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In a previous question, the metric in the frame of reference of a rotating disc takes the following form:

\begin{align} ds^2 &= - dt^2 + dx^2 + dy^2 + dz^2\\ &= -\left(1 - \omega^2 ({x'}^2 + {y'}^2) \right){dt'}^2 + 2 \omega (- y' dx' dt' + x' dy' dt') + {dx'}^2 + {dy'}^2 + {dz'}^2. \end{align}

Or in cylindrical coordinates,

$$ds^2=-(1-r^2\omega^2)dt^2 +2\omega r^2 dt d\theta + r^2d\theta ^2+dr^2+dz^2$$

As discussed in the linked question, the spacetime described by this metric is still flat because it is simply a coordinate transformation of Minkowski space. The curvature of the spacetime does not depend on what coordinates you choose.

However, I'm puzzled why this reasoning does not agree with Ehrenfest's Paradox, which states that the geometry on a rotating disc is non-Euclidean. Ehrenfest's Paradox has been mentioned multiple times on the site (here and here), but most explanations deal with the rigidity of the disc, none of which mention this metric.

I have a few questions:

  • If Minkowski space is always flat regardless of what coordinates you choose, why is the spacetime geometry of the rotating disc in Ehrenfest's Paradox curved?
  • Does this metric above describe the same situation as in Ehrenfest's Paradox? If so, why doesn't the intuitive length contraction effect along the $\theta$-direction appear in the metric above? The metric accounts for time dilation effects, but not for length contraction. If the metric doesn't describe the same situation, please explain why. (From online sources, I think the correct metric is the Langevin-Landau-Lifschitz metric, but I really can't understand the derivations.)
  • Is there a distinction between the length scale on a physical object and with space itself? For example, would the proper length of a stick be the same as the proper distance between two events in empty space on the ends of where the stick would be? If not, I don't see why arguing about the rigidity of the disc has any consequence on the actual geometry of space.
Chang Hexiang
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    The space-time is flat, but that does not mean that the spacelike slice of space-time picked out by a choice of $t$ is flat. We can have a curved surface embedded in flat space. – mike stone Apr 20 '21 at 13:48
  • @mikestone Interesting! Could you elaborate on this? – Chang Hexiang Apr 20 '21 at 14:53
  • I think the wikipedia article explains this. – mike stone Apr 20 '21 at 16:06
  • @mikestone ( and to add) you can check that the whole 4d space time has vanishing Riemann tensor everywhere but a t=const slice that induces a metric on a pure spacelike 3d submanifold has non vanishing Rieamnn Tensor. You don’t need to resort to embedding then. – Shashaank Apr 22 '21 at 05:17
  • @Shashaank A $t=constant$ slice is an embedding of a 3-surface into the flat 4-space. – mike stone Apr 22 '21 at 11:48
  • @mikestone yeh, what I meant was that we just look a the Reimann tensor for both the 4d space time ( vanishing) and 3d space ( non vanishing) without talking about extrinsic curvature – Shashaank Apr 22 '21 at 14:35
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    If I may ask, how do you get the correct $t=constant$ slice to do the embedding? Because obviously if you set $dt=0$ in the metric you just get a Euclidean metric. What is the correct choice of $t$ to set to constant, or $dt$ to set to zero? – Chang Hexiang Apr 22 '21 at 15:12
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    @mikestone I guess you would be the better person to answer to the above comment of the OP in which he forgot to tag any of us. – Shashaank Apr 23 '21 at 05:01

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