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 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.
