# Transverse-traceless gauge: Why the traceless condition?

I'm right now following a course on GR and I arrived to the gravitational waves part. Letting the metric be that of the plane Minkowski space with a small perturbation:

$$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu} \hspace{1cm} (h_{\mu\nu}\ll1),$$

We can write the Einstein's equations as:

$$\Box \bar{h}_{\mu\nu}=-\frac{16 \pi G}{c^4}T_{\mu\nu}$$

Provided that $$\bar h_{\mu\nu}$$ satisfies the harmonic gauge condition: $$\partial^\mu\bar h _{\mu \nu}=0$$.

Here, $$\bar h_{\mu\nu}=h_{\mu\nu}-\frac{1}{2}h\eta_{\mu\nu}$$, where $$h$$ stands for the trace of $$h_{\mu\nu}$$, this is $$h=h^\mu{}_\mu$$. (Related to this question, it would be very useful if someone could explain how that $$\frac{1}{2}h\eta_{\mu\nu}$$ term differs from $$\frac{1}{2}h^\mu{}_\mu\eta_{\mu\nu}=\frac{1}{2}h_{\nu\mu}=\frac{1}{2}h_{\mu\nu}$$, because if this was the case then we would simply have $$\bar h_{\mu\nu}=\frac{1}{2}h_{\mu\nu}$$).

Now, it is clear that in vacuum, Einstein's equations would be $$\Box \bar h_{\mu\nu}=0$$, so we can solve it as $$\bar h_{\mu\nu}=a_{\mu \nu}e^{i k_\sigma x^\sigma}$$, where $$a_{\mu\nu}$$ is a symmetric and constant tensor, the polarization tensor. As it is symmetric, it has 10 independent components, but since $$\partial^\mu\bar h _{\mu \nu}=0$$ (gauge condition), we find that $$k_\mu a^{\mu \nu}=0$$, imposing four extra constraints over the components of the polarization vector.

My doubt arises here, since my professor now states that there are another four constraints that come from the residual gauge when applying an infinitesimal coordinate transformation:

$$x^\mu \rightarrow x'^\mu= x^\mu - \xi^\mu(x)$$

$$h^\mu \rightarrow h'_{\mu\nu}= h_{\mu\nu} + \partial_\mu\xi_\nu(x) + \partial_\nu\xi_\mu(x)$$

Imposing the harmonic gauge condition again: $$\Box \xi^\mu=0$$, we get that $$\xi_\mu=B_{\mu}e^{i \textbf{k} \textbf{x}}$$. Hence, we can choose these $$B_\mu$$ vectors so they cancel out another four components of the polarization tensor, leaving only two free components, each standing for one polarization. This I understand, but my professor expressed these new residual gauge conditions as:

$$a^\mu{}_\mu=0$$ $$a_{0\nu}=0$$

Where the second one I guess stands for the four cancelled components of the polarization tensor, chosen to be those corresponding to the time coordinate. I don't get though where the first "traceless" condition comes from following this argument...

Any help will be much appreciated, thanks!

• In $h^\mu_\mu\eta_{\mu\nu} = h_{\nu\mu}$ you are confusing dummy indices and "true" indices. The left-hand side should be written as $h^\lambda_\lambda\eta_{\mu\nu}$. There is no contraction between the two objects. – NDewolf Mar 30 at 12:24

• You cannot reuse a free index as a dummy index, so $$\frac12h\eta_{\mu\nu}\equiv\frac12h^\alpha{}_\alpha\eta_{\mu\nu}\ne\frac12h_{\nu\mu}$$
• Verify for yourself that the residual gauge transformation acts on the polarisation tensor as $$a_{\mu\nu}\to a_{\mu\nu}+i(k_\mu B_\nu+k_\nu B_\mu-k^\sigma B_\sigma\eta_{\mu\nu})$$. Contracting,
$$\eta^{\mu\nu}a_{\mu\nu}\to \eta^{\mu\nu}a_{\mu\nu}+i(\eta^{\mu\nu}k_\mu B_\nu+\eta^{\mu\nu}k_\nu B_\mu-\eta^{\mu\nu}k^\sigma B_\sigma\eta_{\mu\nu}) \\a^\mu{}_\mu\to a^\mu{}_\mu+i(k^\mu B_\mu+k^\nu B_\nu-4k^\sigma B_\sigma) \\a^\mu{}_\mu\to a^\mu{}_\mu-2i(k^\mu B_\mu)$$
We can set $$k^\mu B_\mu$$ to $$-\frac i2a^\mu{}_\mu$$, making the polarisation tensor traceless, and this imposition is consistent with setting $$a_{0\nu}=0$$.
Something seems suspicious: didn't we use four degrees of freedom in $$B_\mu$$ to impose 5 conditions? No, because one of these is equivalent to setting the wave vector orthogonal to the polarisation tensor, and so is redundant.