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I have a question about the linearized Einstein Field Equations, and in particular about the Newtonian limit. It goes as follows. If one uses the trace-reversed form of the EFE for the 00-component and uses the linearized Ricci tensor, one gets:

$$-\frac{1}{2}\square h_{00} + \partial_{0}V_{0} = \frac{8\pi G}{c^{4}}T_{00}$$

with $V_{0}:= \partial_{\rho}h^{\rho}_{0} - \frac{1}{2}\partial_{0}h$. From here, if one assumes a static spacetime ($\partial_{0}$ annihilates everything) and models the matter field as dust comoving with the reference frame ($T^{\mu\nu} = \delta^{\mu}_{0}\delta^{\nu}_{0}\rho c^{2}$) then approximating to lowest nontrivial order in the metric perturbation one gets the classical Poisson equation for the newtonian potential:

$$-\frac{1}{2}\nabla^{2}h_{00} = \frac{4\pi G}{c^{2}}\rho.$$

Simple enough... but what I don't understand is the normal form of the linearized EFE in this setting (not the trace-reversed form). In that case, it seems that one really needs to choose a gauge in order to obtain the Newtonian limit, since the linearized Einstein tensor contains a lot of undesired terms that complicate everything, yet the approximation of the right hand side remains almost unchanged...

So: trace-reversed form requires no gauge fixing, but normal form seems to require it. Why is this the case? Is the Newtonian limit dependent on some specific gauge condition, and if so, what is the relevance of this fact?

Qmechanic
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Cristián Paris
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  • Could you write down the normal (not trace reversed) form the linearized EFE that you are using? (It is likely to have a mistake.) – TimRias Feb 20 '18 at 10:31
  • With $V_{\mu}:= \partial_{\rho}h^{\rho}{\mu} - \frac{1}{2}\partial{\mu}h$, I have:

    R_{\mu\nu} = -\frac{1}{2}\square\left( h_{\mu\nu} - \frac{1}{2}\eta_{\mu\nu}h\right)

     – Cristián Paris Feb 21 '18 at 14:01
  • With $V_{\mu}:= \partial_{\rho}h^{\rho}{\mu} - \frac{1}{2}\partial{\mu}h$, I have this Einstein tensor:

    $G_{\mu\nu} = -\frac{1}{2}\square h_{\mu\nu} + \frac{1}{2}\left(\partial_{\mu}V_{\nu} + \partial_{\nu}V_{\mu}\right) + \frac{1}{4}\eta_{\mu\nu}\square h - \frac{1}{4}\eta_{\mu\nu}\eta^{\alpha\beta}\left(\partial_{\alpha}V_{\beta} + \partial_{\beta}V_{\alpha}\right)$

     – Cristián Paris Feb 21 '18 at 14:07
  • So linearized EFE:

    $$-\frac{1}{2}\square h_{\mu\nu} + \frac{1}{2}\left(\partial_{\mu}V_{\nu} + \partial_{\nu}V_{\mu}\right) + \frac{1}{4}\eta_{\mu\nu}\square h - \frac{1}{4}\eta_{\mu\nu}\eta^{\alpha\beta}\left(\partial_{\alpha}V_{\beta} + \partial_{\beta}V_{\alpha}\right) = \frac{8\pi G}{c^{4}}T_{\mu\nu}$$

     – Cristián Paris Feb 21 '18 at 14:11
  • So trace reversed form yields: $$ -\frac{1}{2}\nabla h_{00} = \frac{4\pi G\rho}{c^{2}}$$ while non-trace reversed form yields: $$ -\frac{1}{2}\nabla h_{00} + \frac{1}{4} \square h + \frac{1}{4}\eta^{\alpha\beta}\left(\partial_{\alpha}V_{\beta} + \partial_{\beta}V_{\alpha}\right) = \frac{8\pi G\rho}{c^{2}}$$ And I don't know what to do with the last term on the R.H.S. of the second equation... – Cristián Paris Feb 21 '18 at 17:04
  • (assuming static perturbation $h$, so the d'alambertian in the last equation was meant to be a laplacian... this 5 minute edit limit time in the comments is really annoying, sorry for the stream of comments). – Cristián Paris Feb 21 '18 at 17:12
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    (For clarifications of the question it is usually better to edit the original question.) – TimRias Feb 21 '18 at 21:55
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    Thanks for your clarification, I now understand what you are asking. (I originally thought you were asking about the linearize Einstein equation featuring the trace reversed metric perturbation.) – TimRias Feb 22 '18 at 09:01
  • And thanks for your patience with all my follow ups to your comment. Now I just realized that I wrote the gradient instead of the laplacian in some of the comments, but sadly I cannot edit them... – Cristián Paris Feb 22 '18 at 15:21

2 Answers2

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The core of the answer is that the linearized Einstein equation is a coupled set of linear PDEs. A priori, there is no reason to expect that these equations separate in to individual uncoupled equations, even when taking the Newtonian limit. The only that we can expect (because we no the Newtonian limit for the $h_{00}$ component) is that some linear combination of these PDEs, gives an uncoupled equation for $h_{00}$.

Taking this in mind, it is somewhat of a coincidence that when you linearize the field equations with the Ricci tensor and trace reversed Energy-Momentum and take the Newtonian limit, the $00$ component immediately produces the uncoupled equation for $h_{00}$. The situation that you find for the linearization of the field equations with the Einstein tensor and Energy-Momentum, is more in line with the generic expectation.

The two results are not incompatible, they are just different linear combinations of the 10 components of the equation. In particular, starting from both starting points you should be able to find a linear combination of the resulting 10 linear PDEs in the Newtonian limit, that gives you the expected $$-\frac{1}{2}\nabla^{2}h_{00} = \frac{4\pi G}{c^{2}}\rho.$$

TimRias
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  • Thanks a lot, your answer is very illuminating indeed. But in that case, what would be the physical significance of said linear combination? Because according to what you say, fixing a gauge isn't just a matter of convenience, it appears to me now that you need to do it to get the desired Poisson equation for the newtonian potential... – Cristián Paris Feb 22 '18 at 15:18
  • Or maybe not with a gauge, but what I meant was: the procedure that gives you the correct newtonian limit ought to have some physical significance in my opinion... or does it? – Cristián Paris Feb 22 '18 at 15:36
  • The procedure is: Try to isolate an equation for $h_{00}$ from the coupled set of 10 linear PDEs for 10 unknown functions. That you can do this is not a priori given, but when you can the result is unique. In the case of GR you can and the result is the Poisson equation. – TimRias Feb 22 '18 at 20:04
  • Yes, it works! I just posted it for reference. Many thanks. – Cristián Paris Feb 23 '18 at 00:15
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I found the answer. As mentioned by mmeent, a linear combination of the PDEs does the job. Starting from the linearized EFE:

$$ -\frac{1}{2}\square h_{\mu\nu} + \frac{1}{2}\left(\partial_{\mu}V_{\nu}+\partial_{\nu}V_{\mu}\right) + \frac{1}{4}\eta_{\mu\nu}\square h - \frac{1}{4}\eta_{\mu\nu}\eta^{\alpha\beta}\left(\partial_{\alpha}V_{\beta}+\partial_{\beta}V_{\alpha}\right) = \frac{8\pi G}{c^4}T_{\mu\nu}$$

and taking its trace with $\eta^{\mu\nu}$, one finds that:

$$ \frac{1}{2}\eta^{\alpha\beta}\left(\partial_{\alpha}V_{\beta}+\partial_{\beta}V_{\alpha}\right) = \frac{1}{2}\square h - \frac{8\pi G}{c^4}T$$

Replacing this term in the $00$-th component of the linearized EFE, we find that:

$$ -\frac{1}{2}\square h_{00} + \partial_{0}V_{0} - \frac{1}{4}\square h + \left\{ \frac{1}{4}\square h - \frac{4\pi G}{c^4}T \right\} = \frac{8\pi G}{c^4}T_{00}$$

(which in fact is nothing more than the $00$-th component of the trace-reversed linearized EFE!) From here, requiring that the perturbation be static and imposing that $T_{\mu\nu} = \delta^{0}_{\mu}\delta^{0}_{\nu}\rho c^2$ (and thus $T_{00} = \rho c^{2}, T = -\rho c^{2}$), we arrive at the Poisson equation for $h_{00}$:

$$ -\frac{1}{2}\nabla^{2}h_{00} = \frac{4\pi G\rho}{c^2} $$

Cristián Paris
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