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The most straightforward way is to simply take the transverse-traceless (TT) part of $h_{ij}$. The TT part of the metric, denoted $h^{\mathrm{TT}}_{ij}$, contains precisely the two propagating degrees of freedom, which correspond to the two polarizations of gravitational waves. This enables coordinate effects to be removed and exposes the true propagating gravitational waves. It is possible to find a gauge transformation in which the only nonzero part of $h_{\mu\nu}$ is $h^{\mathrm{TT}}_{ij}$. This is known as the TT gauge.

The wavevectors can then be found by taking the Fourier transform of $h^{\mathrm{TT}}_{ij}$. If there is just one single gravitational wave with propagation direction $n^i$, it is possible to find $h^{\mathrm{TT}}_{ij}$ by defining $P_{ij} = \delta_{ij} - n_i n_j$. Then, given $h_{kl}$ in the Lorentz gauge, $$h^{\mathrm{TT}}_{ij} = \left(P_{ik}P_{jl} - \frac{1}{2}P_{ij}P_{kl}\right)h_{kl}.$$

In your example, since there is only one spatial term in your $h_{\mu\nu}$ and it is on the diagonal, it is immediately obvious that its TT part is zero.

The most straightforward way is to simply take the transverse-traceless (TT) part of $h_{ij}$. The TT part of the metric, denoted $h^{\mathrm{TT}}_{ij}$, contains precisely the two propagating degrees of freedom, which correspond to the two polarizations of gravitational waves. This enables coordinate effects to be removed and exposes the true propagating gravitational waves. It is possible to find a gauge transformation in which the only nonzero part of $h_{\mu\nu}$ is $h^{\mathrm{TT}}_{ij}$. This is known as the TT gauge.

The wavevectors can then be found by taking the Fourier transform of $h^{\mathrm{TT}}_{ij}$. If there is just one single gravitational wave with propagation direction $n^i$, it is possible to find $h^{\mathrm{TT}}_{ij}$ by defining $P_{ij} = \delta_{ij} - n_i n_j$. Then, given $h_{kl}$ in the Lorenz gauge, $$h^{\mathrm{TT}}_{ij} = \left(P_{ik}P_{jl} - \frac{1}{2}P_{ij}P_{kl}\right)h_{kl}.$$

In your example, since there is only one spatial term in your $h_{\mu\nu}$ and it is on the diagonal, it is immediately obvious that its TT part is zero.

The most straightforward way is to simply take the transverse-traceless (TT) part of $h_{ij}$. The TT part of the metric contains precisely the two propagating degrees of freedom, which correspond to the two polarizations of gravitational waves. This enables coordinate effects to be removed and exposes the true propagating gravitational waves. In your example, since there is only one spatial term in your $h_{\mu\nu}$ and it is on the diagonal, it is immediately obvious that its TT part is zero.

The most straightforward way is to simply take the transverse-traceless (TT) part of $h_{ij}$. The TT part of the metric, denoted $h^{\mathrm{TT}}_{ij}$, contains precisely the two propagating degrees of freedom, which correspond to the two polarizations of gravitational waves. This enables coordinate effects to be removed and exposes the true propagating gravitational waves. It is possible to find a gauge transformation in which the only nonzero part of $h_{\mu\nu}$ is $h^{\mathrm{TT}}_{ij}$. This is known as the TT gauge.

The wavevectors can then be found by taking the Fourier transform of $h^{\mathrm{TT}}_{ij}$. If there is just one single gravitational wave with propagation direction $n^i$, it is possible to find $h^{\mathrm{TT}}_{ij}$ by defining $P_{ij} = \delta_{ij} - n_i n_j$. Then, given $h_{kl}$ in the Lorentz gauge, $$h^{\mathrm{TT}}_{ij} = \left(P_{ik}P_{jl} - \frac{1}{2}P_{ij}P_{kl}\right)h_{kl}.$$

In your example, since there is only one spatial term in your $h_{\mu\nu}$ and it is on the diagonal, it is immediately obvious that its TT part is zero.

The most straightforward way is to simply take the projection of $h_{\mu\nu}$ onto the transverse-traceless (TT) gauge. The TT part of the metric contains precisely the two propagating degrees of freedom, which correspond to the two polarizations of gravitational waves. This enables coordinate effects to be removed and exposes the true propagating gravitational waves.

The most straightforward way is to simply take the transverse-traceless (TT) part of $h_{ij}$. The TT part of the metric contains precisely the two propagating degrees of freedom, which correspond to the two polarizations of gravitational waves. This enables coordinate effects to be removed and exposes the true propagating gravitational waves. In your example, since there is only one spatial term in your $h_{\mu\nu}$ and it is on the diagonal, it is immediately obvious that its TT part is zero.