# shape of a relativistic wheel

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

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1. As an "eclipse", no. As an "ellipse", yes:

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

It is very easy to prove the above.

Only if the wheel does NOT roll. If it rolls, it looks elliptical because the Terrell effect turns the circle sideways.

While your claim is true for spheres, it is false for circles. Even for a sphere, it is only true if the sphere has a regular, repeated texture, like the soccer ball. If the texture is irregular while the object still appears spherical, its texture will show distortion that is dependent on its rolling speed wrt the observer. The higher the speed, the more severe the distortion.

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3. Maybe, you would need to calculate the kinetic energy for each infinitesimal piece of the semi-circle and sum them up. What does this have anything to do with the fact that relativistic mass has nothing to do with gravitation?

Last edited: Nov 1, 2011

5. ### RJBeeryNatural PhilosopherValued Senior Member

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I don't think you'd need to do that. You agreed that $KE(\theta = 0) <> KE(\theta = \pi)$. It should be obvious that the KE of each half would be different without carrying out the calculations. Note that this is true from any frame OTHER than the co-moving translational one which sees the KE of both halves as equal.

It all comes back to the same point: how can a property which is relative to an observer (such as KE or relativistic mass) affect gravity? It seems to me that gravity would be an absolute property. If we allow a small object to pass by the spinning wheel's center of gravity are we suggesting that the direction toward which it is drawn (up or down) depends upon who is watching the experiment?

Absurd. I hope you can see my confusion.

7. No, I didn't. I said "Maybe". And I added that you need to do the calculations.

How is it "obvious" if you did not do any calculations?

Energy gravitates, this is well known. On the other hand the statement "relativistic mass affects gravity" is a fringe idea that you oft hear from people spouting "if I move very fast wrt a mass, it will collapse into a black hole". Not so.

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8. ### RJBeeryNatural PhilosopherValued Senior Member

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...
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So, you agree that the tangential speeds are unequal but you reserve the possibility that the KEs are the same, at least until you do the maths? Jesus man if you cannot "see" this without doing the calculations you have an unusual blindness.

Anyway, I pose the question to any other forum members here...

9. 1. The KE's are functions of the speeds at EACH point along the circumference.
2. The KE for each semi-circle is the sum of all the elementary KE's.
3. So, the fact that the speed at the top of the circle is different from the speed at the bottom (two isolated points) gives you NO indication as to how the energies for the semi-circles compare.
4. So, you don't have any escape but to do the calculations.
5. You can't make blind claims like the one you've been making.
6. I went ahead and I calculated the kinetic energies of the two semi-circles in an arbitrary frame. They are EQUAL. So, this settles the issue.

Last edited: Nov 1, 2011
10. ### DRZionTheoretical ExperimentalistValued Senior Member

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I agree with RJ, to me it always seemed that
e=mc^2
so, mass=energy, which should have a gravitational effect.
I have read before that this is not the case, and I do not like it. If mass really is energy, how come then doesn't it exert gravity, if it is really just mass. Are there two categories of mass now? Its completely unsatisfying, and I was always taught e=mc^2 is flawless.

How are asymmetrical spokes incorrect? I trust the links I posted before, not to draw out meaningless details. We both agree the spokes will at least appear to be bent out of shape, right?

Relativity is not just visual. There are also the time effects. For instance a train 2 km in length with fit inside a tunnel which is only 1 km long if it is going fast enough. The gates on the tunnel can even be shut at the same time (in the observer's reference frame), because it will not be simultaneous to the people on the train.

If a wheel is rotating and and is at the same time traveling close to the speed of light, it will appear distorted in that there will be more spokes one one half of the wheel. This is scenario is not physically impossible, unlike a wheel which rotates at relativistic velocities.

It will appear distorted (or will it?). If there are lights on the wheel then any sensor will tell you that this light comes from a distorted wheel. If there is any charge on the wheel, any charge sensor will tell you it comes from a distorted wheel. If there is any gravity exerted by the wheel, a gravimeter will tell you that it comes from a distorted wheel!! IE center of mass is not where the axel is!

So, I guess the only thing you can argue now is that the wheel will not look distorted. I don't see how this can be true...
:shrug:

11. ### AlexGLike nailing Jello to a treeValued Senior Member

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4,304
The m in the equation is rest mass. Relativistic mass does not enter into that equation. The e is not relativistic either. The kinetic energy of a moving object is relative to another object.

12. ### MasterovRegistered Senior Member

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728

What we'll see when this (revolving) wheel is fixed relative to the observer?
Plane of rotation of wheel have to on a monitor screen.
_____________________________

PS
My question demonstrate - you have wrong. Have not you?

Last edited: Nov 1, 2011
13. ### prometheusviva voce!Registered Senior Member

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This thread is a perfect demonstration of why relativistic mass is stupid.

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

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What does that equation tell you about the distribution of the spokes?

I tried approaching it by consider the y-value of point of intersection of a spoke with the rim, wanting to compare the times it spends above and below the axis.

Solving for t when $y-r=0$, and assuming $v=\omega r$ (ie using the ground rest frame for a wheel rolling without slipping), we find:
$t_{falling} = nT + \frac{T}{2} + \frac{\gamma vr}{c^2} \\ t_{rising} = nT - \frac{\gamma vr}{c^2} \\ n = 0,1,2,... T = \frac{\gamma 2\pi}{\omega} \mbox {(Time per revolution)}$

So it seems that the rim end of each spoke spends more time above the axel than below, which would be inconsistent with an even distribution.

Next post - two semi-handwaving ways to show that the spokes on the relativistic rolling wheel are curved in the ground rest frame.

Last edited: Nov 1, 2011
15. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
One
The angle of an inertial rigid rod is frame dependent.

Specifically, if a rod is parallel to the x axis and has speed u parallel to the y-axis in frame S, then in frame S' moving at speed v parallel to the x-axis relative to S, the angle $\theta$ of the rod with the horizontal is given by:
$tan \theta = \frac{\gamma u v}{c^2}$

Now, consider a spoke on our rolling wheel.
In the inertial frame of the axle, a spoke segment at distance r from the axle is horizontal and moving with speed $u=\omega r$ as it passes through y=R (the y-coordinate of the axle).

Transforming to the ground frame, we see that as a spoke segment passes through y=R, it forms an angle $\theta$ with the horizontal, which varies along the spoke:

$tan \theta = \frac{\gamma^2 \omega^2 Rr'}{c^2}$
Where r' is the distance from the axle to the spoke segment.

Therefore, the spoke is curved.

Two
Consider the events of the points on a spoke crossing the horizontal in the axle inertial frame.
In the axle inertial frame, the events are simultaneous.
In the ground rest frame, the events are displaced in the direction of the relative frame velocity.
Therefore, the events are not simultaneous in the ground frame.
Therefore, the spoke is curved.

(Yes, I know this one is far from rigorous. If any step looks dodgy, let me know and I'll fill in the blanks.)

I posted the correct equation of the spokes, all you have to do is to use a drawing software that draws them by making t=some constant.

Prove it.

Actually I have proven that the wheel looks elliptical. Did you read my posts?

17. What you really want is to fix t and to draw the resulting implicit curve in the (x,y) space. You will need some sort of drawing package (Mathematica will do the trick) or you can develop your own. The way to do it is to vary x, for example and to evaluate y:

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

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18. See here

19. ### RJBeeryNatural PhilosopherValued Senior Member

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I can't imagine how you're getting this result. The only frame in which the KE of the top and bottom semi-circles should be equal would be the axle's frame, correct? The average velocity of the bottom semi-circle is clearly less than the average velocity of the top one from the frame of the surface upon which is it rolling. You've agree with this, correct? How could you equate their KE's unless the mass is REDUCED in the top semi-circle? Perhaps you could share your calculations? :huh:

20. By calculating and by using proper physics. I can give you the tools such that you too, can calculate.

Math (and proper application of physics) says you are wrong.

"Clearly" to you. Not clearly for people that can actually do the calculations.

21. ### RJBeeryNatural PhilosopherValued Senior Member

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Excellent, please provide me the tools such that I may agree with your conclusions.

22. Gladly, the speed components in the moving frame are:

$v'_x=\frac{V(1+cos(\omega t +\phi))}{1+(V/c)^2 cos(\omega t +\phi)}$

$v'_y=-\frac{V sin(\omega t +\phi)}{\gamma (1+(V/c)^2 cos(\omega t +\phi))}$

Now, you have all the tools to calculate the KE for the two halves of the ellipse.

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

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Am I mistaken that we can discard rotational KE completely since we know they are equal, and we're only concerned with the differential possibly introduced by the translational velocities?

It's a sincere question, Tach. Please give an intellectually sincere response.