] Making Fourier transform by time $t$, we get (using the same notation for the Fourier images) \[\bar{h}_{\mu\nu}(\omega,\mathbf{x})=4G\int d^3 y\; \frac{e^{i\omega (\mathbf{x}-\mathbf{y})}} {|\mathbf{x}-\mathbf{y}|} T_{\mu\nu}(\omega,\mathbf{y}).\] Now we use the approximation that the source is isolated, far away from the observer and its internal motion is non-relativistic. Isolation and slow motion imply that $\omega |\mathbf{x}-\mathbf{y}|\ll 1$ in the region where the integrand is concentrated ($\omega \delta r \sim \delta r/t \sim v \ll c$); combining this with large distance to the observer, we can bring the term $e^{i\omega (\mathbf{x}-\mathbf{y})}/|\mathbf{x}-\mathbf{y}|$ outside the integral, thus obtaining \[\bar{h}_{\mu\nu}(\omega,\mathbf{x}) \approx 4G \frac{e^{i\omega r}}{r} \int d^3 y\; T_{\mu\nu}(\omega,\mathbf{y}),\] where $r$ is the distance to observer. In the Lorenz gauge we only need the spatial components of the metric perturbation, as the rest are related to them via the gauge conditions (raising and lowering of spacial indices changes the sign) \[\partial_\mu \bar{h}^{\mu\nu}=0 \quad\Leftrightarrow\quad \partial_0 \bar{h}_{0\mu} =\partial_{\alpha}\bar{h}_{\alpha\mu},\] or in Fourier space \[h_{0\mu} =\frac{1}{i\omega}\partial_{\alpha}h_{\alpha\mu}.\] Very similar to this, the spatial components of the energy-momentum tensor can be expressed through the $T^{00}$ components, thanks to the energy conservation law, which in the first order takes the familiar form with partial (instead of covariant) derivatives \[\partial_{\mu}T^{\mu\nu}=0, \quad\Leftrightarrow\quad \partial_{\alpha}T_{\alpha\mu}=\partial_0 T_{0\mu}.\] Then for an isolated source, for which the surface integral vanishes, integration by parts twice gives \begin{align*} 0&=\int d^3 y\; \partial_\gamma (y_{\alpha} T_{\beta\gamma}) =\int d^3 y\; T_{\alpha\beta} +\int d^3 y\; y_\alpha \partial_\gamma T_{\beta\gamma} =\int d^3 y\; T_{\alpha\beta} +\int d^3 y\; y_\alpha \partial_0 T_{0\beta},\\ &\Rightarrow\quad \int d^3 y\; T_{\alpha\beta}= -\int d^3 y\; y_\alpha \partial_0 T_{0\beta} =-i\omega \int d^3 y\; y_{\alpha}T_{0\beta};\\ 0&=\int d^3 y\; \partial_\gamma (y_{\alpha}y_{\beta} T_{0\gamma}) =\int d^3 y\; (y_{\alpha}T_{0\beta} +y_{\beta}T_{0\alpha}) +\int d^3 y\; y_\alpha y_{\beta} \partial_\gamma T_{0\gamma}=;\\ &=2\int d^y y_\alpha T_{0\beta} +\int d^3 y\; y_\alpha y_\beta \partial_0 T_{00} =2\int d^y y_\alpha T_{0\beta} +i\omega \int d^3 y\; y_\alpha y_\beta T_{00},\\ &\Rightarrow\quad \int d^3 y\; T_{\alpha\beta} =-i\omega \int d^3 y\; y_{\alpha}T_{0\beta} =-\frac{\omega^2}{2} \int d^3 y\; y_\alpha y_\beta T_{00}. \end{align*} So in the Fourier space \[\bar{h}_{\alpha\beta}(\omega,\mathbf{x}) \approx 2G \frac{e^{i\omega r}}{r} (-\omega^2)\int d^3 y\; y_\alpha y_\beta T_{00}.\] It is conventional to define the quadrupole momentum tensor as \begin{equation} I_{\alpha\beta}(t) =\int d^3 y\; y_\alpha y_\beta T^{00}(t,\mathbf{y}), \end{equation} so that back in coordinate space the final formula is \begin{equation} \bar{h}_{\alpha\beta}=\frac{2G}{r}\; \frac{d^2 I_{\alpha\beta}(t_{ret})}{dt^2}, \quad\text{where}\quad t_{ret}\equiv t-r \end{equation} is the retarded time.