Difference between revisions of "Wave equation"
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<p style="text-align: left;">1) Using the expression for variation of the metric perturbation under gauge transformations $x^\mu {x'}^{\mu}= x^\mu +\xi^\mu$, in the new frame we get | <p style="text-align: left;">1) Using the expression for variation of the metric perturbation under gauge transformations $x^\mu {x'}^{\mu}= x^\mu +\xi^\mu$, in the new frame we get | ||
\begin{align} | \begin{align} | ||
| − | 0&=\ | + | 0&=\partial_{\mu}\bar{h'}^{\mu}_{\nu} |
| − | =\ | + | =\partial_{\mu} |
\big[{h'}^{\mu}_{\nu} | \big[{h'}^{\mu}_{\nu} | ||
-\tfrac12 h' \delta^\mu_\nu\big]=\\ | -\tfrac12 h' \delta^\mu_\nu\big]=\\ | ||
| − | &=\ | + | &=\partial_{\mu}{h'}^{\mu}_{\nu} |
| − | -\tfrac12 \ | + | -\tfrac12 \partial_{\nu}h' |
| − | =\ | + | =\partial_{\mu}({h}^{\mu}_{\nu} |
| − | -\ | + | -\partial^{\mu}\xi_{\nu}-\partial_{\nu}\xi^{\mu}) |
| − | -\tfrac12 \ | + | -\tfrac12 \partial_{\nu}(h-2\partial_{\mu}\xi^{\mu})=\\ |
| − | &=\ | + | &=\partial_{\mu}h^{\mu}_{\nu}-\tfrac12 \partial_{\nu}h |
-\square \xi_\nu\\ | -\square \xi_\nu\\ | ||
| − | &=\ | + | &=\partial_{\mu}\bar{h}^{\mu}_{\nu}-\square \xi_\nu, |
\end{align} | \end{align} | ||
thus | thus | ||
| − | \[\square \xi_{\mu}=\ | + | \[\square \xi_{\mu}=\partial_{\nu}\bar{h}^{\nu}_{\mu}.\] |
2) If the Lorenz gauge is already fixed, it will be preserved by any additional coordinate transformation with $\xi^\mu$ satisfying | 2) If the Lorenz gauge is already fixed, it will be preserved by any additional coordinate transformation with $\xi^\mu$ satisfying | ||
\[\square \xi^{\mu}(x)=0.\]</p> | \[\square \xi^{\mu}(x)=0.\]</p> | ||
Revision as of 14:16, 26 December 2012
Contents
Problem 1: Lorenz (Hilbert, harmonic) gauge
The Lorenz\footnote{By analogy with electromagnetism; note the spelling: Lorenz, not Lorentz.} gauge conditions are \begin{equation} \label{LorenzGauge} \partial_{\mu}\bar{h}^{\mu\nu}=0. \end{equation} Write down the Einstein's equations in the Lorenz gauge.
Hint: They are reduced to the wave equation \begin{equation} \label{WaveEq} \square \bar{h}_{\mu\nu}=\frac{16\pi G}{c^4}T_{\mu\nu}. \end{equation}
Using the general expression for the linearized Einstein tensor, we see that it takes especially simple form \begin{equation} G_{\mu\nu}=\square\bar{h}_{\mu\nu}, \end{equation} thus the Einstein's equations are reduced to the wave equation \begin{equation} \label{WaveEq2} \square \bar{h}_{\mu\nu}=\frac{16\pi G}{c^4}T_{\mu\nu} \end{equation} with the stress-energy tensor as the source (but note $16\pi$ in the numerator). This simple form makes the Lorenz gauge the preferred choice for studying most aspects of gravitational waves.
Problem 2: Lorenz frame
Lorenz frame is the coordinate frame in which the Lorenz gauge conditions (\ref{LorenzGauge}) are satisfied.
1) find the coordinate transformation $x^\mu \to {x'}^{\mu}=x^\mu +\xi^\mu$ from a given frame to the Lorenz frame;
2) is the Lorenz frame unique? what is the remaining freedom for the choice of $\xi^\mu$?
Hint:
1) The equation for $\xi^\mu$ is $\square \xi^\mu
=\partial_{\nu}{\bar{h}^{\nu}}_{\mu}$;
2) no, any $\xi^\mu$ satisfying $\square \xi^{\mu}=0$ preserves the Lorenz gauge.
1) Using the expression for variation of the metric perturbation under gauge transformations $x^\mu {x'}^{\mu}= x^\mu +\xi^\mu$, in the new frame we get \begin{align} 0&=\partial_{\mu}\bar{h'}^{\mu}_{\nu} =\partial_{\mu} \big[{h'}^{\mu}_{\nu} -\tfrac12 h' \delta^\mu_\nu\big]=\\ &=\partial_{\mu}{h'}^{\mu}_{\nu} -\tfrac12 \partial_{\nu}h' =\partial_{\mu}({h}^{\mu}_{\nu} -\partial^{\mu}\xi_{\nu}-\partial_{\nu}\xi^{\mu}) -\tfrac12 \partial_{\nu}(h-2\partial_{\mu}\xi^{\mu})=\\ &=\partial_{\mu}h^{\mu}_{\nu}-\tfrac12 \partial_{\nu}h -\square \xi_\nu\\ &=\partial_{\mu}\bar{h}^{\mu}_{\nu}-\square \xi_\nu, \end{align} thus \[\square \xi_{\mu}=\partial_{\nu}\bar{h}^{\nu}_{\mu}.\] 2) If the Lorenz gauge is already fixed, it will be preserved by any additional coordinate transformation with $\xi^\mu$ satisfying \[\square \xi^{\mu}(x)=0.\]
Problem 3: Wave equation from action
Obtain the Einstein's equation in the Lorenz gauge in the first order by $h$ directly from the action
Problem 4: Vacuum equations in the transverse gauge
Simplify the vacuum Einstein's equations in the transverse gauge (without the Lorenz gauge conditions), assuming vanishing boundary conditions.
Hint:
$\Phi=\Psi=0,\quad w^\alpha =0,\quad \square s_{\alpha\beta}=0$
Substitution of the gauge conditions into the Einstein tensor yields the following. The $00$ equation is \[G_{00}=-2\triangle \Psi =0,\] thus, assuming vanishing boundary condition (at infinity or elsewhere), $\Psi=0$. Then the $0\alpha$ equations take the form \[G_{0\alpha}=\tfrac12 \triangle w_\alpha =0,\] and therefore $w_\alpha =0$ as well. Then the $\alpha\beta$ equations are reduced to \[G_{\alpha\beta}=(\delta_{\alpha\beta}\triangle -\partial_\alpha \partial_\beta)\Phi -\square s_{\alpha\beta}=0.\] Taking the trace, we see that \[\triangle \Phi=0,\] which means that $\Phi=0$ and the remaining non-trivial equation is \[\square s_{\alpha\beta}=0.\] Note, that such representation was only possible in vacuum, i.e. for gravitational wave solutions.
The only difference of the wave equation for gravitational perturbations from the one for the electromagnetic field is in its tensorial nature. The Green's function for the wave equation is known (see e.g. here or here), and the retarded solution, which is usually thought of as the one physically relevant, is
