# Gravitational Waves¶

In the weak field regime of sourceless Einstein’s equation ($$T^{\mu\nu}=0$$), the equation for metric with perturbations is reduced to a wave equation,

$\left( - \frac{\partial^2}{ \partial t^2 } + \nabla^2 \right) \bar h^{\alpha\beta} = 0,$

where $$\bar h^{\alpha\beta}$$ is the trace-reversed perturbation of the metric on top of Minkowski metric background, i.e.,

$\bar h^{\alpha\beta} = h^{\alpha\beta} - \frac{1}{2} \eta^{\alpha\beta} h,$

where $$h^{\alpha\beta} = g^{\alpha\beta} - \eta^{\alpha\beta}$$ and $$h$$ is the trace of metric perturbation $$h^{\alpha\beta}$$.

Trace Reverse

The tensor $$\bar h^{\alpha\beta}$$ is called trace reverse of $$h^{\alpha\beta}$$ for its trace is $$-h$$.

## Gauge¶

To solve the equation we introduce a solution of the form $$\hat h^{\alpha\beta} = A^{\alpha\beta}e^{i k_\mu x^\mu }$$, which simiplifies the equation

$\eta^{\mu\nu} k_{\mu}k_\nu \bar h^{\alpha\beta} = 0.$

To solve the amplitude $$A^{\alpha}$$ we need constraints on it. We can derive that gravitational waves are always null, that is $$k^\mu k_\mu=0$$.

Some of the conditions requires a gauge transformation. In any case, we have the second gauge condition as

$A_{\alpha\beta} U^{\beta} = 0,$

which specifies that $$A_{\alpha\beta}$$ is orthogonal to the vector we chose $$U^{\beta}$$. A practical choice of $$U^\beta$$ is a four velocity. This removes another four degrees of freedom. For illustration purpose, we choose $$U^{\beta} \to ( 1, 0, 0, 0 )$$ since it’s a null vector. The degrees of freedom removed can be visualized as the first rwo and column.

The second one we can think of is a transverse condition,

$A_{\alpha\beta} k^\beta = 0,$

which removes another three degrees of freedom. This specifies that the wave is transverse, i.e., $$A_{\alpha\beta}$$ can not have elements that is in the direction of four wavevector. We specify a wavevector $$k^\beta \to (\omega, 0, 0, \omega )$$, which leads to the removal of the remaining elements of the fourth row and column.

The matrix we have now becomes

$\begin{split}A_{\alpha\beta} \to \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & A_{xx} & A_{xy} & 0 \\ 0 & A_{yx} & A_{yy} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.\end{split}$

The last gauge condition is traceless condition $$A^\alpha_\alpha = 0$$ which also requires the gauge transformation. This condition fixes the phase relations between different spatial directions, that is $$A_{xx} = e^{i\pi} A_{yy} = - A_{yy}$$. This conditions insists that the two directions of distance oscillations should be quadrupole-like, i.e., contracts in one direction (say x) while extend in the other direction (say y).

Slicing

The first two conditions are basically specifying slicings of spacetime.

Physical Significance of Transverse-traceless Gauge

Transverse-traceless gauge is the very gauge that determines a coordinate system that a test particle is stationary in terms of coordinates.

To show this we assume that we have a test particle being stationary initially, i.e., $$U^\alpha\vert_{\tau=0} \to (1,0,0,0)^{\mathrm T}$$.

The particle should travel on geodesics,

$\frac{d}{d\tau} U^\alpha + \Gamma^\alpha_{ \mu\nu } U^\mu U^\nu =0,$

$\frac{ d }{ d\tau } U^\alpha \vert_{\tau = 0} = - \Gamma^\alpha_{00} = 0.$
The same is true for a particle moving in $$z$$ direction. However, the conclusion doesn’t hold for other motions. p