# shape of a relativistic wheel

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

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1. ### przyksquishyValued Senior Member

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Well it's pretty easily understandable given that the whole stress-energy tensor goes into determining the gravitational field: pressure is just the diagonal components of the stress-energy tensor. I was actually puzzling over how this worked some time ago. The result is just that matter going either way through a surface make the same contribution to the momentum flux through that surface (the momentum flux has a velocity dependence that goes as $\sim v^{2}$), so the momentum flux can be non-zero even if the net flow of matter is zero.

3. ### TachBannedBanned

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You claimed that you read it and that you found out error(s). Could you please point them out? Thank you.

Yes, I would like to debate the document with you.

5. ### TachBannedBanned

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Post 215 shows the complete derivation. Do you ask for all the intermediate steps (including the trivial ones) in journal papers?

Let's try again:

$x'=r cos (\theta)$
$y'=r sin (\theta)$

is a STATIONARY circle in the frame comoving with the axle. NOT the one you want for this problem.

$x'=r cos (\theta+\omega t')$
$y'=r sin (\theta+ \omega t')$

is a ROTATING circle in the frame comoving with the axle. THIS is the circle you need for this problem.

It points out the error in your result. You should be getting time-varying results in place of your time-invarying results. This is why you are getting the unphysical answer you got.

Last edited: Nov 7, 2011

7. ### przyksquishyValued Senior Member

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If you were at all following anything I said, you would see that the main issue I have with your derivation is that you seem to be solving the wrong problem in the first place.

Again you show that you haven't even understood the calculation I'm doing. The coordinates $r$ and $\theta$ I introduced are just a general purpose coordinate system that simply make the calculation of a particular integral on a plane more convenient. They aren't the definition of any circle. Fixed $\theta$ doesn't represent a fixed point on the wheel, isn't intended to, and the way I'm using it, absolutely doesn't need to. Only $r = R$ coincides with the outline of the rolling wheel and even that's a matter of convenience rather than necessity.

Last edited: Nov 7, 2011
8. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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I decided to look it up real quick just to refresh my memory and be certain I'm not talking out of my arse. From Wikipedia:

So yes, not only does pressure affect the spacetime curvature and gravity, but tension does as well. I always viewed this type of effect as a simple consequence of the demand that the theory remain self-consistent for arbitrary coordinate systems and reference frames, kind of like how magnetism arises as a necessary consequence when an electric force in a "stationary frame" is viewed from a "moving frame".

The important point: One must be very careful when trying to intuit the workings of gravity, it's nowhere near as simple as the picture Newton put forth. Newton's gravitational laws are only valid in weak gravitational fields generated by masses moving much slower than light speed.

9. ### DRZionTheoretical ExperimentalistValued Senior Member

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I think the answer to 1. must be yes. The light distribution is.. and because gravity travels at the speed of light all the same effects should also apply.

I would hope so ehehehehe

My argument is that, because gravity field travel at the same speed as light (btw, has this been proven?), it should arrive at the observer just like light..

That has a nice ring to it

Hmm.

So, is it very naive to try to argue about what the gravitational effects of a relativistic rolling wheel may appear relative to a stationary observer without knowing the math? I guess any further arguments I make on this subject are more or less shots in the dark.. unless I learn the EFEs .
:roflmao:

I think I may have come up with an even simpler scenario. Just take a massive rod which travels at relativistic speeds skirting a small test particle.

If the mass acts as a gravitational emitter x or 2x in length will change the gravitational attraction of the particle exactly as the test particle passes the centerpoint of the rod. A longer rod of equivalent mass will cause less of a gravitational pull on the test particle at this point... I can prove it with a python program!!

10. ### OnlyMeValued Senior Member

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A bit controversial, but in essence no it has not been proven. There are a number of different conditions to consider. In practice for planetary systems while changes in a gravitational field are theoretically limited to c, the influence of the gravitational field of a star or planet is "essentially" toward the instantaneous position of the gravitational source(s) involved. A little more complicated than that...

For the hypothetical here involved, a propagation velocity of c is likely accurate as a relativistically rolling wheel must be gravitationally considered to be constantly accellerating. That acceleration would represent a change in the field that would propagate out at c. At least that is the way I read it.

See the following FAQ for a better description of the involved conditions and interpretations... http://johanw.home.xs4all.nl/PhysFAQ/Relativity/GR/grav_speed.html

The above link should not be taken as a definitive last word, only as a reference that was easily found via a Google search...

11. ### hardaleeRegistered Senior Member

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“ Originally Posted by DRZion
My argument is that, because gravity field travel at the same speed as light (btw, has this been proven?), it should arrive at the observer just like light.. ”

Gravity fields travel at the speed of light. The gravation has not yet been found but any other speed would mess up GR.

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

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See the scenario in [post=2851724]post 251[/post].

13. ### OnlyMeValued Senior Member

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First, you folks do present some interesting thought experiments!

Pete, in the first link above the post ends with, "So what does GR say?". Speaking in a general sense as to how the motion of the rod or rods and their gravitational fields affects a test mass or particle I am unsure. Just the mention of GR does bring up what may be a side issue involving the question of whether the rod or rods would have an associated frame-dragging effect and whether that would also be involved?

It would seem that were the rods long enough, the frame-dragging effect may even be the dominant "force" on the test mass or particle, at least during a potion of the total interaction.

Is this something that would have already been a consideration? Or does it complicate things even further?

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

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I don't know, but my uneducated suspicion is that frame dragging is necessary to resolve it.

15. ### OnlyMeValued Senior Member

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I think so, but would it not just contribute some element of added angular momentum to the test mass? Where the gravitational field of the rod would actually add an acceleration toward the rod's center of gravity?

16. ### James RJust this guy, you know?Staff Member

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Yes. I will do so in the Formal Debate.

Good.

Please see here to agree to the Formal Debate proposal that I have initiated:

[thread=110880]Proposal: There is no Doppler shift of light reflected from a moving mirror[/thread]

17. ### RJBeeryNatural PhilosopherValued Senior Member

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Oh snap the gauntlet has been thrown down.

I'm curious to see the debate because I hadn't considered the moving mirrors, just the moving wheel. Tach claims that neither will exhibit Doppler shift, of course...

18. ### TachBannedBanned

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Once again, JamesR, you claimed that you found errors in this file. Please point them out or admit that there are no errors.

19. ### RJBeeryNatural PhilosopherValued Senior Member

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Tach, it seems that if you accept his challenge he will be FORCED to either point out the errors or admit that they do not exist. Why not carry this conversation to the other thread?

20. ### James RJust this guy, you know?Staff Member

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Tach:

I have issued a challenge for a Formal Debate to you. I am willing to take your document apart, if necessary, in our Formal Debate.

The link is in post #273, above.

Seeing as you have ignored it, it looks a lot like you're wimping out at this stage. Is that correct?

21. ### TachBannedBanned

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Besides, the file in discussion is instrumental to this thread since, based on it, I am going to prove later that there is no relativistic Doppler effect for the light bouncing off the circumference of the rolling wheel. So, let's have the discussion here, no point in opening another thread, the discussion belongs here. Go ahead, show which equation is wrong and why.

Last edited: Nov 8, 2011
22. ### DRZionTheoretical ExperimentalistValued Senior Member

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I see what you are getting at here - gravitation exerted by a far away star should pull where the star was when its light was emitted, because both gravitation and light travel at c.

It seems like this settles the question through an a priori most people will agree with - gravitation is exerted by where a star appears to be. Anything else would be faster than light signalling.

Absolutely, this becomes clear in light of the above. But then, if you look at the video it wouldn't it mean that the center of mass of the rolling wheel is shifting, without any force causing it?

This is basically what Pete's concurrent thread is about, if I am not mistaken. But, approaching the issue from a gravitational perspective draws the attention more towards newtonian concepts of force and mass, which may or may not be the best approach.

This is great! It brings into picture the gravitational effects which we were discussing, and at the same time it brings to light the contradictions between the two frames without any confusion.

You should send it in a formal letter to a popular science magazine. It is simple and clear enough for anyone to grasp very quickly. I am very curious what the resolution will be. I will definitely have to read up on frame dragging if I am to keep up!

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

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You said you'd like to debate it, Tach...
That means a specific debate, just you and James, in a dedicated thread, with clear rules set down beforehand so neither you nor he can easily obfuscate or claim misunderstanding.

A discussion in this thread wouldn't be a debate... everyone can and will jump in, and it will become yet another tangled sidetrack (and no, doppler shift is not central to the thread at all. It's a sidetrack introduced by a throwaway remark.)

This is your golden opportunity to engage James in a one-on-one arena... why aren't you grabbing it with both hands?

Last edited: Nov 8, 2011