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

...you being clearly one of the biggest offenders. See your latest blunder above. At times, both pete and Neddy Bate have made the incorrect claim that my derivation is valid only for some part of the mirror only. Since then, they have surreptitiously refrained from repeating that false claim.

You have been given plenty of math. Perhaps if you stopped being so offensive and you spent more time attempting to understand the math, you would have been less offensive.

Don't get all twisted in your knickers. Nice picture, is this from your driver license?

What you have consistently misunderstood in the scenario I posted is that, when ANY portion of the wheel approaches the camera (resulting into blueshift) it also recedes from the source (resulting into an equal amount of redshift) and that the two effects cancel each other. You have been very consistent in failing to understand this simple effect. Testimony is the above. Combined with your shrill tone laced with ad-hominems (see above), I decided that it wasn't worth answering until after the debate was over. Now, you have your answer.

 

Log in or Sign up to hide all adverts.

Tbh, I don't think he cares really.

 

Log in or Sign up to hide all adverts.

About the stuff you post, like your claim that there has been on observation of zero spin particles? You are right, absolutely not. Most of the time I ignore trolls who have zero contribution to the threads.

 

Log in or Sign up to hide all adverts.

Please Register or Log in to view the hidden image!

Tach, would you casre to tell me how the above diagram is incorrect?

 

If you use SR's specific math, even in this classic example

Please Register or Log in to view the hidden image!

the Doppler effect is not valid.
Contradicts The Principle of Invariant Light Speed.

 

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.

Please Register or Log in to view the hidden image!

 

1. There is no such thing as "doppler shifts opposite in sign".
2. The statement is true for ANY instance of a surface or collection of surfaces moving parallel with their plane. A rolling wheel is such an EXAMPLE. A mirror moving in its own plane is ANOTHER EXAMPLE.

3. There are no "observers on the rim of the wheel".

 

Where?

I thought about it and there is no such stuff in the issue being debated.

In addition, there are no "opposite signs" for the Doppler effect in the formalism of the problem. Besides, you have made another embarrassing goof, the Doppler effects in cause are definitely NOT "of equal magnitude (and opposite sign) " as you claimed.

Looking at the above goofs, you should definitely think again.

 

Prove it or retract your incorrect claims.

Is this what you think that I claimed? Given that your claim 1 is not even wrong your claim 2 may or may not be true. I vote for the third option: "Hilarious". I really don't understand why you are getting involved in stuff that you have demonstrated to know nothing about.

 

There is no "summing" of Doppler effects. If you bothered to read (and comprehend) you would have seen that the effect is multiplicative. Give it a rest.

 

I said earlier I thought the issue would boil down to one very fundamental error. Now that the debate is over, I will explain.

Tach's scenario is ideal, whereas JamesR's was a classical real-world view. It seems odd to me that none of the discussion (that I read) noted this error.

As soon as I read Tach's paper I was surprised he was assuming the equal pairing that also assumes perfect flatness and perfect measurement of returned light. As soon as JamesR said "radar" I realized he was thinking in terms the real-world systems, which are limited by ambient noise, the signal to noise ratio of the detector, and other effects that would normally not produce a zero just because the object is rolling.

Tach's method relies on a pairing of complementary rays, one advancing, one regressing ,whose sum in doppler is zero. So far so good. But the implications on the mirror are unrealistic, since there exists no surface in the real world that is continuous, perfectly flat in one dimension, and perfectly round in another. Only such a surface would be capable of producing such a pairing.

I thought it was an interesting idea that he advanced and that he should get kudos because it is thought provoking and can be applied in practical applications. For example I can envision a test for surface flatness that simply uses a doppler measurement as he proposes, then as the polishing operation improves, the doppler will diminish to some lower limit, not zero, but who cares.

I was surprised that the audience was so antagonistic. It's great to put an idea to the test, but it was hard to find any remarks directly contradicting his thesis. To me the math looked fine. As did JamesR's. Both are correct as far as I can tell. The fact that they get different results doesn't seem to have anything to do with the very laborious arguments that ensued. One was ideal, one was not. It would have been a lot more interesting and informative if the discipline many of you have in analysis and teaching was expended answering the direct and fundamental question he raised, namely, the sum of infinitesimals having pairs in complementary phases is zero.

 

What does the above have to do with the Doppler effect being discussed? Hint: nothing.
So, you didn't read anything, you are just making stuff up to cover your hilarious goofs.

I admit, I didn't take any "calculas" but I did take "calculus".

 

Tach, are you sure you've carefully considered the possibility that maybe you are wrong, and that the people who are consistently and independently saying the same thing might be right?

No, Tach, we're using different equations with different angle specifications.

The first equation gives the frequency detected by a stationary observer (the mirror) for light from a moving source.
For that equation, \(\phi\) is the angle from the observer-source displacement vector to the source velocity vector.

The second equation gives the frequency detected by a moving observer for light from a stationary source (the mirror).
For that equation, \(\phi\) is the angle from the source-observer displacement vector to the observer velocity vector.

Careful, Tach, you're getting approaches confused.
The approach I took in my diagram uses the mirror rest frame (the one you insist is best).
\(\phi_v\) is the angle between the mirror surface and the observer velocity.
That angle is not zero except at a single point on the wheel (two points actually, except one is on the far side).

That appears to be the relativistic aberration equation, which transforms the apparent angle of light rays for moving observers.
You can't naively apply it to just any angle.

Consider the leading surface of a rolling wheel. The vertical microfacet obviously has a horizontal velocity component, correct?

 

The problem with the wheel rolling parallel to the plane of the light source and camera, is that unless the light source and/or camera are also moving there can be only one instant where the light from the source reflects off the edge of the rim and strikes the camera. And that one unique instant must be when the light path from the source to the wheel is equal to the light path from the wheel to the camera.

It may be that in that one instant there might be no Doppler effect, but in the next instant, even assuming that the camera could still catch the reflected light, the light paths will not be equal, the angles of the two light paths relative to the wheel will no longer be equal and a Doppler effect will become detectable.

The rim of a rolling wheel is not equivalent to a flat mirror in this hypothetical.

Repeatedly Tach has claimed others did not understand him and it was his claim that the original hypothetical, involving a wheel moving "away" from a light source and observer was wrong which led to this whole debate/discussion. It was, is and has been, Tach who could not understand the original hypothetical, or perhaps he has just been arguing for the sake of arguing.

Under almost all conditions involving a moving object, a light source and an observer, a Doppler effect will occur. That is what the original hypothetical two or three threads back involved. And it is in only very narrowly defined cases, where light reflected from a rolling wheel would not be Doppler shifted by the motion of the wheel.

 

We have been over this repeatedly in the PM exchange, "that angle" is zero in all points in EVERY frame attached to EVERY point on the wheel circumference. The same way it is zero at all points in the frame of the axle.

Indeed, how else do you plan to account for the transformation of angles? Especially given the fact that \(\phi_v\) is zero for all points of the circumference as measured in the axle frame?

Not in its own frame. Not in the frame of the axle. In both frames the microfacet moves in its own plane.