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I am reading Einstein's book on relativity and ran into the relativistic rotating disc example, also known as Ehrenfest paradox. What is its resolution?

More specifically, imagine two identical rigid discs with 1m rods lining both their circumferences such that exactly $N$ rods suffice.

Imagine the two identical discs are exactly one on top of the other, and that one of them is still while the other one spins at a relativistic velocity.

The rods of the rotating disc should appear shortened to an observer from the reference frame of the stationary disc.

Imagine we take a snapshot at time $t$ and examine it.

The discs should still exactly coincide since the radius of the spinning disc remains unaltered.

What should we expect to see when we examine the frozen snapshot of the circumference of the two discs?

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    $\begingroup$ related questions $\endgroup$
    – Ghoster
    Commented Jan 15 at 17:48
  • $\begingroup$ > the radius of the spinning disc remains unaltered Here's one problem. Just because the radial lines do not Lorentz contract, it does not mean they do not contract. The disk elements will experience tension or compression forces, and in relativistic theory the disk has to deform. $\endgroup$ Commented Jan 15 at 18:37

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The resolution is that there is no such thing as rigid angular acceleration. You simply cannot have a rigid disk in the situation you describe. The disk must deform.

So what we see in the photograph you mention depends on how the disk deforms. The disk will physically strain as measured by strain gauges attached to the disk.

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  • $\begingroup$ What does rigidity actually have to do with it? Isn't the contraction geometrical? I mean, there isn't any physical stress causing the tangential contraction, is there? $\endgroup$
    – nir
    Commented Jan 15 at 19:47
  • $\begingroup$ @nir yes, there are physical stresses causing physical strains. That is what I stated in my last sentence. This is not the same as inertial-frame length contraction. It is a measurable physical distortion of the material, and it is unavoidable. $\endgroup$
    – Dale
    Commented Jan 15 at 20:36
  • $\begingroup$ But it seems that according to Einstein the length contraction is an inertial-frame length contraction - see page 96 of his relativity book archive.org/details/cu31924011804774/page/n117/mode/2up $\endgroup$
    – nir
    Commented Jan 16 at 21:21
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    $\begingroup$ @nir Einstein showed there that the inertial frame length contraction doesn’t work for rotation. Subsequent authors developed this idea further. In particular, Paul Ehrenfest building on Max Born’s work, and followed by Herglotz and others. $\endgroup$
    – Dale
    Commented Jan 16 at 23:22
  • $\begingroup$ @Dale Does the problem of the rotating disk require treatment according to the General theory of relativity? $\endgroup$ Commented Jan 18 at 23:29
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Imagine the rings of saturn. Try to find a stable rotating configuration in the following preparation idea: Each particle rotates on a circle at constant 3d-speed. Each particle tries to maintain constant distances to its two neighbours on the same circle, and, of course, its distance to the neighbouring inner and outer circle. This poses no difficulty.

Next, two neighbouring particles on the same radius try to circle in constant distance with respect to their board watch readings. This will establish a constant rigid rotation in their local system of reference. Both partners determine the distance at the ticks of their respective clocks, that run on different speeds with respect the global time. But independent of board time speed, the aim is to keep the local position independent wrt. to the angular coordinate $\phi$.

In Newtonian mechanics, with board time synchronous to global time at the axis, this preparation is a model of a rigidly rotating disk with constant angular frequency $\omega(r) =\frac{2\pi}{T(r)} =v(r)/r$.

Now switch representation to flat Minkowski space.

The proper time at radius r is $$\tau(r)= \int \sqrt{dt^2 - r^2 d\phi ^2 } = \int \sqrt{1- r^2 \frac{d\phi^2 }{dt^2}} \ dt=\int \sqrt{1- r^2 \Omega^2} \ dt=\sqrt{1-r ^2\Omega ^2} \ t$$
with $\Omega$ representing the frequency of rotation with respect to time t at the axis at rest.

Appying this formula to the inverses yields the local angular frequency in terms of the global one

$$ \omega(r)^2\ = \ \frac{\Omega(r)^2}{1-\Omega(r) ^2} $$ with the inverse $$\Omega = \frac{\omega }{\sqrt{\omega ^2+1}}$$

Since the stability of such a system in flat Minkowski space depends on the constancy of rigid space coordinate differences at global flat time , $\Omega(r)$ has to be constant, keeping all radial distances between particles rotating on the same radius constant, determined by simultaneity of global time.

Such a model represents a rigidly rotating disk with local time rate slowing with radius, with a limiting radius where local time stands still because $r \Omega = 1$.

The local board time is central in determining the radial forces, and, insofar, determines the radial mass distribution. But the stable configuration is determined in the global system, because the potential fields are of electromagnetic nature, best understood in a global nonrotating system.

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