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 \(-h\).

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,

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

which leads to

\[\frac{ d }{ d\tau } U^\alpha \vert_{\tau = 0} = - \Gamma^\alpha_{00} = 0.\]

The four acceleration is 0 for the test particle. No motion would be detected within the coordinate system.

The same is true for a particle moving in \(z\) direction. However, the conclusion doesn’t hold for other motions. p

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