Pete said:He is now arguing that a moving mirror's surface is necessarily parallel to its velocity.
If that's true, it's no wonder he couldn't grasp my argument in the debate.
Pete said:He is now arguing that a moving mirror's surface is necessarily parallel to its velocity.
It's getting worse. Much worse.
Tach was earlier insisting that in the mirror's rest frame, doppler shift depends on the angle of the mirror's velocity.
I don't know if he's let that go yet.
He is now arguing that a moving mirror's surface is necessarily parallel to its velocity.
Go figure.
It is very simple, really. Let's try a different way, since you either cannot or will not understand any of the other proofs. We agreed that in the frame of the axle each elementary mirror moves parallel to itself while the wheel is spinning. Yes or no?
That said.., if you are speaking of the mirrored surface being the edge of the rim when viewed from 90 degrees to the wheel's rotation, the whole rim edge mirror, remains parallel to itself, while the wheel spins.
Good, try explaining this to pete and scores of the others. BTW, you don't need any viewing from 90 degrees angle.
Good, try explaining this to pete and scores of the others. BTW, you don't need any viewing from 90 degrees angle.
Heh. Yes.
The scenario where we have a section of stainless steel pipe of zero thickness:
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And a stainless steel washer of zero width:
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Both of which display perfect specular reflection, and are moving between a light source and a camera, with some velocity $$ v \leq \omega r$$, while rotating about their C[sub]n[/sub] axis, with some angular velocity $$\omega$$, oriented so that the plane containing their nC[sub]2[/sub] axes also contains the light source and the camera (or. alternatively, oriented so that the vector of motion is perpendicular to the C[sub]n[/sub] axis - same thing, different wording).
And now that the problem has been defined unambiguously, sensible discussion of the two seperate problems, or components of the problem, can begin.
Source
Tach, it seems, is now talking about the second scenario (a washer of zero width), where everything in the debate, and the discussion so far, including the material provided by Tach to support his argument, appears to have been discussing the first scenario (pipe of zero thickness).
That's true. It was however necessisary to be certain that what I was describing was that mirrored edge and no part of the tread surface that would be visible from any other angle.
:roll eyes:You know, if you don't understand things, just ask. This problem has been previously defined in mathematical terms and nowhere was any discussion about thickness. The reflection is off a circle, you only need to follow the drawing. While creating your own drawings might be fun, it doesn't add anything to the description, especially when it is not accompanied by any mathematical formalism.
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The above sketch has only one light source, only one camera, and only one wheel rolling between the two.
Do you notice how the length of the optical path changes from 5.829 to 5.656 between the first and second slides? As the length of the optical path becomes shorter, the frequency of the reflected light must become greater. Surely you must agree with that simple principle?
Then do you see how the length of the optical path changes from 5.656 to 5.829 between the second and third slides? As the length of the optical path becomes longer, the frequency of the reflected light must become lower, surely?
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Tach, would you casre to tell me how the above diagram is incorrect?
What in "valid for only a part of the wheel only" did you not understand? The equations are valid for the entire circumference of the wheel. This is the second time I point this to you. There is only one light source , only one camera and one wheel rolling between the two. The LHS and RHS of the drawing bellow represent two SEPARATE scenarios. This is explained clearly in the text. There is only ONE light source in each scenario.
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Having said that, the same exact formalism applies to BOTH "tube" and "washer". If you did the math accompanying your pretty drawings, you would have found that out.
Isn't it?So, it isn't clear why you have created the "washer" case.
Can you post the math formalisms explaining the difference(s)? That would be a lot more interesting.
Perhaps, later in the day, or later in the week, if I can find time to sit down with a pen and some paper.
Actually, the zero Doppler effect applies not only to the circumference but to the "sidewall" and the spokes as well. I think I have mentioned the spokes before, actually I am quite sure I did. The same exact formalism as the one applied to the rim applies to the other parts of the wheel.
Posting Neddy Bates' images was about the experimental setup, not the formalism.Out of curiosity, why did you post the repeated errors by Neddy Bate? Is there any connection with your formalism?
In an idealized hypothetical, it may apply to any flat and ideally mirrored surface, which is moving parallel to the plane of the light source and camera.
For a flat mirror this will occur of some time equivalent to the length of the mirror in the direction of its motion divided by its velocity.
In the case of a wheel the time that the light reflects between the source and camera is diminished by the reduction in the length of the mirrored surface in the direction of the wheel's motion.
If you assume that the light can reflect off of a wheel that is moving parallel to the plane of the light source and the camera, at any point other than that point midway between the source and camera, the light path between the source and wheel and the light path between the wheel and camera will not be equal..,
and the speed of the wheel relative to the source and the speed of the wheel relative to the camera will also not be equal. And there will in practical terms be a detectable Doppler effect.
If any other result were true, we could not explain the cosmological redshift, as resulting from an expansion of space. Where in this case, the instantaneous expansion of space at the instantaneous location of a photon, affects the wavelength of the photon.
The special case of an ideal mirror with a velocity parallel to the plane of the light source and camera.., may be seen as an exception, but only in an idealized hypothetical thought experiment.
Now the question is, not whether others understand you, but rather do you understand anyone else? If you did there are numerous times your responses should have begun with, "OK I see your point, but what I was referring to...
Posting Neddy Bates' images was about the experimental setup, not the formalism.
I repeat. My posting the other two images is about the experimental setup not the conclusions he reached from that setup.You realize that the claims embedded in his posts/figures have been shown to be false, right?
I repeat. My posting the other two images is about the experimental setup not the conclusions he reached from that setup.
You do realize the difference between the experimental setup, and the conclusions reached based on that experimental setup, don't you?
Really? That's interesting. Not surprising, but still interesting. Because when I look at those three images, I see what appears to be the same experimental setup, with the minor (and frankly irrelevant) illustrative difference that one of them contains an extra light source.Yes, I do. I don't see any "experimental setup"...
The scenario where we have a section of stainless steel pipe of zero thickness, and a stainless steel washer of zero width, both of which display perfect specular reflection, and are moving between a light source and a camera, with some velocity $$ v \leq \omega r$$, while rotating about their C[sub]n[/sub] axis, with some angular velocity $$\omega$$, oriented so that the plane containing their nC[sub]2[/sub] axes also contains the light source and the camera (or. alternatively, oriented so that the vector of motion is perpendicular to the C[sub]n[/sub] axis - same thing, different wording).
Along with the repetition of the experiment, with the pipe and the washer painted, is an accurate summation of your experiment, and that it accurately breaks your experimental setup down into its components - with the proviso that your 'no slip' condition implies $$ v = \omega r$$, rather than $$ v \leq \omega r$$. My time is both precious and expensive, and I have no wish to sit down and perform any form of analysis for you to turn around at the end of it and claim the experimental setup was wrong to begin with.
Let us agree on the setup first.