# Length contraction of rotating wheel

1. May 12, 2014

### Erland

Suppose that a wheel is rotating around its axis in free space with no friction, with no external forces acting on it. In theory, the wheel will rotate forever with constant angular velocity. This velocity is assumed to be small (essentially non-relativistic).

Now, consider an observer O moving with constant (high, relativistic) linear velocity relative to the center of the wheel, in a direction perpendicular to the axis of the wheel.
O will perceive (although not strictly see) the wheel as contracted in the direction of the relative motion, but not contracted in the direction perpendicular to the motion. So O will perceive the wheel as an ellipse instead of a circle.
So accoording to O, a point P at the perimeter of the wheel will move in an elliptic orbit around the moving center, changing its distance to the center of the wheel all the time.

How can then O explain the physics of this motion of the wheel, w.r.t O's own frame of reference? Which forces act on the point P to give it such a strange motion?

2. May 12, 2014

### HallsofIvy

There are no "forces" required to contract a length in the direction of motion.

3. May 12, 2014

### WannabeNewton

This is incorrect. It would be true if the wheel was non-rotating but the rotation actually causes the wheel to take on a much more complicated shape on the simultaneity surfaces of the moving observer's frame. It might be instructive for you to first consider the case of a rod rigidly rotating uniformly with angular velocity $\omega$ about the $z$-axis in a given inertial frame with coordinates $(x,y,z)$: what shape does the rod take on the simultaneity surface at any given instant in the frame of an observer moving along the $x$-axis with constant velocity $v$?

As Halls noted there are no forces involved. All the observer has to appeal to is the relativity of simultaneity.

4. May 12, 2014

### Staff: Mentor

I like this question. It is gives a very strong motivation for using four-vectors. Are you familiar with the concept? I will assume that you are, but if not we can go back and work on that.

Consider a point on the wheel. In the rest frame of the center of the wheel (units where c=1 and r=1) its worldline can be written as: $s=(t,x,y,z)=(t, \cos(\phi+t\omega), \sin(\phi+t\omega),0)$ where $\phi$ identifies a particular point on the wheel. Then, the four-force on that part of the wheel is given by
$$F=m \frac{\partial^2 s}{\partial\tau^2} = -m\frac{\omega^2}{1-\omega^2} (0,\cos(\phi+t\omega),\sin(\phi+t\omega),0)$$
Since this is a four-vector, you can boost it to any other frame using the standard Lorentz transform and the relationship will remain. The boosted forces will be the second derivative wrt proper time of the boosted position.

Last edited: May 12, 2014
5. May 12, 2014

### xox

$$F=m \frac{\partial^2 s}{\partial\tau^2}=m \frac{\partial^2 s}{\partial t^2} (\frac{\partial t}{\partial\tau})^2$$

where $$\frac{\partial t}{\partial\tau}=\frac{1}{\sqrt{1-\omega^2}}$$

So, DaleSpam's answer is at odds with the ones by WannabeNewton and HallsofIvy. The force appears to be non-null. It is true that the direction of the force is perpendicular on the direction of the motion, since $$v= \frac{\partial s}{\partial\tau}=\frac{\omega}{\sqrt{1-\omega^2}} (1,-\sin(\phi+t\omega),\cos(\phi+t\omega),0)$$

Last edited: May 13, 2014
6. May 13, 2014

### Staff: Mentor

No, it is not at odds, they are talking about something else. There is no force required to contract a length. I was deriving the "centripetal" forces necessary to produce the acceleration of points on the tire. Those are two different things, and hence two different quantities.

I agree that no forces are required for length contraction, and I am sure they agree that forces are needed for centripetal motion.

7. May 13, 2014

### xox

Yes, this is why I pointed out that $$F \cdot v=0$$

i.e. the force is perpendicular on the direction of motion. So, the force $$F$$ is not responsible for the length contraction but, there is a force present (obviously).
The elementary element along the circumference gets contracted according to the formula:

$$ds'=ds \sqrt{1-(\omega R/c)^2}$$

where $$ds$$ is the arc element measured in the frame co-rotating with the circumference and $$ds'$$ is the arc length measured in the (inertial) frame attached to the center of the circle.
The proof is not trivial by any stretch of imagination but I can provide it for the interested parties.

How do I get rid of the annoying linefeeds?

Last edited: May 13, 2014
8. Jun 9, 2014

### Erland

I made some Matlab simulations of this, and I find that the shape is perfectly elliptic.

Here is my Matlab code:
---
clear ti
a=linspace(-pi,pi,50);
v=0.9;
w=0.3;
r=0.3;
for i=1:50
tprim=@(t) (t-v*r*cos(t*w+a(i)))./sqrt(1-v^2);
ti(i)=fzero(tprim,0);
end
xprim=(r*cos(ti*w+a)-v*ti)./sqrt(1-v^2);
yprim=r*sin(ti*w+a);
plot(xprim,yprim,'o');
axis equal
---
In this case v=0.9c, the angular frequency is w=0.3 rad/sec, and the radius of the wheel is r=0.3 light-seconds. As you see, the shape looks perfectly elliptic (sorry I can't post the plot), and this doesn't change even if we change the parameters v, w, and r.
Notice that I am not interested in the wheel's accelerated reference frame, only in the frame in which the wheel is rotating with angular frequency w, and the frame moving with constant velocity v relative to this frame.

Please, let me know if there are any errors in my Matlab program.

Thank you all for your replies, anyway.

9. Jun 9, 2014

### WannabeNewton

Hi Erland. When you say "rotating wheel", are you talking about a rotating circle or a rotating circle with spokes such as an actual wheel would have?

Certainly for a rotating circle, we will get an ellipse in the moving frame. This is in fact very easy to verify directly. In frame $S$ the circle is given by, for instance, $x(t,\phi) = \cos(\omega t + \phi), y(t,\phi) = \sin(\omega t + \phi)$ and so in the frame $S'$ which moves along $x$ with velocity $v$, we have $\gamma (x' + vt') = \cos(\omega \gamma t' + \omega \gamma vx' + \phi(\phi')), y' = \sin(\omega \gamma t' + \omega \gamma vx' + \phi(\phi'))$ and so $\gamma^2 (x' + vt')^2 + y'^2 = 1$ which is an ellipse that varies in time in $S'$.

But if you consider an actual wheel (which is what the thread title implies), that is a circle with thin straight rods acting as its spokes, then the situation is, at least ostensibly, different. This is because for a rotating rod in $S$ we have for the rod $y = x\tan \omega t$ so in $S'$ we get $y' = \gamma (x' + vt')\tan (\omega \gamma t' + \omega \gamma vx')$ which is a rather complicated shape compared to the original rod. For example at $t = 0$ this becomes $y' = \gamma x' \tan \omega \gamma vx' \approx \gamma^2 \omega v x'^2$ if $x' \ll 1$ which is a parabola instead of a rod. As such, I don't see a priori why a rotating wheel, which is a combination of a rotating circle and rotating rods, would simply be an ellipse in the moving frame. I may be wrong of course, due to symmetry or otherwise.