Wave equation - Revision history

Igor at 13:05, 15 January 2013

2013-01-15T13:05:13Z

Cosmo All at 03:08, 8 January 2013

2013-01-08T03:08:34Z

Cosmo All: /* Problem 3: Wave equation from action */

2012-12-26T14:17:22Z

‎Problem 3: Wave equation from action

Cosmo All: /* Problem 2: Lorenz frame */

2012-12-26T14:16:47Z

‎Problem 2: Lorenz frame

Cosmo All: /* Problem 2: Lorenz frame */

2012-12-26T14:15:39Z

‎Problem 2: Lorenz frame

Cosmo All: /* Problem 2: Lorenz frame */

2012-12-26T14:15:05Z

‎Problem 2: Lorenz frame

Cosmo All: /* Problem 1: Lorenz (Hilbert, harmonic) gauge */

2012-12-26T14:14:49Z

‎Problem 1: Lorenz (Hilbert, harmonic) gauge

Cosmo All at 14:14, 26 December 2012

2012-12-26T14:14:12Z

Cosmo All at 14:13, 26 December 2012

2012-12-26T14:13:40Z

Cosmo All: Created page with "4"

2012-12-26T12:06:23Z

Created page with "4"

New page

[[Category:Weak field limit and gravitational waves|4]]

@@ Line 9: / Line 9: @@
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 1: Lorenz (Hilbert, harmonic) gauge===
-The Lorenz\footnote{By analogy with electromagnetism; note the spelling: Lorenz, not Lorentz.} gauge conditions are
+The Lorenz${}^{*}$ gauge conditions are
 \begin{equation}
 \label{LorenzGauge}
@@ Line 16: / Line 16: @@
 Write down the Einstein's equations in the Lorenz gauge.
-'''Hint:'''
+${}^{*}$By analogy with electromagnetism; note the spelling: Lorenz, not Lorentz.
 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}
+\end{equation}-->
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
@@ Line 37: / Line 40: @@
    </div>
 </div></div>
@@ Line 50: / Line 51: @@
 ) is the Lorenz frame unique? what is the remaining freedom for the choice of $\xi^\mu$?<br/>
-'''Hint:'''
+<!--'''Hint:'''
-−
 ) The equation for $\xi^\mu$ is $\square \xi^\mu
 	=\partial_{\nu}{\bar{h}^{\nu}}_{\mu}$;<br/>
-) no, any $\xi^\mu$ satisfying $\square \xi^{\mu}=0$ preserves the Lorenz gauge.
+) no, any $\xi^\mu$ satisfying $\square \xi^{\mu}=0$ preserves the Lorenz gauge.-->
 <div class="NavFrame collapsed">
@@ Line 80: / Line 80: @@
    </div>
 </div></div>
@@ Line 96: / Line 94: @@
    </div>
 </div>--></div>
@@ Line 107: / Line 103: @@
 Simplify the vacuum Einstein's equations in the transverse gauge (without the Lorenz gauge conditions), assuming vanishing boundary conditions.
-'''Hint:'''
+<!--'''Hint:'''
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>

← Older revision	Revision as of 03:08, 8 January 2013
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	+
	+	__TOC__

@@ Line 88: / Line 88: @@
 === 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
-<div class="NavFrame collapsed">
+<!-- <div class="NavFrame collapsed">
-<!--  <div class="NavHead">solution</div>
+  <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
      <p style="text-align: left;"></p>
@@ Line 101: / Line 101: @@
 <div id="gw24"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === 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.

@@ Line 59: / Line 59: @@
      <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}
-&=\partialrtial_{\mu}\bar{h'}^{\mu}_{\nu}
+&=\partial_{\mu}\bar{h'}^{\mu}_{\nu}
-	=\partialrtial_{\mu}
+	=\partial_{\mu}
 	\big[{h'}^{\mu}_{\nu}
 		-\tfrac12 h' \delta^\mu_\nu\big]=\\
-&=\partialrtial_{\mu}{h'}^{\mu}_{\nu}
+&=\partial_{\mu}{h'}^{\mu}_{\nu}
-	-\tfrac12 \partialrtial_{\nu}h'
+	-\tfrac12 \partial_{\nu}h'
-	=\partialrtial_{\mu}({h}^{\mu}_{\nu}
+	=\partial_{\mu}({h}^{\mu}_{\nu}
-	-\partialrtial^{\mu}\xi_{\nu}-\partialrtial_{\nu}\xi^{\mu})
+	-\partial^{\mu}\xi_{\nu}-\partial_{\nu}\xi^{\mu})
--\tfrac12 \partialrtial_{\nu}(h-2\partialrtial_{\mu}\xi^{\mu})=\\
+-\tfrac12 \partial_{\nu}(h-2\partial_{\mu}\xi^{\mu})=\\
-&=\partialrtial_{\mu}h^{\mu}_{\nu}-\tfrac12 \partialrtial_{\nu}h
+&=\partial_{\mu}h^{\mu}_{\nu}-\tfrac12 \partial_{\nu}h
 	-\square \xi_\nu\\
-&=\partialrtial_{\mu}\bar{h}^{\mu}_{\nu}-\square \xi_\nu,
+&=\partial_{\mu}\bar{h}^{\mu}_{\nu}-\square \xi_\nu,
 \end{align}
 thus
-\[\square \xi_{\mu}=\partialrtial_{\nu}\bar{h}^{\nu}_{\mu}.\]
+\[\square \xi_{\mu}=\partial_{\nu}\bar{h}^{\nu}_{\mu}.\]
 ) 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>

@@ Line 57: / Line 57: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">\begin{enumerate}
+     <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
-\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}
 &=\partialrtial_{\mu}\bar{h'}^{\mu}_{\nu}
@@ Line 75: / Line 74: @@
 thus
 \[\square \xi_{\mu}=\partialrtial_{\nu}\bar{h}^{\nu}_{\mu}.\]
-\item If the Lorenz gauge is already fixed, it will be preserved by any additional coordinate transformation with $\xi^\mu$ satisfying
+) If the Lorenz gauge is already fixed, it will be preserved by any additional coordinate transformation with $\xi^\mu$ satisfying
-\[\square \xi^{\mu}(x)=0.\]
+\[\square \xi^{\mu}(x)=0.\]</p>
-\end{enumerate}</p>
    </div>
 </div></div>

@@ Line 44: / Line 44: @@
 === Problem 2: Lorenz frame===
 Lorenz frame is the coordinate frame in which the Lorenz gauge conditions (\ref{LorenzGauge}) are satisfied.<br/>
 ) find the coordinate transformation  $x^\mu \to {x'}^{\mu}=x^\mu +\xi^\mu$ from a given frame to the Lorenz frame;<br/>
@@ Line 88: / Line 87: @@
 <div id="gw23"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === 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

← Older revision		Revision as of 14:13, 26 December 2012
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	[[Category:Weak field limit and gravitational waves\|4]]		[[Category:Weak field limit and gravitational waves\|4]]
		+
		+
		+
		+
		+	<div id="gw21"></div>
		+	<div style="border: 1px solid #AAA; padding:5px;">
		+	=== 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}
		+	\pa_{\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}
		+	<div class="NavFrame collapsed">
		+	<div class="NavHead">solution</div>
		+	<div style="width:100%;" class="NavContent">
		+	<p style="text-align: left;">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{WaveEq}
		+	\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.</p>
		+	</div>
		+	</div></div>
		+
		+
		+
		+
		+
		+	<div id="gw22"></div>
		+	<div style="border: 1px solid #AAA; padding:5px;">
		+	=== Problem 2: Lorenz frame===
		+	{
		+	Lorenz frame is the coordinate frame in which the Lorenz gauge conditions (\ref{LorenzGauge}) are satisfied.<br/>
		+	1) find the coordinate transformation $x^\mu \to {x'}^{\mu}=x^\mu +\xi^\mu$ from a given frame to the Lorenz frame;<br/>
		+	2) is the Lorenz frame unique? what is the remaining freedom for the choice of $\xi^\mu$?<br/>
		+
		+	'''Hint:'''
		+
		+	1) The equation for $\xi^\mu$ is $\square \xi^\mu
		+	=\pa_{\nu}{\bar{h}^{\nu}}_{\mu}$;<br/>
		+	2) no, any $\xi^\mu$ satisfying $\square \xi^{\mu}=0$ preserves the Lorenz gauge.
		+
		+	<div class="NavFrame collapsed">
		+	<div class="NavHead">solution</div>
		+	<div style="width:100%;" class="NavContent">
		+	<p style="text-align: left;">\begin{enumerate}
		+	\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}
		+	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
		+	\[\square \xi_{\mu}=\partial_{\nu}\bar{h}^{\nu}_{\mu}.\]
		+	\item If the Lorenz gauge is already fixed, it will be preserved by any additional coordinate transformation with $\xi^\mu$ satisfying
		+	\[\square \xi^{\mu}(x)=0.\]
		+	\end{enumerate}</p>
		+	</div>
		+	</div></div>
		+
		+
		+
		+
		+
		+	<div id="gw23"></div>
		+	<div style="border: 1px solid #AAA; padding:5px;">
		+	=== 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
		+	<div class="NavFrame collapsed">
		+	<!-- <div class="NavHead">solution</div>
		+	<div style="width:100%;" class="NavContent">
		+	<p style="text-align: left;"></p>
		+	</div>
		+	</div>--></div>
		+
		+
		+
		+
		+
		+	<div id="gw24"></div>
		+	<div style="border: 1px solid #AAA; padding:5px;">
		+	=== 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:'''
		+
		+	$\Phi=\Psi=0,\quad w^\alpha =0,\quad
		+	\square s_{\alpha\beta}=0$
		+	<div class="NavFrame collapsed">
		+	<div class="NavHead">solution</div>
		+	<div style="width:100%;" class="NavContent">
		+	<p style="text-align: left;">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
		+	\[G_{\alpha\beta}=(\delta_{\alpha\beta}\triangle
		+	-\pa_\alpha \pa_\beta)\Phi
		+	-\square s_{\alpha\beta}=0.\]
		+	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.</p>
		+	</div>
		+	</div></div>
		+
		+
		+	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. [http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node75.html here] or [http://en.wikibooks.org/wiki/Aeroacoustics/Wave_Equation_and_Green\%27s_function 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}