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Chapter 42 of the Feynman Lectures on Physics claims that a clock "higher up" in a gravitational field will tick more slowly, and that is used to argue that space-time in a constant gravitational field is curved. I think the argument consists of three steps, which can be summarized as follows.

  1. Imagine two clocks, A at the front and B at the back of a rocket ship accelerating forward, emitting light pulses once a second. Because the distance the light has to travel from A to B keeps decreasing, and the distance light has to travel from B to A keeps increasing, the rate of clock A must be slower than that of B.

  2. The equivalence principle implies that the same must be true if the rocket ship is standing vertically on the ground under the pull of earth's gravity.

  3. Since the following rectangle "doesn't close up", spacetime must be curved:

enter image description here

I'm a novice to general relativity, and this raises a few difficulties for me; I'd appreciate your help in sorting them out. Here they are:

A. Clearly, if we start out with flat space and have two observers accelerating keeping a constant displacement from one another, their proper time as a function of the frame time would be the same (the integral expression for the proper time is exactly the same for both observers)

B. Curvature is supposed to be invariant; if it vanishes in one coordinate system, it should vanish in any other. So it is not clear how you can start from flat space, change coordinates, and get curved space. This answer is somewhat related, showing that Rindler metric is flat.

C. The fact BAC and BDC' have the same length, but the rectangle don't close up, needn't imply there's curvature - in "most" coordinate systems for flat space the rectangle will not close up, since the images of the edges will be different wiggly lines (the lines are not geodesics, though maybe they are special in some other way). This is related to a more fundamental difficulty: how can one define the coordinate system of the rocket? (that is, without assuming it is sufficiently small so that we can neglect precisely the kind of effects we're trying to discuss...)

EDIT for (A): As a few people pointed out, one needs to be a bit careful. If their separation in the rest frame is constant, their proper time would be the same - but the rocket contracts as it speeds up, so this is not the case here.

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  • $\begingroup$ I asked the same question once here! $\endgroup$
    – knzhou
    Commented Apr 10, 2023 at 20:48
  • $\begingroup$ Your point A is definitely not a "Clearly". In any given inertial frame the two clocks will be getting closer together due to length contraction. In the non-inertial frame of the rocket the clocks are ticking at different rates. Either way the coordinate acceleration is very much frame dependent, and calculating the proper acceleration is tricky. $\endgroup$
    – Eric Smith
    Commented Apr 10, 2023 at 23:10
  • $\begingroup$ @knzhou your question is indeed related to the difficulty (C) of my question. But note that the answer you accepted does not address the full problem - while it is quite clear that if geodesic rectangles don't close up, that indicates curvature (when there's no torsion) the edges are not geodesics. A geodesic in our frame would correspond to a falling object. So there's in some sense a "deeper" difficulty with the argument. $\endgroup$
    – Just Me
    Commented Apr 11, 2023 at 9:20
  • $\begingroup$ @EricSmith thanks for pointing this out, added a comment explaining my mistake there. $\endgroup$
    – Just Me
    Commented Apr 11, 2023 at 9:23
  • $\begingroup$ @knzhou in this example I think the failure of the rectangle to close does not reflect torsion (vanishes by assumption for the Levi-Civita connection) nor curvature. It is more analogous to trying to walk at a constant speed along a straight path from P to Q and along a curved path from P to Q, and noticing that you get to Q earlier along the straight path... $\endgroup$
    – Just Me
    Commented Apr 11, 2023 at 9:30

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You're right: this spacetime region is flat* and Feynman's argument in section 42-7 is wrong.

(* In reality it's slightly curved, but that curvature isn't responsible for the effects he's talking about; it's just a tiny correction.)

Section 42-8 is mistitled since it's really about motion in flat spacetime (in an accelerating coordinate system), but the calculations are okay. The rest of the chapter looks okay to me.

A. Clearly, if we start out with flat space and have two observers accelerating keeping a constant displacement from one another, their proper time as a function of the frame time would be the same (the integral expression for the proper time is exactly the same for both observers)

You have to be careful here, since in a so-called uniform gravitational field (flat spacetime in Rindler coordinates), the acceleration of an object at rest in the accelerating coordinates varies with height, because Rindler coordinates are analogous to polar coordinates in Euclidean space and the height is the distance from the center. If your accelerating clocks start out at different heights but their acceleration is the same, they can't both remain at rest in the accelerating coordinates and the relationship between them is actually rather complicated. Bell's spaceship paradox is related to this. If they maintain a constant distance then their accelerations are different, so you can't argue by symmetry that they'll remain in sync (and they don't).

B. Curvature is supposed to be invariant; if it vanishes in one coordinate system, it should vanish in any other. [...]

C. The fact BAC and BDC' have the same length, but the rectangle don't close up, needn't imply there's curvature [...]

You're right about all of that.

how can one define the coordinate system of the rocket?

I'm not sure what you're asking, but the rocket effectively is a coordinate system—or at least, it's a reference frame, if you maintain the distinction that reference frames are physical objects and coordinate systems are just numbers. You can use metersticks to assign spatial coordinates to everything on the ship, and you can scatter clocks around to provide the time. If the clocks all measure their own proper time then they will drift out of sync, which doesn't invalidate the coordinate system but does make it inconvenient. You would probably want to tweak their rates instead so that they remain in sync. That gives you Rindler coordinates.

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  • $\begingroup$ Thanks! this really helps clarify the situation. Re. the coordinate system of the rocket - I meant how to synchronize the clocks. Without a precise definition, one can't make sense of diagrams like the one above. $\endgroup$
    – Just Me
    Commented Apr 11, 2023 at 18:56

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