Doubt-Schild's argument about spacetime curvature

Question

Reading MTW Gravitation book, to me is unclear the point related to the Schild's argument about spacetime curvature. It is basically the following:

Consider two observers at rest in the gravitational field of the Earth, one at height $z_1$ and the other at height $z_2>z_1$, in the context of Special Relativity where the frame at rest with the Earth acts as a global Lorentz frame.

The lower observer sends two successive light pulses to the upper observer. This defines four events in spacetime as follows: $E_1$ and $E_2$ are the emissions of the two light pulses by the lower observer, and $R_1$ and $R_2$ are the receptions of the two light pulses by the upper observer. In the spacetime diagram related to the global Lorentz frame, these four events form a parallelogram--it must be a parallelogram because opposite sides are parallel. The lower and upper sides, $E_1E_2$ and $R_1R_2$, are parallel because the two observers are at constant heights; and the light pulse sides, $E_1R_1$ and $E_2R_2$, are parallel because the spacetime is static, so both light pulses follow exactly identical paths--the second is just the first translated in time, and time translation leaves the geometry of the path invariant.

So, in principle, $E_1R_1$ and $E_2R_2$ paths could be not represented as straight lines in the diagram, what is really needed is just the "congruence" between them.

Now my question is: if we admit the above about $E_1R_1$ and $E_2R_2$ paths, it would mean the speed of the light would be not isotropic and constant in the global Lorentz frame violating the really basic feature of a Lorentz frame !

What's wrong in the above argument ?

Answer 1

The clever thing about Schild's argument is that it does not rely on light travelling in straight lines: even if it follows curves, as everything else does under gravity, then you still can't keep SR alive. In particular there's no expectation that things will be isotropic in the presence of gravity (and indeed things aren't, of course), but this can't save you.

Here's a modified version of the experiment which I think makes it clear. I'll explain at the end why the original is better.

Imagine two observers as in his experiment. But now the lower observer has built a very precise machine. What this machine does is, once a second, throw one of a large supply of identical stones at a very accurately known velocity which the upper observer can catch. These stones obviously do not travel in straight lines, because there is gravity. But, because the machine is accurately made, they do all follow the same curves upwards, just translated along the time axis.

Now, the lower observer fires off a bunch of stones, once a second. And the upper observer catches them and measures the times between their arrival. If special relativity is correct, then they will arrive once a second. But they don't: they arrive less frequently than that.

This experiment can be made more practical by accepting that the machine can not be as accurate as you'd like it to be but ensuring that in the long-run it does fire as many stones as there are seconds, and then comparing the long-run rate of arrival with the long-run rate of throwing: after the person on the ground has thrown 100,000 stones they can ask the person high up how long they took to arrive.

But it still depends on all sorts of making-things-very-accurately-ness. If, instead, you fire light at the upper observer then you avoid the accurate-machine problem to a great extent because you know that light always travels at the same speed (relative to you), carries its own clock, and you can generate it at very accurately known frequencies using lasers, say. In particular if you fire a pulse of light which is $n$-cycles long from a source with a very accurately known frequency, then the person at the top can measure how long those $n$ cycles take at their end, and if the times aren't the same then special relativity can't be true, with the only constraint being that all the light took the same path, not whether the path is straight, curved, or whatever.

Answer 2

My point is that I could never understand Schild's argument, for a deeper reason. Maybe explaining my problem I could also help solving your doubt. My counter-argument runs as follows.

Keep mind at the equivalence principle: all what you can experiment in your laboratory fixed on Earth, will also happen in the same way in a spaceship moving in deep space, far from gravitational sources, with its rockets firing and giving it a constant acceleration equal to g. No doubt, however, spacetime around the spaceship is not curved. The problem is only that the physicist in it is naively using a wrong coordinate system, not a Lorentz one (he will likely be using Rindler coordinates, which are exactly suited for a uniformly accelerated frame).

When drawing spacetime diagrams, one must be careful: a diagram is a map of spacetime, just as a geographic map of Earth's surface. And just as there are available many and many cartographic projections, so there are many projections for drawing maps of spacetime. (Of course, they are called "coordinate systems".) But you must not take the map for the real spacetime, and infer from the map properties of the spacetime itself. Or better: you can do, but must know what you are doing...

Of course, it is true that spacetime near Earth is curved, near the spaceship isn't: this can be discovered by a subtler exploration of the maps, involving the effect of the variation of g, or the so-called "tide forces". Gravitational redshift by itself is not enough.