shape of a relativistic wheel

Przyk said "actually, they are very closely related..." he didn't say your sentence was literally incorrect.
So his "correction" (which wasn't a correction) is correct, and needs no correction.
Correct?

Please Register or Log in to view the hidden image!

Consider the [thread=56359]track of a relativistic tank[/thread]...

The animations in that old thread are long gone, and I don't know if I kept them anywhere. Will dig and see...

 

Log in or Sign up to hide all adverts.

Shouldn't it be \(y_2'-R = +R.sin(\omega t')\)?

What sort of transform are you talking about? The formulas are dimensionally wrong. I cannot follow.

Shouldn't it be:

\(-R<x'_1<0\)
\(0<x'_2<R\)?

 

Log in or Sign up to hide all adverts.

No. As the y1 end rises, the y2 end falls.

Transforming from S' to S.
We start with expressions in S' giving y1' and y2' in terms of t', and transform to S to get expressions for y1 and y2 at t=0.

Yes, thanks. Fixing now.

 

Log in or Sign up to hide all adverts.

...meaning that it was irrelevant at best since it did not address the sentence it purported to address.

I have no idea what to look for, it is a long thread with lots of stories and no math.

 

The dimensions are still wrong

 

Well, I suspect it's still relevant to the argument you're having with RJBeery, but whatever. It's not worth splitting hairs.

Sorry, mistake in the tex.

sin\frac{-\omega^2Rx_1'}{c^2}
...needs brackets...
sin(\frac{-\omega^2Rx_1'}{c^2})

Fix'd, thanks.

 

Yeah, nasty thread.
Just consider the concept - a tank, proper length L driving along the road at speed v.

The tank's track runs forward and back along the length of the tank, around infinitesimal driving wheels at each end.

The bottom half of the track is in contact with the road.

In the tank frame, the top and bottom halves of the track are both moving at speed v.

The track's total proper length is \(2\gamma L\).

In the road frame, the bottom half of the track is at rest, and has proper length \(L/\gamma\).


Since track density is proportional to proper length, it is clear that in the road frame less than half the track's mass is in the bottom half.

 

Yes, looks right now.
Except that a problem surfaces: if you put up a "fence" of height R , more than half the spokes will be visible above the "fence". This doesn't seem right. This is where this thread started from.

 

WTF? :bugeye:
James, is that really necessary?
Tach can be an egotistical guy, but he is a capable mathematician as far as I can tell, at least as far as is necessary for SR.

I don't think that pointless antagonizing is going to accomplish anything good.

 

I didn't deny that. I just said they were "closely related" since the total energy and relativistic mass are proportional. I included the clarification "total (kinetic + rest)" for a reason you know. Also in picking this nit you are missing the wider context I posted this "correction" in: the total energy is relevant in GR (more so than just the kinetic energy), and since the total energy and relativistic mass are proportional to one another, the relativistic mass is just as relevant in GR as the total energy is. That is what I was correcting you on.

Also I'm not actually going to say you were completely wrong. It's a point of view issue that's a bit vague because exactly what "relevant" means is a bit vague. As I explained to RJ the gravitational field in GR doesn't depend on just the energy/relativistic mass in a given coordinate system, but also on momentum flux and density. This is why it's possible in GR, as you yourself said earlier, that a mass won't turn into a black hole just because it started moving or you changed frames. In that sense you were perfectly correct when you said the relativistic mass was irrelevant (to the general characteristics of the gravitational field around a dense mass). I'm just saying that you can't deny that the relativistic mass is relevant in GR in exactly the same way and to exactly the same extent as the total energy is, because apart from a factor of c[sup]2[/sup] they're exactly the same thing.

But it's clear that the energy in the top and bottom halves isn't the same. The total energy in each half is given by an integral over the perimeter which goes something like

\(E = \int \mathrm{d}l \, \rho \, k(v)\)​

where \(\rho\) is the linear mass density and \(k(v)\) is the energy per unit mass (either \(\gamma c^{2}\) or \((\gamma - 1) c^{2}\) depending on whether you're interested in the total or just the kinetic energy). Since both these functions are higher throughout the top half of the wheel than in the bottom half (because the mass density is higher, and because the speed is higher and k(v) is a monotonically increasing function in v), and the integrals are over domains of the same length (half the perimeter of the ellipse), the total energy of the top half is necessarily going to be higher than in the bottom half, and you've certainly made an error if you actually did the calculation and found otherwise.

I did, later in the very same post.

So why did you bring it up?

So specifically what's wrong with it? Also the bit about length contraction at the end wasn't the main explanation.

I had to explain a similar issue over a year ago to tsmid in some posts leading to and following [POST=2294731]this[/POST] one. The topic was different (the current in a wire loop), but the issues involved were similar. (Warning: tedium ensues, since the wire loop wasn't the only thing being discussed.)

Come again?


EDIT: 2000[sup]th[/sup] (non-cesspooled) post. Woo.

 

RJ did, this is how it all got started. But , of course, you didn't read the thread before pouncing with both feet (in your mouth).

Except that the claim (by RJ Berry) was that "kinetic energy and relativistic mass are the same thing". A simple math proof shows that they aren't.

So, you are back to stalking. Not very honorable for a moderator.

Nice try, there is no mention about any rigidity anywhere, actually, if you took the trouble to read, you would have understood that this a kinematics problem about how the wheel LOOKS. But, then of course, you needed to jump in with both feet....

Surely you need to stop stalking. You aren't contributing anything. Makes you look bad.

 

I wasn't wrong at all.

Ok,

It depends how you define "top" and "bottom", did you miss the fact that the midline separating the two "halves" is frame dependent, i.e. it is aberrated in the moving frame?

No, the midline does not coincide with the axis of symmetry in the moving frame. You keep missing this point.

 

Woo!
And a lovely mix of science and argumentation it is. The true essence of SciForums.

 

Tach, if you're going to put quote marks around claims that you think someone has made please be accurate. Also, you would do well to check your condescension in the future, particularly when I was (appropriately) open-mindedly questioning your logic while you were authoritatively making pronouncements that were, in fact, wrong.

I said it before: when your maturity level catches up with your math ability you'll make a fine contributor to this forum.

 

Ummm, that's pretty much the point przyk is making. And isn't it a point that you explicitly denied early in the thread?

The axle-frame midline is not the moving frame midline, but that's not relevant to the point to which you were responding, which is about the portion of the wheel above the axle in the moving frame.

 

It isn't. Look at his reaction.

No. There is no mention of aberration in the whole thread except in my latest posts. If you think otherwise, please point out the post.

I am glad that you agree with me on this one.

Sure it is, this is how you separate the "upper half" from the "lower half". The two points where the midline intersects the ellipse determine the limits of your two line integrals for calculating E or KE or whatever. The two "halves" aren't symmetrical in any of the moving frames.

 

Anyway...

Are we all agreed that in the moving frame, the portion of the wheel above the level of the axle has more kinetic energy than the portion of the wheel below the level of the axle?

 

Actually I think what James said is quite... lenient compared to some posters here.

 

...
And this of course cuts to my original question before getting sidetracked:

To which Przyk replied

But the implication here is that the momentum density and/or momentum flux must be reduced in the upper semi-circle in such a way that the total gravitation of each half remains the same? It still doesn't sit well with me.