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

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

Not open for further replies.
1. ### TachBannedBanned

Messages:
5,265
Moderator note: The full, agreed topic of this debate is:

"That that the light reflected off the circular rim of a mirror-like wheel shows zero Doppler effect between the source and the receiver in the case of the wheel rolling without slipping"

Debaters: Tach for the affirmative; James R for the negative.

Standard rules apply.

Tach's opening post follows:

----

I claim that :

"The light reflected of the circular rim of a mirror-like wheel shows zero Doppler effect between the source and the receiver in the case of the wheel rolling without slipping" . The speed may or may not be relativistic, it doesn't really matter.

The formal proof is here

Last edited: Nov 10, 2011

3. ### James RJust this guy, you know?Staff Member

Messages:
38,355
I thank Tach for agreeing to this debate.

I may get to his document in a future post. But for now I will settle for providing a counter-example that is sufficient to win the debate.

Consider a circular wheel with a mirror-like rim, as specified in the topic statement. Now, let us assume that the wheel is rolling without slipping directly towards an observer who is holding something like a police radar gun.

That gun emits light of a particular frequency, $f_s$, measured in the frame of the gun. The light heads out to the wheel along what I will call the x axis. I define the positive x direction as pointing from the wheel to the gun, and the velocity of the wheel as it rolls along this axis is v in the frame of the gun.

The apparent frequency of light observed by the wheel (again, as measured in the frame of the gun) can be shown to be:

$f_w = f_s\sqrt{\frac{1+v/c}{1-v/c}}$

This is a standard relativistic result. I can provide references or proofs if Tach wishes to dispute this equation.

The mirror-like surface of thye wheel rim reflects a frequency that is equal to $f_w$ and this propagates back to the gun. However, the gun now perceives that light as coming from a moving source (i.e. the wheel rim moving towards it), and so the frequency received back at the detector (gun) is:

$f_o = f_w\sqrt{\frac{1+v/c}{1-v/c}} = f_s\left(\frac{1+v/c}{1-v/c}\right)$

Clearly, the received frequency $f_o$ of the reflected light is different from the initial frequency $f_s$, so there is a shift of frequency given by:

$\Delta f = f_o - f_s = f_s\left(\frac{2v/c}{1 - v/c}\right)$

This is the formula for the Doppler shift in the given situation. Note that unless v=0, the Doppler shift is non-zero.

The light reflected off the circular rim of a mirror-like wheel shows non-zero Doppler effect between the source and the receiver in the case of the wheel rolling without slipping, at least in this particular case.

Therefore, Tach's assertion that the Doppler shift is always zero for light reflected from the mirror-like rim of a rolling wheel is proven false.

5. ### TachBannedBanned

Messages:
5,265
Thank you. Despite all your previous claims I have no qualms debating with you.

Actually, ignoring my document is not OK since it provides you (and the others) with the rigorous proof of the thread claim because whatever diversion you took, it does not disprove the information given to you in the opening post, the Doppler effect is still null for the cases described in post 1. You should have read it first and you should have tried to refute the claim shown in the file. Instead , you elected to construct your own, cartoonish version of the exact case.

No, that would not wash since you are not addressing the claim of the OP but rather generating your own, cartoonish version of the scenario.

A radar gun is not a camera, and you just moved the observer and the light source in the same point in space, so you have just moved the goalposts.
You need to consider instead the case shown to you in the detailed writeup, the wheel is either rolling away from the light source and approaching the camera or it is rolling towards the light source and rolling away from the camera. The description explains why there is no net Doppler effect in the way the rolling wheel reflects the light. If you engage in a debate, you need to understand its parameters as outlined in the detailed mathematical description I linked at post 1, not to make up your own description. Your simplistic treatment covers neither of the two cases explained to you in the opening post of the debate.

The above is only partially correct (and certainly incorrect for the case in discussion). You have been given the general equation (formula 1 in the writeup):

$f_{mirror}= \frac{\sqrt{1-(v/c)^2}}{1+(v/c) cos (\phi)} f_{source}$
(1)

valid for a mirror receding from the source. The angle $\phi$ is the angle between the mirror velocity (in our case a point on the circumference of the wheel) and the direction of the incident ray (see the standard description of the relativistic Doppler effect). Only for the PARTICULAR case of $\phi=0$ one gets the formula:

$f_{mirror}=f_{source} \sqrt{\frac{1-(v/c)}{1+(v/c)}}$

Now , looking at the attached figure, one understands immediately that $\phi=0$ only for the two incident rays tangent to the wheel. NEITHER of these two rays ever returns to the camera, let alone to the gun.

At the very WORST, you should have gotten the above and not what you wrote, i.e.

$f_w = f_s\sqrt{\frac{1+v/c}{1-v/c}}$

is clearly wrong. Moreover and unfortunately for your argument, the simplistic formula that you are employing is useless for the case in study clearly described to you in the first post of the debate since $\phi$ ISN'T zero in the general case outlined in the opening post of the debate. This is the FIRST mistake in your argumentation.

Incorrect. Remember that the wheel is rolling, so, the light is NOT propagating back to the source (the gun) but onwards to the camera. See fig. 2 in the writeup (or the picture attached to the bottom of this post). As the wheel (mirror) recedes from the source it also approaches the camera, thus the red-shift due to receding is cancelled out by the blue-shift due to it approaching the camera. This is your SECOND mistake.

This is , again, overly simplistic, since it is valid ONLY for the PARTICULAR case of the angle between the velocity of the mirror and the reflected (secondary) ray is (formula 2 in the writeup):

$f_{camera}=f_{mirror} \frac{1+(v/c)cos(\phi)}{\sqrt{1-(v/c)^2}}$
(2)

For the trivial (and useless) case when $\phi=0$ one recovers the trivial formula:

$f_{camera}=f_{mirror} \sqrt{\frac{1+(v/c)}{1-(v/c)}}=f_{source}$

Turns out that, if you use the CORRECT formulas (1) and (2) for the Doppler effect, you get $f_{camera}=f_{source}$ for all angles $\phi$ and for all speeds $v$. This is quite intuitive since the wheel circumference is either rolling away from the source and towards the camera or vice-versa (see fig 2).

You elected to treat a dumbed down scenario. This is the THIRD and most fatal mistake. Now, in this dumbed down scenario, where you merged the camera and the light source into your radar gun, you will have indeed the trivial situation that the radar gun will get ONLY one ray reflected back (the one that lines up with the center of the wheel). You will not get a picture of the rolling wheel from your radar gun and this is not what the thread is about. It is true, that in this trivial case:

$f_{reflected}=\frac{1+v/c}{1-v/c} f_S$

for wheel approaching the gun and

$f_{reflected}=\frac{1-v/c}{1+v/c} f_S$

for the wheel receding from the gun (using your notation).

Nope, you did not even make the effort to understand the explanation from post 1 and replaced the scenario with your dumbed-down version. You did not address the case described in the opening post. This is why I was so insistent on getting the exact title and the exact description, so you could not move the goal-posts with a dumbed-down version of the scenario.

Luckily, the people from references [1-3] got things right and they show NO Doppler effect in the colors of their rolling wheels. If what you are claiming were true (it clearly ISN'T) the color of the wheels in their animations would have looked severely Doppler shifted. They don't because the authors knew what they were doing.

Actually, it turns out that your claim is easily proven to be in error: I have taken the trouble to understand and address your position whereas you have not bothered to understand the original claim and you have not addressed it. I guess that you have one more shot at doing that....

References:

[1] V.F.Weisskopf, “The Visual Appearance of Rapidly Moving Objects”, Physics Today 13, 24 (1960)
[2] U. Kraus, H. Ruder, D. Weiskopf, C. Zahn, "Was Einstein noch nicht sehen konnte - Visualisierung relativistischer Effekte", Physik Journal, 7/8, (2002)

Last edited: Nov 11, 2011

7. ### James RJust this guy, you know?Staff Member

Messages:
38,355
First, let me deal with Tach's claims that I have made (at least) three mistakes in my first post:

I did not study Tach's link in the first post of this debate. Rather, I gave one specific situation (different from Tach's) that shows a non-zero Doppler shift of light reflected from a rolling wheel.

This is all I had to do to win this debate. I have no obligation to address any scenario that Tach puts forward to show zero Doppler shift. All I need to do to win the debate is to show at least ONE situation in which a Doppler shift occurs. I have done that, and my proof remains unrefuted.

Thus, my "FIRST mistake" is no mistake at all. It may annoy Tach that I produced a situation that shows a Doppler shift. But my proof is unrefuted by Tach at this point. If this is still the case at the end of the debate, then clearly I must be judged the victor.

I do not rely on any figure from Tach's writeup or anywhere else. My first post to this thread is self-contained. The light in my example does propagate back to the source (the gun) - it is projected out from the gun and it reflects directly back from the mirror-like rim of the wheel to the gun. This ought to be quite clear from my first post. Maybe Tach didn't understand it.

Thus, my "SECOND mistake" is also no mistake at all. Maybe Tach thinks that there's no such thing as light that reflects straight back towards a viewer from a mirror surface. If that is the case, I suggest that Tach purchase a pen-light torch and try some experiments in front of his bathroom mirror in the dark.

Tach also makes the silly claim in his post that:

Of course, it is trivial to replace the radar gun in my first post by a camera with a flash mounted next to the lens, so Tach's attempted distinction between a radar gun and a camera is spurious. Perhaps Tach wants to argue that a camera can never take a photo of something straight in front of it using its inbuilt flash. If that is the case, once again I suggest that Tach ought to obtain a camera and try a few experiments himself.

I made my scenario just "dumb" enough so that even Tach should be able to understand it, but just smart enough to win the debate. My example speaks for itself.

Tach goes on:

Obviously, a radar gun has a finite size, and it does not gather "ONLY one ray". Perhaps Tach should review his ray optics. And have enough imagination to realise that replacing the radar gun by a camera makes no meaningful difference to the example I presented.

Thus, my "THIRD" and "fatal" mistake is no mistake at all, and again not fatal to my argument.

Essentially, Tach's complaints about my supposedly "fatal" mistakes amount to a whinge that I considered a scenario different from the one he would prefer to discuss. But this was no mistake. It was a deliberate choice on my part. Only a supreme egotist would assume that he could steer the course of the debate to the extent where his opponent would only consider his preferred talking points and put no rebuttal of his own.

I point out that Tach's document is not a "general case" as he would have us believe. It is a carefully-selected case - as carefully selected as my example.

I note that the whole "rolling wheel" part of this debate is a red herring. Essentially, Tach's claim is that there can be no Doppler shift of light reflected off a mirror. Whether that mirror happens to be attached to the rim of a rolling wheel or is just a plane mirror flying through space is neither here nor there.

Tach's preferred example has a mirror whose surface is parallel to the y-z plane and which moves in the y direction (say). My preferred example has a mirror whose surface is parallel to the y-z plane, but which moves in the x direction. Tach cannot claim that my example is a "dumbed-down scenario" any more than his scenario is "dumbed-down".

Perhaps Tach would like to claim that my scenario is impossible, which would be him making a claim roughly equivalent to the claim that you can never remove the mirror from your bathroom wall. If Tach wants to make this argument, I suggest that he first tries the relevant experiment at home.

---

Loose ends?

Quite clearly, Tach's cases in post 1 are not exhaustive. In particular, my example in post 2 is a case that Tach did not cover.

I point out again that all I need to win this debate is to show ONE case (any case) where there is a Doppler shift of light reflected from a rolling wheel. I have done so.

There are many other cases that show the same thing, of course. For example, suppose we take Tach's preferred example, but allow his "camera" to move relative to the source of the light that is reflected from the wheel. Clearly, this is another case where a Doppler shift will be observed. Again, it is no argument for Tach to cry about how this wasn't what he was thinking of when he agreed to this debate.

Tach's second post comes across as like a child complaining that people must play his game by his rules. But this is an adult debate, not a child's game. I don't have to play by Tach's rules - only by the rules that we agreed to in advance in the Proposal thread for this debate. If Tach can refute my physics, then I will be forced to present an alternative scenario. If he cannot, then I win, regardless of how many cases he can present that show zero Doppler shift.

Here's Tach again advising me how he would like me to debate him:

I repeat that I have no obligation to refute any part of Tach's file to win this debate, unless he claims that it covers all possible cases of light reflected from the rim of a mirror-like wheel. Clearly, as I have shown, it does not.

Terms such as "cartoonish" are a weak attempt to refute my argument by ridicule. I know that smart readers will see straight through that type of childishness and concentrate on the physics.

I notice that Tach did make one claim in his last post that one of my equations was "clearly wrong". It is not clear what is wrong to me, and I don't think it will be "clear" to smart readers either. The onus is on Tach to explain what is wrong, rather than make empty assertions about what is and isn't "clear". If he claims to have found a mistake in my mathematics, he should explain clearly where the mistake lies.

It appears that Tach's intuition is faulty here, as is his understanding of my scenario. My scenario quite clearly has a wheel rolling towards BOTH the source and the camera.

Tach's claim that wheels must always roll towards a source and away from a camera or vice versa is empty, as the shortest application of common sense shows. As I said above, it is equivalent to the claim that you cannot remove a bathroom mirror from the wall with the bathroom light behind you and your eye as the "camera", because the mirror would then be moving towards both the light and the "camera" - something Tach apparently imagines is impossible.

Tach's conclusion from his last post was:

If my claim is "easily proven to be in error" then I look forward to Tach's proof that it is in error in his next post. At this stage, we only have an assertion. We are not obliged to take Tach's word about what is "clear" and "easily proven". In a debate such as this, assertions must be backed up with argument and/or evidence.

8. ### TachBannedBanned

Messages:
5,265
There are two issues with your approach:

1. (The lesser one) You have used a specific case of radar propagation to show that there is Doppler shift for a wheel rolling in the dumbed down case when the emitter and receptor are ONE and the SAME. (the radar gun). Of course, this doesn't cover any of the interesting cases of a wheel rolling between an emitter and a receptor that are DIFFERENT from each other. It is my fault that I allowed the description to be not as tight as necessary, giving you the opportunity to weasel out of dealing with the case that was presented quite clearly in the opening post, a case that you clearly HAVE NOT addressed and you have avoided addressing ever since. The origin of the current debate traces all the way back to your false claim in post 241 that you have found errors in my treatment of the issue of a mirror-like object traveling between a light source and a camera.

2. (The bigger one). Don't be so quick to declare "I win the debate" since there is more than ample evidence in mainstream literature that (for non trivial cases), "there IS evidence of ZERO Doppler shift for a wheel rolling between a source and a camera".

No, this is a gross misrepresentation of what I said: the wheels do not have to always roll BETWEEN the source and the camera, to prove my stance it is SUFFICIENT that they roll this way just in the example outlined in the opening argument in this thread, an example that you have not taken the time to either understand, let alone to refute. This is all I claim and it is the sufficient condition to prove the statement in my OP correct, thus disproving your stance.

No, I understood your scenario all right, this is why I made the effort to play it back to you while outlining the differences in the assumptions between the case I described to you and the dumbed-down case you insist on presenting. While I made the effort to understand your scenario (it is quite trivial, really), you never made the effort to understand mine, so, in effect, you never responded to the claims in the OP. What you also fail to mention, is that if we cast your scenario into the case when the mirror moves between the light source and the camera (see references [4,5]), say away from the light source and towards the camera, you get a NULL Doppler effect, as explained to you multiple times even before this debate started. Here is again the explanation why there is no Doppler effect in this scenario:

$f_{mirror}=\sqrt{\frac{1-v/c}{1+v/c}} f_{source}$ (redshift due to the mirror moving away from the source)
$f_{camera}=\sqrt{\frac{1+v/c}{1-v/c}} f_{mirror}$ (blueshift due to mirror moving towards the camera)

Conclusion: $f_{camera}=f_{source}$, clearly contradicting your strawman argument. If you don't agree with that, I suggest you consult references [1-5]. take for example reference [3], despite of the fact that the speeds of different components of the wheel vary from point to point between 0 and $2v$, the color of the wheel is uniform, i.e. there is no Doppler shift. And there shouldn't be, as explained in the math above....

Now, evidence for the above has been provided to you much earlier for translation motion, in this file, in the other thread, and, again, for the case of rolling motion, in the current thread. In NEITHER cases, did you address the issue, let alone refute it. You have been provided with direct evidence that , in fact, there is NO Doppler effect in the cases described in this thread (and in the "rolling wheel" thread) for light being reflected off moving object (translation or rolling motion). You never addressed either.

References

[4] H . Bateman, "The reflexion of light at an ideal plane mirror moving with a uniform velocity of translation" ,Phil. Mag., (18) (1909) p.890
[5] W.Pauli, "Theory of Relativity" , 1958, Pergamon Press, p.95 (top)

Last edited: Nov 14, 2011
9. ### James RJust this guy, you know?Staff Member

Messages:
38,355
This debate ended prematurely. James R did not reply within the previously agreed time limit, and Tach refused an extension of the time limit by 12.5 hours.