How Does a Plane Wave Affect the Metric in a Weak Gravitational Field?

[Silviu]

Homework Statement

Show that a plane wave with $A_{xy}=0$ (see below) has the metric $ds^2=-dt^2+(1+h_+)dx^2+(1-h_+)dy^2+dz^2$, where $h_+=A_{xx}sin[\omega(t-z)]$

Homework Equations

$h_{\mu \nu}$ is small perturbation of the Minkowski metric i.e. in the space now $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. If we go in TT gauge, and consider a plane traveling in the z direction $k = (\omega,0,0,\omega)$, we have $h_{\mu \nu}^{TT}=A_{\mu \nu}^{TT}exp(ik_\alpha x^\alpha)$, with $A_{\mu \nu}^{TT} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & A_{xx} & A_{xy} & 0 \\ 0 & A_{xy} & -A_{xx} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
This is from Schutz page 205 Second edition

The Attempt at a Solution

So as $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ and only $h_{xx}$ and $h_{yy}$ are non-zero, we have $h_{xx}=A_{xx}e^{i\omega(z-t)}$. For $dt^2$ and $dz^2$ it is obvious that they remain the same but for $dx^2$ for example, the coefficient would be $1+A_{xx}e^{i\omega(z-t)}$. How do I get from this to $1+A_{xx}sin[\omega(t-z)]$? Thank you!

 

[mark726]

Hello,

Thank you for your post. To show that the given plane wave has the metric $ds^2=-dt^2+(1+h_+)dx^2+(1-h_+)dy^2+dz^2$, we can start by writing out the metric in terms of the Minkowski metric and the perturbation $h_{\mu \nu}$:

$$ds^2 = \eta_{\mu \nu}dx^\mu dx^\nu + h_{\mu \nu}dx^\mu dx^\nu$$

Next, we can substitute in the values for $h_{\mu \nu}$ given in the homework statement:

$$ds^2 = \eta_{\mu \nu}dx^\mu dx^\nu + A_{\mu \nu}^{TT}e^{ik_\alpha x^\alpha}dx^\mu dx^\nu$$

Since we are considering a plane wave in the z-direction, we can simplify this further by setting $k_\alpha x^\alpha = \omega(z-t)$:

$$ds^2 = \eta_{\mu \nu}dx^\mu dx^\nu + A_{\mu \nu}^{TT}e^{i\omega(z-t)}dx^\mu dx^\nu$$

Now, we can use the TT gauge condition to simplify the perturbation further. In TT gauge, the perturbation satisfies $h_{0\mu} = 0$ and $h_{ij} = 0$, where $i,j=1,2,3$. This means that the only non-zero components of the perturbation are $h_{xx}$ and $h_{yy}$. Substituting this into the equation above, we get:

$$ds^2 = \eta_{\mu \nu}dx^\mu dx^\nu + \begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & h_{xx} & h_{xy} & 0 \\
0 & h_{xy} & -h_{xx} & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}e^{i\omega(z-t)}dx^\mu dx^\nu$$

Since $h_{xy} = A_{xy}e^{i\omega(z-t)} = 0$, we can ignore it in the above equation. This leaves us with