shape of a relativistic wheel

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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.
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?
:p
I don't think this is a valid explanation, especially the bit about length contraction generating higher density in the upper half.

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

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

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

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


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

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


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:
$$-1 < x_1' < 0 \\
0 < x_2' < 1$$

Shouldn't it be:

$$-R<x'_1<0$$
$$0<x'_2<R$$?
 
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Shouldn't it be $$y_2'-R = +R.sin(\omega t')$$?
No. As the y1 end rises, the y2 end falls.
What sort of transform are you talking about? I cannot follow.
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.

Shouldn't it be:

$$-R<x'_1<0$$
$$0<x'_2<R$$?
Yes, thanks. Fixing now.
 
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)

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



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...

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

The dimensions are still wrong
 
...meaning that it was irrelevant at best since it did not address the sentence it purported to address.
Well, I suspect it's still relevant to the argument you're having with RJBeery, but whatever. It's not worth splitting hairs.

The dimensions are still wrong
Sorry, mistake in the tex.

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

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

As usual, this was another attempted ego trip by Tach, and as usual Tach got caught out again.

Nothing przyk said was incorrect.

And, of course, kinetic energy is one of the components of so-called "relativistic mass". Nobody claimed that relativistic mass is the same as kinetic energy. All przyk pointed out - correctly - was that the two are closely related.

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.

This is another usual Tach ploy. Almost invariably, he claims to be able to do various calculations and so on, but when pushed can never produce them.

True but there is nothing about rigidity in this problem.

Here Tach completely missed the point - again, not uncommon with him.

The mass distribution in the different halves of a rolling wheel is not equal in different frames, precisely because the wheel is not rigid.

Tach has been corrected again and he doesn't even realise it.

Maybe Tach should spend more time learning physics and less time preening himself.
 
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.

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.
 
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As usual, this was another attempted ego trip by Tach, and as usual Tach got caught out again.
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.
 
Yes, I heard. The sentence that you attempted to correct was "Kinetic energy is not relativistic mass".
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.

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.
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.

So prove it, assertions don't count as proofs.
I did, later in the very same post.

True but there is nothing about rigidity in this problem.
So why did you bring it up?

I don't think this is a valid explanation, especially the bit about length contraction generating higher density in the upper half.
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.)

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
Come again?


EDIT: 2000[sup]th[/sup] (non-cesspooled) post. Woo.
 
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Nobody claimed that relativistic mass is the same as kinetic energy.

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).


All przyk pointed out - correctly - was that the two are closely related.

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.





This is another usual Tach ploy. Almost invariably, he claims to be able to do various calculations and so on, but when pushed can never produce them.

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






The mass distribution in the different halves of a rolling wheel is not equal in different frames, precisely because the wheel is not rigid.

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....



Maybe Tach should spend more time learning physics and less time preening himself.

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

I wasn't wrong at all.


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.

Ok,

But it's clear that the energy in the top and bottom halves isn't the same.

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?


and the integrals are over domains of the same length (half the perimeter of the ellipse),

No, the midline does not coincide with the axis of symmetry in the moving frame. You keep missing this point.
 
Tach said:
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.
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.
 
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?
Ummm, that's pretty much the point przyk is making. And isn't it a point that you explicitly denied early in the thread?

No, the midline does not coincide with the axis of symmetry in the moving frame.
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.
 
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Ummm, that's pretty much the point przyk is making.

It isn't. Look at his reaction.

And isn't it a point that you explicitly denied early in the thread?


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.




The axle-frame midline is not the moving frame midline,

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


but that's not relevant to the point to which you were responding.

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?
 
Pete said:
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?
...
And this of course cuts to my original question before getting sidetracked:
RJBeery said:
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?
To which Przyk replied
Przyk said:
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.
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.
 
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