Transverse traceless gauge

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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 transverse traceless (TT) gauge is then \begin{equation} \partial_\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.

Problem 1: Transverse traceless (TT) gauge

Rewrite the gauge conditions for the TT and Lorenz gauge

1) in terms of scalar, vector and tensor decomposition; 2) 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.

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 $t,x,y,z$) \begin{equation} h_{\mu\nu}=\begin{pmatrix} 0&0&0&0\\ 0&h_{+}&h_{\times}&0\\ 0&h_{\times}&-h_{+}&0\\ 0&0&0&0\end{pmatrix} 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)&= \begin{pmatrix} 0&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&0\end{pmatrix} h_{+}(t-z) +\begin{pmatrix} 0&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&0\end{pmatrix}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: \[\bar{h}_{\mu\nu} =A_{\mu\nu}e^{i\,k_{\lambda}x^\lambda}, \quad k^\mu k_\mu =0.\]

1) 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};\\ 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} 2) What is the transformation to the Lorenz gauge for arbitrary gravitational wave in vacuum?

Problem 4: Curvature of a plane wave

Consider the plane-wave solution, in which \[R_{\mu\nu\rho\sigma} =C_{\mu\nu\rho\sigma}e^{ik_{\lambda}x^{\lambda}}.\] 1) Using the Bianchi identity, show that all components of the curvature tensor can be expressed through $R_{0\alpha0\beta}$;
2) 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;
3) in the TT gauge (denoted by the superscript $TT$) \[R_{0\alpha 0\beta} =-\tfrac{1}{2}\partial_0^2 g_{\alpha\beta}^{TT};\]