# Discussion: Zero Doppler effect for reflected light from a rolling wheel

Discussion in 'Formal debates' started by James R, Nov 10, 2011.

1. ### EmilValued Senior Member

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This double effect it would have to appear gradually, depending on the quality of the mirror? http://www.delstar.com/mechanical-polishing.html
I believe that the same change in frequency for "mirror" and for Diffuse reflection

Also, I am unclear with the color of a moving object .
If it were illuminated by a light with one frequency, then the frequency is changed when reflected. Right.
But if it is illuminated with "white" light?

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3. ### AlexGLike nailing Jello to a treeValued Senior Member

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There is no 'white light'. White light is the combination of every colored wavelength.

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5. ### EmilValued Senior Member

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Yes, this is also my position. For that I used "...".

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7. ### PeteIt's not rocket surgeryModerator

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In the frame of the mirror, the incident and reflected frequencies are always the same.
Tach uses the angle-dependent formula to show that the result is the same regardless of the mirror-source direction.

The angle dependent doppler formulae Tach uses give:
1. The incident frequency in the rest frame of the mirror, given:
• the source frequency in the rest frame of the source,
• the angle in the mirror frame between the mirror-to-source displacement vector and the source velocity,
2. And the receiver frequency, given:
• the reflected frequency in the rest frame of the mirror,
• the angle in the mirror frame between the mirror-to-receiver displacement vector and the receiver velocity.
Note that the two formulae are reciprocal given the same angles. The angles are the same if the source and receiver are stationary relative to each other and the velocity vector is parallel to the mirror's surface.
And that is the situation Tach is insisting on.

Yeah, I was wrong about that part.

Rule 23 of Tachtical warfare:
If you find that you have given an incorrect solution to a problem, redefine the problem so that you are correct.

Works well with Rule 14:
Any statement or argument that apparently refutes your position is obviously nonsense and can be ignored.

And both are logical consequence of Rules 1 and 2:
1 - You are always right.
2 - If you are ever wrong, see Rule 1.

*Disclaimer*
The above 'rules' are satire - a humorous strawman, not to be taken too seriously.
This disclaimer brought to you by the Department of the Bleeding Obvious

Last edited: Nov 13, 2011
8. ### PeteIt's not rocket surgeryModerator

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It seems that Tach is now arguing that there is sometimes no doppler shift for light reflected off the rim of the rolling wheel.

He's kind of right - for source and receiver appropriately positioned, there will indeed be a single point on the surface of the wheel with no doppler shift.

9. ### PeteIt's not rocket surgeryModerator

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The 'finish' of a surface describes how good a mirror it is.
A 'glossy' finish will reflect more and absorb or scatter less.
A 'matt' finish will reflect less and absorb or scatter more.

A perfect matt surface of a single pure color will not reflect any light, and will absorb all incident frequencies except one.

The light scattered off such an object is always the same frequency, regardless of the motion of the illuminating source.

So doppler-shifted 'white light' could still be white-light, right?
(Within certain practical limitations)

10. ### RJBeeryNatural PhilosopherValued Senior Member

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The issue isn't whether there is a high polish to the mirror. The free electrons in a mirror allow ALL frequencies to be absorbed and re-emitted, and they could do this even with a rough surface...but electrons of most atoms are different and can only absorb/emit specific frequencies. Therefore if an object were a particular color and illuminated with another frequency it would actually appear black.

There would also be a lesser color change associated with the heating of these objects by this light source, but I'm less familiar with that and it also isn't "Dopplar related".

11. ### EmilValued Senior Member

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Yes.
But for moving receiver?
I understand the Doppler effect is because of the relative speed between the "source" (reflectance) and receiver.
For this single frequency, Doppler effect does not exist?
This is also my opinion.

12. ### EmilValued Senior Member

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I think in this case it is easier to use wave.
The model particles cause problems for the "mirror".
Namely: the direction of the photon re-issued, which here is very precise.

13. ### RJBeeryNatural PhilosopherValued Senior Member

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Actually...the particle model works just fine. Imagine the photon bouncing just like a ball off the mirror.

Using the ball analogy, I throw a ball to you with velocity V, and you swing a bat and hit it back with velocity -V. Now if I run towards you with velocity Y and throw the ball with velocity V, it reaches you with a velocity V+Y and returns to me at -(V+Y). However, I am still travelling at Y so from my perspective the ball returns with a velocity -(V+2Y). This is the mirror doubling the blueshift (in this case, Y is the blueshift).

Now you drop the bat and just catch the ball before throwing it back at -V each time. When I run at you and throw the ball it returns to me at -(V+Y). This is the colored object reflecting constant-frequency light, and the blueshifting only occurs once.

The analogy works but I'm not feeling very eloquent tonight. Perhaps a bit less bourbon..or perhaps a bit more!

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14. ### EmilValued Senior Member

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Cheers,
I looked much a source where the photon direction re issued depending on the direction photon absorbed, and I have not found.
(Not depending on the electron bombardment)

15. ### AlphaNumericFully ionizedModerator

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JamesR's posts in the debate thread seem to be mirroring many of my own thoughts, mostly about how there's clearly some break down of communication, either between Tach and everyone else or just between Tach's ears. I think we can nail this in the head even quicker, by considering the generalisation of my previous post.

Tach says in a general emitter/wheel/observer setup the rotation of the wheel has no effect on the frequencies. In other words you cannot tell whether a wheel is rotating or not by looking at it, in the absence of other information. Now this I would be inclined to agree with. So without loss of generality you can say "The wheel isn't spinning at all". Now you just have a circular mirror. Clearly it will give a Doppler shift to light due to its axle's relative motion to the emitter and observer, just like a normal object would (the mirror just happens to reflect all light). Tach can't have it both ways, if there's no shift in the axle's frame there'll be a shift in other frames.

So you can only measure translational motion, not rotational, from a perfect mirror by looking at the light bouncing off it in the absence of other information. However, the no slip condition explicitly links the translational motion and the rotational, ie $v = \omega R$. Thus if you can measure one you know the other and you can certainly measure the translational motion of a mirror in the right setup, like a speed camera.

I think that's what Tach has been referring to in his citations, people showing how the alteration of frequencies is independent of the rotational motion of a wheel just hanging in space. If some other property is rotation dependent then you might be able to measure that. In the rolling wheel example using the rotational invariance you cannot tell whether the wheel coming at you with speed v is rolling with angular velocity $\omega$ or whether it's just 'gliding' along the frictionless ground.

Tach's document is sufficient to prove that the rotational motion doesn't alter the frequency, because you boost into the wheel's axle frame, do Tach's calculations, and you're done. However, as I demonstrated in my last post, the boost and unboost will not undo one another, they compound one another!

A few questions remain. Does Tach understand this? Has he just been communicating his claims incorrectly? If so what's the roll (no pun intended) of the no slip condition in all of this, which explcitly links translational motion and rotational?

It's actually not a bad explanation (yes, yes, the world is going to end, I gave you a compliment).

16. ### hardaleeRegistered Senior Member

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To Cool Hand Luke: "What we have here is a failure to communicate."

That what this has turned into.

As an engineer, the first thing I do is to define the problem. That dosen't appear to have been done well enough in this debate.

17. ### Aqueous Idflat Earth skepticValued Senior Member

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Despite all that has been said in all three threads, there is very basic concept that has not been brought "to light". When the dust settles, I would like to offer it as the simplest proof that one of the arguments is flawed.

18. ### RJBeeryNatural PhilosopherValued Senior Member

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LOL!
Holy sh*t!
Perhaps I was wrong when I said Tach provided little value to the forum. We're all suffering a common tragedy and it's bringing us closer together!

19. ### EmilValued Senior Member

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A question.
1) Radar in the police car that is stationary. A car is coming at 60 km / h. Radar shows 60 km / h.
2) Radar in the police car which travel at 70 km / h. From the opposite direction a car is coming at 60 km / h.
Show radar 130 km / h?

20. ### RJBeeryNatural PhilosopherValued Senior Member

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Depending on the model, the radar gun is intelligent enough to correct for the motion of the patrol car. See here for more info.

Actually...now that the subject has been raised...perhaps Tach is right after all and now we know the true reason for purchasing chrome rims. Zero Doppler against a mirrored wheel means no more speeding tickets!

Last edited: Nov 13, 2011
21. ### hardaleeRegistered Senior Member

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Is the emission of the reflected photon instantaneous? I would think not. If it is not, a shift would happen, however small if the mirror was moving.

22. ### TrippyALEA IACTA ESTStaff Member

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This whole argument is absurd (I know I should be more polte, but I'm disinclined to be so.

It doesn't matter if the wheel is stationary (but rotating), or moving relative to the observer, any observer will see part of the wheel moving towards them, and part of the wheel moving away from them - the only thing that changes his how much of the wheel is doing what.

Here's a real world example:

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It's a doppler image of the sun.
Notice how the doppler image is red shifted on one side, and blue shifted on the other?

It doesn't even matter if we're observing a distant wheel (so that the light we're observing is effectively the integral over the entire surface) - if we observe, for example, a distant wheel being illuminated by monochromatic light, we would observe a spreading - the FWHH bandwidth, for example would be wider on the reflected source than a direct source. If we consider a distant wheel illuminated by a source with a thermalized continuum emission spectrum (eg an incandescent light bulb) illuminating a distant wheel, then we'd observe the peak of the spectrum to flatten, and the spectrum itself to widen. Finally, if we consider a distant wheel illuminated by a flat-white (mmm yes please) spectrum, we'd see an unexpected excess of light at either end, and a deficit of light in the middle.

The whole argument boils down to what is essentially a modern eample of Zeno's paradoxes, and if we apply the logic that has been espoused, wheels wouldn't work at all.

23. ### Neddy BateValued Senior Member

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Yes, that is what Tach seems to be doing. It is trivial to prove there is no Doppler shift for that special case, because the optical path length does not change over time.