Transverse traceless gauge - Revision history

Igor at 13:44, 15 January 2013

2013-01-15T13:44:13Z

Cosmo All at 03:08, 8 January 2013

2013-01-08T03:08:42Z

Cosmo All at 01:04, 8 January 2013

2013-01-08T01:04:32Z

Cosmo All at 14:42, 26 December 2012

2012-12-26T14:42:08Z

Cosmo All at 14:41, 26 December 2012

2012-12-26T14:41:36Z

Cosmo All: /* Problem 3: Curvature of a plane wave */

2012-12-26T14:31:46Z

‎Problem 3: Curvature of a plane wave

Cosmo All: /* Problem 3: Plane wave TT gauge transformation */

2012-12-26T14:30:20Z

‎Problem 3: Plane wave TT gauge transformation

Cosmo All: /* Problem 1: Transverse traceless (TT) gauge */

2012-12-26T14:26:42Z

‎Problem 1: Transverse traceless (TT) gauge

Cosmo All at 14:25, 26 December 2012

2012-12-26T14:25:38Z

Cosmo All at 14:25, 26 December 2012

2012-12-26T14:25:13Z

Show changes

@@ Line 56: / Line 56: @@
 \[k^{\alpha}s_{\alpha\beta}=0.\]
 ) Transversality implies $h_{\mu 0}=0$, Lorenz gauge condition that $h_{\mu z}=0$, so the only non-zero components are $h_{xx},h_{yy},h_{xy},h_{yx}$. Of those only two are independent due to symmetry $h_{xy}=h_{yx}$ and tracelessness $h_{xx}+h_{yy}=0$.
-\end{enumerate}</p>
+</p>
    </div>
 </div></div>
@@ Line 62: / Line 62: @@
+So any plane-wave solution with $k^{\mu}=(\omega,0,0,k)$ in the $z$-direction in the TT gauge has the form (in coordinates $t,x,y,z$)
 So any plane-wave solution with $k^{\mu}=(\omega,0,0,k)$ in the $z$-direction in the TT gauge has the form
 \begin{equation}
-h_{\mu\nu}=\bordermatrix{
+h_{\mu\nu}=\begin{pmatrix}
- 	~&t&x&y&z\cr
+&0&0&0\\
-	t&0&0&0&0\cr
+&h_{+}&h_{\times}&0\\
-	x&0&h_{+}&h_{\times}&0\cr
+&h_{\times}&-h_{+}&0\\
-	y&0&h_{\times}&-h_{+}&0\cr
+&0&0&0\end{pmatrix}
-	z&0&0&0&0}e^{ikz-i\omega t},
+e^{ikz-i\omega t},
 \end{equation}
 or more generally any wave solution propagating in the $z$ direction can be presented as
 \begin{align}
 h_{\mu\nu}(t,z)&=
-	\bordermatrix{
+	\begin{pmatrix}
- 	~&t&x&y&z\cr
+&0&0&0\\
-	t&0&0&0&0\cr
+&1&0&0\\
-	x&0&1&0&0\cr
+&0&-1&0\\
-	y&0&0&-1&0\cr
+&0&0&0\end{pmatrix} h_{+}(t-z)
-	z&0&0&0&0}h_{+}(t-z)
+	+\begin{pmatrix}
-	+\bordermatrix{
+&0&0&0\\
- 	~&t&x&y&z\cr
+&0&1&0\\
-	t&0&0&0&0\cr
+&1&0&0\\
-	x&0&0&1&0\cr
+&0&0&0\end{pmatrix}h_{\times}(t-z)=\\
-	y&0&1&0&0\cr
-	z&0&0&0&0}h_{\times}(t-z)=\\
 	&=e^{(+)}_{\mu\nu}h_{+}(t-z)
 		+e^{(\times)}_{\mu\nu}h_{\times}(t-z).
@@ Line 179: / Line 176: @@
 	=-\frac{1}{2i}{A^{\lambda}}_{\lambda}.
 \end{equation}
-Plugging this into the gauge conditions, one can see that the first one (tracelessness) is obeyed immediately, and the second one (transversality) after a little bit of more algebra.
+Plugging this into the gauge conditions, one can see that the first one (tracelessness) is obeyed immediately, and the second one (transversality) after a little bit of more algebra.</p>
 ) The general first-order vacuum solution of Einstein's equations in the Lorenz gauge can be presented through its spatial Fourier transform (temporal part is integrated over due to fixed dispersion relation)
 \[h_{\mu\nu}(x)=\int d^{3}k h_{\mu\nu}(\mathbf{k})
 	e^{ik_{\lambda}x^{\lambda}},\quad\text{with}\quad
@@ Line 189: / Line 187: @@
 	e^{ik_{\lambda}x^{\lambda}}.\]
 Working with individual plane-wave solutions is equivalent to working in the full Fourier space.
-\end{enumerate}</p>
+</p>
    </div>
 </div></div>

← Older revision	Revision as of 03:08, 8 January 2013
Line 20:	Line 20:


−	+	__TOC__

@@ Line 4: / Line 4: @@
-From now on we consider only vacuum solutions. Suppose we use the Lorenz gauge. As shown above, we still have the freedom of coordinate transformations with $\square \xi^\mu =0$, which preserve the gauge.  So, let us choose some arbitrary (timelike) field $u^\mu$ and, in addition to the Lorenz gauge conditions, demand that the perturbation is also transverse $u^{\mu}\bar{h}_{\mu\nu}=0$ with regard to it plus that it is traceless $\bar{h}^{\mu}_{\mu}=0$. Then $h_{\mu\nu}=\bar{h}_{\mu\nu}$ and we can omit the bars. In the frame of observers with 4-velocity $u^\mu$ the full set of conditions that fix the \emph{transverse traceless (TT) gauge} is then
+From now on we consider only vacuum solutions. Suppose we use the Lorenz gauge. As shown above, we still have the freedom of coordinate transformations with $\square \xi^\mu =0$, which preserve the gauge.  So, let us choose some arbitrary (timelike) field $u^\mu$ and, in addition to the Lorenz gauge conditions, demand that the perturbation is also transverse $u^{\mu}\bar{h}_{\mu\nu}=0$ with regard to it plus that it is traceless $\bar{h}^{\mu}_{\mu}=0$. Then $h_{\mu\nu}=\bar{h}_{\mu\nu}$ and we can omit the bars. In the frame of observers with 4-velocity $u^\mu$ the full set of conditions that fix ''the transverse traceless (TT) gauge'' is then
 \begin{equation}
 \partial_\mu {h^{\mu}}_{\nu}=0,\quad

@@ Line 199: / Line 199: @@
 <div style="border: 1px solid #AAA; padding:5px;">
-=== Problem 3: Curvature of a plane wave===
+=== Problem 4: Curvature of a plane wave===
 Consider the plane-wave solution, in which
 \[R_{\mu\nu\rho\sigma}

← Older revision		Revision as of 14:41, 26 December 2012
Line 18:		Line 18:
	* either the wave vector is null $k^{\lambda}k_{\lambda}=0$, which roughly translates as that gravitational waves propagate with the speed of light,		* either the wave vector is null $k^{\lambda}k_{\lambda}=0$, which roughly translates as that gravitational waves propagate with the speed of light,
	* or $h_{\mu\nu}=0$, which means that in any other (non-TT) coordinate frame, in which metric perturbation is non-zero, it is due to the oscillating coordinate system, while the true gravitational field vanishes.		* or $h_{\mu\nu}=0$, which means that in any other (non-TT) coordinate frame, in which metric perturbation is non-zero, it is due to the oscillating coordinate system, while the true gravitational field vanishes.
−	~~\end{itemize}~~

@@ Line 204: / Line 204: @@
 \[R_{\mu\nu\rho\sigma}
 	=C_{\mu\nu\rho\sigma}e^{ik_{\lambda}x^{\lambda}}.\]
-\begin{enumerate}
+) Using the Bianchi identity, show that all components of the curvature tensor can be expressed through $R_{0\alpha0\beta}$;<br/>
-\item Using the Bianchi identity, show that all components of the curvature tensor can be expressed through $R_{0\alpha0\beta}$;
+) Show that in the coordinate frame such that $k^\mu =(k,0,0,k)$ is directed along the $z$-axis the only possible nonzero components are $R_{0x0x}$, $R_{0y0y}$ and $R_{0x0y}$, obeying $R_{0x0x}=-R_{0y0y}$, leaving only two independent non-zero components;<br/>
-\item Show that in the coordinate frame such that $k^\mu =(k,0,0,k)$ is directed along the $z$-axis the only possible nonzero components are $R_{0x0x}$, $R_{0y0y}$ and $R_{0x0y}$, obeying $R_{0x0x}=-R_{0y0y}$, leaving only two independent non-zero components;
+) in the TT gauge (denoted by the superscript $TT$)
-\item in the TT gauge (denoted by the superscript $TT$)
 \[R_{0\alpha 0\beta}
 	=-\tfrac{1}{2}\partial_0^2 g_{\alpha\beta}^{TT};\]
-\end{enumerate}
-[based on Padm p403].
+<!--[based on Padm p403].-->
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">\begin{enumerate}
+     <p style="text-align: left;">1) The covariant derivatives in the  Bianchi identity
-\item The covariant derivatives in the  Bianchi identity
 \[R_{abcd;e}+R_{abec;d}+R_{abde;c}=0\]
 in the first order can be replaced by partial derivatives. Then for a plane-wave solution
@@ Line 229: / Line 227: @@
 	R_{0\beta 0\gamma}
 	+\big(\text{three terms restored by symmetry}\big).\]
-\item From vacuum Einstein equations
+) From vacuum Einstein equations
 \[0=R^{\lambda \mu\lambda \nu}
 	=R_{0\mu 0\nu}-R_{\alpha\mu\alpha\nu},\]
@@ Line 240: / Line 238: @@
 The $00$ equation translates to the wave being traceless, and the $0\beta$ equation to it being transverse: in the chosen frame
   \[R_{0x0x}+R_{0y0y}=0,\qquad R_{0z0\alpha}=0.\]
-\item In the TT gauge $g_{00}$ and $g_{0\alpha}$ are zero, so in the first order
+) In the TT gauge $g_{00}$ and $g_{0\alpha}$ are zero, so in the first order
 \[R_{0\alpha 0\beta}=\frac{1}{2}\big(
 	\partial_0 \partial_\beta g_{0\alpha}
@@ Line 247: / Line 245: @@
 	-\partial_\alpha \partial_\beta g_{00}\big)
 	=-\tfrac12 \partial_0^2 g_{\alpha\beta}.\]
-As the curvature tensor is gauge invariant, the last relation allows one to calculate the metric components in the TT gauge $g_{\alpha\beta}^{TT}$ without the prior gauge fixing.
+As the curvature tensor is gauge invariant, the last relation allows one to calculate the metric components in the TT gauge $g_{\alpha\beta}^{TT}$ without the prior gauge fixing.</p>
-\end{enumerate}</p>
    </div>
 </div></div>

@@ Line 139: / Line 139: @@
 	=A_{\mu\nu}e^{i\,k_{\lambda}x^\lambda},
 	\quad k^\mu k_\mu =0.\]
-\begin{enumerate}
-\item Show that the TT gauge is fixed by the coordinate transformation $x\to x+\xi$ with
+) Show that the TT gauge is fixed by the coordinate transformation $x\to x+\xi$ with
 \begin{align}
 \xi_{\mu}&=B_{\mu}e^{ik_\lambda x^\lambda};\\
@@ Line 152: / Line 152: @@
 	l^{\mu}=(\omega,-\mathbf{k}).
 \end{align}
-[Padm. p \textsection 9.3, p.403 (wrong signs!), also see  MTW Ex.35.1 for similar formulation; \emph{the solution can probably be derived if we introduce the null frame $k$ and $l$ (the other two spacial basis vectors are not needed) and look for solution in the form
+<!--[Padm. p \textsection 9.3, p.403 (wrong signs!), also see  MTW Ex.35.1 for similar formulation; ''the solution can probably be derived if we introduce the null frame $k$ and $l$ (the other two spacial basis vectors are not needed) and look for solution in the form $B=C_1 k+ C_2 l +C_{3} w$.'']-->
-$B=C_1 k+ C_2 l +C_{3} w$.}]
+) What is the transformation to the Lorenz gauge for arbitrary gravitational wave in vacuum?
-\item What is the transformation to the Lorenz gauge for arbitrary gravitational wave in vacuum?
-\end{enumerate}
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">\begin{enumerate}
+     <p style="text-align: left;">1) The gauge transformation for $\bar{h}_{\mu\nu}$ and therefore $A_{\mu\nu}$ is
-\item The gauge transformation for $\bar{h}_{\mu\nu}$ and therefore $A_{\mu\nu}$ is
 \[A_{\mu\nu}\to A_{\mu\nu}'
 	=A_{\mu\nu}-2\partial_{(\mu}\xi_{\nu)}
@@ Line 184: / Line 181: @@
 \end{equation}
 Plugging this into the gauge conditions, one can see that the first one (tracelessness) is obeyed immediately, and the second one (transversality) after a little bit of more algebra.
-\item The general first-order vacuum solution of Einstein's equations in the Lorenz gauge can be presented through its spatial Fourier transform (temporal part is integrated over due to fixed dispersion relation)
+) The general first-order vacuum solution of Einstein's equations in the Lorenz gauge can be presented through its spatial Fourier transform (temporal part is integrated over due to fixed dispersion relation)
 \[h_{\mu\nu}(x)=\int d^{3}k h_{\mu\nu}(\mathbf{k})
 	e^{ik_{\lambda}x^{\lambda}},\quad\text{with}\quad
@@ Line 202: / Line 199: @@
 <div id="gw28></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 3: Curvature of a plane wave===
 Consider the plane-wave solution, in which

@@ Line 28: / Line 28: @@
 === Problem 1: Transverse traceless (TT) gauge===
 Rewrite the gauge conditions for the TT and Lorenz gauge
-\begin{enumerate}
-\item in terms of scalar, vector and tensor decomposition;
+) in terms of scalar, vector and tensor decomposition;
-\item in terms of the metric perturbation for the plane wave solution
+) in terms of the metric perturbation for the plane wave solution
 \[h_{\mu\nu}=h_{\mu\nu}e^{ik_{\lambda}x^\lambda}
 	=h_{\mu\nu}e^{i\omega t-ikz}\]
 with wave vector $k^{\mu}=(\omega,0,0,k)$ directed along the $z$-axis.
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">\begin{enumerate}
+     <p style="text-align: left;">1) The two algebraic conditions are
-\item The two algebraic conditions are
 \begin{align}
 &=u^{\mu}h_{\mu\nu}=h_{0\nu}=(\Phi,-\mathbf{w});\\
@@ Line 58: / Line 56: @@
 Thus the only remaining non-trivial component of the perturbation is the tensor one (the traceless part) $s_{\alpha\beta}$, and it is transverse, in the sense that for a plane wave solution $s_{\alpha\beta}\sim e^{i\,k_{\mu}x^{\mu}}$ the metric perturbation is transverse with respect to the wave vector:
 \[k^{\alpha}s_{\alpha\beta}=0.\]
-\item Transversality implies $h_{\mu 0}=0$, Lorenz gauge condition that $h_{\mu z}=0$, so the only non-zero components are $h_{xx},h_{yy},h_{xy},h_{yx}$. Of those only two are independent due to symmetry $h_{xy}=h_{yx}$ and tracelessness $h_{xx}+h_{yy}=0$.
+) Transversality implies $h_{\mu 0}=0$, Lorenz gauge condition that $h_{\mu z}=0$, so the only non-zero components are $h_{xx},h_{yy},h_{xy},h_{yx}$. Of those only two are independent due to symmetry $h_{xy}=h_{yx}$ and tracelessness $h_{xx}+h_{yy}=0$.
 \end{enumerate}</p>
    </div>
@@ Line 101: / Line 99: @@
 <div id="gw26"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 2: Two polarizations===
 Show that $e^{(\times)}_{\mu\nu}$ and $e^{(+)}_{\mu\nu}$ transform into each other under rotation by $\pi/8$

@@ Line 6: / Line 6: @@
 From now on we consider only vacuum solutions. Suppose we use the Lorenz gauge. As shown above, we still have the freedom of coordinate transformations with $\square \xi^\mu =0$, which preserve the gauge.  So, let us choose some arbitrary (timelike) field $u^\mu$ and, in addition to the Lorenz gauge conditions, demand that the perturbation is also transverse $u^{\mu}\bar{h}_{\mu\nu}=0$ with regard to it plus that it is traceless $\bar{h}^{\mu}_{\mu}=0$. Then $h_{\mu\nu}=\bar{h}_{\mu\nu}$ and we can omit the bars. In the frame of observers with 4-velocity $u^\mu$ the full set of conditions that fix the \emph{transverse traceless (TT) gauge} is then
 \begin{equation}
-\pa_\mu {h^{\mu}}_{\nu}=0,\quad
+\partial_\mu {h^{\mu}}_{\nu}=0,\quad
 h_{0\mu}=0,\quad {h^{\mu}}_{\mu}=0.
 \end{equation}
@@ Line 48: / Line 48: @@
   Then the remaining Lorenz condition is reduced to
 \begin{align*}
-&0=\pa_0 h^0_\nu +\pa_\alpha h^\alpha_\nu ;\\
+&0=\partial_0 h^0_\nu +\partial_\alpha h^\alpha_\nu ;\\
 \nu=0:\quad& 0=0;\\
-\nu=\beta:\quad&0= \pa_{\alpha}{s^{\alpha}}_{\beta},
+\nu=\beta:\quad&0= \partial_{\alpha}{s^{\alpha}}_{\beta},
 \end{align*}
 so it takes the form
 \begin{equation}
-\pa_{\alpha}{s^{\alpha}}_{\beta}=0.
+\partial_{\alpha}{s^{\alpha}}_{\beta}=0.
 \end{equation}
 Thus the only remaining non-trivial component of the perturbation is the tensor one (the traceless part) $s_{\alpha\beta}$, and it is transverse, in the sense that for a plane wave solution $s_{\alpha\beta}\sim e^{i\,k_{\mu}x^{\mu}}$ the metric perturbation is transverse with respect to the wave vector:
@@ Line 163: / Line 163: @@
 \item The gauge transformation for $\bar{h}_{\mu\nu}$ and therefore $A_{\mu\nu}$ is
 \[A_{\mu\nu}\to A_{\mu\nu}'
-	=A_{\mu\nu}-2\pa_{(\mu}\xi_{\nu)}
+	=A_{\mu\nu}-2\partial_{(\mu}\xi_{\nu)}
-		+\eta_{\mu\nu}\pa_{\lambda}\xi^\lambda,\]
+		+\eta_{\mu\nu}\partial_{\lambda}\xi^\lambda,\]
 so
 \begin{equation}
@@ Line 212: / Line 212: @@
 \item in the TT gauge (denoted by the superscript $TT$)
 \[R_{0\alpha 0\beta}
-	=-\tfrac{1}{2}\pa_0^2 g_{\alpha\beta}^{TT};\]
+	=-\tfrac{1}{2}\partial_0^2 g_{\alpha\beta}^{TT};\]
 \end{enumerate}
 [based on Padm p403].
@@ Line 245: / Line 245: @@
 \item In the TT gauge $g_{00}$ and $g_{0\alpha}$ are zero, so in the first order
 \[R_{0\alpha 0\beta}=\frac{1}{2}\big(
-	\pa_0 \pa_\beta g_{0\alpha}
+	\partial_0 \partial_\beta g_{0\alpha}
-	+\pa_\alpha \pa_0 g_{0\beta}
+	+\partial_\alpha \partial_0 g_{0\beta}
-	-\pa_0 \pa_0 g_{\alpha\beta}
+	-\partial_0 \partial_0 g_{\alpha\beta}
-	-\pa_\alpha \pa_\beta g_{00}\big)
+	-\partial_\alpha \partial_\beta g_{00}\big)
-	=-\tfrac12 \pa_0^2 g_{\alpha\beta}.\]
+	=-\tfrac12 \partial_0^2 g_{\alpha\beta}.\]
 As the curvature tensor is gauge invariant, the last relation allows one to calculate the metric components in the TT gauge $g_{\alpha\beta}^{TT}$ without the prior gauge fixing.
 \end{enumerate}</p>
    </div>
 </div></div>