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

[Aqueous Id]

The vertical microfacet obviously has a horizontal velocity component, correct?

Hi Pete.

I see an implicit assumption that is not correctly applied to Tach's scenario.

When you say "microfacet" it implies that the reflector is not ideal, as Tach's scenario implies.

Under Tach's scenario, there need not be any "facet" of a forward velocity. And in the ideal case he presents, the microfacet has no capacity to reflect anything.

Reflection, in the real scenario, will only occur at some scale that subdivides the mirror surface to a order of magnitude in Angstroms. Below that there is no interaction between light and the mirror material.

Yet Tach is below that threshold in his theoretical treatment, or else he is just at the threshold. He is using rays. He is decomposing the light source into infinitesimals in a theoretical sense. He is assuming a perfect reflector which is in reality impossible.

You can pick any microfacet you want that you assume has a forward velocity, and he can just as well posit that there is no facet, it's perfectly round, and the tangent is an infinitely thin line, not a facet, and it will reflect nothing. A zero dimensional line is advancing with forward velocity, but what of it?

So when you say "facet" you are slipping into the real world, quantized, molecular, and a huge disparity opens between your premise and his.

 

[Pete]

Hi Pete.

I see an implicit assumption that is not correctly applied to Tach's scenario.

When you say "microfacet" it implies that the reflector is not ideal, as Tach's scenario implies.

"microfacet" is a word that Tach invented and applied to his scenario himself.

I think you're on the wrong track with ideal vs real-world scenarios.

Everyone, as far as I can tell, is assuming ideal optics, with lossless
and noiseless reflections and transmissions.

 

[Aqueous Id]

Regardless of noise and losses which are part of the disparity, his premise remains at great disparity with you folks.

His surface is perfectly round, perfectly continuous, and perfectly reflective, isn't it? Under those conditions, his rays cancel each other in phase. The net phase is zero. No phase, no doppler.

 

[Kittamaru]

Question - can such a scenario exist in the real world...?

 

[Trippy]

So, a wheel rolling between a source and a receiver does not exhibit Doppler effect of equal magnitude and opposite signs in logarithmic space?

Stop twisting my words.

You're being dishonest, and you know it.

 

[Trippy]

Question - can such a scenario exist in the real world...?

Yes, and the available evidence from comparable scenarios contradicts Tach's assertions.

 

[Kittamaru]

Yes, and the available evidence from comparable scenarios contradicts Tach's assertions.

No no, I mean the "perfectly round, perfectly reflective, and perfectly continuous" that Aqueuous was talking about - can that really exist?

 

[Tach]

I explained that the aberration formula applies to the apparent direction of light rays. You're claiming without support that it applies to arbitrary angles.
And I've given you a specific counterexample that you seem to be avoiding:[

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

Incorrect. One of the benefits of dealing with you is that it affords multiple , different ways of proving that there is indeed zero Doppler effect of a circular wheel rolling between a source and a receiver. In the above, though you claim having done the computations in the microfacet(s) frame of reference(s) you actually keep thinking in the frame of reference of the source/receiver. No matter, I will explain in a different way.

In frame S (the axle frame) an observer measures the angle $$\phi_v$$ as $$tg(\phi_v)=\frac{y}{x}$$. In any other frame S" (like the ground frame)moving with speed $$V$$ wrt S, the angle is $$tg(\phi'_v)=\frac{y'}{x'}=\frac{1}{\gamma(V)}.tg( \phi_v)$$
Since we agreed that $$\phi_v=0$$ for ALL circumference points in frame S, it follows that $$\phi'_v=0$$ for ALL points on the circumference in the ground frame S', thus contradicting your long held belief that $$\phi'_v=0$$ ONLY for two points (the ground contact point and its polar opposite). Come to think of this, this is quite intuitive, the microfacets move tangent to a circle in the axle frame and tangent to the ellipse in the ground frame. This is enough to prove that the two angles in the calculation of the Doppler shift are equal, resulting into the cancellation of the red and blue shift.

So, let me me clearer:
In my diagram, $$\phi_v$$ is the angle between the mirror surface and the camera velocity, not the mirror velocity.

Fine, see above, the angle $$\phi_v$$ is zero for all points on the circumference.

Once again, you might like to answer this simple question:
Consider the leading surface of a rolling wheel. The vertical microfacet obviously has a horizontal velocity component, correct?

In the frame of the ground, certainly. I showed you the exact equations long ago, in the "rolling wheel" thread. So does the orientation of the microfacet, so the velocity and the orientation have a zero degree angle between them (see above). Which brings us back to the zero Doppler shift.

 

[Tach]

Question - can such a scenario exist in the real world...?

Perfectly reflective mirrors do not exist but, interestingly enough, this shortcoming does not affect the result.


Perfectly circular wheels do not exist. This affects the result of the scenario.

We were talking about a theoretical scenario. The argument is not about the realism but about the equations describing the theory. It is already known (for more than 100 years) that the case of translation motion already exhibits zero Doppler effect. The next shoe to drop is the realization that this is also true for a (circular) wheel rolling between a source and the camera. The emotions are running very high in the opposition campus

 

[RJBeery]

To the extent that Tach refers to this as his work, he is a fraud. In addition, he has misinterpreted aspects of the very work he borrowed.

 

[OnlyMe]

Perfectly reflective mirrors do not exist but, interestingly enough, this shortcoming does not affect the result.


Perfectly circular wheels do not exist. This affects the result of the scenario.

We were talking about a theoretical scenario. The argument is not about the realism but about the equations describing the theory. It is already known (for more than 100 years) that the case of translation motion already exhibits zero Doppler effect. The next shoe to drop is the realization that this is also true for a (circular) wheel rolling between a source and the camera. The emotions are running very high in the opposition campus

This is the second time you have referred to a proof of this, 100 years in the past. Since it was far beyond the technology of the day to experimentally prove this at that time.., and perhaps even today in a laboratory, you must be referring to a mathematical proof....., a horse of quite a different color. A mathematical proof can provide the basis for a hypothesis, a model, a theory, even a discussion of logic. But without conclusive supporting observation, it cannot rise to the level of proving anything more than, a hypothesis, a model, a theory, or discussion of logic.

Cite the experimental proof, from 100 years ago. I have been under the impression that the Doppler effect for light post dates that claim and thus could not be any part of an experimental proof at that time.

It seems you have a great deal of difficulty in separating mathematical proofs and logic from experience, observation and that which has been experimentally confirmed and/or proven.

 

[Pete]

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

Incorrect.

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

In the frame of the ground, certainly.

Contradicting yourself is a bad sign.

In the above, though you claim having done the computations in the microfacet(s) frame of reference(s) you actually keep thinking in the frame of reference of the source/receiver.

The diagram and explanation I posted is in the microfacet frame. I'm also able to think in the ground frame and axle frame when appropriate.

In frame S (the axle frame) an observer measures the angle $$\phi_v$$ as $$tg(\phi_v)=\frac{y}{x}$$. In any other frame S" (like the ground frame)moving with speed $$V$$ wrt S, the angle is $$tg(\phi'_v)=\frac{y'}{x'}=\frac{1}{\gamma(V)}.tg( \phi_v)$$

Please define which angle you're talking about. You appear to be referring to a different $$\phi_v$$ than is defined in my diagram.

Come to think of this, this is quite intuitive, the microfacets move tangent to a circle in the axle frame and tangent to the ellipse in the ground frame.

It seems you've given up thinking altogether. A simple application of common sense immediately shows that the microfacets do not move tangent to the ellipse in the ground frame.
Immediately next to the ground contact point, for example, the almost horizontal facets move almost vertically.
The vertical facets at the leading and trailing extremities of the ellipse have a horizontal compenent to their velocity.

So, let me me clearer:
In my diagram, is the angle between the mirror surface and the camera velocity, not the mirror velocity.

Fine, see above, the angle $$\phi_v$$ is zero for all points on the circumference.

Try again.
The angle between the all the mirror surfaces and the camera velocity is not zero in any reference frame.

So does the orientation of the microfacet, so the velocity and the orientation have a zero degree angle between them (see above).

Rubbish. The orientation of the leading microfacet is vertical, otherwise it would not be leading.

 

[Pete]

I majorly edited the previous post in the twenty minutes after first posting it.
Apologies to anyone who read it in between.

 

[Neddy Bate]

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 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?

 

[Tach]

The angle between the all the mirror surfaces and the camera velocity is not zero in any reference frame.

Who told you anything about the angle between the camera a velocity and and the mirror surface being zero?
The relevant angle is the angle between the plane of the facet and its velocity (once again, see H. Bateman and W. Pauli). You , of all people, know about that. This is the angle we have been arguing about all along. If this angle is zero, then there is no Doppler effect off the moving window. The rim of a circular, rolling wheel is one such example. So is a mirror whose velocity is embedded in its plane.

The vertical facets at the leading and trailing extremities of the ellipse have a horizontal compenent to their velocity.

The nice thing about your stubbornness is that it allows me to provide fresh rebuttals to your errors. Here is the final one:

Theorem: Given a function defined parametrically as :

$$x=f(t)$$
$$y=g(t)$$

where f and g are continuous with continuous first order derivatives and $$\frac{df}{dt}\ne 0$$ and $$\frac{dg}{dt}\ne 0$$ then any elementary mirror moving tangent to the above curve has the same direction as its velocity along the curve in any point. This means that any ray of light reflected off the elementary mirror as it traces the curve between a light source and a receiver exhibits a zero Doppler effect. This is a generalization of the previous discussions about Doppler effect produced by moving mirrors, now you have been given the generalization to arbitrary trajectories. Thank you for your persistence.

Particular case: Any point on a circular wheel rolling without slipping along a straight line describes a cycloid. The cycloid satisfies the conditions set above (with the exception of one point) so, the Doppler effect as explained above is null.


Proof:

The tangent to any point $$x_0,y_0$$ of the curve has the equation:

$$y-y_0=\frac{dy}{dx}(x-x_0)=\frac{dy/dt}{dx/dt}(x-x_0)$$

Thus:

$$\frac{y-y_0}{dy/dt}=\frac{x-x_0}{dx/dt}$$

On the other hand:

$$v_x=\frac{dx}{dt}$$
$$v_y=\frac{dy}{dt}$$

The above proof is frame-independent and you already know what happens when the elementary mirrors move parallel to their own planes (for the people not familiar with the outcome, see the paper by H.Bateman and the chapter in the Pauli book). It was nice interacting with you despite your occasional rudeness.Thank you for providing the occasion to prove you wrong so many different ways.

 

[Tach]

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?

You have made this same error several times before, the outcome has nothing to do with "length of the light path". Pete will explain this to you.

 

[Tach]

This is the second time you have referred to a proof of this, 100 years in the past.

If the proof stands , then you should make the effort to learn it.
Since it clearly stands, you will need to learn the math that allows you to understand it.

 

[Neddy Bate]

You have made this same error several times before, the outcome has nothing to do with "length of the light path". Pete will explain this to you.

If the total length of the optical path changes, that is equivalent to the light source moving relative to the camera. This is basic stuff. For example, if the total optical path becomes shorter, that is equivalent to the light source getting closer to the camera.

However, I am not surprised that you disagree with this basic concept.

 

[OnlyMe]

If the proof stands , then you should make the effort to learn it.
Since it clearly stands, you will need to learn the math that allows you to understand it.

Tach, I know the difference between a mathematical proof and physics.

I come from a deep family background in mathematics and I studied mathematics and physics, in the late 60s. A mathematical proof can be rigorous and logically compelling and still not represent any description of the "real" world or "physics".

I asked for the reference or citation, because I was interested in seeing what you were referring to.

 

[Tach]

If the total length of the optical path changes, that is equivalent to the light source moving relative to the camera. This is basic stuff.

You sure about that?

For example, if the total optical path becomes shorter, that is equivalent to the light source getting closer to the camera.

However, I am not surprised that you disagree with this basic concept.

What in "the mirror moves away from the light source and towards the camera" do you not understand? The total length of the light path may vary from point to point and from wheel position to wheel position, it has nothing to do with the resultant effect.