# Length contraction: a question about tramwheels

1. Jun 14, 2009

### Rasalhague

This isn’t an example from a textbook, just something that occured to me, so there may be any number of mistaken assumptions in it. I must have gone wrong somewhere, but I don't know where, and I'd appreciate any advice.

Imagine one of those Swiss high speed trams that wreaked so much havoc in early 20th century Berne. In the tram’s rest frame, the time between successive contacts of a point on the rim of a wheel with the rail is the proper time (timelike interval) between these events. Imagine that this point on the rim of the wheel marks the rail wherever it touches it. In the rail’s rest frame, these contact events, each marking one complete revolution of the wheel, will be further apart in time by the Lorentz factor

$$\frac{1}{\sqrt[]{1 - \beta^{2}}} = cosh\left(artanh\; \beta \right)$$

where $$\beta$$ is the speed of the tram in the rail's rest frame, as a fraction of c. In the rail’s rest frame, these events are clearly further apart in space than they were in the tram’s rest frame, where they were no distance at all apart.

Let $$\ell$$ = the length of rail that passes under the tramwheel in the tram’s rest frame in one revolution, and $$\ell'$$ be the length of rail over which the tram passes in the rest frame of the rail.

$$\ell = 2 \pi r = \beta \tau$$

What is the distance between marks in the rail’s rest frame?

$$\ell' = \beta t'$$
$$\ell = \beta t \Rightarrow \beta = \frac{\ell}{t}$$
$$t' = t\gamma$$
$$\therefore \ell' = \frac{\ell}{t} \cdot t \gamma$$

$$\Rightarrow \ell' = \ell \gamma$$

Suppose that the tram is moving at three fifths of c relative to the rail, and that in the tram’s rest frame, 4 light metres (henceforth metres, m) of time pass between contacts of the point on the rim of the wheel with the rail. I calculate that 5m of time pass in the rail’s rest frame between these events. Multiplying the proper time 4m by

$$sinh\left(artanh\; \beta \right)$$

I find that, in the rail’s rest frame, 3m of space separate each contact event from the next. Now I want to know how long this distance is in the tram’s rest frame. It’s the distance between two moving points (a constant distance apart in any individual frame), so it should be contracted. I divide 3m by the Lorentz factor:

$$\frac{3m}{cosh\left(artanh\; \beta \right)} = 2.4m$$

But now I think: the distance of the marks in either frame ought to be the circumference of the wheel in that frame.

In the tram’s rest frame, the length of ground that passes under the wheel equals twice pi times the radius of the wheel in the tram’s rest frame, supposing the wheel to be circular in this frame. From this I calculate the radius, which will also be the semimajor axis of the ellipse-shaped wheel of the tram in the rail’s rest frame, since the wheel is only contracted in the direction of the tram’s motion in that frame.

$$\\2.4 = 2 \pi r \\\Rightarrow r = a = 2.4/2\pi = 1.2/\pi \approx 0.38$$

The semiminor axis of the wheel in the rail’s rest frame will be this length divided by the Lorentz factor.

$$b = \frac{1.2}{\pi\; cosh\left(artanh\; \frac{3}{5} \right)} \approx 0.31$$

To find the perimeter of the ellipse, I used the equation

$$4a \;E\left(1 - \frac{b^{2}}{a^{2}} \right)$$

where E is the complete elliptic integral of the second kind. From Wolfram Alpha, I get:

$$4 \cdot \; 1.2\pi \cdot \; E\left(1 - {cosh^{-2}\left(artanh \frac{3}{5}\right) \right) = 2.1\dot{6}$$

What am I doing wrong; why do I get such a different value when I try to calculate the length in this way?

Also puzzling me: the distance between the events (and hence between the marks left on the rail) should be less in the tram's rest frame, shouldn't it, as it is in the first result, by analogy with the example given in several books I've seen of a muon created in the upper atmosphere which travels far enough to reach the ground, the distance to the ground being less in the muon's rest frame than it is in the rest frame of the Earth. But if the wheel is contracted to an ellipse in the rail's rest frame, wouldn't that mean the wheel has a shorter perimeter in that frame, and which suggests that the marks would be closer together in the rail's rest frame than in the tram's rest frame. What if there were two identical trams moving along the same rail and marking it in the same way, one moving at relativisticly significant speed along the rail, the other moving very slowly? Will the marks they make be equally spaced, and if they differ, whose will be closer together? What tenuous intuition I have has deserted me on this one.

Often when I've been confused, the solution turns out to depend on the relativity of simultaneity, but I'm not sure how that would feature here, given that each contact of the point on the rim with the rail is a spacetime cooincidence.

All suggestions welcome!

2. Jun 14, 2009

### JesseM

Yes, those numbers look right, although it's a little confusing to refer to the time it takes light to move a metre as just a "metre", better to use an abbreviation like lm for "light-metres".
I'm not familiar with the use of the sinh function in SR, but that number isn't right so you must be using the wrong equation. If it takes 4 lm of time between marks in the rail frame, then since the train is moving at 0.6c in this frame, the distance between marks in the rail frame would have to be (4 lm)*(0.6 m/lm) = 2.4 meters.
Should be 2.4/1.25, or 1.92 meters.
That should certainly be true in the tram frame where the wheels are round, though it obviously doesn't still work in the rail frame where the shape of the wheel is distorted into an oval shape with a smaller circumference (because the semi-major axis is the same as the radius of the wheel but the semi-minor axis is Lorentz-contracted), and yet the distance the wheel covers between marks is larger. To think about why it doesn't work in the rail frame, you should consider why it is that we believe the distance a rolling wheel travels in one rotation is equal to its circumference. I think one way of approaching it might be considering the limit of a "rolling" rigid equilateral polygon as the number of sides go to infinity--by "rolling" I mean that it balances on a corner, then falls flat on an edge, the raises up on the next corner (which does not change positions as long as it is touching the ground), then falls flat on the next edge, and so on. Since each corner lands one edge-length apart from the last one, it's not hard to see why after a full rotation the distance the polygon has traveled is equal to the sum of all its edges, i.e. its circumference. But now consider what happens if we no longer require that the polygon remain rigid, so the distance between a given pair of corners is constantly changing as the polygon rotates, and the speed at which each corner rotates is non-uniform too (both of which would be true for nearby points on a rotating wheel in SR). In this case, there's no longer any obvious reason why the distance the polygon moves in one rotation should be equal to its circumference at a single instant.

3. Jun 15, 2009

### Rasalhague

Okay.

Well, 4 lm of time was the interval between events in the tram's rest frame (where they happen in the same place, so this should also be the value of the invariant spacetime interval between the events):

So, by that logic, 2.4 m of space should be the distance between marks in the tram's rest frame, shouldn't it? Since the rail is moving in the tram's rest frame, it will be contracted in the tram's rest frame, so the marks should be further apart in the rail's rest frame, specifically 2.4 m * gamma = 4 sinh(artanh(3/5)) = 4 * beta * gamma = 4 * (3/5) * 1/(sqrt(1 - 9/25)) = 3 m.

(The sinh function just corresponds to beta * gamma, the Lorentz factor multiplied by the positive or negative speed difference between frames as a fraction of c[/c].)

In the rail's rest frame, where the events don't happen in the same place, their separation has a time component of 5 lm and a space component of 3 m. This corresponds to a timelike invariant spacetime interval of sqrt(|5^2 - 3^2|) = 4 lm, which matches the stated value.

Ah, thanks, I hadn't thought of that! Good explanation. I wonder how we'd go about calculating the distance rolled by such a non-rigid wheel. My intuition is telling me that ought to be even smaller than the distance rolled by a rigid ellipse, but it must be mistaken... The wheel is squashed and yet somehow making marks further apart than a physically identical wheel with negligible speed in the rail's rest frame? Hmm, the path of the rotating point on the wheel of the tram travelling at 0.6 c will still be a cycloid won't it, albeit contracted in the direction the tram is going?

Is it correct to say that the faster the tram travels in the rail's rest frame, the slower its wheels turn? I guess this accords with the fact that events ahead of it along the rail which are simultaneous in the rail's rest frame with a particular marking event will be further in the tram's past, the faster it goes (the faster the rail moves under it), although I find it hard to visualise.

4. Jun 15, 2009

### JesseM

Ah OK, I misread the original scenario, thought that was the time interval in the rail frame. So if the time is 5 lm in the rail frame, then since the tram is moving at 0.6c, the marks will be 3 m apart in the rail frame, and they'll be 2.4 m apart in the tram's frame, just as you said.
The calculation would be difficult, you might approach it by imagining that in the tram's rest frame the wheel has a bunch of spokes at equal angles, and then you could try to figure out how the spokes rotate as a function of time in the rail frame, which would also tell you how the angle between spokes varies at any given instant as a function of their position on the oval-shaped wheel--they'd presumably be more spread out in some regions and more bunched up in others. If you look at the point where the wheel is touching the ground, and look at the arc-length of the section of the wheel between the two nearest spokes on either side, then since every other pair of spokes should be about that far apart when they reach the point of being nearest the ground, I think you could multiply this arc-length by the total number of spokes and get the approximate distance the wheel moves in one full rotation in the rail frame, with the distance getting perfectly accurate in the limit as you let the number of spokes approach infinity. Definitely not 100% sure this is right, though!
Well, the speed of that point will vary somewhat because you have to do relativistic addition of velocities, so it might be more complex although it would probably still look cycloid-esque...
Well, let's work backwards from the fact that the marks are always 2.4 m apart in the tram's frame...in that case if the speed of the tram relative to the tracks was 0.8c instead of 0.6c, then the marks would be 2.4/0.6 = 4 m apart in the rail frame...so if the time between marks is T lm in the rail frame, we have (T lm)*(0.8 m/lm) = 4 m, giving T = 5 lm. So interestingly in this case the time for a rotation in the rail frame seems to be exactly the same as in the previous example where the tram was moving at 0.6c! This doesn't work for all speeds though, for example if the speed is 0.9c, the marks are 2.4/sqrt(1 - 0.9^2) = 5.506 m apart in the rail frame, so the time would be 5.506/0.9 = 6.12 lm, so here the rotation is slower. In general if the speed of the tram is V in units of m/lm, the marks are 2.4/sqrt(1 - V^2) in the rail frame, and the time between marks in the rail frame is 2.4/(V*sqrt(1 - V^2)) = 2.4/sqrt(V^2 - V^4). We can figure out where the minimum of this function is by taking the derivative and setting it equal to zero--from this derivative calculator I get that the first derivative of this function is $$\frac{-2.4*(V - 2V^3)}{(V^2 - V^4)^{3/2}}$$, so we have V - 2V^3 = 0, or 2V^2 = 1, or V = sqrt(1/2) = 0.707c, which would mean the fastest possible time between marks in the rail frame is 2.4/0.5 = 4.8 lm.