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

[RJBeery]

I just got another PM from Tach wherein he is insisting that the colored wheel would also show no Doppler shifting. I'm taking the liberty of posting the PM here under the presumption that he isn't doing so because he's adhering to the formal debate rules...

Repeating the same nonsense doesn't make it right

If it didn't appear to be a such a time-consuming and maddening endeavor I'd be happy to show Tach why he's wrong in another formal debate...

Tach, if you'd like to me simply post in this thread why you're mistaken, I will on one condition: you PM me acknowledging in advance that you will publicly admit your error (if I am able to show it) rather than try to move the goal posts, rewrite history, or obfuscate the topic. I currently acknowledge I will do the same thing.

 

[Trippy]

MODERATOR NOTE:
Five off topic posts deleted.

Tach, this is now the second time I've reminded you, do not post in this thread unless it's to raise an administrative issue regarding your debate with James R.

RJ Beery asked you to respond to him in PM, and that would have been the appropriate medium to do so (and remains so). If you want to have a formal debate with RJ Beery, and to set the ground rules - including start date, then do so either via PM, or set up a debate proposal thread.

 

[Neddy Bate]

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?

I'm glad you mentioned the two light sources shown in Tach's diagram. I wasn't sure why it was drawn that way either.

I was PM'ed by Tach that my interpretation of his stance is correct, btw.

Excellent. So now we know that the two light sources are meant to represent incident light from all directions.

But as it stands I declare that the paper is correct, if confusing

I declare the paper to be incomplete to a fault. The paper neglects the myriad light rays reflecting off the wheel which are Doppler shifted.



Here is a modified version of his diagram:

The yellow lines represent light rays of their natural color in the reference frame of the camera. The red and blue lines represent light rays which are are Doppler redshifted and blueshifted, respectively. (I assumed the wheel is rolling from left to right). Tach's paper does not address those Doppler shifted light rays, and that is what James R's rebuttal is based on.

 

[RJBeery]

Cripes, NeddyBate, I didn't consider the circle to be the wheel, parallel to the screen...I don't know what your source is for that picture but if it was from Trippy I believe it to be in error and only confusing the thread. The original "proof" is posted here. In this paper you can imagine (from my interpretation as declared correct by Tach himself) that the wheel is "coming out of the screen" and rolling along the path. The picture you posted would have light rays moving through the wheel itself in order to reflect from the inner rim...:bugeye:

 

[Neddy Bate]

Cripes, NeddyBate, I didn't consider the circle to be the wheel, parallel to the screen...I don't know what your source is for that picture but if it was from Trippy I believe it to be in error and only confusing the thread. The original "proof" is posted here. In this paper you can imagine (from my interpretation as declared correct by Tach himself) that the wheel is "coming out of the screen" and rolling along the path. The picture you posted would have light rays moving through the wheel itself in order to reflect from the inner rim...:bugeye:

I thought I was modifying the diagram from the original paper, but it looks like I was unwittingly modifying Trippy's version of the diagram. Apparently Trippy was the one who added the circles which make it look like the light rays pass through the wheel. I didn't add those.

In my diagram, I only added the colored lines. Those light rays do not pass through the wheel, and they are reflecting off the outer rim. Those rays are Doppler shifted.

 

[Trippy]

...but if it was from Trippy...

Nothing to do with me.

 

[RJBeery]

Nothing to do with me.

Trippy, can you please explain where you got the picture with the circles? I think the author of that picture shifted the debate from a wheel with mirrored sides to a wheel with mirrored rims. This is not what Tach's original paper proposed. It might be argued that at some point he ALSO announced that the mirrored rims would be impervious to the effects of Doppler shifting but that's a separate issue as far as I'm concerned.

 

[Trippy]

The picture you posted would have light rays moving through the wheel itself in order to reflect from the inner rim...:bugeye:

Apparently Trippy was the one who added the circles which make it look like the light rays pass through the wheel. I didn't add those.

The image in my post is TACH'S OWN IMAGE.

It is the attachment labled Fig5final.jpg in his first post (the third post in the debate thread)

Tachs Post

The image has nothing to do with me - it's all Tach, and if it's confusing the thread... :shrug:

 

[Trippy]

Trippy, can you please explain where you got the picture with the circles? I think the author of that picture shifted the debate from a wheel with mirrored sides to a wheel with mirrored rims. This is not what Tach's original paper proposed. It might be argued that at some point he ALSO announced that the mirrored rims would be impervious to the effects of Doppler shifting but that's a separate issue as far as I'm concerned.

I think my previous post covers it.

 

[Neddy Bate]

Below is a NEW sketch which avoids the problem of light rays passing through the wheel. This sketch clearly shows red and blue Doppler-shifted rays reflecting off the outer rim of the rolling mirror. This new sketch also shows the Non-Doppler-shifted rays that Tach described in his paper. Those rays occur in the special case when the object is located midway between the light and the camera.

But Tach's paper only considered the Non-Doppler-shifted case, and his argument seemed to be that those are the only types of rays which reflect off the rolling mirror. If that is what he was claiming, then it is clearly not correct, as James R argued in his rebuttal.

 

[AlphaNumeric]

Those rays occur in the special case when the object is located midway between the light and the camera.

Yep, both Pete and I even went through the pretty simple mathematics. Thus far Tach hasn't replied to PMs pointing it out from me and it seems RJ is having his PMs ignored, yet Tach is quite happy to try to post here to say how we're all wrong and he'll show it.

I can't help but notice the parallel in his behaviour when compared to Chinglu. Both demand mathematics, stating they've proof of their claims. When the mathematics is presented they just say "Wrong!" and then demand more mathematics, each time challenging people to provide mathematics which they will then disprove, when in fact they just ignore it.

Come on Tach, if you're interested in talking about it while you wait for JamesR clearly more than one person is trying to talk to you via PM. What's the problem? If we have a misunderstanding, as you claim, then explain how we've misunderstood the setup. If there's a problem with the mathematics then please tell me, I'll be happy to pass it on here in the thread so we can discuss it.

Hell, you could at least reply to my query about how the colours of these wheels are never stated to be anything other than a simple way of labelling which is which in each picture. Where on the page does it say the colours are included in any possible Doppler shift illustration?

Alternatively you could just be a man and admit your mistake.

 

[Emil]

Below is a NEW sketch which avoids the problem of light rays passing through the wheel. This sketch clearly shows red and blue Doppler-shifted rays reflecting off the outer rim of the rolling mirror. This new sketch also shows the Non-Doppler-shifted rays that Tach described in his paper. Those rays occur in the special case when the object is located midway between the light and the camera.

But Tach's paper only considered the Non-Doppler-shifted case, and his argument seemed to be that those are the only types of rays which reflect off the rolling mirror. If that is what he was claiming, then it is clearly not correct, as James R argued in his rebuttal.

If you do the analysis as if it were a laser beam, this can lead to error.
In my opinion, it is easier to analyze the situation using the wave model.
Light source looks like this:

Put a wheel here and see how the light reflects, if the wheel does not rotate.
After that see the difference if the wheel rotates.

 

[James R]

I realise that this post is about 12.5 hours after the deadline by which I should have posted. If Tach wishes to end the debate here, then we can close this thread. I will then post in the related Discussion thread instead. Otherwise, with Tach's agreement, we each get one more post after this one to end the debate.

I have taken a long time to reply mosty due to being incredibly busy with real-world work over the past week. I have missed the deadline due to being ill over the past few days. I realise that neither of these is an excuse - only a reason.

The other reason I have been slow in replying is that I think this is a more complicated matter than Tach would have us believe. In particular, I have been looking over Tach's "file" cited in the opening post of this thread, and I have some issues with it. At this stage, not having had a chance to do the relevant derivations myself, I merely have questions rather than claims of outright error. Perhaps Tach will clarify in his next post (if there is one).

It seems to me that Tach's "file" is not his own work, but is ripped largely from an article by A. Sfarti titled "The correct representation of objects moving at relativistic speeds" and published in the Romanian Journal of Physics (Vol 55, No.3-4, pp 265-73, 2010). The article is freely available online. The diagrams and formulae therein are uncannily similar to what Tach claims is his own work.

What worries me about Tach's file (which is what I will concentrate on rather than the article it seems to be copied from) is that the frames of reference are not clear (at least not to me). In particular, Tach uses the fact that the angle of reflection from the mirror (wheel) is equal to the angle of incidence. It seems to me that this would only be true in the frame of the mirror, and not in the source/detector frame. In the source/detector frame, the angles will NOT be equal, due to the aberation of light. And if the angles are not equal, then it seems to me that the source frequency cannot possibly equal the detected frequency, even according to Tach's formulae (assuming they are correct - and I also reserve judgment on that at this point). The formula clearly have an angular dependence, and if the angles concerned are not equal, then we will see a frequency (Doppler) shift, even in the situation that Tach prefers.

At this point, I present no proof of this. I merely wish to highlight the issues I see right now, in the hope that Tach can make his case watertight in his last post of the debate. If I have the time and good health, perhaps I will be able to resolve the questions I have raised in my final post of the debate. Otherwise, they may have to wait until some time after the debate is over.

All of this is academic to the debate itself, since I have already technically "won" with my first post. Nevertheless, I'd like to leave as few loose ends as possible, if possible.

In the interests of tying up all the loose ends, I'd also like to take Tach up on his generous offer to post the relevant derivation of the relativistic Doppler effect (including reflection off a moving mirror) from Pauli's book, cited in Tach's previous post.

 

[Reiku]

Wrong, I have already explained (at least twice), that all I needed to do is to show that there EXIST valid cases when there is no Doppler shift. This is the point that you, AN, etc. failed to understand. It is good to see that you finally understood the cases explained in the file.

I've seen you more wrong than you are right.

Explain yourself.




.......................................................In fact don't. It will inexorably lead to a thread like this.

 

[Tach]

I realise that this post is about 12.5 hours after the deadline by which I should have posted. If Tach wishes to end the debate here, then we can close this thread. I will then post in the related Discussion thread instead. Otherwise, with Tach's agreement, we each get one more post after this one to end the debate.

You have had more than 4 days to post your rebuttal. Moreover, you have had more than 14 days to prove the false claims of finding errors in my file (post 241). At this point, you have not done either and I do not see how giving you extra time will produce anything of any value. Please lock the thread, no more posts.

What worries me about Tach's file (which is what I will concentrate on rather than the article it seems to be copied from) is that the frames of reference are not clear (at least not to me). In particular, Tach uses the fact that the angle of reflection from the mirror (wheel) is equal to the angle of incidence. It seems to me that this would only be true in the frame of the mirror, and not in the source/detector frame.

This was explained to you (and pete) several times already. The calculations are done in the frame of the mirror. The mirror acts a receiver for the ray from the source and acts as a source for the ray reflected towards the camera.

In the source/detector frame, the angles will NOT be equal, due to the aberation of light. And if the angles are not equal, then it seems to me that the source frequency cannot possibly equal the detected frequency, even according to Tach's formulae (assuming they are correct - and I also reserve judgment on that at this point).

The calculations are not done in the source/detector frame, they are done in the mirror frame. This is the third time I explain this.

In the source/detector frame, the angles will NOT be equal, due to the aberation of light.

This is a hilarious error that shows what happens when you shoot from the hip and you don't do any calculations.
Let's have two angles that are equal in a frame S: $$\theta_1=\theta_2$$

In frame S', the angles transform into $$\theta'_1$$ and $$\theta'_2$$

The aberration formula tells you that $$cos(\theta'_1)=\frac{cos (\theta_1)-v/c}{1-v/c . cos(\theta_1)}$$ and $$cos(\theta'_2)=\frac{cos (\theta_2)-v/c}{1-v/c . cos(\theta_2)}$$

So, you can easily see that $$cos(\theta'_1)=cos(\theta'_2$$

with the trivial conclusion $$\theta'_1=\theta'_2$$.

I am quite sure that you'll try to wiggle out of your above goof by claiming that $$\theta'_1=2 \pi -\theta'_2$$. To preclude that, think about how you can have that in the case of trivial reflection when the angles are less than $$\pi/2$$

In the interests of tying up all the loose ends, I'd also like to take Tach up on his generous offer to post the relevant derivation of the relativistic Doppler effect (including reflection off a moving mirror) from Pauli's book, cited in Tach's previous post.

Trouble is, if you did your homework you would have found that the angles are equal.

At this point, I present no proof of this.

You had ample time to produce a proof. Had you done the calculations, rather than simply shooting from the hip starting two weeks ago and lying that you "found errors" in my calculations, you would have found out that your gut feel on the issue is incorrect.

In the interests of tying up all the loose ends, I'd also like to take Tach up on his generous offer to post the relevant derivation of the relativistic Doppler effect (including reflection off a moving mirror) from Pauli's book, cited in Tach's previous post.

I have cited this reference several days ago for your benefit. You mean that all this time you couldn't find the time to find it , especially in the context that it is a classical text refuting your opinions? You should have tried to find it first time. I will post it first thing after I return from skiing.

All of this is academic to the debate itself, since I have already technically "won" with my first post. Nevertheless, I'd like to leave as few loose ends as possible, if possible.

No, you have not "won" since you failed to address ANY of the explanations in the OP even after two weeks of working on this issue. You have actually lost but you will never admit it. The above explanations are the last ones you will receive on the subject. Now, please lock the thread.

 

[AlphaNumeric]

Mister has quoted Tach showing his dishonesty. Tach's now claiming he wasn't making a carte blanche statement but only that some part of the mirror for some configuration wouldn't show a Doppler shift.

Tach, you say a bunch of us failed to understand. Why didn't you explain it? Why didn't you clearly state it? Why didn't you say the moment I gave my example "Oh you misunderstood, I'm saying that it's possible, not that it always happens." Instead you do nothing but "You're wrong, shut up whining".

This I view as you back peddling and I think you realise it because you deleted your post, but Mister quoted you before you realised how your post would be viewed. You're doing as people have said you have done in the past, realising you're mistaken and trying to reinterpret what you were claiming to avoid admitting your mistake. It's extremely dishonest.

 

[James R]

Moderator note:

The debate has now ended. James R defaulted on the time limit by 12.5 hours, and Tach refused to allow any extension.

---

Posts 133 and 135 of the current thread would have been part of the debate if Tach had allowed it to continue. However, with Tach having vetoed that, I have moved them here.

 

[James R]

The Debate is over, so both Tach and I may comment freely in this thread now.

I note that the Debate was conducted by Tach in a spirit of bad faith from the start. To not allow me a lattitude of 12.5 hours is petty, to say the least. I expected nothing better from Tach, however.

I do not intend to respond immediately to Tach's last post. Perhaps I will reply at some future date, if I can be bothered. I owe Tach nothing, and frankly I have more important things to do with my time right now than extend any courtesies to Tach. I am (hopefully) recovering from a nasty illness, and I don't need his shit.

I guess Tach realised that he couldn't win the debate fair and square, and so he pulled the plug the first chance he got.

 

[Emil]

My reasoning:
A light source illuminates a toothed wheel .
The toothed wheel rotates clockwise.
Light source looks like this:

Light source illuminates half of the toothed wheel.

Tooth on the extreme "right" approaches the light, and by reflection there is Doppler effect.
Tooth on the extreme "left" is moving away from light, and by reflection there is Doppler effect.
The tooth of the toothed wheel which is in the middle (on the axis of the light source and the center toothed wheels) neither moving away or approaching the light source and there is no Doppler effect.

Now imagine a lot of small teeth on the toothed wheel . Almost we can talk of roughness.
The phenomenon is like the previous one but the light that is reflected is much weaker (reflecting surface is much smaller).

When roughness disappears so it is a mirror, light reflected from moving parts is zero.

So there is no Doppler effect from a wheel that spins only if the surface is mirror.

Oops ... I did this reasoning for a wheel that rotates but does not approaching.
If the wheel is running toward the light then obviously there is Doppler effect.
If this is the case for me so obvious that there is Doppler effect, that I quit the subject.

 

[Tach]

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:
$$\phi_s = \phi_v + \pi - \theta $$

You are very close to understanding the issue but, because you are so fixated on proving me wrong, you end up with an incorrect understanding.

$$\phi_o = \phi_v + \theta$$

This is incorrect, you need to calculate the angles in a consistent way. You calculated $$\phi_s$$ as the angle between the light ray and the velocity $$v$$, you need to do the same thing when calculating $$\phi_o$$. The correct expression is $$\phi_o=\pi-\theta-\phi_v$$

This is the smaller of the two errors in your post (the larger one follows below) but it has a bearing on the deriving the correct result.

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):
$$f_{observed} = f_{source} \ \frac{1 +\beta \cos(\phi_v + \theta)}{1 +\beta\cos(\phi_v - \theta)}$$

Because of the above mistake , you end up with the incorrect formula. The correct formula is:

$$f_{observed} = f_{source} \ \frac{1 -\beta \cos(-\phi_v + \theta)}{1 -\beta\cos(\phi_v + \theta)}$$

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

.

In a few PM's I explained to you, through several different methods why $$\phi_v=0$$. Here is another explanation : in the frame of the axle, $$\phi_v=0$$. Then, in ANY other frame (including the frame of the ground, the one that you insist in doing the calculations despite of being absolutely the WORST choice for such calculations) you will have $$\phi'_v=0$$. The proof is trivial:

$$cos(\phi'_v)=\frac{cos(\phi_v)-v/c}{1-v/c. cos(\phi_v)}=1$$.

Now, when you plug in $$\phi_v=0$$ into the corrected general expression of the Doppler effect, you will get the exact result I've been telling you all along, zero shift.

In other words , in ALL frames, the microfacets composing the wheel rim move parallel with themselves. This means , as you well know, that the Doppler effect is null (see the book by Pauli and the paper by Bateman). It is interesting to note that you too, could have derived the correct result had you not been so fixated in proving me wrong. After all, you have been given all the tools...