Rotating Rod in Special Relativity

Question

Statement of the Problem

In frame $F'$ with co-ordinates $(t',x',y',z')$, a straight rod rotates in the $x',y'$ plane with angular velocity $\omega ′$ about one of its ends. The fixed end is located at the spatial origin in $x'$ which is $x'=y'=z'=0$. At time $t′ = 0$ the rod lies along the positive $x′$ axis. Frame $F′$ moves with constant velocity $v$ in the $x$-direction with respect to frame $F$. The origins of the two frames coincide at time $t′ = t = 0$ (i.e. the points $(t,x,y,z)=0$ and $(t′,x′,y′,z′)=0$ are the same). Also, there is no spatial rotation between the two frames. Find an equation which gives the shape of the rod in frame $F$ at time $t = 0$. In other words, find an equation for $y$ as a function of $x$.

First Thoughts

It's clear to me that the rod won't be straight in $F$. Points closer to the origin have a lower velocity in $F'$, so those points move with different velocities than points further from the origin, in a nonlinear way.

However, I'm not sure how to encode this mathematically within the framework of special relativity and thus ultimately get the equation that describes the shape of the rod in $F$. I considered taking a small element $dl'$ of the rod, and modifying its shape according to length contraction in the direction of the relative velocity between the frames, but that doesn't seem to be leading anywhere. It's also possible that one could use the relativistic velocity addition formula to get the velocity of some point on the rod, but that doesn't seem to lead to the necessary function, either.

Any help is greatly appreciated.

Answer 1

The problem talks of a straight rod. It's also obvious that rotation is assumed uniform. You say

It's clear to me that the rod won't be straight in F.

$\let\g=\gamma \let\om=\omega$ I can't see how you say that. But for a moment let's stay in F$'$. There rod is straight and uniformly rotating, so that you may parametrize its points by $s$, the distance from the fixed end as measured in F$'$ (going from 0 to - say - $a$). We'll have $$\eqalign{ x' &= s \cos\om t' \cr y' &= s \sin\om t'.\cr} \tag1$$

No rigidity assumption is needed: our problem is strictly kinematical. It's true that the rod will be stretched - its total length is greater than in its rest condition, and furthermore the stretch isn't uniform, because acceleration increases with $s$. But in course of time stretching remains unaltered and the same applies to the distance $s$ of any point of the rod from the fixed end. In short, eqs. (1) describe a circular uniform motion for every point of rod in F$'$.

The problem's text exactly states the relationship between F and F$'$. The relevant coordinate transformations are $$\eqalign{ x' &= \g\,(x - vt/c) \cr y' &= y \cr t' &= \g\,(t - vx/c^2).\cr} \tag2$$ where, as usual, $\g=1/\sqrt{1 - v^2/c^2}$. (Note that rod's rotation doesn't enter into this transformation - it only concerns coordinates of the same event as viewed by both frames.)

It remains for you to enter (2) into (1), then eliminate $s$ to find $y$ as a function of $x$ and $t$. If you look at that equation for a fixed $t$ you have a "snapshot" of the rod as seen from F. It's far form a straight line, which means your intuition was right.

Answer 2

Firstly, are you happy with handling kinematics of accelerated point-particles in special relativity (Rindler coordinates and all that)? If not, start there.

If you are ok with it, see Born rigidity https://en.wikipedia.org/wiki/Born_rigidity.

Answer 3

The question seems to be trying to emphasise the mixing of space and time - t' will depend on x when t=0 (which of course itself depends on x' and t', oh dear).

A little playing around considering a single part of the rod at distance $l$ from origin in s':

$x' = l \cos(\omega t')$
$y' = l \sin (\omega t')$

$t = \gamma(t' + \frac{x' v}{c^2})$ but since t=0 ... $$t' = -\frac{v}{c^2} l \cos(\omega t')$$ This can be solved numerically to give specific t' and x' combinations - I'm not 100% sure what the interpretation of multiple solutions would be since a single point of the rod cannot appear to be in multiple places simultaneously in any frame.

Provided $\omega l << c$ then we can make the small angle approximation (to see this, let $\omega t' = k$) so $t' \approx \frac{v}{c^2}l$

Using $x' = \gamma (x - vt)$ and $y = y'$ the small angle approximation leads us to $$ x = \frac{1}{\gamma}l \\ y = -\frac{v}{c^2} l^2 \omega$$

Where $l$ varies parametrically. In the case that $wl << c$ this is basically a straight line.