shape of a relativistic wheel

**Tach** · 11-01-11, 11:07 AM

Originally Posted by RJBeery

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?

Nope, you have the tools to calculate the total kinetic energy without "discarding" anything.

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

You are being insulting implying that I have given you insincere answers.

**RJBeery** · 11-01-11, 11:31 AM

The total KE of the wheel? Why would that do us any good? I'm confused, Tach, please walk me through the calculation for each semi-circle.

**Tach** · 11-01-11, 11:39 AM

Originally Posted by RJBeery

The total KE of the wheel? Why would that do us any good? I'm confused, Tach, please walk me through the calculation for each semi-circle.

No, I gave you all the tools, don't try to get me to do the calculations for you. It is time for you to make some effort and to stop demanding that someone else does the work for you. Only this way you'll learn physics.
I'll give you another hint, I have given you the speed components for each element of the circle. The infinitesimal kinetic energy for each element of the circle is $d_{KE}=dm_0c^2(\frac{1}{\sqrt{1-(v/c)^2}}-1)$ where $dm_0$ is the rest mass of the infinitesimal circle element. Time for you to roll your sleeves and start calculating.

**RJBeery** · 11-01-11, 11:50 AM

If you were being sincere you'd just explain what the hell it is you're calculating. I won't have time to dig into this until after work tonight.

What I can tell you without doing the calculations, though, is that there is no element on the bottom semi-circle whose velocity is greater than any element on the top semi-circle from the frame of the surface on which the circle is moving. Can you understand why I'm puzzled?

**Tach** · 11-01-11, 11:59 AM

Originally Posted by RJBeery

If you were being sincere you'd just explain what the hell it is you're calculating. I won't have time to dig into this until after work tonight.

Then stop whining and do so.

What I can tell you without doing the calculations, though, is that there is no element on the bottom semi-circle whose velocity is greater than any element on the top semi-circle from the frame of the surface on which the circle is moving. Can you understand why I'm puzzled?

You are considering only the x component of the wheel, you are not considering the y component. Calculate the speed from the two components I gave you.

**RJBeery** · 11-01-11, 12:29 PM

Originally Posted by Tach

You are considering only the x component of the wheel, you are not considering the y component. Calculate the speed from the two components I gave you.

No, for any given radius there is not an element in the bottom semi-circle whose velocity is greater than an element in the top semi-circle [from the ground frame]; this includes y components. This should be OBVIOUS, yet what you're saying is bizarre; if you won't show me your calculations I cannot understand what you're talking about.

**Tach** · 11-01-11, 12:37 PM

Originally Posted by RJBeery

No, for any given radius there is not an element in the bottom semi-circle whose velocity is greater than an element in the top semi-circle [from the ground frame]; this includes y components.

How do you know? You haven't done any calculations.

This should be OBVIOUS,

Well, this is what happens when you aren't doing any calculations, you take as "obvious" what is incorrect.

yet what you're saying is bizarre; if you won't show me your calculations I cannot understand what you're talking about.

Try doing the calculations yourself for a change, you have been given $v'_x$ and $v'_y$ , calculate $v'$ as a function of $\phi$ .

**Tach** · 11-01-11, 01:17 PM

Originally Posted by Pete

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.

This is an interesting approach. One can use the same approach to calculate the behavior of a PAIR of spokes that are at opposition (angle of separation is $\pi$ in the frame comoving with the axle). The question that needs answering is: do the two spokes form a "U" shape (as in the picture) or do they form an "S" shape?

**RJBeery** · 11-01-11, 05:21 PM

Tach I'm sorry man but I'm just not seeing how you're analyzing the KE of each semi-circle. Whatever you're doing is way different from my (Newtonian, although it shouldn't matter) approach, which would be to consider the wheel to be rotating around it's point of contact on the surface S, and calculating the KE in purely rotational terms. Since the CoM of the top half of the wheel is $(2*4R)/(3\pi)$ further from the axis of rotation than the CoM of the bottom half, it will have a higher moment of inertia and therefore higher KE.

**AlphaNumeric** · 11-01-11, 06:01 PM

Originally Posted by Tach

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.

As I said, my predictive text on my phone is rubbish.

Originally Posted by Tach

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

Yes, if it rolls things get more complicated.

Originally Posted by Tach

While your claim is true for spheres, it is false for circles.

If you're restricting the motion, rotation and observer all to be in the same plane then you're talking about spheres, just of the 1 dimensional kind. If you're considering a general point of view with relative motion not necessarily in the same plane as the rotation and orientation as the circle then no, things are not pleasant.

Originally Posted by Tach

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.

Obviously, as you're explicitly breaking the spherical symmetry to a discrete subgroup.

**Tach** · 11-01-11, 06:25 PM

Originally Posted by RJBeery

Tach I'm sorry man but I'm just not seeing how you're analyzing the KE of each semi-circle. Whatever you're doing is way different from my (Newtonian, although it shouldn't matter) approach, which would be to consider the wheel to be rotating around it's point of contact on the surface S, and calculating the KE in purely rotational terms. Since the CoM of the top half of the wheel is $(2*4R)/(3\pi)$ further from the axis of rotation than the CoM of the bottom half, it will have a higher moment of inertia and therefore higher KE.

Since you don't want to try to do any work and insist in talking (unfounded) generalities, I'll show you.
In the frame S comoving with the axle, the energy E and the momentum p are equal for the upper and lower half of the circle:

$E_{upper}=E_{lower}$
$p_{upper}=p_{lower}$

I think that you have no problem with that , right?

In any other frame S', moving with speed V wrt S', the energy-momentum transforms according to SR:

$E'=\gamma(E+pV)$

Therefore:

$E'_{upper}=\gamma(E_{upper}+p_{upper}V)$
$E'_{lower}=\gamma(E_{lower}+p_{lower}V)$

It follows immediately that :

$E'_{upper}=E'_{lower}$

Now, kinetic energy , KE , differs from total energy E by an invariant:

$KE=E-m_0c^2$

so:

$KE'_{upper}=KE'_{lower}$

...which is what I told you many posts ago. So, your intuition has been wrong all along.

**przyk** · 11-01-11, 08:25 PM

Originally Posted by Tach

Since you don't want to try to do any work and insist in talking (unfounded) generalities, I'll show you.
In the frame S comoving with the axle, the energy E and the momentum p are equal for the upper and lower half of the circle:

$E_{upper}=E_{lower}$
$p_{upper}=p_{lower}$

Er, no, this is wrong. For starters you should have $p_{\mathrm{upper}} = - p_{\mathrm{lower}}$ : the total momenta for the upper and lower halves of the wheel are directed in opposite directions and have opposite signs. Second, you can't apply the Lorentz transformation in the way you're doing it because it only applies to the total energy and momentum of invariantly defined systems. This isn't the case here: the Lorentz transformation does not map the upper half of the wheel in one frame uniquely to the upper half of the wheel in another. It doesn't even preserve the amount of matter in each half: in the frame in which the wheel is rolling and the ground is at rest, most of its mass is actually concentrated in the top half.

You can, however, transform the stress-energy tensor in the standard way, and then integrate over the top and bottom halves separately.

**przyk** · 11-01-11, 08:30 PM

Originally Posted by Tach

Kinetic energy is not relativistic mass.

Actually they're very closely related. Total (kinetic + rest) energy and the so-called "relativistic mass" are proportional to one another and differ only by a factor of c².

**przyk** · 11-01-11, 08:42 PM

Originally Posted by RJBeery

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?

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. Also the way general relativity is formulated means the relationship between gravity and energy/momentum isn't as straightforward as in Newtonian gravity, and it's absolutely not clear just from the Einstein field equation that more energy will always mean more gravity in all circumstances. Apparently, the rolling wheel is a counterexample.

**Tach** · 11-01-11, 09:39 PM

Originally Posted by przyk

Actually they're very closely related. Total (kinetic + rest) energy and the so-called "relativistic mass" are proportional to one another and differ only by a factor of c².

$KE=m_0c^2(\gamma-1)$
$m_{relativistic}=m_0 \gamma$
$\frac{KE}{m_{relativistic}}=c^2 \frac{\gamma-1}{\gamma}$

They obviously aren't the same thing, nor do they differ by a constant factor.
The main point was that gravity doesn't depend on relativistic mass.

**Tach** · 11-01-11, 09:50 PM

Originally Posted by przyk

Er, no, this is wrong. For starters you should have $p_{\mathrm{upper}} = - p_{\mathrm{lower}}$ : the total momenta for the upper and lower halves of the wheel are directed in opposite directions and have opposite signs.

That is a good point, I did this approach as a (mistaken) backup since I couldn't get RJBerry to calculate the total energy of the upper and lower halves of the circle via a simple integral.

It doesn't even preserve the amount of matter in each half

What makes you say that? The wheel is rigid, so why would a frame change move matter between the two halves? This is not a system of loose particles.

**~~Reiku~~** · 11-01-11, 10:24 PM

Originally Posted by Tach

That is a good point, I did this approach as a (mistaken) backup since I couldn't get RJBerry to calculate the total energy of the upper and lower halves of the circle via a simple integral.

What makes you say that? The wheel is rigid, so why would a frame change move matter between the two halves? This is not a system of loose particles.

Ah, passing the buck because someone wouldn't answer your question.

**Pete** · 11-01-11, 10:34 PM

Originally Posted by Tach

This is an interesting approach. One can use the same approach to calculate the behavior of a PAIR of spokes that are at opposition (angle of separation is $\pi$ in the frame comoving with the axle). The question that needs answering is: do the two spokes form a "U" shape (as in the picture) or do they form an "S" shape?

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.

**przyk** · 11-01-11, 10:37 PM

Originally Posted by Tach

$KE=m_0c^2(\gamma-1)$
$m_{relativistic}=m_0 \gamma$
$\frac{KE}{m_{relativistic}}=c^2 \frac{\gamma-1}{\gamma}$

They obviously aren't the same thing, nor do they differ by a constant factor.
The main point was that gravity doesn't depend on relativistic mass.

Er, did you read what I said? The total energy (which is really more relevant in GR than just the kinetic energy) and relativistic mass are proportional to one another. The total energy is given by $E = \gamma m_{0} c^{2}$ , and the relativistic mass by $m = \gamma m_{0}$ , so you find that total energy and relativistic mass are just related by $E = m c^{2}$ .

Originally Posted by Tach

That is a good point, I did this approach as a (mistaken) backup since I couldn't get RJBerry to calculate the total energy of the upper and lower halves of the circle via a simple integral.

First, don't you think it's a bit silly complaining that RJ didn't calculate the integral when you've just admitted you couldn't be bothered doing it yourself?

Second, it's clear even without doing any calculation that RJ is actually right. The reason is that the speeds of wheel elements in the top and bottom halves of the wheel aren't just different: they are consistently higher in the top half than in the bottom half. Also as stated previously, most of the mass is also concentrated in the top half, which will only further increase the energy in the top half.

What makes you say that? The wheel is rigid

I think one of the nice things illustrated by the relativistic rolling wheel is the generally well known fact that perfectly rigid systems are impossible in relativity.

so why would a frame change move matter between the two halves?

It's just a manifestation of the relativity of simultaneity effect: imagine two wheel elements A and B on opposite sides of the wheel in the frame comoving with the axle. When wheel element A crosses up into the top half, element B on the opposite side simultaneously crosses down into the bottom half. In the frame where the wheel is rolling and the ground is at rest, these events are no longer simultaneous. If you think about it or work it out, you find in that frame that element A crosses up into the top half before B crosses down into the bottom half. The result is that wheel elements tend to lag in the top half. This also plays well with the fact that you expect the mass density to be higher in the top half anyway, since it's moving faster and length contraction is locally more pronounced there.

**Tach** · 11-01-11, 10:51 PM

Originally Posted by przyk

Er, did you read what I said? The total energy (which is really more relevant in GR than just the kinetic energy) and relativistic mass are proportional to one another. The total energy is given by $E = \gamma m_{0} c^{2}$ , and the relativistic mass by $m = \gamma m_{0}$ , so you find that total energy and relativistic mass are just related by $E = m c^{2}$ .

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.

First, don't you think it's a bit silly complaining that RJ didn't calculate the integral when you've just admitted you couldn't be bothered doing it yourself?

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.

Also as stated previously, most of the mass is also concentrated in the top half, which will only further increase the energy in the top half.

So prove it, assertions don't count as proofs.

I think one of the nice things illustrated by the relativistic rolling wheel is the generally well known fact that perfectly rigid systems are impossible in relativity.

True but there is nothing about rigidity in this problem.

It's just a manifestation of the relativity of simultaneity effect: imagine two wheel elements A and B on opposite sides of the wheel in the frame comoving with the axle. When wheel element A crosses up into the top half, element B on the opposite side simultaneously crosses down into the bottom half. In the frame where the wheel is rolling and the ground is at rest, these events are no longer simultaneous. If you think about it or work it out, you find in that frame that element A crosses up into the top half before B crosses down into the bottom half. The result is that wheel elements tend to lag in the top half. This also plays well with the fact that you expect the mass density to be higher in the top half anyway, since it's moving faster and length contraction is locally more pronounced there.

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