# shape of a relativistic wheel

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

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1. ### DRZionTheoretical ExperimentalistValued Senior Member

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To further pursue a topic from another thread -

Say there is a wheel, traveling in one direction through space very close to the speed of light. It is traveling, relative to an observer, just behind a very very, infinitely, long strip of paper. The strip of paper is positioned so that it covers half of the wheel from the axle down, hence only half of the spokes are visible and the wheel travels so that the observer will only see one half of it for an extended period of time.

Next to the observer is a very bright light source. This light source illuminates the strip of paper and the half of the wheel. But, the wheel is rotating so that it looks lorentz contracted:

So, how many spokes of the wheel does the observer see? How many are illuminated? Is there any reason why there shouldn't be 6 illuminated spokes poking from behind the strip of paper?

3. 4

4

Yep, the reason is that there are 4 spokes uncovered. The strip of paper appears rotated away from the direction of motion (check the Terrell-Penrose effect) , so ALL observers see the SAME number of spokes uncovered (4). Where are you going with this?

Last edited: Oct 31, 2011

5. ### DRZionTheoretical ExperimentalistValued Senior Member

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Oh wow.

TP effect:

this is earth-shattering

Because the wheel is travelling slower than light, light coming from some parts of the object will reach the observer at the same time as other parts. Warning, this video may give you vertigo:

Well, the strip of paper won't be moving so it won't be rotated.. another complaint I have is regarding the video above. In the video, the green wheel is the 'stationary' wheel. What would happen if half of this wheel was covered up? I would still see just 4 spokes? 2 spokes? 6 spokes? The video is supposed to account for Terrell-Penrose rotation, post #6.

I guess I am just trying to probe for problems with relativity.
Related question: in an uncovered wheel, how would this affect the gravitational effects of the wheel? Would more than half of the gravity be emanating from one side rather than the other?

Last edited: Oct 31, 2011

7. ### MasterovRegistered Senior Member

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728
What we'll see when this (revolving) wheel is fixed relative to the observer?

8. This is not how you do you it.
Huh? The wheel is round, so it is symmetric. The fact that the wheel appears to look like an ellipse has no effect.

9. Look at the green wheel.

10. ### MasterovRegistered Senior Member

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728
No-no. No from a side.

Plane of rotation of wheel have to on a monitor screen.

11. Why?

12. ### DRZionTheoretical ExperimentalistValued Senior Member

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1,046
Right, but I am not getting at the fact that it appears to look like an ellipse. I am referring to the part that the spokes are bent out of shape. Most of the spokes are on one side of the wheel, and they have mass, so one side of the wheel should be exerting more gravity. This would mean the center of mass shifts as the wheel accelerates, and not in the direction of acceleration.

13. The picture you posted is wrong, the spokes are distributed symmetrically.
The equation of the spokes is:

$\frac{\gamma(x-vt)}{y-r}=tan(\omega \gamma (t-vx/c^2)+\phi_i)$

where $\phi_i=i \frac{ 2 \pi}{8} , i=0,1,2,...7$

No, it doesn't, the picture showing the spokes displaced asymmetrically is wrong.
Even IF the picture was right (it isn't) this doesn't change anything, the picture is just a (not so good) SIMULATION of the PHOTOGRAPHIC process , it has nothing to do with the physical shape of the wheel, that stays symmetric, the spokes do not move from one half of the wheel to the other half.

Last edited: Nov 2, 2011
14. ### DRZionTheoretical ExperimentalistValued Senior Member

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Ah, math is a laugh. I'll get to it someday.

You contradict yourself and I never posted a picture with symmetrical spokes. What are you looking at?

So what it look like has no relevance to what it really is? Its crazy enough that the wheel will get distorted for observers and not the inhabitants; but you're saying it doesn't even distort in reality, just visually?

15. If you don't know math, you will never learn physics.

I didn't claim that you posted a picture with symmetrical spokes, quite the opposite, you posted an incorrect picture showing asymmetrical spokes, I am telling you this the second time so I am not contradicting myself.

Just visually. I do not understand the rest of the word salad you posted.

16. ### RJBeeryNatural PhilosopherValued Senior Member

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Tach, I think what DrZion is confused about (and something I've never been clear about either) is the fact that on a rotating wheel whose translational velocity is V the tangential point touching the surface has a velocity of zero relative to that surface while the very top of the wheel is moving at a substantially higher velocity than V. If we want to claim that actual length contraction is a popular misunderstanding of Relativity which does not actually exist we still must wrestle with the fact that the top half of the wheel should have a relativistic mass greater than the bottom half, correct?

17. Yes, the tangential speeds for $\phi=0$ and $\phi=\pi$ are unequal and NO, the "relativistic mass" DOES NOT intervene in ANY gravitational force or, for that reason, in ANYTHING to do with theory of gravitation. So, it is a non issue.

18. ### rpennerFully WiredValued Senior Member

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4,833
The wheel rotates rigidly, and at constant angular velocity about the hub. Every instance of the wheel is governed by r, the radius of the wheel, u, the velocity of the rim, and t, the time at the hub.
In the non-rotating inertial frame associated with the position and movement of the hub, every point on the wheel is described by its angular position, θ, and its distance from the hub, l. For points on the rim, l = r. For points on the k-th spoke, $\theta = \frac{2 \pi k}{n}$.

Putting this all together, we have the following family of worldlines:
$\textrm{Wheel} \left( r, u; \theta, \ell, t \right) = \begin{pmatrix} \ell \cos \left( \theta + \frac{ u t }{ r } \right) \\ \ell \sin \left( \theta + \frac{ u t }{ r } \right) \\ 0 \\ t \end{pmatrix}, \quad \textrm{Rim}\left( r, u ; \theta, t \right)= \textrm{Wheel} \left( r, u ; \theta, r, t \right), \quad \textrm{Spoke}\left( r, u, n ; k, \ell, t \right) = \textrm{Wheel} \left( r, u ; \frac{ 2 \pi k }{n}, \ell, t \right)$

We can speak of relative coordinates of θ, l and t:
$\Delta \textrm{Wheel} \left( r, u; \theta_1, \ell_1, t_1 ; \theta_0, \ell_0, t_0\right) = \begin{pmatrix} \ell_1 \cos \left( \theta_1 + \frac{ u t_1 }{ r } \right) - \ell_0 \cos \left( \theta_0 + \frac{ u t_0 }{ r } \right) \\ \ell_1 \sin \left( \theta_1 + \frac{ u t_1 }{ r } \right) - \ell_0 \sin \left( \theta_0 + \frac{ u t_0 }{ r } \right) \\ 0 \\ t_1 - t_0 \end{pmatrix}$ with associated Lorentz invariant: $c^2 (\Delta \tau)^2 = c^2 ( t_1 - t_0)^2 - \ell_0^2 - \ell_1^2 + 2 \ell_0 \ell_1 \cos \left( \theta_1 - \theta_0 + \frac{u}{r} \left( t_1 - t_0 \right) \right) = c^2(\Delta t)^2 - (\Delta \ell)^2 \cos^2 \frac{\Delta \theta + \frac{u}{r} \Delta t}{2} - (\Sigma \ell)^2 \sin^2 \frac{\Delta \theta + \frac{u}{r} \Delta t}{2}$

Lorentz boosting in the x-direction by the amount v, we get $\begin{pmatrix} x' \\ y' \\ z' \\ t' \end{pmatrix} = \textrm{Wheel}' \left( r, u; \theta, \ell, t \right) = \Lambda \textrm{Wheel}\left( r, u; \theta, \ell, t \right) = \begin{pmatrix} \left(\cosh \tanh^{-1} \frac{v}{c} \right) \ell \cos \left( \theta + \frac{ u t }{ r } \right) + \left(\sinh \tanh^{-1} \frac{v}{c} \right) c t \\ \ell \sin \left( \theta + \frac{ u t }{ r } \right) \\ 0 \\ \left(\cosh \tanh^{-1} \frac{v}{c} \right) t + \left(\sinh \tanh^{-1} \frac{v}{c} \right) \frac{\ell}{c} \cos \left( \theta + \frac{ u t }{ r } \right) \end{pmatrix}$ but this does not let you talk about the shape yet, since x', y' and z' are written in terms of planes which are slices of constant t not constant t'.
Solving $t' = \left(\cosh \tanh^{-1} \frac{v}{c} \right) t + \left(\sinh \tanh^{-1} \frac{v}{c} \right) \frac{\ell}{c} \cos \left( \theta + \frac{ u t }{ r } \right) = \frac{c^2 t + \ell v \cos \left( \theta + \frac{u t }{r} \right) }{c \sqrt{c^2-v^2}}$ for a general expression for t in terms of t' looks hopeless.

/// Ran out of time

19. The spoke equation is given in post 10. The rim equation is much nicer:

$\frac{(x'- V t')^2}{(r/\gamma)^2}+\frac{(y'-r)^2}{r^2}=1$

Last edited: Nov 1, 2011
20. ### RJBeeryNatural PhilosopherValued Senior Member

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To be clear, I'm not claiming that I know the answer one way or the other. Just the opposite, in fact: I'm claiming open-minded, curious ignorance. That being said, I find your assertion that relativistic mass does not affect gravity as being overly absolute. Is this well established?

http://arxiv.org/abs/gr-qc/9909014

21. Yes, this is basic GR.

I have no idea why you are citing this paper, it has nothing to do with the subject. Kinetic energy is not relativistic mass. Energy is known to gravitate.

Last edited: Nov 1, 2011
22. ### AlphaNumericFully ionizedRegistered Senior Member

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6,699
The wheel will not appear like an eclipse. There is a funny combination of light delay effects which h means the wheel will remain looking like a circle. Its named after Penrose and someone else. Length contraction happens but visually a second opposite effect conspires to cancel it in the case of circles and spheres, not anything else. Its to do with how the boost is like a complex rotation in time. Ill Google when I am not using a phone to post.

/edit

Terrel, that is the guy. Voice recognition Googling rocks, even if my predictive text is dumb as a post.

Last edited: Nov 1, 2011
23. ### RJBeeryNatural PhilosopherValued Senior Member

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Wouldn't the "ground frame" calculate a differential in kinetic energy between the wheel halves?