The linearized Einstein field equations in the Lorenz gauge (with $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$ and $\bar h_{\mu \nu}=h_{\mu \nu}-\frac{\eta_{\mu \nu}}{2}h$) are given by: $$ \Box \bar h_{\mu \nu}=-\frac{16\pi G}{c^4}T_{\mu \nu}$$ This has the general solution: $$\bar h_{\mu \nu}=\frac{4\pi G}{c^4}\int \frac{T_{\mu \nu}(\vec{r}_s,t-R/c)}{R} d^3\vec r_s+\phi_{\mu \nu}$$ Where $R=|\vec r-\vec r_s|$ and $\phi_{\mu \nu}$ is any tensor that satisfies $\Box \phi_{\mu \nu}=0$. I am guessing that any arbitrary choice of $\phi_{\mu \nu}$ will not lead to a solution $\bar h_{\mu \nu}$ that is in the Lorenz gauge, and may even change the curvature. My question is therefore: for a situation where no other sources are present (or ever have been) how do we go about choosing $\phi_{\mu \nu}$ such that we are still in the correct gauge. I have seen people take $\phi_{\mu \nu}=0$ is this always allowed?


After some research I am unsure we are allowed to simply add on $\phi_{\mu \nu}$ since this would change our boundary conditions and our boundary conditions have already been predetermined by our use of Green's function. If this the case then we appear to have lost all gauge freedom in our solution. I.e. $$\bar h_{\mu \nu}=\frac{4\pi G}{c^4}\int \frac{T_{\mu \nu}(\vec{r}_s,t-R/c)}{R} d^3\vec r_s$$ only holds for a specific gauge (since I can plug in a $T_{\mu \nu}$ and get a unique $\bar h_{\mu \nu}$). Yet I have seen people use this formula to derive the quadrapole formula in the $TT$-gauge. I my edit here is correct, where is the gauge freedom in this expression that allows us to produce a solution in the $TT$-gauge?


2 Answers 2


The linearized theory is invariant up to the gauge changes $$\bar{h}_{\mu\nu} \to \bar{h}_{\mu\nu} + 2 \xi_{(\mu,\nu)} - \xi^\alpha_{\;,\alpha} \eta_{\mu\nu}.$$ By working in the Lorenz gauge $\bar{h}_{\mu\nu}^{\;\;\;,\nu} =0$ we still have a residual gauge freedom in the form of $\xi^\mathrm{R}_\nu$ which satisfy the condition $$(2 \xi^\mathrm{R}_{(\nu,\mu)}- \xi^{\mathrm{R}\alpha}_{\;\;\;,\alpha} \eta_{\mu\nu})^{,\nu}=0$$ which leads to the simple condition $$\Box \xi^\mathrm{R}_\nu = 0.$$ You can see that this also implies $$\Box (2 \xi^\mathrm{R}_{(\nu,\mu)}- \xi^{\mathrm{R}\alpha}_{\;\;\;,\alpha} \eta_{\mu\nu}) = 0$$ and your residual gauge freedom in the retarded potential formula is exactly $\phi_{\mu\nu} = 2 \xi^\mathrm{R}_{(\nu,\mu)}- \xi^{\mathrm{R}\alpha}_{\;\;\;,\alpha} \eta_{\mu\nu}$.

The reason why some people would write the retarded potential formula with $\phi_{\mu \nu} = 0$ is probably because they assume the reader understands that it is valid only up to the residual gauge freedom. At later points, this gauge is changed to shift to the TT gauge (i.e. $\phi_{\mu \nu}$ becomes nonzero).

As for the question of boundary conditions, we can indeed fix $\phi_{\mu\nu} = 0$ by assuming that the matter content is isolated and finite, and by requiring $\bar{h}_{\mu \nu} \to 0$ at infinity and $\Box \xi_\nu = 0$ everywhere. This is because $\xi_\nu$ can then be just superpositions of plane waves.

The funny thing, however, is that in the use of TT gauge to derive the quadrupole formula we also have $\bar{h}_{\mu \nu} \to 0$ at infinity but this gauge is different from the "Green-function gauge" $\phi_{\mu\nu} = 0$. We can do that because in that case we are using a gauge transformation which has $\Box \xi^\mathrm{R}_\nu \neq 0$ at the coordinate origin. I.e., $\bar{h}_{\mu \nu}^{\;\;\;,\nu} = 0$ in the TT gauge is valid only away from the origin and this is achieved by $\xi^\mathrm{R}_\mu$ being set equal to something like the electromagnetic quadrupolar radiation field.

The only case in which the TT gauge can be applied globally in the whole space-time along with the Lorentz condition is the case of a planar gravitational wave. In other cases such as the gravitational field of an isolated matter source, you will always have to leave out coordinate patches to apply that gauge choice.

  • $\begingroup$ So the boundary conditions do not force $\phi_{\mu \nu}=0$ but simply restrict it to take the form $ 2 \xi^\mathrm{R}_{(\nu,\mu)}- \xi^{\mathrm{R}\alpha}_{\;\;\;,\alpha} \eta_{\mu\nu}$? $\endgroup$ Mar 23, 2017 at 15:35
  • $\begingroup$ @Quantumspaghettification I tried to answer your question in an edit. What you can still do is to violate the gauge conditions in regions you are not interested in. $\endgroup$
    – Void
    Mar 23, 2017 at 17:32
  • $\begingroup$ @Void Hello, I have a couple of questions after reading your answer. Firstly, what is the sub-gauge where $\phi_{\mu \nu} = 0$ called? Secondly, let's say we want to go from $\phi_{\mu \nu} = 0$ to the TT gauge using a residual gauge transformation. We can use either (1) $\bar{h}_{\mu\nu} \to \bar{h}_{\mu\nu} + 2 \xi^R_{(\mu,\nu)} - \xi^{R\alpha}_{\;,\alpha} \eta_{\mu\nu}$ (for some known $\xi^R$) or (2) the projection operator $\bar{h}_{ij} = (P^k_i P^l_j - \frac{1}{2} P_{ij} P^{kl}) \bar{h}_{kl}$. Is there any difference between the two? $\endgroup$ Feb 27, 2021 at 10:18
  • 1
    $\begingroup$ @VincentThacker I would probably call it something like the "Green's-function gauge", or "wave-generation gauge", even though I have never seen a specific name for it. As for the projection, in principle you should show that you are able to find a $\xi^R_\mu$ that does the same thing as the projection operator (which you are, but you should demonstrate it directly). $\endgroup$
    – Void
    Feb 27, 2021 at 17:09

The classical reference for the multipole formalism is Thorne (1980); have a look through section V in particular (page 21). This link might also help.

The point is that the TT-portion of the metric is the radiation field part of the perturbation. The $\textit{total}$ perturbation $\boldsymbol{\bar{h}}$ can have portions of it which are not radiative, depending on the source, which you don't care about if you are doing GW analysis. You could do this analysis for a time-independent source, for example, and the linearised metric perturbation will be non-zero since $\boldsymbol{T}\neq0$, but you won't have any radiation; $d E_{\text{GW}}/ dt = 0$. This is more important when you want to think about post-Newtonian or non-linear radiation fields.

The Lorenz gauge is nothing more than a choice of fancy coordinates -- no information is lost by using it. It just simplifies the algebra. The expression you have written indeed relies on these coordinates (and your boundary conditions, which are Dirichlet for this particular Green's function you have written). Other terms might enter in a different coordinate system, but the content is the same.

Hopefully this answers (at least some of) your question.


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