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

Is that sufficiently clear?

I think I understand, but I understood the original wheel about 1200 posts and a few threads back.

So, to know for sure we have have to wait until Tach.., responds.

So, to know for sure we have have to wait until Tach.., responds.
If you ask me, Tach seems in imminent danger of imploding into a singularity of emo-rage.

So when considering the situation of a rolling mirrored toiletpaper roll, because the camera is always going to see the reflection off a tangential plane that is co-moving with the axel, but not co-rotating with the wheel, we can model any infinitessimal part of the surface of the wheel that connects the raypath from the light source to the ray path from the camera as if it were a flat mirror moving with some velocity v, paralell to the ground, with some angle to the ground, and the velocity v is the velocity of the axel added to the component of the tangential velocity that is paralell to the ground.
I believe looking at it in terms of an infinite number of mirrors is just fine. The problem is that if the mirror's orientation is anything but zero wrt its velocity there will be a detectable Doppler shift. The only place the orientation angle is zero is at the top and bottom of the wheel, which would show no Doppler.

...or perhaps this is your position as well?

...or perhaps this is your position as well?

I know you're not asking me, but YES! That is what everyone has been trying to explain to Tach.

Yeah I welcome your thoughts as well Neddy Bate. I'm glad to see Tach is consistently wrong...

I believe looking at it in terms of an infinite number of mirrors is just fine. The problem is that if the mirror's orientation is anything but zero wrt its velocity there will be a detectable Doppler shift. The only place the orientation angle is zero is at the top and bottom of the wheel, which would show no Doppler.

...or perhaps this is your position as well?

This is my position as well - this represents the 'special case' that I have readily admitted right from my first comment on the matter.

In the case of the mirror at the bottom of the cycle, it's because the mirror is stationary, in the case of the top of the cycle, it's because the red shift and the blue shift are equal and inverse, not because it is stationary.

Heh. Yes.

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).

And now that the problem has been defined unambiguously, sensible discussion of the two seperate problems, or components of the problem, can begin.

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My feeling is that no one here disagrees on anything, but perhaps I should verify:

1. The mirrored cylinder will exhibit Doppler effects everywhere except the very special (infinitesimal) case of top/bottom
2. The mirrored washer would exhibit Doppler effects nowhere
3. The matte versions of either would exhibit Doppler effects everywhere except, again for very special cases