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In the weak field limit of general relativity (with matter described by a perfect fluid consisting only of dust), we have the following correspondences:

  • $00$-component of the Einstein field equations (EFEs) $\leftrightarrow$ Poisson's equation $\nabla^2\phi = 4\pi G \rho$;
  • spatial components of the geodesic equation $\leftrightarrow$ force equation $\vec{g} = -\vec{\nabla}\phi$,

where $\phi$ is the Newtonian gravitational potential. My question is about the other components of the EFEs and geodesic equation. In the several textbooks I have consulted these are not worked out or discussed. The remaining $0$-component of the geodesic equation reduces nicely to $0=0$ and hence does not add anything. Similarly for the mixed $0i$-components of the EFEs. But the $ij$-components of the EFEs do not seem to reduce to a triviality. In fact, we obtain an equation of the form (schematically)

$$\sum \partial_i\partial_j h_{\mu\nu} = 0,$$

where the sum represents the fact that there are several of these terms with permuted indices (some of which are contracted over). This equation constrains the spatial and mixed components $h_{ij}, h_{0i}$ in terms of $h_{00} = 2 \phi/c^2$. Does this represent anything physical? Since the $h_{ij}, h_{0i}$ components do not enter in the (approximate) geodesic equation for massive particles in the Newtonian limit, the equation has no bearing on the movement of (massive) test particles, at least. I'm wondering whether it is still relevant in any way.

Inzinity
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  • The "Newtonian limit" normally means that because all particles move at speeds much less than $c$, the components of $h$ other than the 00 component are so much smaller than the 00 component that we can just treat them as 0. The same goes for all terms in $T$ other than the 00 component. That means you just get 0 = 0 for the other 15 components of the EFE. If there's some other Newtonian limit where the question is interesting, you should explain what you mean by Newtonian limit. – Brian Bi Mar 20 '24 at 11:29

1 Answers1

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The "Newtonian limit" is not the same as linearized gravity. The former is a subset of the latter.

In the Newtonian limit, the assumptions are:

  1. The spacetime is considered to be a perturbation of flat spacetime.
  2. The metric is static.
  3. The velocities are small compared to the speed of light.

This is different from linearized gravity, which uses only the first assumption. In the Newtonian limit, it is indeed true that the only relevant component is the 00 component due to the two additional assumptions. It is used to check that GR indeed reproduces Newtonian gravity under the three assumptions.

In full linearized gravity, the spatial components of the metric perturbation contain (among other things) the propagating degrees of freedom. This means that we get gravitational waves in vacuum, which is a huge difference compared to Newtonian gravity alone. The linearized Einstein field equations are $$\partial_\alpha \partial_\mu\bar{h}_\nu^{\;\alpha} + \partial_\alpha \partial_\nu\bar{h}_\mu^{\;\alpha}-\Box^2 \bar{h}_{\mu\nu} - \eta_{\mu\nu} \partial^\alpha \partial^\beta \bar{h}_{\alpha\beta} = 16\pi G T_{\mu\nu}.$$

In addition, for photons and test particles at relativistic velocities, the spatial terms of the metric perturbation do affect the geodesics as well. This is used to calculate the bending of light by the Sun.

Vincent Thacker
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  • You say that only the 00-component (of the Ricci tensor, I guess?) is nonzero in the Newtonian limit. How do you arrive at that conclusion? Assumptions 1,2,3 do not seem to imply this, at least it's not clear to me how they would. As I stated in my question, I calculate the $R_{ij}$ components and they do not vanish in general. – Inzinity Mar 21 '24 at 17:04
  • @Inzinity The third assumption implies that spatial components of velocity are negligible compared to temporal components. So among the Christoffel symbols, only $\Gamma^\mu_{00}$ needs to be considered. The same argument can be used to neglect all components of $T$ except $T_{00}$. Finally the second assumption implies that $\Gamma^\mu_{00} = (-1/2)g^{\mu\sigma}\partial_\sigma g_{00}$. This is in Carroll's Spacetime and Geometry, section 4.1. – Vincent Thacker Mar 21 '24 at 18:24
  • Thanks for the suggestion but as I see it, the fact that that spatial components of velocity are negligible compared to temporal components does not allow you to neglect Christoffel symbols with spatial lower indices in the Ricci tensor. It only allows you to do that in the geodesic equation, as only then those terms are multiplied by the spatial components of the velocity. Sean Carroll section 4.1 does not say anything on this this if I'm not mistaken and just only considers the $R_{00}$ component without even mentioning the $R_{ij}$ components. – Inzinity Mar 22 '24 at 13:07
  • @Inzinity No. The assumption implies that temporal intervals are much larger than spatial ones (i.e. $\mathrm{d}x^i \ll \mathrm{d}t$). So the 00 component of h is much larger than the other components. All of Newtonian gravitation occurs in the 00 component alone. As Brian mentioned above, the other components of the EFE are 0 = 0. – Vincent Thacker Mar 22 '24 at 16:13
  • Can you explain how you get from the low velocity assumption to the conclusion that the 00 component of h is much larger than the other ones? – Inzinity Mar 22 '24 at 20:19
  • @Inzinity The explanation is already in my previous comment. The point is that in order to derive Newtonian gravity, only the 00 component is needed. I never claimed that the Ricci tensor is always zero. In case it isn't explicit enough, the answer to your original question is no, they are not relevant in the Newtonian limit because the velocities are small compared to c. Only when we consider full linearized gravity are they needed, such as gravitoelectromagnetism, gravitational waves, light bending, etc. – Vincent Thacker Mar 22 '24 at 22:05
  • @Inzinity See https://physics.stackexchange.com/a/460930, https://physics.stackexchange.com/q/371990, https://physics.stackexchange.com/a/704193, etc. – Vincent Thacker Mar 22 '24 at 22:17
  • @Inzinity I have updated the answer to better reflect this. – Vincent Thacker Mar 23 '24 at 01:16
  • I appreciate it, but your answer still does not answer my original question. – Inzinity Mar 24 '24 at 13:12
  • @Inzinity What exactly is your question? As currently stated, it has been answered completely. The answer is "No" and I have explained multiple times why. – Vincent Thacker Mar 24 '24 at 13:29