# shape of a relativistic wheel

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

Not open for further replies.
1. ### James RJust this guy, you know?Staff Member

Messages:
30,376
I looked at Tach's document. The guy seems to go off on a math frolic at a moment's notice, but without at any time engaging his brain.

What an unnecessarily complicated mess that document is! And wrong besides.

This is usual for Tach. Overcomplication. Introduction of irrelevancies. An over-reliance on math untempered by any physical intuition. Top it with a hefty does of blindness to his own errors and you get the mess that is Tach's attempts at physics.

3. ### przyksquishyValued Senior Member

Messages:
3,151
Personally I've gotten sick of Tach's blatant denial, deliberate obfuscation, and "demands" others do all his work for him. Since the weekend has started for me and I haven't gone skiing, here's how one can calculate the energies in the top and bottom halves of the wheel, using the approach I suggested in my [POST=2848590]very first post in this thread[/POST]: transforming the stress-energy tensor (cue charges that Tach had days to do this, and failed). Note that by "top" and "bottom" halves I mean what everyone with the possible exception of Tach (who seems to like keeping this ambiguous) means: everything above the axle, or above the strip of paper mentioned in the OP. So specifically, to calculate the energy of the top half, I'll just be integrating the energy density everywhere above the axle, which in the coordinates I'll be using, means integrating everywhere over z > 0.

Before tansforming the stress-energy tensor, we actually need something to transform, so...

Stress-energy tensor in the axle frame

I'll map this coordinate system with the "primed" coordinates x'[sup]0[/sup] = t', x'[sup]1[/sup] = x', and x'[sup]2[/sup] = z'. The idea is that the wheel will be moving in the x direction in the ground frame, and z (= z') is the vertical coordinate, with the axle at the origin, the ground at z = -R, and the top of the wheel at z = +R.

Let the wheel have a linear total (kinetic + rest) energy, or "relativistic mass" (since I'll be setting c = 1) density of $\rho$ and be rotating in the axle frame with angular frequency $\omega$, so the velocity of a point on edge of the wheel is given by $v_{x} = \omega z$, $v_{z} = -\omega x$.

The Stress-energy tensor is a unified way of describing energy, momentum, and momentum flux well suited to relativity (because the components of this tensor transform in a simple way under general Lorentz transformations). The $T^{00}$ component is the energy density in the frame under consideration, $T^{0i} = T^{i0}$ is linear momentum, and the components $T^{ij} = T^{ji}$ represent momentum flux. In our case, these components are given by
\begin{align} T'^{00} \,&=\, \rho \delta(r - R) \,, \\ T'^{10} \,&=\, \rho \omega z \delta(r - R) \,, \\ T'^{20} \,&=\, - \rho \omega x \delta(r - R) \,, \\ T'^{11} \,&=\, \rho \omega^{2} z^{2} \delta(r - R) \,, \\ T'^{21} \,&=\, - \rho \omega^{2} x z \delta(r - R) \,, \\ T'^{22} \,&=\, \rho \omega^{2} x^{2} \delta(r - R) \,, \end{align}​
where $r^{2} = x'^{2} + z'^{2}$, and the Dirac delta is included because I'm considering an infinitely thin wheel rim as an idealisation. Integrating $T'^{00}$, one can check that the total energy is $2 \pi R \rho$ as one would expect.

Transformation to ground frame

The energy density is the new component $T^{00}$ in the ground frame, related to the axle frame by the (inverse) Lorentz boost
\begin{align} t \,&=\, \gamma (t' + vx') \,, \\ x \,&=\, \gamma (x' + vt') \,, \\ z \,&=\, z' \,. \end{align}​
In general we could consider any boost of any velocity, though for the ground frame, $v = \omega R$.

In order to transform the stress-energy tensor, we have to transform the components, which transform according to $T^{\mu\nu} = \Lambda^{\mu}_{\;\rho} \Lambda^{\nu}_{\;\lambda} T'^{\rho\lambda}$ where ($\Lambda^{\mu}_{\;\rho}$ are the matrix elements of the Lorentz boost above), and re-express the resulting quantity in terms of the ground frame coordinates. In our case, re-expressing things in terms of the ground frame coordinates just means re-expressing $r$ in the Dirac delta as $r^{2} = \gamma^{2} (x - vt)^{2} + z^{2}$, And we're only interested in calculating the transformed energy density, given in terms of non-zero contributions by
$T^{00} \,=\, \bigl(\Lambda^{0}_{\;0}\bigr)^{2} T'^{00} \,+\, 2\Lambda^{0}_{\;1} \Lambda^{0}_{\;0} T'^{10} \,+\, \bigl(\Lambda^{0}_{\;1}\bigr)^{2} T'^{11} \,.$​
Substituting and simplifying, one finds
$T^{00} \,=\, \gamma^{2} \rho (1 + v \omega z)^{2} \delta(r - R)$​
with $r^{2} = \gamma^{2} (x - vt)^{2} + z^{2}$. Notice that we already see the asymmetry appear in the energy distribution, since $(1 + v \omega z)^{2}$ is higher in the top half of the wheel (where z is positive) than in the bottom half (z negative). But of course that won't be enough for Tach, so...

Energy of the upper and lower halves

This means calculating the inegral of $T^{00}$ seperately over z >0 (for the upper half) and z < 0 (for the lower half):
\begin{align} E_{\mathrm{upper}} \,&=\, \int_{-\infty}^{\infty} \mathrm{d}x \int_{\;0}^{\infty} \mathrm{d}z \, T^{00} \,, \\ E_{\mathrm{lower}} \,&=\, \int_{-\infty}^{\infty} \mathrm{d}x \int_{-\infty}^{0} \mathrm{d}z \, T^{00} \,. \end{align}​
The easiest way to calculate these is to express things in terms of $r$ and an angular variable $\theta$ defined such that
\begin{align} \gamma(x - vt) \,&=\, r \cos(\theta) \,, \\ z \,&=\, r \sin(\theta) \,. \end{align}​
Then the measure changes to $\mathrm{d}x \mathrm{d}z = \frac{r}{\gamma} \mathrm{d}r \mathrm{d} \theta$. For $E_{\mathrm{upper}}$, one finds
\begin{align} E_{\mathrm{upper}} \,&=\, \int_{0}^{\infty} \frac{r}{\gamma} \mathrm{d}r \int_{0}^{\pi} \mathrm{d}\theta \, T^{00} \\ \,&=\, \int_{0}^{\infty} \frac{r}{\gamma} \mathrm{d}r \int_{0}^{\pi} \mathrm{d}\theta \, \gamma^{2} \rho \, \bigl(1 \,+\, v \omega r \sin(\theta)\bigr)^{2} \, \delta(r - R) \\ \,&=\, \gamma R \rho \int_{0}^{\pi} \mathrm{d}\theta \, \bigl(1 \,+\, v \omega R \sin(\theta)\bigr)^{2} \,. \end{align}​
Getting a result is just a matter of evaluating the integral, which isn't difficult. In general,
$\int \mathrm{d}\theta \, \bigl( 1 \,+\, \alpha \sin(\theta) \bigr)^{2} \,=\, \Bigl( 1 \,+\, \frac{\alpha^{2}}{2} \Bigr) \theta \,-\, 2 \alpha \cos(\theta) \,-\, \frac{\alpha^{2}}{4} \sin(2\theta) \,+\, C \,.$​
Using this, the final result is
$E_{\mathrm{upper}} \,=\, \gamma \pi R \rho \Bigl( 1 \,+\, \frac{\omega^{4} R^{4}}{2} \Bigr) \,+\, 4 \gamma \omega^{2} R^{3} \rho \,,$​
where I've finally substituted in $v = \omega R$. Similarly, for the lower half, the result is
$E_{\mathrm{lower}} \,=\, \gamma \pi R \rho \Bigl( 1 \,+\, \frac{\omega^{4} R^{4}}{2} \Bigr) \,-\, 4 \gamma \omega^{2} R^{3} \rho \,.$​
Clearly, $E_{\mathrm{upper}} > E_{\mathrm{lower}}$ except in the case where $\omega = 0$, with a difference of $E_{\mathrm{upper}} - E_{\mathrm{lower} = 8 \gamma \omega^{2} R^{3}$.

All this just to show something that was pretty much self evident to everyone except Tach right from the beginning: in the ground frame, more of the energy associated with the rolling wheel is concentrated above the axle (and strip of paper mentioned in the OP) than below.

Cue Tach's last ditch attempt to redefine the problem...

Last edited: Nov 5, 2011

5. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,136
Impressive work, but Tach doesn't want to see the math. After tiring of his demands for it I gave him a Newtonian analysis. (As an aside, I'm curious whether some adjustments would equate our solutions sans the gamma component, but I don't have time until later)
His response was a casual dismissal with no other comment beyond referring to his previous logic.
I'm done with this thread. The consensus appears to be that the energy differential would exist even though the paradox would not occur, but no one really knows why or how the paradox would be avoided.

7. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
Unfortunately, Tach's blockade has prevented a proper exploration.
But like I said before, I suspect we need someone with a reasonable understanding of GR to address it.

8. ### przyksquishyValued Senior Member

Messages:
3,151
I'm not sure I see what the issue is: normally the point with a paradox is that someone presents an argument that leads to apparently paradoxical conclusions. "Resolving" the paradox just means examining the argument and identifying something wrong with it. For example, we resolve the twin paradox by pointing out that the relativistic time dilation formula doesn't necessarily apply in an accelerating frame. We don't normally find it necessary to give a full blown derivation of gravitational time dilation.

In the case of the rolling wheel, the paradox relies on an assumption that more energy and momentum will always lead to stronger gravitational effects. This is completely unsupported. There is no justification for this assumption. So the paradox raised in this thread has no foundation to it.

9. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
It hasn't been formally supported, but it's intuitively compellingly.
The upper half of the wheel has more mass, and more energy... so intuitively, it has a stronger gravitational field than the lower half.

The naive conclusion is that the resulting force on small test mass near the axle will have a vertical component.

10. ### przyksquishyValued Senior Member

Messages:
3,151
True, my calculation is pretty irrelevant anyway, since it's clear the real issue is that Tach isn't solving the same problem as the rest of us are in the first place. Posting a calculation will only mean he can't distract from this anymore with his favourite "hasn't done the calculation" charge.

In retrospect I'm having second thoughts about the validity of the calculation I posted. Some half-hearted looking into the stress energy tensor (of eg. a rotating disc) suggests that the momentum flux terms should actually be zero. This seems really odd to me, but it would simpify the final result I posted down to
\begin{align} E_{\mathrm{upper}} \,&=\, \gamma \pi R \rho \,+\, 4 \gamma \omega^{2} R^{3} \rho \,, \\ E_{\mathrm{lower}} \,&=\, \gamma \pi R \rho \,-\, 4 \gamma \omega^{2} R^{3} \rho \,. \end{align}​
This actually seems like a more sensible result, since the total energy $E_{\mathrm{upper}} + E_{\mathrm{lower}}$ in this case is $\gamma 2 \pi R \rho$, which is just the Lorentz transform of the total energy $2 \pi R \rho$ in the axle frame.

I may post an update to this later if I get this sorted out one way or the other. Of course, comments from anyone with more experience dealing with the relativistic stress energy tensor are always welcome.

11. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,136
My apologies, przyk. I guess my definition of a resolved paradox is stricter, but I'm certainly not going to ask that anyone carry out calculations beyond my understanding just to satisfy my own curiosity. This was really the only thing I saw you mention on the subject:
And, to me, this was akin to saying that a particular perpetual motion machine MUST have a flaw in it because otherwise the 1st and/or 2nd Laws of Thermodynamics would be violated...without actually discovering the flaw.

Anyway, don't take my comments as doubt (I'm certain you're right), but it's very surprising to me that there can be cases where more energy does not always translate to more gravitation.

12. ### Pincho PaxtonBannedBanned

Messages:
2,387
I know why, but I'm not allowed to say in here, and no matter which forum section I post it in, I don't have the maths to prove it. Let's just say that you should be thinking of an opposite effect, and leave it at that.

13. ### przyksquishyValued Senior Member

Messages:
3,151
I also posted some clarifications [POST=2848867]here[/POST] and [POST=2849970]here[/POST], in case they got drowned out in the flurry of "Tach vs. the World" posts. Basically to recap, there's a contrast between the way Newtonian gravity and GR are formulated with regard how easy it is to visualise how things will work out. Newtonian gravity is very direct: it says that if there's a mass $m$ somewhere in the universe, it will cause other masses in the universe to accelerate toward it at a rate given by $g = Gm/r^{2}$. From that it's easy to see that more mass means more acceleration. GR, by contrast, isn't formulated in such a direct way. In GR a measure of the curvature of spacetime is related to the whole stress-energy tensor via the Einstein field equation (EFE)
$R_{\mu\nu} \,-\, \frac{1}{2}R g_{\mu\nu} \,=\, 8 \pi G T_{\mu\nu} \,.$​
Then the curvature ends up causing geodesic trajectories in spacetime to converge (or in general even diverge). Unlike with Newtonian gravity, it's generally not easy to visualise, just from the form of the EFE, what the end result is going to be, partly because the EFE is a local, differential equation. There is no analogue of the Newtonian equation directly giving how the behaviour of a mass in one place will depend on the presence of another a long distance away.

We know that under non-relativistic conditions (so where relativistic mass is essentially the same as rest mass) and low mass density we recover Newtonian gravity as an approximation, but even that isn't obvious (it's not an especially difficult result to obtain, but I wouldn't call it trivial either). We also have a few exact solved cases and approximations, and those are our main source of intuition for GR predictions. That's why I say it's not clear a priori that more energy will always mean more gravitation. Momentum also appears in the stress-energy tensor, and there's no particular reason momentum shouldn't decrease gravitational influences, for instance.

Basically, the point I'm making is that because of the difficulty of visualising what you'll get from the EFE, you shouldn't approach GR with any expectations either way, except that under classical conditions you'll get behaviour approximately similar to Newtonian gravity.

Not exactly. A better analogy would be showing that electromagnetism and mechanics have associated energy conservation laws, and then concluding that a proposed perpetual motion machine based only on electromagnetism and mechanics is impossible.

For a more helpful answer that might help get my point of view across: suppose for a moment you did plan to derive the gravitational field produced by the rolling wheel from the Einstein field equation. The problem with this is that by far the easiest way to do this - and the way any sane physicist would want to do it - would be to calculate the gravitational field around the wheel in the frame co-moving with the axle, and then apply a boost to it. This is perfectly rigorous, because the EFE is formulated in a way that makes it manifestly coordinate system independent. Then you get a rolling wheel and a gravitational field around it that can't be "paradoxical" by definition, because you derived it by transforming the field around the wheel in the axle frame. Then what?

14. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
Is this scenario simple enough to properly analyse with GR?

Consider two infinite straight thin parallel rods of uniform density $\rho$ (mass per unit proper length), in constant relative inertial motion such that they remain parallel and separated by proper distance 2L.

Choose an inertial (x,y,t) reference frame S, such that:
• The upper rod is positioned at y=L, and has velocity v parallel to the x-axis.
• The lower rod is positioned at y=-L, and has velocity -v parallel to the x-axis.
S' (the upper rod rest frame) has velocity v parallel to the x-axis relative to S.
S'' (the lower rod rest frame) has velocity -v parallel to the x-axis relative to S.

In S, the density of both rods is $\gamma\rho$.
In S', the upper rod has density $\rho$, and the lower rod has density $\gamma'\rho$.
In S'', the lower rod has density $\rho$, and the upper rod has density $\gamma'\rho$.
Where $\gamma' = (1 + v^2/c^2)/\gamma^2$

So, three questions:
1. What is the acceleration of a test mass A at rest in S at (x,y,t)=(0,0,0)?
2. What is the acceleration of a test mass B with velocity v along the x-axis?
3. What is the acceleration of a test mass C with velocity -v along the x-axis?

By symmetry, A must have zero acceleration, and the proper accelerations of B and C should be equal and opposite.

Working in the rest frames of the test masses, the naive use of SR + newtonian gravity suggests paradoxical answers:
• In S, all test masses have zero acceleration
• In S', the test masses accelerate in the negative y direction
• In S'', the test masses accelerate in the positive y direction

Clearly, SR + newtonian gravity isn't an adequate model in this scenario.
So what does GR say?

Last edited: Nov 7, 2011
15. ### OnlyMeValued Senior Member

Messages:
3,914
Pete, where did the acceleration come from? The hypothetical was set up with velocities of v and -v for the two rods. Did I miss something simple..?

16. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
The acceleration is that of a test mass between the rods, due to the gravitational fields of the rods.

17. ### OnlyMeValued Senior Member

Messages:
3,914
O.K. Missed that.

18. ### TachBannedBanned

Messages:
5,265
I am still skiing, so I have borrowed someone else's computer for a fe minutes. The velocity profile is NOT static, it is clearly time-variable. See post 39 for the correct expressions of the speed in the ground frame.

19. ### TachBannedBanned

Messages:
5,265
Well you should try it sometimes, this is how I pay for my tuition (by teaching skiing and windsurfing to rich kids).

This is , of course, false. I have calculated the total energy and posted the results in post 215. I have shown that the energies oscillate and that their averages are equal. Please keep that in mind for what is to follow.

Here is where you introduce your error. I have shown earlier in post 109, that the correct parametrization is (I am using your notation):

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

The $\omega$ is KEY, it signifies the ROTATION. Dropping it is not an option since it results into your incorrect final result.
So, in reality:

$x=\gamma (x'+Vt')=\gamma(r cos(\omega t' +\theta)+Vt')$
$z=z'+r=r(1+sin(\omega t'+\theta)$

as shown early on, in post 39 where I used the above in order to calculate the components of the speed in the ground frame.

Now, you can do your integrals and, to your surprise, you will see the oscillating terms in $sin (\omega t')$ replacing your non-oscillating terms, exactly as I have already shown in post 215.

...except that the correct integral is:

$\int \mathrm{d}\theta \, \bigl( 1 \,+\, \alpha \sin(\theta + \omega t') \bigr)^{2} \$

No, the correct final result needs to have the oscillating terms, in powers of $sin (\omega t')$, as shown in post 215 earlier.

This was the mathematical rebuttal. Now, for the physics one:
Imagine that the experiment were done in a spaceship far away from any gravitational body. Then, an observer comoving with the axle will observe a test body as floating at the center of the wheel. By contrast, an observer on the spaceship ground, by virtue of your calculations, will observe the test body as being pulled up by virtue of your claim that $E_{upper}>E_{lower}$.

I expect a very vitriolic denial of any wrongdoing and an even more vitriolic set of personal attacks. Luckily, I won't see them since I will be skiing...

Last edited: Nov 7, 2011
20. ### TachBannedBanned

Messages:
5,265
It is wrong? We are talking about this document. The equations are numbered , so I would appreciate you pointing the exact equation you think it is wrong and explaining why do you think it is wrong. I decided to extend my skiing stay, so take your time, I will be away from a computer for the rest of next week.

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

Messages:
5,357
Assuming I recall correctly, if for example you were to calculate the gravitational field inside a ball of gas, even the pressure of the gas has an effect on the resulting spacetime curvature and gravity. Not very intuitive at all, most definitely a case of "shut up and calculate" before trying to draw or guesstimate any conclusions.

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

Messages:
30,376
Tach:

Pete already pointed out what is wrong with it.

Do you want to take the matter to a formal debate?

Do you actually assert that there is no Doppler shift from a moving mirror?

I'm not very inclined to waste my time chasing your mistakes, because I know that you won't appreciate my educating you. But if you agree to a Formal Debate on the topic, at least we'll have a permanent record showing some of your mistakes. After that, I probably won't bother much with you again.

23. ### przyksquishyValued Senior Member

Messages:
3,151
False. You haven't shown anything. You have merely asserted results with no derivation. In any case, as we keep telling you over and over again, and as you keep ignoring over and over again, from the little you say about what you're actually doing it's clear that your main problem is that you aren't even considering the same problem everyone else is:
This is not how the rest of us are defining the separation between the two halves. As I clearly stated in the introduction of the very post you are replying to, I am defining the top part of the wheel as the part above the axle and strip of paper, and in the region where z > 0, in the ground frame.

It's funny how you keep quietly ignoring this.

What are you talking about? I'm just calculating an integral over a 2D plane. There is no such thing as an "incorrect" parameterisation. There are only parameterisations that are more or less convenient for the particular integral I want to calculate. It is a standard mathematical procedure to reparameterise an integral under an arbitrary change of coordinates.

Your charge of an "error" here is completely made up. You wanting to do the calculation a particular way does not make every other way wrong.

No, it would signify that the coordinate system is rotating, and I have no particular reason to use a rotating coordinate system.

You haven't even understood the calculation I'm doing.

Nope, just a few minutes needed pointing out that your rebuttals are completely made up and unsupported. Apparently, you don't know how to reparameterise an integral.

Last edited: Nov 7, 2011