# I Questions about a spinning disk/ring and Lorentz boosts

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1. Jul 22, 2016

### Juxtaroberto

I'm trying to understand the relativistically spinning disk within the framework of SR (if that is even possible). I thought to first simplify the problem by considering a spinning ring/annulus, but I don't know if my analysis is correct.

I imagined a spinning ring of radius R, spinning at an angular velocity of ω, and I wrote down an expression for its 4-velocity. I then used u⋅u=-c^2 to normalize it. This gave me an expression for the Lorentz factor due to rotation, which I called γ_ω.

Can I then premultiply this 4-velocity by the Lorentz matrix (for linear motion in the x direction) to find the 4 velocity of an annulus spinning at angular velocity ω and moving at some speed v w.r.t some inertial observer? Or are there issues due to the fact that a rotating annulus is noninertial? The center of the annulus is inertial, right? So as long as I have the basic parametrization for a circle in polar coordinates, which parametrizes the circle with respect to the center, I should be fine, right?

I'm asking because I tried this, and then tried to get the 3-velocity from the 4-velocity by u_x/u_t=dx/dt, and in this expression both the γ_v and γ_ω terms cancel out, γ_v being the Lorentz factor coming from the linear motion w.r.t. to an inertial observer, i.e. the typical (1-v^2/c^2)^(-1/2). This seems wrong, because I expect there to be length contraction due to the spinning, but then further length contraction due to its linear motion w.r.t. an inertial observer. The dy/dt term contains the γ_v term only, and I can't figure out if that makes sense or not.

I'm working on this problem because I was trying to suss out how the geometry of the annulus would change, because if it is both spinning and moving linearly, then one side of the annulus will be moving forward more quickly than the opposite side w.r.t. to the inertial observer, so what would happen there? How would the Einstein velocity addition mess with the geometry of the annulus? Or, being a rigid body (I'm aware that the definition of rigidity needs to be amended in SR) would the annulus slow down to match the speed of its slowest part? Can anyone share any insight on this?

Last edited: Jul 22, 2016
2. Jul 22, 2016

### Simon Bridge

If you google for "relativistic wheels" you will get a lot of literature on the subject, as well as other peoples attempts at the same thing.
Of note: http://www.spacetimetravel.org/rad/rad.html
... and the Ehrenfest paradox.

3. Jul 22, 2016

### Staff: Mentor

It is. But it would help if you would post the actual formulas you came up with.

4. Jul 22, 2016

### Juxtaroberto

Alright. I just didn't want to make it look like a homework post.

For the 4-velocity of the circle/annulus, I came up with γ_ω*[c, -Rωsin(ωt), Rωcos(ωt), 0].

Using u⋅u=-c^2 I found that γ_ω=(1-[Rω/c]^2)^(-1/2). So far, this is just a spinning circle of radius R, centered at (0,0).

I then tried to perform a Lorentz boost by premultiplying this 4-vector by the Lorentz matrix. I thought this would work because when I was taking senior electrodynamics, we had a homework assignment where we had to derive the Einstein velocity addition formula via the matrix premultiplication of 4-vectors. Basically, what I did there is what I did here, the only difference is that all motion was assumed to be inertial in that homework assignment, whereas here I have a spinning circle.

The matrix I used is (I apologize, I'm not sure how to format matrices here):
[γ_v, -(γ_v)*v/c, 0, 0
-(γ_v)*v/c, γ_v, 0, 0
0, 0, 1, 0
0, 0, 0, 1]

Here, γ_v=(1-[v/c]^2)^(-1/2).

My goal was to first describe a spinning circle, and then view it from a frame of reference moving at velocity v in the x direction, thus the observer would see the spinning disk moving in the -x direction relative to him.

This yields the 4-vector: u=[(γ_ω)(γ_v)c+(γ_ω)(γ_v)*(v/c)*ωRsin(ωt), (γ_ω)(γ_v)v-(γ_ω)(γ_v)ωRsin(ωt), (γ_ω)*Rωcos(ωt), 0].

The order of the components above is such that u=[u_t, u_x, u_y, u_z].

So, then, to find dx/dt I divided (u_x)/(u_t). I found dy/dt similarly. Based on what I have, though, dx/dt loses dependence on γ_ω and γ_v, because those terms cancel out. I feel like they should be there, because the velocity of a point on the wheel around the wheel depends on how the wheel has been length contracted both due to rotation and due to linear motion w.r.t. the inertial observer, doesn't it?

5. Jul 22, 2016

### Staff: Mentor

You also have to transform $t$, since it appears in the arguments of the sine and cosine. In other words, those arguments need to be transformed as well; they will end up being functions of both $t'$ and $x'$ in the new frame.

That won't work as it stands because you haven't transformed $t$, as above. You have to transform the arguments of the sines and cosines, and then calculate $dx'/dt'$ and $dy'/dt'$ as you describe.

6. Jul 22, 2016

### Juxtaroberto

Do I use the Lorentz transformation formulas to do this? Meaning, t'=γ(t-vx/c^2), etc.? Which γ would this be, then?

7. Jul 22, 2016

### Staff: Mentor

Yes. You are transforming from one inertial frame to another; that means everything that is frame-dependent, which includes the coordinates themselves, needs to be transformed. The same Lorentz transformation applies to all of them.

8. Jul 23, 2016

### DrGreg

See our LaTeX Primer for all maths processing on this website.

9. Jul 23, 2016

### Juxtaroberto

So that means I replace t by what it equals in terms of t' and x'? Is it really as simple as that?

Meaning, take the first component of my 4-velocity. Does it change into (γ_ω)(γ_v)c+(γ_ω)(γ_v)*(v/c)*ωRsin(ω*(γ_v)[t'+vx'/c^2])?

That then leads me to another question: how do I check that this makes sense? What does this predict about a disk spinning at a relativistic velocity, then being made to move linearly with respect to an inertial observer, also at a relativistic velocity? What happens to the geometry of this object? Would one side appear to spin more quickly due to the Einstein velocity addition formula? Are there any simulations out there? I've been trying to find answers to this question but nowhere have I found both a wheel that's rotating and moving through space (I've seen it moving on the ground, but I'm looking for motion through empty space, not a wheel where one point is always stationary with respect to the ground).

10. Jul 23, 2016

### Juxtaroberto

Thank you, I'll definitely take a look at this.

11. Jul 23, 2016

### Staff: Mentor

Since that's what a Lorentz transformation does, yes.

Yes, except that I think there should be a minus sign on the $vx'/c^2$ term.

Since all you're doing is changing frames, no actual physical predictions will change. Changing frames doesn't change anything about the disk itself.

What I think you're really asking is what the disk's motion will look like in the new frame. To find that you have to work out the math. Intuition is probably not a good guide here; for example, your intuitive questions don't appear to take into account relativity of simultaneity, which is a key factor in this scenario. (It shows up, for example, in the fact that the arguments of the sines and cosines have both $t'$ and $x'$ in them in the new frame.)

12. Jul 23, 2016

### Juxtaroberto

Could you elaborate on this point a little further? How does the relativity of simultaneity affect the answers to the questions I'm making? Or, perhaps, how have I assumed simultaneity in the way I worded the question?

13. Jul 23, 2016

### Staff: Mentor

Let's go back to post #4, where you gave the 4-velocity of the annulus in the original frame:

First note that this is not the 4-velocity of every point on the annulus; it's only the 4-velocity of one particular point--the one that is at $(x, y) = (1, 0)$ at time $t = 0$. If we consider the point at, for example, $(x, y) = (0, 1)$ at time $t = 0$, the 4-velocity of this point must be $\gamma_\omega \left[ c, - R \omega \cos ( \omega t ), - R \omega \sin ( \omega t ) \right]$, because this point at time $t = 0$ will be moving to the left (the $-x$ direction), not up (the $+y$ direction). We can re-express this using standard trig formulas as $\gamma_\omega \left[ c, - R \omega \sin ( \omega t + \frac{\pi}{2} ), R \omega \cos ( \omega t + \frac{\pi}{2} ) \right]$. More generally, for a point on the annulus which is at angle $\theta$ at time $t = 0$ (where $\theta$ is measured counter-clockwise starting from $(x, y) = (1, 0)$), the 4-velocity will be

$$u = \gamma_\omega \left[ c, - R \omega \sin ( \omega t + \theta ), R \omega \cos ( \omega t + \theta ) \right]$$

Now, when we transform this into the new frame, we will obtain

$$u' = \gamma_\omega \gamma_v \left[ c + \frac{v}{c} R \omega \sin \left( \omega \gamma_v ( t' + \frac{v x'}{c^2} ) + \theta \right), - R \omega \sin \left( \omega \gamma_v ( t' + \frac{v x'}{c^2} ) + \theta \right) - v, R \omega \cos \left( \omega \gamma_v ( t' + \frac{v x'}{c^2} ) + \theta \right) \right]$$

What is the meaning of $\theta$ here? Is it the angular position of the chosen point on the annulus at $t' = 0$? No. At the very least, it is the angular position of the chosen point on the annulus at $t' + vx'/c^2 = 0$, or $t' = - vx'/c^2$. But this depends on $x'$ as well as $t'$, and $x'$ for a given point on the annulus is changing with $t'$, so figuring out at exactly which value of $(x', t')$ a given point on the annulus will be at angle $\theta$ is even more complicated. All of this is because of relativity of simultaneity: it makes the coordinate dependence of the 4-velocity of a given point on the annulus much more complicated than it is in the original frame.

To try to separate out the different coordinate dependencies, I would recommend using trig formulas for the sines and cosines of sums to convert each 4-velocity component's formula to a product of sines and cosines, where each factor is a sine or cosine of a function of either $t'$ alone or $x'$ alone. That will make it easier to see what things look like at a given instant of time in the new frame (for example $t' = 0$).

14. Jul 23, 2016

### Juxtaroberto

Oh wow, I hadn't even thought of that. Thanks!

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