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

[RJBeery]

Moderator note: The participants for this debate are Tach for the affirmative and RJBeery for the negative. Agreed parameters for the debate can be found in the Proposal thread.

[thread=111038]Discussion thread[/thread]
[thread=111009]Proposal thread[/thread]

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There is no Doppler shift off a matte wheel rolling between a source and the receiver.

The above statement is false, and I will demonstrate why. The apparent basis for this belief by Tach is here, and it is this paper that I will critique and dismantle.

There are no technical or mathematical errors in the paper if we restrict the analysis to mirrored wheels. The title of the paper, "The color of objects rolling at relativistic speeds", does present a problem though because it exposes the author's belief that this would extend to non-mirrored objects.

The difference is that mirrors are specular and have the ability to reflect all frequencies of light, whereas a matte wheel is restricted to reflecting a particular one. In the scenario below we'll examine a green wheel which emits a frequency of 600 THz (in it's axle frame). [Note: We'll examine a sliding wheel rather than a rolling one; a sliding wheel will exhibit uniform Doppler effects while a spinning one will exhibit varied Doppler effects across its surface due to the varying relative speeds between the light source, a point on the wheel, and the camera. Since the paper argues that no Doppler exists for any movement velocity V, it would also apply to each point on the wheel whether its relative movement was rotational in nature or not. Therefore, proving Doppler effects of a sliding wheel is equivalent to proving Doppler effects on a rotating one.]

Here we see the wheel receding from the light source and moving towards the camera.

$$f_0$$ = frequency emitted by light source
$$f_{wheel}$$ = frequency emitted by wheel
$$f_{s'}$$ = frequency perceived by the camera

$$f_{wheel} = \frac{f_0\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}cos\phi_{1}$$​

This represents the red-shifting of $$f_{0}$$ due to the relative movement between the light source and the wheel.

$$f_{s'} = \frac{f_{wheel}(1+\frac{v}{c}cos\phi_{1})}{\sqrt{1-\frac{v^2}{c^2}}}$$​

This represents the blue-shifting of $$f_{wheel}$$ due to the relative movement between the wheel and the camera.

Combining them cancels out the Doppler effect such that $$f_{s'}=f_0$$. I'm sure you agree with all of this, Tach, because I'm basically just parroting your paper. (I'm doing this mostly because I've always needed an excuse to learn TEX )

Anyway, your mistake is in neglecting the optical nature of a colored, matte wheel. Unlike a mirror, the wheel would only reflect light of 600THz in the frame of the wheel, not light of 600THz in the frame of the light source. In this case:

$$f_{wheel}$$=600THz so

$$\frac{f_0\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}cos\phi_{1}$$=600THz​

Solving for $$f_{0}$$...

$$f_0 = \frac{f_{wheel}(1+\frac{v}{c}cos\phi_{1})}{\sqrt{1-\frac{v^2}{c^2}}}$$​

Arbitrarily picking

$$\frac{v}{c} = .5$$ and $$\phi_1 = \frac{\pi}{4}$$​

We get

$$f_0 = \frac{600THz(1+.3535)}{.75} = $$1082THz​

Since we have already established that $$f_{s'}=f_0$$, we conclude that

$$f_{s'} = $$1082THz​

Q.E.D.

 

[Tach]

The difference is that mirrors are specular and have the ability to reflect all frequencies of light, whereas a matte wheel is restricted to reflecting a particular one.

The fact that a matte wheel reflects only one frequency (more correctly it reflects only a subset of the incident frequencies) is irrelevant. The reflected frequencies are subjected to the same exact laws in either case. In other words, the frequency reflected by the matte surface has been "redshifted" when the wheel rolled away from the light source only to be blueshifted by the same amount in its approach to the camera. The fact that a spectrum of frequencies has been absorbed by the wheel in the process is irrelevant. Herein lies RJBeery's error.

In the scenario below we'll examine a green wheel which emits a frequency of 600 THz (in it's axle frame). [Note: We'll examine a sliding wheel rather than a rolling one; a sliding wheel will exhibit uniform Doppler effects while a spinning one will exhibit varied Doppler effects across its surface due to the varying relative speeds between the light source, a point on the wheel, and the camera.

You are compounding your errors: You have already been shown that , for the scenario of a wheel rolling between a source and a receiver, there is NO Doppler effect.

Since the paper argues that no Doppler exists for any movement velocity V, it would also apply to each point on the wheel whether its relative movement was rotational in nature or not.

No, it doesn't. You need to understand what has been explained in prior threads. In a nutshell, there is no Doppler effect in the case of surfaces moving in their own plane between a source and a receiver stationary wrt each other. This is equally true for matte or mirror like surfaces. You would need to read the references on this subject (Pauli, Bateman). They are quite old but correct.

Combining them cancels out the Doppler effect such that $$f_{s'}=f_0$$.

Congratulations, you contradicted your first sentence of the thread, i.e. you contradicted yourself.

Anyway, your mistake is in neglecting the optical nature of a colored, matte wheel. Unlike a mirror, the wheel would only reflect light of 600THz in the frame of the wheel, not light of 600THz in the frame of the light source.

In the frame of the wheel (the frame used for performing all the calculations that you copiedfrom me) , the correct application of Maxwell equations (see for example Jackson) show that the frequency of the reflected light is equal to the frequency of the incident light. This is basic electrodynamics, if you want to write your own fringe theories, feel free. On the other hand, if you want to do the correct calculations, using the basic physics gives you zero Doppler shift.
Now, in a moving frame, like the frame of the light source the two frequencies are STILL equal, so you have no valid argument.

We get

$$f_0 = \frac{600THz(1+.3535)}{.75} = $$1082THz​

Since we have already established that $$f_{s'}=f_0$$, we conclude that

$$f_{s'} = $$1082THz​

This is the second time when your conclusion contradicts the original claim of your debate.

 

[RJBeery]

In a nutshell, there is no Doppler effect in the case of surfaces moving in their own plane between a source and a receiver stationary wrt each other. This is equally true for matte or mirror like surfaces.

I'm agreeing with you completely, and I concur with the math in your paper. I agree that there is no net Doppler effect between a the primary source and receiver, and my extension of your work as illustrated above shows this. $$f_0 = f_{s'}$$, just as you have been claiming in other threads.

However, we are discussing a Doppler shift off of the matte wheel (i.e. the secondary light source) as described in the debate topic, not a Doppler shift of light between the primary light source and receiver. The IS a Doppler shift off of the green matte wheel. The wheel is green, yet the camera would see it as blue-shifted, just as common sense would dictate. I think it's fair (nay, ironic) to ask for a mathematical proof if you claim otherwise.

 

[Tach]

I'm agreeing with you completely, and I concur with the math in your paper. I agree that there is no net Doppler effect between a the primary source and receiver, and my extension of your work as illustrated above shows this. $$f_0 = f_{s'}$$, just as you have been claiming in other threads.

Good.

However, we are discussing a Doppler shift off of the matte wheel (i.e. the secondary light source) as described in the debate topic, not a Doppler shift of light between the primary light source and receiver.

No, I was very careful this time (after the experience with JamesR) not to allow any wiggle room in the definition of the debate. We are discussing "The absence of the doppler effect for a matte wheel rolling between a source and a receiver", remember?
The matte wheel acts like a reflector, not like a "secondary light source". It is simply reflecting frequencies in a selective manner, that is all.

The IS a Doppler shift off of the green matte wheel. The wheel is green, yet the camera would see it as blue-shifted, just as common sense would dictate. I think it's fair (nay, ironic) to ask for a mathematical proof if you claim otherwise.

This is of course, false. The wheel is NOT "green", it simply reflects frequencies in the green spectrum preferentially (in your case you have declared green to be the ridiculous frequency of 600THz but no matter). The wheel will not reflect ANY frequency incoming from the source UNLESS the speed of the wheel wrt the source is such that the incoming frequency is not in the 600THz spectrum. So, in your case, the source emits 1082 THz frequency, the wheel , due to its motion receives it downshifted to 600THz and reflects it towards the camera. The camera receives it back as upshifted to the original 1082 THz. No Dopler shift.

 

[RJBeery]

A few points...

1. The wheel is "rolling between" the source and receiver. Standard English.
2. You claim that the wheel isn't a secondary light source now, but in your paper you wrote "the object acts as a secondary light source" in the opening sentence.
3. The wheel is not green? I'd be curious to know how you would define a green object if my description does not suffice.

Changing English grammar rules and the very definition of words is a pretty desperate tactic, Tach.

You referred to "your" math, in a possessive sense, and I see that Aqueous Id has called for others to "give credit where credit is due". However, when you agreed to this debate, one thing it did was verify to me that THIS ISN'T EVEN YOUR WORK. I had been confounded how a UC Berkeley Physics student could put so much work into a paper with such a glaring hole (i.e. the topic of this debate), so I read the ORIGINAL paper in its entirety...guess what? The author of that paper got it right. It was YOU that misinterpreted HIS paper, then you ran off onto the internet to make pronouncements about your misinterpretation!

HERE is the paper you've been pulling your ideas from. Page 5, section 3 contains the section that you've been touting. Starting at page 6, however...

Combining (3.4) and (3.5) we conclude that there is no frequency shift due to the relative motion between the object and the camera for any angle ' [0, ] φ ∈ π . The above reasoning applies to mirror-like objects that reflect the light coming from the source without changing its frequency. This type of reflection is called specular in computer graphics. There is a second type of objects that absorb the incoming spectrum of frequencies and re-emit a selected subset of frequencies. This type of reflection, characteristic to matte objects is called diffuse. In the case of diffuse reflection, the moving body acts as a secondary light emitter and the perceived frequencies follow the relativistic Doppler effect equations (see Fig. 6)[...]When the object is aligned with the camera objective ( ' / 2 φ = π ), we will detect a blue shift consistent with the transverse Doppler effect

Perhaps I should contact A. Sfarti in California and ask him or her to chime in on the subject?

 

[Tach]

A few points...

1. The wheel is "rolling between" the source and receiver. Standard English.
2. You claim that the wheel isn't a secondary light source now, but in your paper you wrote "the object acts as a secondary light source" in the opening sentence.

1. 
   
   I did not write any paper, what are you babbling about?
   
   

   [*]The wheel is not green?

   
   Correct, you need to learn proper physics, this has nothing to do with grammar. So far, you managed to get all your physics wrong. Which is not a surprise.

 

[RJBeery]

That response is devoid of a substance. Please explain the source of your original calculations, and why they appear to be identical to the Sfarti paper. Ten seconds of comparison makes your plagiarism obvious and another ten seconds makes your misinterpretation of the author's conclusion regarding matte wheels obvious as well. Or shall I accept your non-response as a concession?

 

[Tach]

That response is devoid of a substance. Please explain the source of your original calculations, and why they appear to be identical to the Sfarti paper. Ten seconds of comparison makes your plagiarism obvious and another ten seconds makes your misinterpretation of the author's conclusion regarding matte wheels obvious as well. Or shall I accept your non-response as a concession?

So, once you have been shown to have contradicted yourself and your position in the OP twice, not once, you switched to personal attacks?
Please answer the direct challenges (posts 2 and 4) to your errors. This may prove difficult since you already contradicted your debate stance twice in a single post.
For your information, I do not use the author's stance on matter surfaces. He/she is talking about raytracing ( a computer graphics process, not a physics one). In raytracing mirror-like surfaces are treated as in physics but matte surfaces are NOT. The reason is that matte surfaces reflect light in ALL directions, making a raytracer an untractable process due to the infinite number of rays that would need chasing. This is why in a raytracer program rays cast from the eye are traced back all the way to the light source for mirror-like objects and ONLY back to the object for matte ones. This is not the case in physics, where rays of light are traced forward all the way from light source to object to camera for both matte and mirror-like objects, as physics dictates. If you don't understand the basics, as you have demonstrated repeatedly, you should not engage in debates.
In this thread we are debating the physics , not the computer graphics simplification of the problem. You obviously have a zero understanding of both.
Now, that I explained to you the source of your confusion, please answer the challenges (posts 2 and 4) or admit that you have been wrong all along.

 

[RJBeery]

The fact that I agree with your calculations for mirrored wheels does not mean there is a contradiction in my stance, therefore there is nothing to address. There is no Doppler shift between the primary light source and the camera, we both agree on that, and this is not what we're debating. We are debating Doppler shift off a matte wheel. You wrote the debate topic, and I accepted it. You wrote the rules, and I accepted them. There is NO ROOM for misinterpretation of the debate topic:

There is no Doppler shift off a matte wheel rolling between a source and the receiver
​

Cannot be interpreted as

There is no Doppler shift between a source and the receiver off a matte wheel rolling ​

...due to the fact that we are both speaking English:bravo:

Now: Define a green wheel, then calculate whether that moving wheel would appear green to the camera. Show your math, as it stipulated in the rules YOU defined and outlined in the proposal thread. Do this, or concede.:truce:

 

[Tach]

Now: Define a green wheel, then calculate whether that moving wheel would appear green to the camera.

It will not, you have shown that yourself in your ill-fated OP.

Show your math, as it stipulated in the rules YOU defined and outlined in the proposal thread.

But this is not what you did: you defined the "green wheel" to be a reflecting surface at the (ridiculous) frequency of 600Thz. When illuminated with the (ridiculous) frequency of 1082Thz while rolling away from the source you concluded that the wheel has the 1082Thz "color". This means three things:

1. The wheel DOES NOT appear green (600Thz) on the camera.
2. Instead it appears to be the "color" of the source , i.e. 1082Thz.
3. Therefore, there is no Doppler shift off the matte wheel .


So, you lost, straight from the start. This is what happens when you don't know physics.

 

[RJBeery]

General direction of our disagreement:

It is very easy to prove, I can upload a file that shows all the details. Before I do, think about this The light from the source arrives at an object receding from it redshifted. The same light is reflected by the object approaching the camera, so it will be blueshifted by the same exact amount. Net effect: zero Doppler shift

BTW, Tach, this "zero Doppler shift" is only valid for mirrors. An object such as a colored, spoked wheel would indeed appear Doppler shifted. Your comment is therefore irrelevant and in the context of this thread, false.

did your armchair teach you that as well? you got to stop taking lessons from it , it is giving you bad information

A green wheel will emit green light, period. Move the wheel relative to a light source and you're only shifting the average frequency of the illuminating white light. As long as the wheel continues to be illuminated by the proper (green) frequency of light it will continue to emit that green light, and absorb the rest.

The apparent color of the wheel is indeed dependent on the the movement of the observer (i.e. camera) relative to the wheel. Conclusion: the objects under consideration in this thread do appear Doppler shifted.

Nope, because the frequency (green , in this case) is equally blueshifted when reaching the object and redshifted (by the same amount) when arriving at the camera, as I explained in the file that pete had so much trouble understanding.

MOST RELEVANT QUOTES:

An object such as a colored, spoked wheel would indeed appear Doppler shifted.

Nope

$$f_{wheel}$$=600THz
[...]
$$f_{s'}$$=1082THz
Q.E.D.

The wheel DOES NOT appear green (600Thz) on the camera

In the end, you agree with me and have been proven to now see the (green) light. Recorded history is a bitch isn't it?:thumbsup:

 

[Tach]

Here we see the wheel receding from the light source and moving towards the camera.

Conclusion (final post, according to the rules): The amusing thing is that you kept proving yourself wrong throughout the thread. See the picture above? You show the light source emitting a ray in the ultra-violet range (the ridiculous 1082THz). Do you also see the frequency of light arriving at the camera? Let me help you, it is the SAME ultra-violet at 1082Thz. So, what the camera records is the color of the source, i.e. no Doppler shift, exactly as in the case of mirror-like objects.
See how you prove yourself wrong? You've exhausted all your six posts proving yourself wrong. You did that without failing in every each one of them.

 

[RJBeery]

Yeah, Tach, and if my grandmother had tires she would be a trolleybus!
:roflmao:
DEBATE IS OVER.

 

[Trippy]

MODERATOR NOTE

Four off topic posts deleted. If you're going to post in the formal debates subforum, you would do well to familiarize yourself with the rules. Further infringements may result in warnings and or bans being issued.

Addendum, thread closed, per request, and per rules the rules.