# shape of a relativistic wheel

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Really? You've invested all this time arguing about the mass distribution, and now you're suddenly just not interested?

Why?

You agreed that this is an accurate diagram of the rolling wheel in the ground frame at t=0, right?

Sigh. Because what you are looking at is the geometric locus of all points whose light transit time from the point on the object to the camera is equal within an error of less than half the shutter time. It is a rendering, no spokes have moved around, ok?

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I hesitate to start another sidetrack, but feel free to post your proof.

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,

Tach, the rim of the green wheel is moving at 0.93c relative to the camera. Figure out the scale.
Hint - this is a simulation, not a practical camera recording, and it's not real-time (unless those wheels are light-seconds wide)

Why do you refuse to do the calculation? Because it proves you wrong? The difference in light transit time for the two examples I gave you are $$1.6 . 10^{-10}$$ and $$1.6 . 10^{-11}$$. Find a camera with a shutter aperture of half that.

Sigh. Because what you are looking at is the geometric locus of all points whose light transit time from the point on the object to the camera is equal within an error of less than half the shutter time. It is a rendering, no spokes have moved around, ok?
No, Tach, it's a diagram graphing your equation for the spokes in the ground frame.
You agreed to this back in post 168.

Don't be confused.
spacetime.org movies = camera simulation.
quickmath diagram = your spoke equation.

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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,

Why do you refuse to do the calculation? Because it proves you wrong? The difference in light transit time for the two examples I gave you are $$1.6 . 10^{-10}$$ and $$1.6 . 10^{-11}$$. Find a camera with a shutter aperture of half that.
It's a simulation, Tach.
The wheels are light seconds wide, the camera is light seconds away.
Do the calculation.

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

The wheels are light seconds wide,

LOL

quickmath diagram = your spoke equation.

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

You could learn some manners, especially when you are asking for something.
Beams and specks spring to mind, but I apologize for the abruptness of my request.
What, you can't comprehend the idea of a simulation spanning light-seconds?

my equation.......graphed for t=constant.
Correct.
Here it is:

At t=0, your equation demonstrates that there are more spokes above y=R than below.

Let it be, I will not answer any more nonsense on this subject.
As you wish. The conclusion is clear.

Beams and specks spring to mind, but I apologize for the abruptness of my request.

What, you can't comprehend the idea of a simulation spanning light-seconds?

I do, I know a very good professor who, when confronted with such nonsense, answers: "if my grandmother had tires she would be a trolleybus".

Correct.
Here it is:

At t=0, your equation demonstrates that there are more spokes above y=R than below.

As you wish. The conclusion is clear.

No, there aren't, the raytracer program renders more spokes in the upper half, no spokes have really "migrated". It is easy to disprove that, put a strip of paper covering the lower half of the wheel and you'll only see half of the spokes. So, no spokes have migrated in the upper half. If you paid attention to one of my earlier posts to you , you would have thought about this. Physics always trumps nonsense.

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No, there aren't, the raytracer program renders more spokes in the upper half, no spokes have really "migrated".
Tach, we're not talking about a raytracer program here, this is a simple graphing of your equation.
It is easy to disprove that, put a strip of paper covering the lower half of the wheel and you'll only see half of the spokes.
Your handwaving will get you nowhere.
Physics always trumps nonsense.
Maths always trumps handwaving.

The equation of the spokes is:

$$\frac{\gamma(x-vt)}{y-r}=tan(\omega \gamma (t-vx/c^2)+\phi_i)$$

where $$\phi_i=i \frac{ 2 \pi}{8} , i=0,1,2,...7$$
Do you think your equation is nonsense?
This is its graphical representation with r = 2, v = 0.8c, t = 0:

Do you think this graphical representation is nonsense?
If so, then please graph it yourself and show the results.

Do you think this graphical representation is nonsense?
If so, then please graph it yourself and show the results.

So, you did not understand the argument. Forget about the renderer/ camera/ etc. Put a strip of paper covering the lower half of the wheel. How many spokes will you see: 4 , 5 or 6?

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,

You describe an object approaching the camera and receding from the lightsource. That's a window, not a mirror.

Try again, for a mirror that is not between the light source and the camera.

You describe an object approaching the camera and receding from the lightsource. That's a window, not a mirror.

Try again, for a mirror that is not between the light source and the camera.

No, it isn't worth it, you failed to recognize a moving mirror. This is specifically the case occurring in raytracers, object BETWEEN the light source and the camera (in general, the light source is at infinity).
I know what you are talking about, the case you are talking about does not exist in raytracers.

Tach said:
No, it isn't worth it, you failed to recognize a moving mirror. This is specifically the case occurring in raytracers, object BETWEEN the light source and the camera (in general, the light source is at infinity).
I know what you are talking about, the case you are talking about does not exist in raytracers.
Firstly, Tach, you're claim wasn't limited to raytracers:
Tach said:
I can prove this quite easily, there is no Doppler shift of a moving mirror.
Secondly, in your analysis the light passes though the object without changing direction, so of course there is no doppler shift, and the object is obviously not a mirror.

If light reflects off an object, then there is a doppler shift.
Try again - light source behind the camera, moving object in front of the camera.
Light reflects off the object onto the camera.
Easy.

Actually, I don't even know why I'm pushing this. Your claim is so ludicrously false that I shouldn't have taken the bait in the first place.

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So, you did not understand the argument. Forget about the renderer/ camera/ etc. Put a strip of paper covering the lower half of the wheel. How many spokes will you see: 4 , 5 or 6?

Your own equation shows that at t=0, there are 5 spokes above the paper.

What we see is a different question, addressed by the spacetime.org videos which we were discussing until you decided your grandma was a trolleybus.

Your own equation shows that at t=0, there are 5 spokes above the paper.

What we see is a different question, addressed by the spacetime.org videos which we were discussing until you decided your grandma was a trolleybus.

You are reverting to being an ass.

Count them, Tach:

Remember, this is your equation for the spokes at t=0.
No raytracing. The actual spoke positions at t=0.

Firstly, Tach, you're claim wasn't limited to raytracers

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

Secondly, in your analysis the light passes though the object without changing direction,

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.

so of course there is no doppler shift, and the object is obviously not a mirror.

Unfortunately you can't read a very simple figure.

If light reflects off an object, then there is a doppler shift.
Try again - light source behind the camera, moving object in front of the camera.
Light reflects off the object onto the camera.
Easy.

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

Count them, Tach:

Remember, this is your equation for the spokes at t=0.
No raytracing. The actual spoke positions at t=0.

No raytracing, just a piece of paper between you and the wheel. You are 40 years old, if there is any chance that you will become a physicist, you need to start trying to understand what is being explained to you. This starts with being willing to learn rather than being stuck on being right at any cost.

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