# Discussion: There is no Doppler shift off a matte wheel rolling between a source and the receiver

Discussion in 'Formal debates' started by RJBeery, Nov 19, 2011.

1. ### RJBeeryNatural PhilosopherValued Senior Member

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General discussion thread for the topic listed in the title, "There is no Doppler shift off a matte wheel rolling between a source and the receiver". This is a belief held by Tach and will be argued against by myself.

Last edited by a moderator: Nov 21, 2011

3. ### Neddy BateValued Senior Member

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Why not make the color of the wheel "matte white"? That way we can study a general case, instead of another special case.

5. ### RJBeeryNatural PhilosopherValued Senior Member

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Thanks...anyway as I mentioned the matte white wheel could indeed appear to be white regardless of the wheel's speed, given the proper light source. You have to think about what we mean by "white" though. Does it reflect UV and IR as would a mirror? Because that will determine what light source is needed. In other words, Doppler shift COULD occur but a human eye might be oblivious to it because the white wheel would continue to look white. Think of a matte wheel as a potential band-pass filter, whereas a mirror filters nothing.

Does that make sense?

[EDIT: below is my initial response to Neddy Bate posted in the debate thread, copied over so I can delete it from the other]

Last edited: Nov 20, 2011

7. ### Neddy BateValued Senior Member

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Yes that makes sense. I now agree with your choice of using a wheel that only reflects light rays which are monochromatic, (in your case green), according to the wheel's own reference frame.

8. ### TrippyALEA IACTA ESTStaff Member

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I disagree with this:

9. ### Neddy BateValued Senior Member

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I also disagree with the claim that a rolling wheel "would NOT show any Doppler shifting whatsoever," regardless of whether the wheel is mirrored or matte white. I only meant to agree with RJBeery's choice of using a monochromatic (green) wheel rather than a white wheel. I do think the light source should be the full EM spectrum, so that it can illuminate the green wheel as it rolls at any speed.

10. ### RJBeeryNatural PhilosopherValued Senior Member

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Well, only because it isn't really related to the debate topic, I presume I'm permitted to interject in this thread (unless Tach has a problem with it). I have to admit that I haven't been following everyone's arguments in the other thread, Trippy, but perhaps you could give me a summary (or at least a link to the best summarizing post in the other thread) of your objection; the reason is because I'm prepared to agree with and defend Tach's position* on the movement of a mirrored wheel - as much as it pains me - and I'm curious to know where everyone else's confusion lies, if it indeed exists at all.

* Please note that this only applies to a particular position; one which he PM'ed me as being the CORRECT position. I have since seen the discussion drift into other areas (mirrored rims, single source, rotating mirror, relative movement between camera and source, etc). Basically, the position I concur with is anything derived from the fact that a camera viewing the Universe through a mirror which is travelling on a plane parallel to its orientation would not detect any Doppler shifting of the Universe whatsoever.

11. ### TrippyALEA IACTA ESTStaff Member

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I'm sticking with the first part of this: http://sciforums.com/showpost.php?p=2857607&postcount=105
The bit before the addendum, and the more reading I do , the more convinced I become that I am correct.

12. ### RJBeeryNatural PhilosopherValued Senior Member

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In that post are you considering a mirrored rim or mirrored sides of a wheel? Those circles really confused the direction of that thread. Also, what do you mean by "simultaneously moving toward and away from the light source"?

13. ### TrippyALEA IACTA ESTStaff Member

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Rim.

Those circles were Tach's, as presented in the debate thread. Perhaps you should consider the significance of that, if the evidence Tach presented to support his case confused the direction of the discussion of his case.

Exactly what it sounds like.

Consider the scenario where the light source us to the right (or in the 3'oclock position relative to) of a stationary wheel (for simplicities sake) and the wheel is rotating clockwise about its axel. In the frame of the light source (or in the rest frame) the top of the wheel is moving towards the light source, and the bottom of the wheel is moving away from it, with the 3 oclock position neither moving towards or away from the observer.

Now consider exactly the same setup, only this time the axel has a translational velocity towards the light source such that $v<r\omega$. This instance there will be some point on the rim of the wheel that appears to be stationary, because the instaneous velocity is at a sufficient angle that its horizontal component is equal, but opposite to the motion of the wheel.

In the case that $v=r\omega$ (at which point we're dealing with what is physically equivalent to a wheel rolling without slip), that part of the wheel is located at the 6 o'clock position, and obviously for $v>r\omega$ all points on the wheel appear to have some velocity in the direction of the motion of the axel.

My contention was/is that irrespective of whether we're talking about specular reflection, diffuse reflection, or emission, then in the special case of a wheel rolling without slip, the part of the wheel that is in contact with the surface it is rolling along shows no doppler shift, but the rest of it shows red shift, or blue shift, depending on the relative direction of motion of the wheel.

I understand Tach's claim that he has made, and I even understand the physical principles of Tach's claim, however, I have yet to convince myself of the universality of it within specular reflection, let alone beyond that. And I will not be bullied into accepting it.

14. ### RJBeeryNatural PhilosopherValued Senior Member

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As suspected, this is a case of us arguing different scenarios. The rim would cause noticeable Doppler shift, whether matte or mirrored, because the reflective surface orientation is not parallel with its velocity (except at the single point which you correctly pointed out). Analyze the light path length when the wheel is closest to either the source or camera and compare it to the light path length when the wheel is equidistant from both. A change in light path length over time = Doppler shift.

15. ### RJBeeryNatural PhilosopherValued Senior Member

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Also, Trippy, I believe both sides concur that the debate is over, and the thread can be closed. The unstoppable force of truth has collided with Tach's immovable inability to accept it.

16. ### TrippyALEA IACTA ESTStaff Member

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Are we? I thought we were discussing the presence or absence of doppler shit of a matte painted wheel rolling between a light source and an observer without slipping?

17. ### RJBeeryNatural PhilosopherValued Senior Member

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Yes, but discussing the RIM or the SIDE of the wheel makes all the difference for the reason I stated (i.e. that determines whether or not the reflective surface orientation is parallel with its velocity). The rolling aspect of the wheel makes the SIDE of the wheel necessarily parallel with its velocity. And I was actually asking about your stance in the other thread, where mirrors were being discussed.

18. ### TrippyALEA IACTA ESTStaff Member

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Okay - I'm mildly confused now.

As I understood, and understand it, based on the discussions, and based on Tachs diagrams, and Tachs commentary on other diagrams, we were talking about this:

With a polished or painted surface in place of a tyre, rather than any properties of the hub, or spokes. Correct?

19. ### OnlyMeValued Senior Member

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A couple of threads back (I think) the example included a spoked wheel and involved light reflected from the tread surface. Tach challenged that but shifted the discussion to the edge of the rim of that tread surface.

The paper he linked to was actually using a flat mirror instead of a rolling wheel.

It has seemed to me that it has been very difficult to get everyone on the same page as to what surface was being discussed.

20. ### TrippyALEA IACTA ESTStaff Member

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Well, based on the diagram in the paper, and the diagram used in the debate, and the discussions on the diagrams in the other thread, not to mention the fact that Tach not once critiscized the placement of the mirror on the wheel, only relative positions, and wheel(s), camera(s), and light source(s), I had come to the conclusion that we were talking about the tread surface IE in this image:

Not the Hub.
Not the white-wall surface.
But the surface that comes into direct physical contact with the ground.

And I thought I had been fairly explicit, and fairly clear that:
A. That was what I was discussing. and
B. That that was what was being discussed.

Now, If I'm wrong on that, that's fine, I can handle that, mea culpa, but if that's the case, it's only because Tach has been unclear, or muddied the waters, or shifted the goal posts. And if that's the case, given some of the things he's had to say in the other thread he owes several people apologies.

Last edited: Nov 21, 2011
21. ### RJBeeryNatural PhilosopherValued Senior Member

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Agreed, OnlyMe! I've only made claims about the side of an essentially flat mirror or wheel. Trippy, I'll be happy to discuss anything with you but to this point I've not made any comments about the RIM of a wheel.

Actually in this debate I could've dismantled Tach in another way but I wasn't prepared for him to misinterpret the *@#\$(!! debate topic (which he wrote) in the way that he did. The fact that $f_0$=$f_{s'}$ is only true because I chose the point where the angle of incidence equals the angle of reflection; this is required for mirrors but not for matte surfaces. I didn't think it mattered though because it sufficed to show that $f_{wheel}$ didn't match either of them, which is all I cared about proving. I could easily redo the analysis where all 3 values differ.

22. ### RJBeeryNatural PhilosopherValued Senior Member

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I don't think you're wrong at all. In my heart I think Tach took his initial result which he pilfered from A. Sfarti, misinterpreted it to extend to matte surfaces, and then continued to extend it to the RIM of the wheel as well. I frankly give a shit about what Tach thinks or when he thought it, but I am grateful to have finally learned how to use TeX!! I also believe I have a firm grasp on Doppler effects coming from a rolling wheel, mirrored or not, and will be happy to discuss it with you in a civil manner (which I have no doubt is what would occur).

23. ### TrippyALEA IACTA ESTStaff Member

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I understood this to be because he was considering the tangential surface.

Consider an unobtainium rolling mirrored toilet paper roll with zero thickness

Every point on the surface has a tangent.

In the case of a stationary mirrored toilet paper roll, every point on the (curved) surface has a tangent.
The reflection from any point on the surface can be computed by considering the light ray to be reflected by a mirror lying within the tangent plane. Essentially, for all intents and purposes, for calculating refelection from a mirrored toilet paper roll, the toilet paper roll can be considered to be an extruded n-gon, where n is infinitely large - hence the comment I made about infinitessimals, and infinitessimal surfaces.

So when considering the situation of a rolling mirrored toiletpaper roll, because the camera is always going to see the reflection off a tangential plane that is co-moving with the axel, but not co-rotating with the wheel, we can model any infinitessimal part of the surface of the wheel that connects the raypath from the light source to the ray path from the camera as if it were a flat mirror moving with some velocity v, paralell to the ground, with some angle $\theta$ to the ground, and the velocity v is the velocity of the axel added to the component of the tangential velocity that is paralell to the ground.

Hence, I saw no contradiction between the diagram with the circles, and the diagram in the document - the two situations appeared (to me) to be physically equivalent (hence my confusion from whence stemmed the confusion).

Is that sufficiently clear?