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

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

1. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
No.

You need to apologize.

And you need to learn how to use the Forums search features.

For example:
http://sciforums.com/showpost.php?p=2801445&postcount=139
http://sciforums.com/showpost.php?p=2799295&postcount=59
http://sciforums.com/showpost.php?p=2789182&postcount=7
http://sciforums.com/showpost.php?p=2753684&postcount=77
http://sciforums.com/showpost.php?p=2733451&postcount=243
http://sciforums.com/showpost.php?p=2659983&postcount=4
http://sciforums.com/showpost.php?p=2651343&postcount=43
http://sciforums.com/showpost.php?p=2341324&postcount=55
http://sciforums.com/showpost.php?p=2249027&postcount=82

While specific asthetics may vary, I have always made it explicitly clear when I am posting as a moderator, which leaves us with you and your absurd assumptions.

Kindly apologize, and move on.

3. ### Aqueous Idflat Earth skepticValued Senior Member

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6,152
Wow. Rough crowd. I'm still waiting for the shoe to drop.

If one of them is wrong, what is the most fundamental error?

I'm seeing this from a different perspective than what most of you are seeing. I see something very basic that hasn't really come up yet.

5. ### TrippyALEA IACTA ESTStaff Member

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I wonder if it's the same thing I've seen.

7. ### KittamaruAshes to ashes, dust to dust. Adieu, Sciforums.Valued Senior Member

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13,938
Would you two care to enlighten those of us not so well versed in this station?

8. ### TrippyALEA IACTA ESTStaff Member

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10,890
Well, from my perspective, it boils down to this.

The thing is that wave mechanics allows us to treat a solid surface as a series of point sources - reflection of an incident wave from a wall, for example, can be modled as a series of very small, very closely spaced point sources.

Given that a wheel is just a wall bent into a circle, and with it ends joined together, we can model a wheel as a series of very small, very closely spaced point sources rotating around a common center.

The upshot of all of this is that when you think about a wheel in these terms, there is no part of the wheel - no point source, that is is simultaneously moving towards and away from the observer. Whether any 'infinitessimal' part of the wheel is moving towards, or away from the observer depends on where in its cycle it is.

However, the reflected spectrum, be it specular or diffuse, is the sum, or the integration of the reflected light from each infinitessimal part of the wheel, which is where my first post on the matter comes into it. Basically, because wave mechanics allows us to treat a surface as the sum of an infinite amount of infinitessimal point sources, and because as the wheel rotates, none of the point sources is simultaneously moving towards and away from the source of light (or the camera), then Tach's assertions fall over at the first step.

Or at least, that's part of it anyway (from my perspective).

I'm also reasonably confident that these considerations:

Fail to take into account the phase angle (camera - object - light source).

Let's assume that Tach's right, for a minute, that with the setup illustrated that there is no net redshift because the redshift of the reflected light is of the same magnitude as the incident light, what does that actually mean?

Well, what it means is that multiple observers might not be able to agree on the colour of the object. In the situation on the left an observer in front of the object might not be able to detect any blue shift, but an observer behind it will be able to detect redshift. And in the situation on the right, an observer behind it might not be able to detect any red shift, but an observer in front of it will be able to detect the blue shift. Neither or which contradicts anything I have said.

I word it like that, because at this point I'm not willing to cede that there are any observers that will observe no reshift in either setup.

But then, what do I know, I'm just an environmental chemist, I should stick to water quality, right?

Last edited: Nov 17, 2011
9. ### TachBannedBanned

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5,265
I was apologizing. I am apologizing again.
On a different note, you may want to rethink your explanation from post 105.

10. ### TrippyALEA IACTA ESTStaff Member

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10,890
MODERATOR NOTE
The leeway that is granted is granted for debators to post in the discussion thread is for administrative issues only. Please refrain from posting in this thread to discuss the matter of the debate.

11. ### EmilValued Senior Member

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2,801
Oops....It's complicated than how I thought at first.
My post predecessor does not sufficiently accurate.
There are several cases.

Light source on the wheel.
Case 1.
All perimeter is a light source.
Looks like this

where the source rotates. There is no Doppler effect.
Case 2.
A point light source on the perimeter.
Looks like this

except that the light source has a rotating motion instead of linear. There is Doppler effect.

Source of light is not on the wheel.
Case 3.
Spherical light source. Illuminate half of the wheel (180 degrees).Resembles case 1. There is no Doppler effect.
Case 4.
Light source is a laser beam. Illuminates a single point on the perimeter.
Resembles case 2 but still is different.
Two consecutive laser pulses, generates a light source by refraction, with the same coordinates, although they are two different points on the wheel. There is no Doppler effect. (Doppler effect to exist, the laser must generate a light source as in case 2.)

The analogy with the mirror.

The drawing is a little confusing because the bulb suggests a spherical light source,
but is designed as a laser beam.

If a spherical light source that illuminates the whole mirror, then the secondary light source (mirror) move. There is Doppler effect.

If a laser beam illuminates a single point on the mirror, the secondary light source (reflected) do not move even if the mirror moves. There is no Doppler effect.

Conclusion: Doppler effect to exist, there must be a light source (direct or reflected) that moves.

12. ### RJBeeryNatural PhilosopherValued Senior Member

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That isn't true Emil. Imagine the mirror to be infinite in size, moving and spinning. The fact is, you could not detect its movement (in the very special case referred to above). The camera would see a stationary light source.

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

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10,167
One difficulty with deciphering Tach's argument is that it's not clear what reference frame he is working in.

His diagrams are clearly set in the source/observer rest frame, but he [post=2850227]claimed earlier (post 209)[/post] to be exclusively working in the mirror rest frame, and his equations are only correct if $\phi_1$ is an angle in the mirror rest frame.

And in this line:
...he is using another frame, the inertial rest frame of the wheel's axle, so he's using the wrong velocity in his equations.

Anyway...
Well, Tach appears to be relying on the special condition of the source and observer being in different places such that the wheel is approaching one and receding from the other.
But, given Tach's other special condition, that the facet velocity relative to the observer is parallel to the facet surface, then it can still work even if the source and observer are together.

If the light source and observer are together, the ray must strike the mirror at right angles in the observer's rest frame. So yes, the mirror can not be both approaching or receding in that rest frame.

But in the instantaneous inertial reference frame of the facet (which is the frame that Tach implicitly uses for his equations, though not his diagram), the observer is approaching as the ray is emitted, and receding as the ray is detected:

Bear in mind that this is strictly reflected light, not scattered light, and that Tach is relying on the special condition of having the mirror aligned with it velocity relative to the observer and light source.
Under those conditions, you can't have observers in different directions - the light ray can only be reflected in one specific direction.

Last edited: Nov 17, 2011
14. ### RJBeeryNatural PhilosopherValued Senior Member

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4,215
That paper is clear as mud. Why are there two pictures? The camera is stationary but the light source is moving?? Or is the idea that the light source is ambient and incident from all directions?

What a minute...if that's what Tach is thinking then his "special case" remains a special case, but would be extended to the entire trajectory of the mirror rather than at a single point of its path; that would explain a lot, because I knew Tach was stubborn and immature but I didn't think he was a barking lunatic. Either way, it's only valid when the mirror's translational direction is parallel with its surface orientation.

Edit: Actually, this condition may be implied by the "rolling wheel" aspect of the mirror...if we consider the wheel to be a circled mirror, for simplicity, and we're asking whether or not a camera viewing the universe through this mirror would see any color-shifting, the answer is actually NO. Interpreted in this manner, Tach's stance is correct!

Last edited: Nov 18, 2011
15. ### EmilValued Senior Member

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2,801
In this case you are right.
But imagine now, for the case 2, a tiny mirror.
I have not questioned how we can detect. Only if there Doppler effect.

16. ### AlphaNumericFully ionizedRegistered Senior Member

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6,702
Tach, my post was in reference to your silence via pm. You ignored my questions and haven't addressed my mathematics, despite previously whining i hadn't done any. If through mirror doesn't alter the frequency in its axle rest frame then it alters it in the grounds frame. i have given the mathematics. Rather than repeat your rapid and typical tactic of just saying everyone but you is wrong actually show it. I addressed one of your references explicitly, pointing out the website with the coloured wheels doesn't say the colors are physical. It appears to just be a visual labelling. The boosts alter the spokes and circumference. You haven't addressed any of these and yet you are here complained James isn't addressing you. What a joke!

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

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10,167
I was saving this for after the debate, but I'm getting bored.

Tach's diagram is in the wrong reference frame for his equations, it is for a special case where the mirror-observer relative velocity is parallel to the mirror surface, and it is unnecessarily complicated by having two separate cases where one is sufficient.

This diagram is clearer and more general:

In the rest frame of the mirror:
V is the velocity of the source and the observer.
$\theta$ is the angle of incidence and reflection.
$\phi_v$ is the angle between the mirror surface and V.
$\phi_s$ is the angle between the mirror-source displacement vector and V.
$\phi_o$ is the angle between the mirror-observer displacement vector and V.
Note that:
$\phi_s = \phi_v + \pi - \theta \\ \phi_o = \phi_v + \theta$

Now, let's figure the doppler shift.
$\beta$ = |V|/c
$\gamma = 1/\sqrt{1 - \beta^2$
$f_{source}$ is the frequency of the light ray emitted by the source in the source rest frame.
$f_{mirror}$ is the frequency reflected by the mirror in the mirror rest frame.
$f_{observed}$ is the frequency detected by the observer
(Note that in Tach's document, $f_o$ is the source frequency, and $f_s'$ is the observed frequency. Not sure why he chose that convention.)

We can now correctly use the relativistic doppler equations used in Tach's document:

$f_{mirror} = f_{source}/\gamma(1+\beta \cos\phi_s)$

$f_{observed} = f_{mirror} \gamma (1+\beta \cos\phi_o)$

Note that these equations work for any angles. You don't need separate cases for approaching/receding, this is covered by allowing the angles to vary from zero to $2\pi$.

Combining and manipulating the equations, I end up with this (please check if you're keen):
$f_{observed} = f_{source} \ \frac{1 +\beta \cos(\phi_v + \theta)}{1 +\beta\cos(\phi_v - \theta)}$

Note that for the special case of $\phi_v = 0, \pi$ which Tach relies on, the result is $f_{observed} = f_{source}$.

If anyone is really keen, it might be interesting to go even more general and allow the source and observer to have different velocities, and enumerate all the special cases where the total doppler shift is zero.

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

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10,167
That's what I thought Tach was doing (see [post=2854243]post 20[/post], but as przyk pointed out, Tach is clearly referring to the outer 'tread' of the wheel in his document.

His mistake is in using mirror's velocity relative to the axle (which is indeed parallel to the surface) to figure the doppler shift when he obviously needs to use the velocity relative to the source/observer.

It is true (as Alphanumeric and probably others have noted) that there is no doppler shift for source/observer at rest in the axle rest frame.

19. ### TrippyALEA IACTA ESTStaff Member

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I'm not sure how this affects anything I have said, to be honest - after all, when you take phase angle into account, which I mentioned later.

I'm struggling with this for a number of reasons, none of which have anything to do with my level of understanding of Physics.

Now stop and consider the whole picture.

For example, while any given light ray might only be able to follow one path, what's the upper limit on possible light rays?

I think one of the reasons why I'm struggling with this is that you seem to be taking issue with what I've said (for lack of a better way of putting it), but you're saying the same things I've said - even down to the existence of special conditions where there is no doppler shift.

It's one of the reasons I told Tach he needed to pay closer attention to what I've actually said rather than knee-jerk reacting to what he thinks I've said.

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

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10,167
Sorry, I misunderstood what you were thinking, putting it in the wrong context.

21. ### TrippyALEA IACTA ESTStaff Member

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Accepted - I'm willing to cop some flack for any ambiguity. Some times 1000 words just isn't enough. :3

22. ### RJBeeryNatural PhilosopherValued Senior Member

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4,215
Well, there are many things going on here. People are debating the paper, what they think the paper is saying, what they think the debate topic is, and each other. :argue:

I was PM'ed by Tach that my interpretation of his stance is correct, btw. The problem (for me) was in the computer graphics ray-tracing aspect of the paper, where the primary ray moves from the camera to the light-source; in Physics optics ray-tracing that I've seen the discussion begins with the photon source. Because of this confusion I had presumed a single light-source...which made the paper's conclusion basically trivial and uninteresting (see post #23).

I have to note that it's possible that Tach convinced himself a mirrored wheel would not show a Doppler shift under any circumstances after reading (or authoring?) the paper. Perhaps he thought it would extend to a mirrored rim, who knows? But as it stands I declare that the paper is correct, if confusing, and the debate topic is unfortunately too ambiguous to declare a winner. :shrug:

23. ### RJBeeryNatural PhilosopherValued Senior Member

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I'd better add, however, that Tach's veracity is limited to the claim that the paper itself is correct. He's still flat wrong that a colored rolling wheel would not show Doppler shifting, which I pointed out long ago in the relativistic wheel thread.