# shape of a relativistic wheel

Discussion in 'Physics & Math' started by DRZion, Oct 31, 2011.

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1. ### TachBannedBanned

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Nope, you have the tools to calculate the total kinetic energy without "discarding" anything.

You are being insulting implying that I have given you insincere answers.

3. ### RJBeeryNatural PhilosopherValued Senior Member

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The total KE of the wheel? Why would that do us any good? I'm confused, Tach, please walk me through the calculation for each semi-circle.

5. ### TachBannedBanned

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No, I gave you all the tools, don't try to get me to do the calculations for you. It is time for you to make some effort and to stop demanding that someone else does the work for you. Only this way you'll learn physics.
I'll give you another hint, I have given you the speed components for each element of the circle. The infinitesimal kinetic energy for each element of the circle is $d_{KE}=dm_0c^2(\frac{1}{\sqrt{1-(v/c)^2}}-1)$ where $dm_0$ is the rest mass of the infinitesimal circle element. Time for you to roll your sleeves and start calculating.

7. ### RJBeeryNatural PhilosopherValued Senior Member

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If you were being sincere you'd just explain what the hell it is you're calculating. I won't have time to dig into this until after work tonight.

What I can tell you without doing the calculations, though, is that there is no element on the bottom semi-circle whose velocity is greater than any element on the top semi-circle from the frame of the surface on which the circle is moving. Can you understand why I'm puzzled?

8. ### TachBannedBanned

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Then stop whining and do so.

You are considering only the x component of the wheel, you are not considering the y component. Calculate the speed from the two components I gave you.

9. ### RJBeeryNatural PhilosopherValued Senior Member

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No, for any given radius there is not an element in the bottom semi-circle whose velocity is greater than an element in the top semi-circle [from the ground frame]; this includes y components. This should be OBVIOUS, yet what you're saying is bizarre; if you won't show me your calculations I cannot understand what you're talking about.

10. ### TachBannedBanned

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How do you know? You haven't done any calculations.

Well, this is what happens when you aren't doing any calculations, you take as "obvious" what is incorrect.

Try doing the calculations yourself for a change, you have been given $v'_x$ and $v'_y$ , calculate $v'$ as a function of $\phi$.

Last edited: Nov 1, 2011
11. ### TachBannedBanned

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This is an interesting approach. One can use the same approach to calculate the behavior of a PAIR of spokes that are at opposition (angle of separation is $\pi$ in the frame comoving with the axle). The question that needs answering is: do the two spokes form a "U" shape (as in the picture) or do they form an "S" shape?

12. ### RJBeeryNatural PhilosopherValued Senior Member

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Tach I'm sorry man but I'm just not seeing how you're analyzing the KE of each semi-circle. Whatever you're doing is way different from my (Newtonian, although it shouldn't matter) approach, which would be to consider the wheel to be rotating around it's point of contact on the surface S, and calculating the KE in purely rotational terms. Since the CoM of the top half of the wheel is $(2*4R)/(3\pi)$ further from the axis of rotation than the CoM of the bottom half, it will have a higher moment of inertia and therefore higher KE.

13. ### AlphaNumericFully ionizedRegistered Senior Member

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As I said, my predictive text on my phone is rubbish.

Yes, if it rolls things get more complicated.

If you're restricting the motion, rotation and observer all to be in the same plane then you're talking about spheres, just of the 1 dimensional kind. If you're considering a general point of view with relative motion not necessarily in the same plane as the rotation and orientation as the circle then no, things are not pleasant.

Obviously, as you're explicitly breaking the spherical symmetry to a discrete subgroup.

14. ### TachBannedBanned

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Since you don't want to try to do any work and insist in talking (unfounded) generalities, I'll show you.
In the frame S comoving with the axle, the energy E and the momentum p are equal for the upper and lower half of the circle:

$E_{upper}=E_{lower}$
$p_{upper}=p_{lower}$

I think that you have no problem with that , right?

In any other frame S', moving with speed V wrt S', the energy-momentum transforms according to SR:

$E'=\gamma(E+pV)$

Therefore:

$E'_{upper}=\gamma(E_{upper}+p_{upper}V)$
$E'_{lower}=\gamma(E_{lower}+p_{lower}V)$

It follows immediately that :

$E'_{upper}=E'_{lower}$

Now, kinetic energy , KE , differs from total energy E by an invariant:

$KE=E-m_0c^2$

so:

$KE'_{upper}=KE'_{lower}$

...which is what I told you many posts ago. So, your intuition has been wrong all along.

Last edited: Nov 1, 2011
15. ### przyksquishyValued Senior Member

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Er, no, this is wrong. For starters you should have $p_{\mathrm{upper}} = - p_{\mathrm{lower}}$: the total momenta for the upper and lower halves of the wheel are directed in opposite directions and have opposite signs. Second, you can't apply the Lorentz transformation in the way you're doing it because it only applies to the total energy and momentum of invariantly defined systems. This isn't the case here: the Lorentz transformation does not map the upper half of the wheel in one frame uniquely to the upper half of the wheel in another. It doesn't even preserve the amount of matter in each half: in the frame in which the wheel is rolling and the ground is at rest, most of its mass is actually concentrated in the top half.

You can, however, transform the stress-energy tensor in the standard way, and then integrate over the top and bottom halves separately.

16. ### przyksquishyValued Senior Member

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Actually they're very closely related. Total (kinetic + rest) energy and the so-called "relativistic mass" are proportional to one another and differ only by a factor of c[sup]2[/sup].

17. ### przyksquishyValued Senior Member

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The short answer is that gravity doesn't just depend on energy. It depends on the whole stress-energy tensor, which also includes momentum density and momentum flux. Also the way general relativity is formulated means the relationship between gravity and energy/momentum isn't as straightforward as in Newtonian gravity, and it's absolutely not clear just from the Einstein field equation that more energy will always mean more gravity in all circumstances. Apparently, the rolling wheel is a counterexample.

18. ### TachBannedBanned

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$KE=m_0c^2(\gamma-1)$
$m_{relativistic}=m_0 \gamma$
$\frac{KE}{m_{relativistic}}=c^2 \frac{\gamma-1}{\gamma}$

They obviously aren't the same thing, nor do they differ by a constant factor.
The main point was that gravity doesn't depend on relativistic mass.

19. ### TachBannedBanned

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That is a good point, I did this approach as a (mistaken) backup since I couldn't get RJBerry to calculate the total energy of the upper and lower halves of the circle via a simple integral.

What makes you say that? The wheel is rigid, so why would a frame change move matter between the two halves? This is not a system of loose particles.

Last edited: Nov 2, 2011

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21. ### PeteIt's not rocket surgeryRegistered Senior Member

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Consider the worldlines of the rim ends of opposite spokes, horizontal at t'=0:

$x_1' = -R.cos(\omega t') \\ y_1'-R = R.sin(\omega t')$

$x2' = R.cos(\omega t') \\ y_2'-R = -R.sin(\omega t')$

Transforming and solving for y at t=0, we get:
$y_1-R = R sin(\frac{-\omega^2Rx_1'}{c^2}) \\ y_2-R = -R sin(\frac{-\omega^2Rx_2'}{c^2})$

I can't get an exact expression for x1' and x2' in terms of t, but I think it's clear that at t=0:
$-R < x_1' < 0 \\ 0 < x_2' < R$

This means that at t=0:
$y_1-R > 0 \\ y_2-R > 0$

And since the axis is at:
$y_0 - R = 0$
...the pair of spokes must be U-shaped.

Last edited: Nov 2, 2011
22. ### przyksquishyValued Senior Member

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3,203
Er, did you read what I said? The total energy (which is really more relevant in GR than just the kinetic energy) and relativistic mass are proportional to one another. The total energy is given by $E = \gamma m_{0} c^{2}$, and the relativistic mass by $m = \gamma m_{0}$, so you find that total energy and relativistic mass are just related by $E = m c^{2}$.

First, don't you think it's a bit silly complaining that RJ didn't calculate the integral when you've just admitted you couldn't be bothered doing it yourself?

Second, it's clear even without doing any calculation that RJ is actually right. The reason is that the speeds of wheel elements in the top and bottom halves of the wheel aren't just different: they are consistently higher in the top half than in the bottom half. Also as stated previously, most of the mass is also concentrated in the top half, which will only further increase the energy in the top half.

I think one of the nice things illustrated by the relativistic rolling wheel is the generally well known fact that perfectly rigid systems are impossible in relativity.

It's just a manifestation of the relativity of simultaneity effect: imagine two wheel elements A and B on opposite sides of the wheel in the frame comoving with the axle. When wheel element A crosses up into the top half, element B on the opposite side simultaneously crosses down into the bottom half. In the frame where the wheel is rolling and the ground is at rest, these events are no longer simultaneous. If you think about it or work it out, you find in that frame that element A crosses up into the top half before B crosses down into the bottom half. The result is that wheel elements tend to lag in the top half. This also plays well with the fact that you expect the mass density to be higher in the top half anyway, since it's moving faster and length contraction is locally more pronounced there.

23. ### TachBannedBanned

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Yes, I heard. The sentence that you attempted to correct was "Kinetic energy is not relativistic mass". So, I had to correct your correction, since kinetic energy is indeed not relativistic mass.

First off, I did it, so you are off base. I wasn't going to just write it down for him, I gave him the tools to write it himself.

So prove it, assertions don't count as proofs.

True but there is nothing about rigidity in this problem.

I don't think this is a valid explanation, especially the bit about length contraction generating higher density in the upper half.
A much simpler proof is that the midline in the ground frame is no longer horizontal but inclined (due to relativistic aberration), so it no longer coincides with the axis of symmetry of the ellipse.