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I had a discussion with someone about the motion of a photon in the field of an infinite massive plane. The metric of the spacetime above and below the plate is:

$$ds^2=e^{2g|z|}(-dt^2+dx^2+dy^2)+dz^2$$

From this metric, we can deduce that a photon starting out horizontally, parallel to the plate, will stay parallel. On the other hand, a time-like particle will fall to the plate.

Does this mean that the equivalence principle doesn't hold here, or is it just an "adjusted" equivalence principle?

Il Guercio
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    The trajectories of massless and massive particles are always different, not just in this case. – Javier Aug 27 '23 at 14:38
  • Yes, but both move towards the center of gravity in the same amount wrt to r. Like in an accelerated rocket both move towards the floor of the rocket. But in this case, only time-like particles fall to the ground. A photon starting out parallel to the plate will stay pararllel, while a massive particle drops down. – Il Guercio Aug 27 '23 at 15:14
  • @Javier - It's not the mass but the speed - if you take a testparticle with limit v→c you get the same geodesic as with a photon with v=c, although it is numerically better to calculate it with ds²=0 and v=c since the proper time of a particle in the limit gets very small while the affine parameter of a photon doesn't, but the trajectory is de facto de same (for example the geodesics of Neutrinos in the limit v→c is not different than for photons with v=c). The trajectories (along y) are velocity (along x) dependend though, but that's not limited to the speed of light but continuously. – Yukterez Aug 27 '23 at 19:48

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Il Guercio asked: "Does this mean that the equivalence principle doesn't hold here?"

In your metric of the infinite plane (which is metric 5 in here) you have a cosmological constant with density ρ=-3g²/8/π and p=-ρ, see here at Output 9, so a parallel light ray will indeed stay parallel but you can't compare that to the Newtonian plane which doesn't need a Λ for a time independend infinite plane metric as you do in relativity.

So in your scenario you don't even have a vacuum around the plane, but expanding space (without that the infinite plane would collapse under its own weight and increase in density like in a 2-dimensional big crunch).

By the way, a light ray grazing the earth in the x-direction also falls with d²y/dT²=(2v²+1)g, which is 3x faster than an object at rest, see here, but that also doesn't violate the equivalence principle, see here. On the infinite plane the difference is in the other direction, but the principle is the same.

The equivalence principle only holds for small length and time scales, whereas the speed of light covers a relatively large length scale over a short time scale. That is always the case, not only on the infinite plane, but also on the round earth or sun. If the equivalence priciple was all there is it would have been easier to calculate the deflection of light around the sun and give just the Newtonian value, which it doesn't.

Yukterez
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  • Why should the plate contract to a black hole without space expansion? The metric I give in my question is a static metric. – Il Guercio Aug 27 '23 at 19:51
  • @Il Guercio - You have put in your static metric tensor, meaning the metric independend of t. That metric tensor automatically leads to an energy tensor with ρ=-p=-3g²/8/π everywhere outside of the plane, if you force a vacuum around the plane in the energy tensor the metric tensor you get will no longer be independent of t but lead to a contraction of the plane (since it is infinite is stays infinite though, but the density will increase), so you can't have it both ways at once, at least not in general relativity. – Yukterez Aug 27 '23 at 19:54
  • In the PDF I linked above it says on page 6: "One can give a physical motivation for the appearance of pressures or tensions (i.e. negative pressures) in the static, general relativistic solutions: a matter source with only a plane of mass-energy – T00 ∝ δ(z) – is not stable, but will collapse under its gravitational self-attraction. To have a static configuration one must stabilize the mass-energy density by pressures/tensions (and in the case of the “brane” world metric by a cosmological constant)." – Yukterez Aug 27 '23 at 20:01
  • What then is the physical motivation? If the mass is uniformly distributed, nothing will happen. You could use a thick infinite slab of polystyrene. – Il Guercio Aug 27 '23 at 20:03
  • Why not, if you have an infinite universe with uniformly distributed matter in 3 dimensions it would also collapse, that's why Einstein invented the cosmological constant. Same here, but with the plane spanning 2 dimensions instead of a 3 dimensional volume. – Yukterez Aug 27 '23 at 20:06
  • An infinite 3d universe with mass uniformly distributed in it would not collapse, I think. Why should it? – Il Guercio Aug 27 '23 at 22:35