# shape of a relativistic wheel

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

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1. ### TachBannedBanned

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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?

Last edited: Nov 4, 2011

3. ### TachBannedBanned

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5,265
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.

5. ### PeteIt's not rocket surgeryRegistered Senior Member

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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.

Last edited: Nov 4, 2011

7. ### PeteIt's not rocket surgeryRegistered Senior Member

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It's a simulation, Tach.
The wheels are light seconds wide, the camera is light seconds away.
Do the calculation.

8. ### TachBannedBanned

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5,265
You could learn some manners, especially when you are asking for something.
See here.

LOL

9. ### TachBannedBanned

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

10. ### PeteIt's not rocket surgeryRegistered Senior Member

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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?

11. ### PeteIt's not rocket surgeryRegistered Senior Member

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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.

12. ### TachBannedBanned

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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".

13. ### TachBannedBanned

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5,265
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.

Last edited: Nov 4, 2011
14. ### PeteIt's not rocket surgeryRegistered Senior Member

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Tach, we're not talking about a raytracer program here, this is a simple graphing of your equation.
Your handwaving will get you nowhere.
Maths always trumps handwaving.

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.

15. ### TachBannedBanned

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5,265
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?

16. ### PeteIt's not rocket surgeryRegistered Senior Member

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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.

17. ### TachBannedBanned

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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.

18. ### PeteIt's not rocket surgeryRegistered Senior Member

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

Last edited: Nov 4, 2011
19. ### PeteIt's not rocket surgeryRegistered Senior Member

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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.

20. ### TachBannedBanned

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You are reverting to being an ass.

21. ### PeteIt's not rocket surgeryRegistered Senior Member

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Count them, Tach:

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

22. ### TachBannedBanned

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5,265
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.

23. ### TachBannedBanned

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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.