Difference between revisions of "Wave equation"

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Line 10: Line 10:
 
\begin{equation}
 
\begin{equation}
 
\label{LorenzGauge}
 
\label{LorenzGauge}
\pa_{\mu}\bar{h}^{\mu\nu}=0.
+
\partial_{\mu}\bar{h}^{\mu\nu}=0.
 
\end{equation}
 
\end{equation}
 
Write down the Einstein's equations in the Lorenz gauge.
 
Write down the Einstein's equations in the Lorenz gauge.
Line 51: Line 51:
  
 
1) The equation for $\xi^\mu$ is $\square \xi^\mu
 
1) The equation for $\xi^\mu$ is $\square \xi^\mu
=\pa_{\nu}{\bar{h}^{\nu}}_{\mu}$;<br/>
+
=\partial_{\nu}{\bar{h}^{\nu}}_{\mu}$;<br/>
 
2) no, any $\xi^\mu$ satisfying $\square \xi^{\mu}=0$ preserves the Lorenz gauge.
 
2) no, any $\xi^\mu$ satisfying $\square \xi^{\mu}=0$ preserves the Lorenz gauge.
  
Line 60: Line 60:
 
\item 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
 
\item 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&=\partial_{\mu}\bar{h'}^{\mu}_{\nu}
+
0&=\partialrtial_{\mu}\bar{h'}^{\mu}_{\nu}
=\partial_{\mu}
+
=\partialrtial_{\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}
+
&=\partialrtial_{\mu}{h'}^{\mu}_{\nu}
-\tfrac12 \partial_{\nu}h'
+
-\tfrac12 \partialrtial_{\nu}h'
=\partial_{\mu}({h}^{\mu}_{\nu}
+
=\partialrtial_{\mu}({h}^{\mu}_{\nu}
-\partial^{\mu}\xi_{\nu}-\partial_{\nu}\xi^{\mu})
+
-\partialrtial^{\mu}\xi_{\nu}-\partialrtial_{\nu}\xi^{\mu})
-\tfrac12 \partial_{\nu}(h-2\partial_{\mu}\xi^{\mu})=\\
+
-\tfrac12 \partialrtial_{\nu}(h-2\partialrtial_{\mu}\xi^{\mu})=\\
&=\partial_{\mu}h^{\mu}_{\nu}-\tfrac12 \partial_{\nu}h
+
&=\partialrtial_{\mu}h^{\mu}_{\nu}-\tfrac12 \partialrtial_{\nu}h
 
-\square \xi_\nu\\
 
-\square \xi_\nu\\
&=\partial_{\mu}\bar{h}^{\mu}_{\nu}-\square \xi_\nu,
+
&=\partialrtial_{\mu}\bar{h}^{\mu}_{\nu}-\square \xi_\nu,
 
\end{align}
 
\end{align}
 
thus
 
thus
\[\square \xi_{\mu}=\partial_{\nu}\bar{h}^{\nu}_{\mu}.\]
+
\[\square \xi_{\mu}=\partialrtial_{\nu}\bar{h}^{\nu}_{\mu}.\]
 
\item If the Lorenz gauge is already fixed, it will be preserved by any additional coordinate transformation with $\xi^\mu$ satisfying
 
\item If the Lorenz gauge is already fixed, it will be preserved by any additional coordinate transformation with $\xi^\mu$ satisfying
 
\[\square \xi^{\mu}(x)=0.\]
 
\[\square \xi^{\mu}(x)=0.\]
Line 118: Line 118:
 
and therefore $w_\alpha =0$ as well. Then the $\alpha\beta$ equations are reduced to
 
and therefore $w_\alpha =0$ as well. Then the $\alpha\beta$ equations are reduced to
 
\[G_{\alpha\beta}=(\delta_{\alpha\beta}\triangle
 
\[G_{\alpha\beta}=(\delta_{\alpha\beta}\triangle
-\pa_\alpha \pa_\beta)\Phi
+
-\partial_\alpha \partial_\beta)\Phi
 
-\square s_{\alpha\beta}=0.\]
 
-\square s_{\alpha\beta}=0.\]
 
Taking the trace, we see that
 
Taking the trace, we see that

Revision as of 14:14, 26 December 2012



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}



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.



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:

$\Phi=\Psi=0,\quad w^\alpha =0,\quad \square s_{\alpha\beta}=0$


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

\begin{equation} \bar{h}_{\mu\nu}(t,\mathbf{x})=\frac{1}{4\pi}\int \frac{d^{3}x'}{|\mathbf{x}-\mathbf{x'}|}\cdot \frac{16\pi G}{c^4}T_{\mu\nu} \Big(t-\frac{|\mathbf{x}-\mathbf{x'}|}{c}, \mathbf{x'}\Big). \end{equation}