# shape of a relativistic wheel

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This is what my writeup is about, does not include trivial cases.

Err no, you didn't understand a simple picture, see the angle of reflection equal to the angle of incidence? See the normal N to the object? No? Look again.

Unfortunately you can't read a very simple figure.

Yes, this is the case you failed to understand. Better luck next time.

Ha. My mistake, I should should have paid more attention.
Nevermind, I'll have a look and find your mistakes for you.

OK, easy mistake to make - in general, the angle of incidence is equal to the angle of reflection only in the rest frame of the mirror.

Formalism to come, if I have time.

This is of course false, the Terrell effect is just as present, you can't "wish it away".

No, the Terrell Penrose effect will still be present but it won't as precisely affect the shape of the track as a perfectly circular wheel does. AlphaNumeric as well as the videos show that when the spinning (not rolling) wheel is in the plane of the monitor it will not appear distorted. This will not be the case for the tank tread because of its non-symmetrical shape.

In other words, there is no proof.

Not if they aren't backed up by solid math. There isn't any backing up this claim.

Tach, if there is no meaning to proofs made of words, then what is the point of talking? If we cannot make sense of anything said without mathematical symbols, then how can we even begin to have a discussion on relativity? If you're correct then we have been spouting gibberish for the past 160 posts!

So, what gravitational effects will it have on a moving observer?

If the observer is moving at the same speed as the wheel's tangential velocity, the wheel will still look distorted.

Actually , the wheel looks distorted, I even gave you the precise equation, it is a family of ellipses moving at speed $$V$$. What is still in debate, it the form of distortion of the spokes, my equations clearly differe from Pete's. I think that I pointed out the source of error in Pete's equations.

Actually the movies show the distortion, all of them.

Yes, but they look differently distorted from different angles! Pete has pointed this out in a later post, quoted here at the end.

Maybe 60 years ago this view was valid, not today.

Hmmmmmmmm.. I don't even know enough about it to make a statement. The way I understand, it is now something like an 'electron cloud' ... too bad, because orbiting electrons would provide a possibility to test this experimentally.

The tank is not a practical example either, as excellerating the entire tank to c - (1 m/s) is itself problematic..., beyond our current technology.

Particle accellerators should be a good alternative, relativistic velocities are attainable, in that case. But I don' see how it fits the current hypothetical.

Particle accelerators accellerate particles and ions, bare nuclei.

Has there been any test involving whole atoms at relativistic velocities? Something like the LHC would likely strip the electrons away. Would seem to have had to happen in a linear accelerator. Even then I would expect the atom to become unstable, torn apart by the EM fields of the accelerator.

I suppose if there were relativistic tanks I would have known already.

I know that accelerators have to work with ions because they use strong electric fields to force these to move. Do they use completely bare nuclei? (and I am too sleepy now to check) Also, isn't it possible that a hydrogen atom with an electron attached will form when particles smash together? This would still conserve charge.

Can you show a fixed wheel which rotates in the plane of the monitor screen?

The animations given earlier are the best I can provide, but these not only account length contraction but other effects such as terrell rotation and ray tracing. This wheel looks 100% normal.

most of its mass is actually concentrated in the top half
I can disprove the redlined sentence , before I do it, I am asking you to provide your proof. This is the claim that you made in post 52 that it is at the origin of our disagreement. Provide your proof (since you made the claim) and I will provide my counter-proof.

How can you disprove it? We agree that visually, at least, most of the spokes will be above a line drawn in the rest frame. Mass will intuitively be found in the same place.

No, I've seen it before. For example, it was mentioned in [post=2385907]DRZion's first Sciforums thread[/post] in 2009.
My point is that you are now disagreeing with the same material you used to support your argument a year ago.

Yep, and this is why I started this thread too.
Chinglu brought up an objection which was just too much like my earlier objection - that there will be no way to determine the course of events using relativity. Minds work alike, and they can fail alike as well. I still have to make up my mind about it.

There isn't "more mass", The raytraced image of the spokes makes them look curved as in the picture.

A simple test proves this: place a strip of paper between the wheel and the light source or between the wheel and the eye and you'll see only half the spokes. No spokes have "wandered" in the half of the wheel peeking over the fence.

Even so, if only 4 spokes are present above the edge of the paper, they will still be bunched up!

I still don't see a contradiction.
• The videos are rendering visual effects, they are showing what a camera would actually record (except for light intensity changes and doppler shift color changes).
• In the camera rest frame the rolling wheel spokes are distorted while the spinning wheel spokes are straight..
• Remember Teller-Penrose - at a distance, the light-delay shape distortion of an approaching object is approximately opposite to the length-contraction shape distortion.
• With the spinning wheel in the video, there is no length contraction distortion, only light-delay distortion.

Did you notice that the spinning wheel distortion is reversed when looking at it from the other side?

This adds even more complexity to the issue. Because now the spokes will be shifting around as the observer moves around. Hence, no force is acting on the spinning wheel, but as the observer moves around the spokes will appear to migrate from one portion of the wheel to another, and quite likely shift the center mass of the wheel.

There are three questions:

1) How does the y-axis mass-energy distribution of the rolling wheel depend on our reference frame?
2) What frame-dependent gravitational effects are implied by the answer to 1)?

Question 1) is being thrashed out.
Tach says the answer should be no, but the calculations so far suggest (to me at least) that the answer is yes, that in the ground frame there is more mass-energy above the fence than below.

The second and third questions haven't been addressed with any kind of formalism.

I think that the reason Tach is disagreeing with przyk, RJBeery, and myself about the first question is that he thinks that would imply that the answer to the third question is yes.

But I'm sure we can all agree that if the answer to the third question turns out to be 'yes', then we're doing something wrong.

These are good questions! I am too tired to think clearly today, I will try tomorrow though.

Ha. My mistake, I should should have paid more attention.
Nevermind, I'll have a look and find your mistakes for you.

You will never become a physicist at this rate. If you are closed - minded and you can't follow someone else's bona fide proof, how will you ever learn?

I should hope not, or this medical degree would be a real waste of time!

I'm going to have to quit anyway. I have exams next week, and this has just been a big exercise in procrastination.

Also, it's beginning to feel like I'm kicking a puppy attacking my shoe...

So I give up, Tach. Just take the goddam shoe!

OK, easy mistake to make - in general, the angle of incidence is equal to the angle of reflection only in the rest frame of the mirror.
Last post for me...

More correctly, the angle of incidence equals the angle of reflection only in reference frames where the mirror's velocity is parallel to the reflecting surface.

I should hope not, or this medical degree would be a real waste of time!

Good, stick to medicine. Hopefully you are more open minded when it comes to treating people

OK, easy mistake to make - in general, the angle of incidence is equal to the angle of reflection only in the rest frame of the mirror.

Yes, this IS the frame used in the derivation. The ONLY frame.

Formalism to come, if I have time.

If you get a different result from zero shift, you got the wrong result.

Just one more... (I must be an addict)

OK, Tach.
In your diagram, you have the object moving at velocity V.
Since you said you're working in the object rest frame, I'll call that a typo and pretend that V is attached to the camera and the light source.
Also, you have the primary ray pointing at the camera instead of the source.
And you don't need separate scenarios for approaching and receding... it's OK to let $$\phi$$ vary from 0 to 2pi.

Now, $$\phi$$ is the angle between V and the primary ray, which I think you've correctly used in your formula for $$f_{mirror}$$.

But you have used $$\phi$$ again in your formula for $$f_{s'}$$.
This is not right.
In the $$f_{s'}$$ equation, you should be using $$\phi + 2\theta$$, the angle between V and the secondary ray.

But don't take my word for it... ask your professor. I'm sure they will be happy to help you with the doppler shift from a moving mirror.

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Easily. Can you stop dodging and provide a mathematical proof? Once you do it, I'll show you why it's false. Don't worry, unlike you I will deliver the math.

Just like you delivered the math on the proof you said you had that there was not more mass above the fence than below it?

Ha!

At this point, it is quite clear you don't even understand the issue you're discussing. For a start, you're confusing the actual coordinate locations of the spokes in different frames with the image that those spokes make on some distant detector or camera. Why you decided to introduce a distant camera at all is a bit of a mystery, since the issue is whether the wheel has more mass above the fence than below it - an issue you pretended you had solved earlier in the thread.

I am not going to touch this one with a ten foot pole.

No surprises there. True to your usual form, the more wrong you are proved, the more you back away. In the end, you resort to blatant denial. In the past, I've even seen you go back and edit previous posts you made to hide your errors. It's as if you can't admit your many mistakes even to yourself. They must be obliterated from your memory for the protection of your extra-large but oh-so-fragile ego.

It is very easy to prove, I can upload a file that shows all the details.

Yeah. Sure. 48 hours (or is it 72?) after you said you had a simple proof that the mass above the fence was always equal to the mass below the fence in all frames, we have yet to see anything from you.

I'm sure your imaginary "file" will be as forthcoming.

You could learn some manners, especially when you are asking for something.

The man who looks down like a self-made god on everybody else apparently has quite a different standard when it comes to the treatment he expects from others.

This must adversely affect your daily life. I wonder if you'll grow up at some point. I must say, the chances aren't looking good right now, from what I can see.

my equation.......graphed for t=constant. Let it be, I will not answer any more nonsense on this subject.

How convenient for you. When your own work refutes your arguments, suddenly you just want to forget the whole thing.

You will never become a physicist at this rate. If you are closed - minded and you can't follow someone else's bona fide proof, how will you ever learn?

Get out your Doppler-shifted mirror and take a good hard look at yourself in light of your comment, Tach.

Surely, hypocrisy is one of the most face-palming traits one can have.

Err, no. $$\rho \, k(v)$$ is NOT larger in the top half than in the bottom half . I'll give you a chance to remedy your proof, calculate the energy density $$k(v)$$.
It shouldn't be necessary - the energy density (per unit mass) is a strictly monotonically increasing function in v. If the speed v is higher, so is the energy. Your not going to deny that the speeds are consistently higher in the top half of the wheel, are you? :bugeye:

If you do it correctly, you will find out that it oscillates with time, sometimes it is larger for the upper half, sometimes it is higher for the lower half of the wheel.
No, here it's clear you're trying to redefine what the discussion is about again. The "top half" of the wheel is the part of it above the axle (and the strip of paper mentioned in the OP). Yes, this is "frame dependent". No, that doesn't make it "invalid".

It shouldn't be necessary - the energy density (per unit mass) is a strictly monotonically increasing function in v. If the speed v is higher, so is the energy.

Actually, it IS necessary. So, please do the calculation, I have given you (and JamesR, who's still contributing nothing, except stalking) all the tools to do this simple calculation.

Just one more... (I must be an addict)

OK, Tach.
In your diagram, you have the object moving at velocity V.
Since you said you're working in the object rest frame, I'll call that a typo and pretend that V is attached to the camera and the light source.

No, the speed V is attached to the object. "Working in the object rest frame" means using the relativistic Doppler equations expressed in angles measured in the rest frame of the object, once for the ray coming from the source, when the object acts as a target and the second time for the reflected ray , when the object acts as a source. You obviously did not take time to understand the writeup.

Also, you have the primary ray pointing at the camera instead of the source.

Yes, this is standard raytracing , if you don't understand the notions, just ask.

Now, $$\phi$$ is the angle between V and the primary ray, which I think you've correctly used in your formula for $$f_{mirror}$$.

Good, you got this part.

But you have used $$\phi$$ again in your formula for $$f_{s'}$$.
This is not right.
In the $$f_{s'}$$ equation, you should be using $$\phi + 2\theta$$, the angle between V and the secondary ray.

Nope, you did not get this part, see above. The angle between the speed V and the secondary ray and the angle between the speed V and the primary ray are the SAME ($$\phi$$) due to geometric symmetry. Definitely not the $$\phi + 2\theta$$ you are claiming. Claiming such a thing shows that you don't even understand the basics of the relativistic Doppler effect.

But don't take my word for it... ask your professor. I'm sure they will be happy to help you with the doppler shift from a moving mirror.

The error is yours. Again. If you don't understand things, just ask, I will be more than happy to explain.

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From post 127:

"So, the only valid approach left is to integrate $$\gamma=\frac{1}{\sqrt {1-(v/c)^2}}$$ over the ever varying "halves" of the ellipse. In another post I showed the functions $$v_x=v_x(V,\phi,t)$$ and $$v_y=v_y(V,\phi,t)$$. If you do the calculations, the integrand has the very nice expression $$\gamma=\frac{1+(V/c)^2 cos (\omega t+\phi)}{1-(V/c)^2}$$. "

Now, all that is left is finding out the integration limits. Do you think you can do that , James? All by yourself? Instead of wasting your life stalking, try to roll up your sleeves and do some science for a change?

So, James, 24 hours after you were given all the tools to calculate a simple thing, you are still not able to perform the calculation, yet you have the time to stalk all the posts for imagined errors. You know, if you used a fraction of the time you are using for stalking and trolling to make the calculation instead you would have found a very interesting fact:

The total energies for the halves of the wheel are:

$$E_1=\frac{1}{1-(V/c)^2}(\pi-2(V/c)^2 sin(\omega t))$$
and
$$E_2=\frac{1}{1-(V/c)^2}(\pi+2(V/c)^2 sin(\omega t))$$

So, contrary to all claims by parties that did lift a finger to do any calculations, waved their arms and flapped their wings, it turns out that:

$$E_1>E_2$$ for $$(2k+1)\pi<\omega t<2k \pi$$
$$E_1<E_2$$ for $$2k \pi<\omega t<(2k+1) \pi$$

On average, the energies are the same:

$$<E_1>=<E_2>$$

The same goes for the energy density, contrary to all unsubstantiated claims that przyk has been making. Had he done the calculations instead of waving his arms , he would have found out the above.

JamesR to Tach said:
In the past, I've even seen you go back and edit previous posts you made to hide your errors.
Searching for "Last Edited By Tach" in this thread: 31 results

What's quite interesting is to note when the edit times are made after there are follow-up comments by other posters.

Tach said:
Actually, it IS necessary. So, please do the calculation, I have given you (and JamesR, who's still contributing nothing, except stalking) all the tools to do this simple calculation.
The last time in this very thread you made this "all the tools to do it yourself" comment you were shown to have done the calculation wrong.

Just like you delivered the math on the proof you said you had that there was not more mass above the fence than below it?

You really, really need to stop stalking. If you used a fraction of the time you are using stalking my posts for doing some real science, you would be learning new things.

Yeah. Sure. 48 hours (or is it 72?) after you said you had a simple proof that the mass above the fence was always equal to the mass below the fence in all frames, we have yet to see anything from you.

I gave you all the tools to do the calculations yourself, had you spent 5 minutes, much less than it took you to compose this post, you would have gotten the result.

I'm sure your imaginary "file" will be as forthcoming.

Since you couldn't do a simple calculation even in the context of all the materials being done for you, you have the results posted for you, ready made. I am sure that the next question coming from you is going to be : "how did you get that?".

Get out your Doppler-shifted mirror and take a good hard look at yourself in light of your comment, Tach.

It isn't my fault that Pete managed to misunderstand every basic notion from the explanation. He imagined errors where there were none.

Surely, hypocrisy is one of the most face-palming traits one can have.

Stop stalking. Stop trolling. They are unbecoming traits for a moderator. If you can't stick to science , get a different job.

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Actually, it IS necessary.
No it's not. It is not always necessary to calculate two quantities to show that one is larger than the other. In this case it simply follows from the fact that energy increases strictly monotonically with speed.

So, please do the calculation, I have given you (and JamesR, who's still contributing nothing, except stalking) all the tools to do this simple calculation.
This calculation that's so simple you never do it?

Tach said:
It is very easy to prove, I can upload a file that shows all the details. Before I do, think about this The light from the source arrives at an object receding from it redshifted. The same light is reflected by the object approaching the camera, so it will be blueshifted by the same exact amount. Net effect: zero Doppler shift
BTW, Tach, this "zero Doppler shift" is only valid for mirrors. An object such as a colored, spoked wheel would indeed appear Doppler shifted. Your comment is therefore irrelevant and in the context of this thread, false.

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