Gravitational waves and matter

13.6.1

Problem 13.6.1: Particle motion in TT gauge

Consider a plane wave with "$+$" polarization.
1) Write down the equations of motion for a non-relativistic particle; what happens to particles at rest?
2) Find the geodesic deviation from the Raychaudhuri equation and show that the result is the same;

solution

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.

13.6.2

Problem 13.6.2: Proper distance variation

Find the variation of proper distance between two particles at rest in the TT frame in the presence of a plane gravitational wave with the "$+$" polarization. Show that a ring of test particles placed initially at rest in the plane orthogonal to the wave vector will be distorted into an ellipse with its axes directed along the polarization tensor's main axes and oscillating with the wave's frequency.

solution

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 \[l_{x}\approx r\cos\phi \cdot (1+\tfrac{h}{2}e^{i \omega t}),\quad l_{y}\approx r\sin\phi \cdot (1-\tfrac{h}{2}e^{i \omega t}).\]

13.6.3

Problem 13.6.3: Particle motion in the proper frame

Derive the equation of motion and geodesic deviation in the proper frame of one of the particles.

13.6.4

Problem 13.6.4: Euclidean spatial coordinates

Consider the plain gravitational wave with "$+$" polarization, propagating in the $z$ direction \[h_{\alpha \beta }(t,z)=e_{\alpha\beta}^{(+)}f(t-z)\] and introduce new coordinates in the $(x,y)$ plane at $z=0$: \[X=\left( 1+\frac{1}{2}f \right)x, \quad Y=\left( 1-\frac{1}{2}f \right)y\] The coordinates $(x,y)$ of test particles do not change with time, but $(X,Y)$ do.
Show, that the distance between the particles can be calculated in the first order by the wave's amplitude in the $(X,Y)$ coordinates using the Euclidean metric

solution

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.

13.6.5

Problem 13.6.5: Metric far from a non-relativistic source

Show that far from an isolated non-relativistic system of small enough mass the spacial components of the metric perturbation in the long-wave approximation are \[\bar{h}_{\alpha\beta}(t,\mathbf{r}) =\frac{2}{r}\frac{d^2 I_{\alpha\beta}(t_r)}{dt^2},\] where $t_{r}=t-r$ is retarded time and \[I_{\alpha\beta}=\int d^3 x\; x^\alpha x^\beta T_{00}(t,\mathbf{x})\] is the second mass moment of the system.

solution

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.