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

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".

Yet, you always avoid any mathematical proof.

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

[4] H . Bateman, "The reflexion of light at an ideal plane mirror moving with a uniform velocity of translation" ,Phil. Mag., (18) (1909) p.890
[5] W.Pauli, "Theory of Relativity" , 1958, Pergamon Press, p.95 (top)

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.

As the total light path gets shorter, the camera sees the light source as getting closer. That makes the light source appear blue-shifted.

As the total light path gets longer, the camera sees the light source as getting farther away. That makes the light source appear red-shifted.

Really basic stuff.

As the total light path gets shorter, the camera sees the light source as getting closer.

Err, no. The distance between the camera and the light source is fixed. Why don't you let me an pete finish the discussion?

Err, no. The distance between the camera and the light source is fixed. Why don't you let me an pete finish the discussion?

Pete is free to reply to you whenever he wants.

5.829 > 5.656 < 5.829

QED

Yet, you always avoid any mathematical proof.

Its been forty years Tach. I did other things in life and though I can follow most of what you youngsters present, it is work and I don't always catch the errors myself. I no longer have a working knowledge. I'd do still have a fair conceptual understanding of at least that portion that applies to reality. I never was interested in pure mathematical logic.

[4] H . Bateman, "The reflexion of light at an ideal plane mirror moving with a uniform velocity of translation" ,Phil. Mag., (18) (1909) p.890
[5] W.Pauli, "Theory of Relativity" , 1958, Pergamon Press, p.95 (top)

Thanks, I'll look at both if possible. Fair certain if I don't have then Pauli book getting dusty on a shelf I will find it somewhere. The Bateman reference I will have to try the net...

Pete is free to reply to you whenever he wants.

5.829 > 5.656 < 5.829

QED

The path of every point on the rolling rim describes a cycloid. As such, every point while receding from the light source it approaches the camera. It would be good if you tried understanding math instead of painting pretty (and irrelevant) pictures.

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

Of course not! But it doesn't matter - it was an ideal scenario he propped up from the outset of his paper. If there was to be any argument against it on those grounds, it should have been advanced up front.

Obviously many of you are gifted. But so is Tach!

Who told you anything about the angle between the camera a velocity and and the mirror surface being zero?
You did:
Tach said:
Pete said:
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 is zero for all points on the circumference.
I really don't think you're paying attention.
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.
You're definitely not paying attention.
Once more:
In my diagram, $$\phi_v$$is the angle between the mirror surface and the camera velocity in the rest frame of the mirror facet.
In that frame, the facet is at rest (obviously), so it is meaningless to talk about the angle of its velocity.

Tach said:
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.
Counterexample:
An element of a vertical mirror moving horizontally.

You are confusing the path of a mirror element (the spatial component of its worldine) with the shape of the mirror surface.
They are two different curves.

The element's orientation is tangent to the surface shape.
The element's velocity is tangent to the path.

For the rolling wheel, the surface shape is ellipsoid, while the path is cycloid.
Different curves, different tangents.

Why does this even need explaining? I thought you were a thinker?

You did:

I really don't think you're paying attention.

You're definitely not paying attention.
Once more:
In my diagram, $$\phi_v$$is the angle between the mirror surface and the camera velocity in the rest frame of the mirror facet.

Yet, the discussion is about the fact that the angle between the mirror surface and its velocity. This angle being zero is what causes the null Doppler effect.

In that frame, the facet is at rest (obviously), so it is meaningless to talk about the angle of its velocity.

But you are talking about the wrong angle, Few ideas but fixed.

Counterexample:
An element of a vertical mirror moving horizontally.

Bzzt, you weren't paying attention, in your rush to find fault. The mirror follows the path described by the parametric representation of the trajectory. New tangent, pay attention.

You are confusing the path of a mirror element (the spatial component of its worldine) with the shape of the mirror surface.

Not at all, you don't understand what I wrote.

The element's orientation is tangent to the surface shape.
The element's velocity is tangent to the path.

The element orientation follows the trajectory of the point, as such it represents the tangent to the trajectory , not to the mirror surface. I have changed the explanation, try following it. The reflection always comes of ONE point. You can consider the point as part of the tangent to the rim (as in the past explanations that you have not understood) or as part as the tangent to the point trajectory (as in the new explanation that you still don't understand).

For the rolling wheel, the surface shape is ellipsoid, while the path is cycloid.
Different curves, different tangents.

I have replaced the tangent to the rim (the tangent to the ellipse, NOT ellipsoid) with the tangent to the trajectory (the cycloid). The text was quite clear , try engaging brain before mouth.

Why does this even need explaining? I thought you were a thinker?

Yes, I engage brain before mouth. How about you?

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Tach, you don't even seem to be trying to understand anything I say, and are confusing separate discussions.
Let's take a step back.

Discussion 1 - the diagram I presented in post 114:

Do you agree that the diagram I drew represents a mirror facet's instantaneous rest frame?
Do you understand that in that diagram, V is the velocity of the light source and camera?

-------------------------------------------------------------------------------------

Discussion 2 - the angle between facet velocity and facet orientation in the ground frame.

Do you agree that in the ground frame, the shape of the mirror is an ellipse?
Do you agree that in the ground frame, the orientation of each facet is tangent to the ellipse?
Do you agree that in the ground frame, the trajectory of each facet is cycloid?
Do you agree that the instantaneous velocity of each facet is tangent to the cycloid?

Tach, you don't even seem to be trying to understand anything I say, and are confusing separate discussions.
Let's take a step back.

I don't confuse any discussions. Since you have been so obtuse in accepting the explanations, I have given you different explanations for the same problem.

Discussion 2 - the angle between facet velocity and facet orientation in the ground frame.

Do you agree that in the ground frame, the shape of the mirror is an ellipse?

Yes, I showed you that long ago.

Do you agree that in the ground frame, the orientation of each facet is tangent to the ellipse?

I showed you that long ago. Not only that, I showed you the exact equations of the speed of any point on the circumference. Not only that, I have shown you that in any frame of reference, the angle between the velocity of any facet and its plane is zero.

Do you agree that in the ground frame, the trajectory of each facet is cycloid?
Do you agree that the instantaneous velocity of each facet is tangent to the cycloid?

Yes, I just showed you that in the previous post. I have also showed you that the angle between the velocity of the facet as it describes the cycloid (or any other trajectory that satisfies the conditions of the theorem in the post) and its plane is zero.

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

I have defined the angle $$\phi_v$$ as an arbitrary angle, call it $$\phi$$ if it confuses you. In the particular case I have explained to you, it is the angle between the velocity of the mirror and its plane, both measured in the frame S of the axle. As such, as already explained (and agreed upon) $$\phi=0$$. In the frame of the ground S', $$\phi'=0$$, courtesy of the way angles transform between the two frames .

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.

You will need to use math instead of insults in order to prove your point.
The equation of the ellipse, as given to you earlier on is here. The parametric representation of the ellipse is:

$$x'=\gamma(V)(rsin(\omega t + \theta)+Vt)$$
$$y'=r(1+cos(\omega t + \theta))$$

I am quite sure that I have posted the above several times already.

The components of the velocity are given a few posts down the page, in post 39. It is easy to verify that the components of the velocity line up with the components of the vector tangent to the ellipse, i.e. the plane of the microfacet.

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

Yes, you have repeated this true but irrelevant statement in order to further your point several times already. What matters is not the respective components, what matters is , when taken together, the velocity components line up with the tangent to the ellipse.

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

Same nonsense repeated multiple times doesn't form a valid argument.

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Tach, I really think that you are being awfully illogical.

Look, we agree that in the ground frame:
The orientation of each facet is tangent to the ellipse.
The instantaneous velocity of each facet is tangent to the cycloid.

It is obvious that these two tangents are not parallel:

The angle between them is obviously not zero.
Agreed?

Therefore the orientation of each facet is not parallel to its velocity.

I have defined the angle $$\phi_v$$
No Tach, I defined $$\phi_v$$ in post 114:
In the rest frame of the mirror:
V is the velocity of the source and the observer[camera].
$$\phi_v$$ is the angle between the mirror surface and V.​

I don't think you've ever read that post properly.

If you want to talk about a different angle, please be explicit and give it a different label.
As such, as already explained (and agreed upon) $$\phi=0$$. In the frame of the ground S', $$\phi'=0$$, courtesy of the way angles transform between the two frames .
You haven't yet shown an understanding of how to properly transform the angle between a surface and an arbitrary velocity vector.
Want me to post the correct transformation (including derivation) for you?

It is obvious that these two tangents are not parallel:

The angle between them is obviously not zero.

They are two DIFFERENT explanations. This is the third time I am correcting your misconception.

When you're comparing trajectory to orientation, you need both.

The tangent to the cycloid gives the trajectory, not orientation.
The tangent to the ellipse gives the orientation, not the trajectory.

Tach... maybe, just maybe, all these other people saying you are wrong... they might have a point here mate... just sayin, ya know?

Oh, this is still going. I don't have time to read the last half dozen pages or so so can someone tell me if Tach's managed to reconcile his specific case 'proof' with the blanket statement the discussion originally started with? Or is he still claiming somehow he isn't wrong, despite the blanket statement from the debate obviously being falsified?

Tach can't reconcile his specific-case proof with his grandiose general claim. The best he can do is to go off on tangents and generally try to obfuscate, in the hope that people will forget his original over-reach.

Oh, this is still going. I don't have time to read the last half dozen pages or so so can someone tell me if Tach's managed to reconcile his specific case 'proof' with the blanket statement the discussion originally started with? Or is he still claiming somehow he isn't wrong, despite the blanket statement from the debate obviously being falsified?

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.