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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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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.

In summary, the metric perturbation produced by a given (static) configuration is identical for both regimes. It's just that in the Newtonian limit, the only relevant component is the 00 component. This isn't surprising at all since the Newtonian limit is designed precisely to reproduce Newtonian gravity, which is based on a scalar potential. If the other components mattered, it won't exactly be a "Newtonian limit" anymore.

Vincent Thacker
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