Non-Euclidean geometry of a rotating cylinder

Question

I am reading Ta-Pei Cheng's book "Relativity, Gravitation and Cosmology" and having some difficulties with question 6.3. It basically asks to work out the spatial distance in a rotating cylinder to show that the Euclidean relation between the circumference and radius is violated. In the solutions he starts by defining coordinates $$(ct, r, \phi, z) \text{ in the lab frame}$$ and $$(ct, r_0, \phi_0, z) \text{ for an observer on the rotating disk}$$ related by $$r = r_0 \text{, } \phi = \phi_0 + \omega t\text{.}$$ The line element in terms of coordinates at rest with respect to the observer on the rotating disk is (ignoring $z$ coordinate) $$ds^2 = -c^2dt^2 + dr_0^2 + r_0^2d\phi_0^2.$$ By substituting $$d\phi_0 = d\phi - \omega dt$$ he rewrites $ds^2$ in terms of the lab coordinates, identifies the metric elements and uses a previously derived result to eventually work out that $$dl^2 = dr^2 + \frac{r^2d\phi^2}{1 - (\omega r/c)^2}$$ "showing clearly length contraction of the circumference, but not the radius."

Using the above result you get that the circumference $C$ is $$C = \gamma(\omega r)2\pi r.$$ How is this a contraction if this is greater than $2\pi r$?

Also, why he uses the same time in both the lab frame and the observer on a disk frame? Isn't $dt_{\text{on a disk}} =\gamma(\omega r) dt_{\text{lab}}$? I am clearly misunderstanding something but I just can't figure out what.

Answer 1

$\let\g=\gamma \let\om=\omega \def\ns#1#2{#1_{\mathrm{#2}}}$ I don't know the book you are quoting and can't exactly figure what is said in the book from your short summary. What I can tell you with certainty is that the problem is an old one, being known, in its first formulation, as Ehrenfest paradox (1909). Einstein used it in his 1916 paper on GR and also in the ensuing book The Meaning of Relativity, to show that a gravitational field warps spacetime.

However there were unclear points in Einstein's argument and since then there has been a big number of papers on the subject, till to recent times. I'm not sure the issue is definitely solved. (I must be cautious as I didn't follow the relevant literature in the last years. E.g. I didn't know the paper you quoted in your update.) This is to say that you should not worry if you have difficulties understanding the problem: you're not alone!

To be more specific:

How is this a contraction if this is greater than $2\pi r$?

I try to interpret as follows. The book is saying the circumference length in the lab frame is contracted wrt length in its rest (rotating) frame.

Also, why he uses the same time in both the lab frame and the observer on a disk frame? Isn't $\ns{dt}{on\ a\ disk} = \g(\om r)\,\ns{dt}{lab}$?

Really not. A point on the disk is moving wrt to lab with speed $\om r$. Then time ratio is the other way round: $$\ns{dt}{lab} = \g(\om r)\,\ns{dt}{on\ disk}$$ if with $\ns{dt}{on\ disk}$ you mean proper time of a point of disk.

With the substitution given in the book we have $$ds^2 = -(c^2 - \om^2 r^2)\,dt^2 + dr^2 + r^2 d\phi_0^2 + 2\,\om\,r\,d\phi_0\,dt.$$ A point fixed on disk has $dr=0$, $d\phi_0=0$ and its proper time is $dt/\g$, where $dt=\ns{dt}{lab}.$

Is there any argument justifying this choice?

I don't know the author's arguments. Consider that in principle you are free to use coordinates of your choice. Then you may search for a physical interpretation. In our case introducing $\phi_0$ is reasonable, as you may integrate to $$\phi = \phi_0 + \om\,t$$ which you can read in the lab frame: a point having fixed $\phi_0$ and $r$ coordinates is moving in the lab of uniform circular motion with angular velocity $\om$.

As to $t$ coordinate it may be justified a posteriori by the above reasoning. It has however some counterintuitive properties, discussed in Cook's paper. I warn you that there are in that paper several points I don't agree with, and other where I don't like his mathematical approach. But all this goes beyond your question.

Answer 2

$(ct, r_0, \phi_0, z) \text{ for an observer on the rotating disk}$

I think you misread or mistranslated this part, or the book is in error. These are the coordinates of non rotating inertial observer at the centre of the disk.

$ds^2 = -c^2dt^2 + dr_0^2 + r_0^2d\phi_0^2$

We can replace $ds^2$ with $-c^2 \tau^2$ where $\tau$ is the proper time of the observer that is rotating with the disk. See Wikipedia so:

. $-c^2 \tau^2 = -c^2dt^2 + dr_0^2 + r_0^2d\phi_0^2$

$\frac{d\tau^2}{dt^2} = 1 - \frac{dr_0^2}{c^2 dt^2} - \frac{r_0^2 d\phi_0^2}{c^2 dt^2}$

For motion at a constant radius $dr_0 =0$ and since angular velocity $\omega = d\phi/ dt$ and tangential velocity $v = r_0 \omega$ we can write:

$\frac{d\tau^2}{dt^2} = 1 - \frac{r_0^2 \omega^2}{c^2} = 1 - \frac{v^2}{c^2}$

$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}} $

Looks familiar? This is the concept of "Instantaneous Comoving Observer" at work. The time dilation of a clock following a circular path and experiencing centripetal acceleration is exactly the same as an inertial clock moving in a straight line, when measured in a small enough spatial and time interval. The clock rotating at radius $r_0$ is ticking slower than the non rotating clock of the inertial observer. I.e. the rotating clock and the lab frame clock (if the lab frame is the inertial frame in this case - you did not make it clear) are not ticking at the same rate.

Using the above result you get that the circumference C is $C=\gamma(ωr) 2 \pi r$. How is this a contraction if this is greater than 2πr ?

It not the circumference of the disk as whole that is contracting. It is the rulers of the observer riding with the rim. Because her rulers are contracted, she can fit more of them around the perimeter and and the circumference appears larger to her.