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.
13.5.1
Problem 13.5.1: 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.
solution
1) The two algebraic conditions are
\begin{align}
0&=u^{\mu}h_{\mu\nu}=h_{0\nu}=(\Phi,-\mathbf{w});\\
0&=h^{\mu}_{\mu}=\Phi-6\Psi,
\end{align}
so we get in the new frame
\[\Phi=\Psi=0,\quad \mathbf{w}=0.\]
Then the remaining Lorenz condition is reduced to
\begin{align*}
&0=\partial_0 h^0_\nu +\partial_\alpha h^\alpha_\nu ;\\
\nu=0:\quad& 0=0;\\
\nu=\beta:\quad&0= \partial_{\alpha}{s^{\alpha}}_{\beta},
\end{align*}
so it takes the form
\begin{equation}
\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:
\[k^{\alpha}s_{\alpha\beta}=0.\]
2) 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 $h_{xx}+h_{yy}=0$.
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.
13.5.2
Problem 13.5.2: Two polarizations
Show that $e^{(\times)}_{\mu\nu}$ and $e^{(+)}_{\mu\nu}$ transform into each other under rotation by $\pi/8$
solution
The two-dimensional transformation matrix $\Phi$ for rotation by $\pi/4$ and the polarization tensors are
\[\Phi=(x^{\mu}_{,\nu})=\frac{1}{\sqrt{2}}
\begin{pmatrix}1&1\\-1&1 \end{pmatrix},\quad
(e^{(+)}_{\mu\nu})
=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad
(e^{(\times)}_{\mu\nu})
=\begin{pmatrix}0&1\\1&0\end{pmatrix}.\]
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
\begin{align*}
\Phi\Phi e^{(+)}&=\frac12 \begin{pmatrix}1&1\\-1&1 \end{pmatrix}
\begin{pmatrix}1&1\\-1&1 \end{pmatrix}
\begin{pmatrix}1&0\\0&-1 \end{pmatrix}
=-\begin{pmatrix}0&+1\\1&0 \end{pmatrix}
=-e^{(\times)},\\
\Phi\Phi e^{(\times)}&=\frac12 \begin{pmatrix}1&1\\-1&1 \end{pmatrix}
\begin{pmatrix}1&1\\-1&1 \end{pmatrix}
\begin{pmatrix}0&+1\\1&0 \end{pmatrix}
=+\begin{pmatrix}1&0\\0&-1 \end{pmatrix}
=+e^{(+)}.
\end{align*}
13.5.3
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?
solution
1) The gauge transformation for $\bar{h}_{\mu\nu}$ and therefore $A_{\mu\nu}$ is
\[A_{\mu\nu}\to A_{\mu\nu}'
=A_{\mu\nu}-2\partial_{(\mu}\xi_{\nu)}
+\eta_{\mu\nu}\partial_{\lambda}\xi^\lambda,\]
so
\begin{equation}
A_{\mu\nu}\to A_{\mu\nu}'
=A_{\mu\nu}-2ik_{(\mu}B_{\nu)}
+i\eta_{\mu\nu}k_{\lambda}B^\lambda,
\end{equation}
and the TT gauge conditions take the form
\begin{align}
0&={{A'}^{\mu}}_{\mu}={A^{\mu}}_{\mu}+2ik^\mu B_\mu;\\
0&={A'}_{0\;\mu}=A_{0\mu}-ik_0 B_{\mu}
-ik_\mu B_0 +i\delta^0_\mu \; k_\lambda B^\lambda.
\end{align}
Also remember that Lorenz gauge implies $k_{\mu}A^{\mu\nu}=0$, thus $k^{\alpha}A_{\alpha\mu}=-\omega A_{0\mu}$, and therefore
\begin{equation}
l^{\mu}k_{\mu}=2\omega^2,\quad
A_{\mu\nu}l^{\mu}=2\omega A_{0\mu},\quad
k^{\lambda}B_{\lambda}
=-\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.
2) 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)
\[h_{\mu\nu}(x)=\int d^{3}k h_{\mu\nu}(\mathbf{k})
e^{ik_{\lambda}x^{\lambda}},\quad\text{with}\quad
k^\mu k_\mu =0,\quad\Leftrightarrow\quad
k_{0}^{2}\equiv \omega^2 =\mathbf{k}^2.\]
Then the TT gauge will be fixed by the coordinate fransformation
\[\xi^\mu =\int d^{3}k B^{\mu}[h_{\mu\nu}(\mathbf{k})]
e^{ik_{\lambda}x^{\lambda}}.\]
Working with individual plane-wave solutions is equivalent to working in the full Fourier space.
13.5.4
Problem 13.5.4: 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 $TT$)
\[R_{0\alpha 0\beta}
=-\tfrac{1}{2}\partial_0^2 g_{\alpha\beta}^{TT};\]
solution
1) 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
\[k_{e}R_{abcd}+k_{d}R_{abec}+k_{c}R_{abde}=0.\]
By setting $\{eabcd\}=\{0\alpha 0\beta \gamma\}$ we get ($k_0\equiv \omega \neq 0$ for a null vector)
\[R_{\alpha 0\beta \gamma}
=-\frac{k_\beta}{\omega}R_{\alpha 0\gamma 0}
+\frac{k_\gamma}{\omega}R_{\alpha 0\beta 0},\]
and by setting $\{eabcd\}=\{0\alpha \beta \gamma\delta \}$ and using the previous result, that
\[R_{\alpha\beta\gamma\delta}=
\frac{k_\alpha k_\gamma}{\omega^2}
R_{0\beta 0\gamma}
+\big(\text{three terms restored by symmetry}\big).\]
2) From vacuum Einstein equations
\[0=R^{\lambda \mu\lambda \nu}
=R_{0\mu 0\nu}-R_{\alpha\mu\alpha\nu},\]
expressing everything through $R_{0\alpha 0\beta}$, we get
\begin{align}
00:&\quad R_{0\alpha 0\alpha}=0;\\
0\beta:&\quad k_{\alpha} R_{0\alpha 0\beta}=0;\\
\beta\gamma:&\quad 0=0.
\end{align}
The $00$ equation translates to the wave being traceless, and the $0\beta$ equation to it being transverse: in the chosen frame
\[R_{0x0x}+R_{0y0y}=0,\qquad R_{0z0\alpha}=0.\]
3) In the TT gauge $g_{00}$ and $g_{0\alpha}$ are zero, so in the first order
\[R_{0\alpha 0\beta}=\frac{1}{2}\big(
\partial_0 \partial_\beta g_{0\alpha}
+\partial_\alpha \partial_0 g_{0\beta}
-\partial_0 \partial_0 g_{\alpha\beta}
-\partial_\alpha \partial_\beta g_{00}\big)
=-\tfrac12 \partial_0^2 g_{\alpha\beta}.\]
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.