Difference between revisions of "Transverse traceless gauge"

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[[Category:Weak field limit and gravitational waves|5]]
 
[[Category:Weak field limit and gravitational waves|5]]
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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
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\begin{equation}
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\pa_\mu {h^{\mu}}_{\nu}=0,\quad
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h_{0\mu}=0,\quad {h^{\mu}}_{\mu}=0.
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\end{equation}
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Simplest solutions of the vacuum wave equation are plane waves
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\[h_{\mu\nu}=h_{\mu\nu}e^{ik_{\lambda}x^\lambda}.\]
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Indeed, substitution into $\square h_{\mu\nu}=0$ yields
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\[k^{\lambda}k_{\lambda}\cdot h_{\mu\nu}=0.\]
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Thus
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* either the wave vector is null $k^{\lambda}k_{\lambda}=0$, which roughly translates as that gravitational waves propagate with the speed of light,
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* 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.
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\end{itemize}
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<div id="gw25"></div>
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<div style="border: 1px solid #AAA; padding:5px;">
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=== Problem 1: Transverse traceless (TT) gauge===
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Rewrite the gauge conditions for the TT and Lorenz gauge
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\begin{enumerate}
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\item in terms of scalar, vector and tensor decomposition;
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\item in terms of the metric perturbation for the plane wave solution
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\[h_{\mu\nu}=h_{\mu\nu}e^{ik_{\lambda}x^\lambda}
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=h_{\mu\nu}e^{i\omega t-ikz}\]
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with wave vector $k^{\mu}=(\omega,0,0,k)$ directed along the $z$-axis.
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\end{enumerate}
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<div class="NavFrame collapsed">
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  <div class="NavHead">solution</div>
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  <div style="width:100%;" class="NavContent">
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    <p style="text-align: left;">\begin{enumerate}
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\item The two algebraic conditions are
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\begin{align}
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0&=u^{\mu}h_{\mu\nu}=h_{0\nu}=(\Phi,-\mathbf{w});\\
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0&=h^{\mu}_{\mu}=\Phi-6\Psi,
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\end{align}
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so we get in the new frame
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\[\Phi=\Psi=0,\quad \mathbf{w}=0.\]
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Then the remaining Lorenz condition is reduced to
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\begin{align*}
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&0=\pa_0 h^0_\nu +\pa_\alpha h^\alpha_\nu ;\\
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\nu=0:\quad& 0=0;\\
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\nu=\beta:\quad&0= \pa_{\alpha}{s^{\alpha}}_{\beta},
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\end{align*}
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so it takes the form
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\begin{equation}
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\pa_{\alpha}{s^{\alpha}}_{\beta}=0.
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\end{equation}
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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:
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\[k^{\alpha}s_{\alpha\beta}=0.\]
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\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 $h_{xx}+h_{yy}=0$.
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\end{enumerate}</p>
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  </div>
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</div></div>
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So any plane-wave solution with $k^{\mu}=(\omega,0,0,k)$ in the $z$-direction in the TT gauge has the form
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\begin{equation}
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h_{\mu\nu}=\bordermatrix{
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~&t&x&y&z\cr
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t&0&0&0&0\cr
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x&0&h_{+}&h_{\times}&0\cr
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y&0&h_{\times}&-h_{+}&0\cr
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z&0&0&0&0}e^{ikz-i\omega t},
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\end{equation}
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or more generally any wave solution propagating in the $z$ direction can be presented as
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\begin{align}
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h_{\mu\nu}(t,z)&=
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\bordermatrix{
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~&t&x&y&z\cr
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t&0&0&0&0\cr
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x&0&1&0&0\cr
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y&0&0&-1&0\cr
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z&0&0&0&0}h_{+}(t-z)
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+\bordermatrix{
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~&t&x&y&z\cr
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t&0&0&0&0\cr
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x&0&0&1&0\cr
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y&0&1&0&0\cr
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z&0&0&0&0}h_{\times}(t-z)=\\
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&=e^{(+)}_{\mu\nu}h_{+}(t-z)
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+e^{(\times)}_{\mu\nu}h_{\times}(t-z).
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\end{align}
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Here $h_{+}$ and $h_\times$ are the amplitudes of the two independent components with linear polarization, and $e^{(\times)}_{\mu\nu},e^{(+)}_{\mu\nu}$ are the corresponding polarization tensors.
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<div id="gw26"></div>
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<div style="border: 1px solid #AAA; padding:5px;">
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=== Problem 2: Two polarizations===
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Show that $e^{(\times)}_{\mu\nu}$ and $e^{(+)}_{\mu\nu}$ transform into each other under rotation by $\pi/8$
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<div class="NavFrame collapsed">
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  <div class="NavHead">solution</div>
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  <div style="width:100%;" class="NavContent">
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    <p style="text-align: left;">The two-dimensional transformation matrix $\Phi$ for rotation by $\pi/4$ and the polarization tensors are
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\[\Phi=(x^{\mu}_{,\nu})=\frac{1}{\sqrt{2}}
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\begin{pmatrix}1&1\\-1&1 \end{pmatrix},\quad
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(e^{(+)}_{\mu\nu})
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=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad
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(e^{(\times)}_{\mu\nu})
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=\begin{pmatrix}0&1\\1&0\end{pmatrix}.\]
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In matrix notation the second rank tensor transformation law is $e'=\Phi\Phi e$, and acting twice on each of the polarization vectors we have
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\begin{align*}
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\Phi\Phi e^{(+)}&=\frac12 \begin{pmatrix}1&1\\-1&1 \end{pmatrix}
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\begin{pmatrix}1&1\\-1&1 \end{pmatrix}
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\begin{pmatrix}1&0\\0&-1 \end{pmatrix}
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=-\begin{pmatrix}0&+1\\1&0 \end{pmatrix}
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=-e^{(\times)},\\
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\Phi\Phi e^{(\times)}&=\frac12 \begin{pmatrix}1&1\\-1&1 \end{pmatrix}
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\begin{pmatrix}1&1\\-1&1 \end{pmatrix}
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\begin{pmatrix}0&+1\\1&0 \end{pmatrix}
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=+\begin{pmatrix}1&0\\0&-1 \end{pmatrix}
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=+e^{(+)}.
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\end{align*}</p>
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  </div>
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</div></div>
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<div id="gw27></div>
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<div style="border: 1px solid #AAA; padding:5px;">
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=== Problem 3: Plane wave TT gauge transformation===
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Consider the plane wave solution of the wave equation in the Lorenz gauge:
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\[\bar{h}_{\mu\nu}
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=A_{\mu\nu}e^{i\,k_{\lambda}x^\lambda},
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\quad k^\mu k_\mu =0.\]
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\begin{enumerate}
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\item Show that the TT gauge is fixed by the coordinate transformation $x\to x+\xi$ with
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\begin{align}
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\xi_{\mu}&=B_{\mu}e^{ik_\lambda x^\lambda};\\
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B_{\lambda}
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&=-\frac{A_{\mu\nu}l^\mu l^\nu}
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{8i \omega^4}k_\lambda
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-\frac{A^{\mu}_{\mu}}{4i\omega^2}l_\lambda
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+\frac{1}{2i\omega^2}A_{\lambda\mu}l^\mu;\\
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&\text{where}\quad
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k^\mu=(\omega,\mathbf{k}),\quad
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l^{\mu}=(\omega,-\mathbf{k}).
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\end{align}
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[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
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$B=C_1 k+ C_2 l +C_{3} w$.}]
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\item What is the transformation to the Lorenz gauge for arbitrary gravitational wave in vacuum?
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\end{enumerate}
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<div class="NavFrame collapsed">
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  <div class="NavHead">solution</div>
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  <div style="width:100%;" class="NavContent">
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    <p style="text-align: left;">\begin{enumerate}
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\item The gauge transformation for $\bar{h}_{\mu\nu}$ and therefore $A_{\mu\nu}$ is
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\[A_{\mu\nu}\to A_{\mu\nu}'
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=A_{\mu\nu}-2\pa_{(\mu}\xi_{\nu)}
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+\eta_{\mu\nu}\pa_{\lambda}\xi^\lambda,\]
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so
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\begin{equation}
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A_{\mu\nu}\to A_{\mu\nu}'
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=A_{\mu\nu}-2ik_{(\mu}B_{\nu)}
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+i\eta_{\mu\nu}k_{\lambda}B^\lambda,
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\end{equation}
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and the TT gauge conditions take the form
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\begin{align}
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0&={{A'}^{\mu}}_{\mu}={A^{\mu}}_{\mu}+2ik^\mu B_\mu;\\
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0&={A'}_{0\;\mu}=A_{0\mu}-ik_0 B_{\mu}
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-ik_\mu B_0  +i\delta^0_\mu \; k_\lambda B^\lambda.
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\end{align}
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Also remember that Lorenz gauge implies $k_{\mu}A^{\mu\nu}=0$, thus $k^{\alpha}A_{\alpha\mu}=-\omega A_{0\mu}$, and therefore
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\begin{equation}
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l^{\mu}k_{\mu}=2\omega^2,\quad
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A_{\mu\nu}l^{\mu}=2\omega A_{0\mu},\quad
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k^{\lambda}B_{\lambda}
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=-\frac{1}{2i}{A^{\lambda}}_{\lambda}.
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\end{equation}
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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.
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\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)
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\[h_{\mu\nu}(x)=\int d^{3}k h_{\mu\nu}(\mathbf{k})
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e^{ik_{\lambda}x^{\lambda}},\quad\text{with}\quad
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k^\mu k_\mu =0,\quad\Leftrightarrow\quad
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k_{0}^{2}\equiv \omega^2 =\mathbf{k}^2.\]
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Then the TT gauge will be fixed by the coordinate fransformation
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\[\xi^\mu =\int d^{3}k B^{\mu}[h_{\mu\nu}(\mathbf{k})]
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e^{ik_{\lambda}x^{\lambda}}.\]
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Working with individual plane-wave solutions is equivalent to working in the full Fourier space.
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\end{enumerate}</p>
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  </div>
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</div></div>
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<div id="gw28></div>
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<div style="border: 1px solid #AAA; padding:5px;">
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=== Problem 3: Curvature of a plane wave===
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Consider the plane-wave solution, in which
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\[R_{\mu\nu\rho\sigma}
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=C_{\mu\nu\rho\sigma}e^{ik_{\lambda}x^{\lambda}}.\]
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\begin{enumerate}
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\item Using the Bianchi identity, show that all components of the curvature tensor can be expressed through $R_{0\alpha0\beta}$;
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\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;
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\item in the TT gauge (denoted by the superscript $TT$)
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\[R_{0\alpha 0\beta}
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=-\tfrac{1}{2}\pa_0^2 g_{\alpha\beta}^{TT};\]
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\end{enumerate}
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[based on Padm p403].
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<div class="NavFrame collapsed">
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  <div class="NavHead">solution</div>
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  <div style="width:100%;" class="NavContent">
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    <p style="text-align: left;">\begin{enumerate}
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\item The covariant derivatives in the  Bianchi identity
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\[R_{abcd;e}+R_{abec;d}+R_{abde;c}=0\]
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in the first order can be replaced by partial derivatives. Then for a plane-wave solution
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\[k_{e}R_{abcd}+k_{d}R_{abec}+k_{c}R_{abde}=0.\]
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By setting $\{eabcd\}=\{0\alpha 0\beta \gamma\}$ we get ($k_0\equiv \omega \neq 0$ for a null vector)
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\[R_{\alpha 0\beta \gamma}
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=-\frac{k_\beta}{\omega}R_{\alpha 0\gamma 0}
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+\frac{k_\gamma}{\omega}R_{\alpha 0\beta 0},\]
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and by setting $\{eabcd\}=\{0\alpha \beta \gamma\delta \}$ and using the previous result, that
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\[R_{\alpha\beta\gamma\delta}=
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\frac{k_\alpha k_\gamma}{\omega^2}
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R_{0\beta 0\gamma}
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+\big(\text{three terms restored by symmetry}\big).\]
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\item From vacuum Einstein equations
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\[0=R^{\lambda \mu\lambda \nu}
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=R_{0\mu 0\nu}-R_{\alpha\mu\alpha\nu},\]
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expressing everything through $R_{0\alpha 0\beta}$, we get
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\begin{align}
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00:&\quad R_{0\alpha 0\alpha}=0;\\
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0\beta:&\quad k_{\alpha} R_{0\alpha 0\beta}=0;\\
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\beta\gamma:&\quad 0=0.
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\end{align}
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The $00$ equation translates to the wave being traceless, and the $0\beta$ equation to it being transverse: in the chosen frame
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\[R_{0x0x}+R_{0y0y}=0,\qquad R_{0z0\alpha}=0.\]
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\item In the TT gauge $g_{00}$ and $g_{0\alpha}$ are zero, so in the first order
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\[R_{0\alpha 0\beta}=\frac{1}{2}\big(
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\pa_0 \pa_\beta g_{0\alpha}
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+\pa_\alpha \pa_0 g_{0\beta}
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-\pa_0 \pa_0 g_{\alpha\beta}
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-\pa_\alpha \pa_\beta g_{00}\big)
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=-\tfrac12 \pa_0^2 g_{\alpha\beta}.\]
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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.
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\end{enumerate}</p>
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  </div>
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</div></div>

Revision as of 14:25, 26 December 2012



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 h_{0\mu}=0,\quad {h^{\mu}}_{\mu}=0. \end{equation}

Simplest solutions of the vacuum wave equation are plane waves \[h_{\mu\nu}=h_{\mu\nu}e^{ik_{\lambda}x^\lambda}.\] Indeed, substitution into $\square h_{\mu\nu}=0$ yields \[k^{\lambda}k_{\lambda}\cdot h_{\mu\nu}=0.\] Thus

  • 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.

\end{itemize}



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; \item in terms of the metric perturbation for the plane wave solution UNIQ-MathJax58-QINU with wave vector UNIQ-MathJax10-QINU directed along the UNIQ-MathJax11-QINU-axis. \end{enumerate}



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{ ~&t&x&y&z\cr t&0&0&0&0\cr x&0&h_{+}&h_{\times}&0\cr y&0&h_{\times}&-h_{+}&0\cr z&0&0&0&0}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{ ~&t&x&y&z\cr t&0&0&0&0\cr x&0&1&0&0\cr y&0&0&-1&0\cr z&0&0&0&0}h_{+}(t-z) +\bordermatrix{ ~&t&x&y&z\cr t&0&0&0&0\cr x&0&0&1&0\cr 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). \end{align} Here $h_{+}$ and $h_\times$ are the amplitudes of the two independent components with linear polarization, and $e^{(\times)}_{\mu\nu},e^{(+)}_{\mu\nu}$ are the corresponding polarization tensors.



Problem 2: Two polarizations

Show that $e^{(\times)}_{\mu\nu}$ and $e^{(+)}_{\mu\nu}$ transform into each other under rotation by $\pi/8$



Problem 3: Plane wave TT gauge transformation

Consider the plane wave solution of the wave equation in the Lorenz gauge: \[\bar{h}_{\mu\nu} =A_{\mu\nu}e^{i\,k_{\lambda}x^\lambda}, \quad k^\mu k_\mu =0.\] \begin{enumerate} \item Show that the TT gauge is fixed by the coordinate transformation UNIQ-MathJax31-QINU with \begin{align} \xi_{\mu}&=B_{\mu}e^{ik_\lambda x^\lambda};\\ B_{\lambda} &=-\frac{A_{\mu\nu}l^\mu l^\nu} {8i \omega^4}k_\lambda -\frac{A^{\mu}_{\mu}}{4i\omega^2}l_\lambda +\frac{1}{2i\omega^2}A_{\lambda\mu}l^\mu;\\ &\text{where}\quad k^\mu=(\omega,\mathbf{k}),\quad 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 $B=C_1 k+ C_2 l +C_{3} w$.}] \item What is the transformation to the Lorenz gauge for arbitrary gravitational wave in vacuum? \end{enumerate}



Problem 3: Curvature of a plane wave

Consider the plane-wave solution, in which \[R_{\mu\nu\rho\sigma} =C_{\mu\nu\rho\sigma}e^{ik_{\lambda}x^{\lambda}}.\] \begin{enumerate} \item Using the Bianchi identity, show that all components of the curvature tensor can be expressed through UNIQ-MathJax39-QINU; \item Show that in the coordinate frame such that UNIQ-MathJax40-QINU is directed along the UNIQ-MathJax41-QINU-axis the only possible nonzero components are UNIQ-MathJax42-QINU, UNIQ-MathJax43-QINU and UNIQ-MathJax44-QINU, obeying UNIQ-MathJax45-QINU, leaving only two independent non-zero components; \item in the TT gauge (denoted by the superscript UNIQ-MathJax46-QINU) UNIQ-MathJax67-QINU \end{enumerate} [based on Padm p403].

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