Moderator note: The participants for this debate are

[thread=111038]Discussion thread[/thread]

[thread=111009]Proposal thread[/thread]

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The above statement is false, and I will demonstrate why. The apparent basis for this belief by Tach is here, and it is this paper that I will critique and dismantle.

There are no technical or mathematical errors in the paper if we restrict the analysis to

The difference is that mirrors are specular and have the ability to reflect

Here we see the wheel receding from the light source and moving towards the camera.

Combining them cancels out the Doppler effect such that $$f_{s'}=f_0$$. I'm sure you agree with all of this, Tach, because I'm basically just parroting your paper. (I'm doing this mostly because I've always needed an excuse to learn TEX )

Anyway, your mistake is in neglecting the optical nature of a colored, matte wheel. Unlike a mirror, the wheel would only reflect light of 600THz in the frame of the wheel, not light of 600THz in the frame of the light source. In this case:

Solving for $$f_{0}$$...

Arbitrarily picking

Q.E.D.

**Tach**for the affirmative and**RJBeery**for the negative. Agreed parameters for the debate can be found in the Proposal thread.[thread=111038]Discussion thread[/thread]

[thread=111009]Proposal thread[/thread]

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**There is no Doppler shift off a matte wheel rolling between a source and the receiver**.The above statement is false, and I will demonstrate why. The apparent basis for this belief by Tach is here, and it is this paper that I will critique and dismantle.

There are no technical or mathematical errors in the paper if we restrict the analysis to

*mirrored wheels*. The title of the paper, "The*of objects rolling at relativistic speeds", does present a problem though because it exposes the author's belief that this would extend to non-mirrored objects.***color**The difference is that mirrors are specular and have the ability to reflect

*all*frequencies of light, whereas a matte wheel is restricted to reflecting a*particular*one. In the scenario below we'll examine a green wheel which emits a frequency of 600 THz (in it's axle frame). [Note: We'll examine a*sliding*wheel rather than a rolling one; a sliding wheel will exhibit uniform Doppler effects while a spinning one will exhibit varied Doppler effects across its surface due to the varying relative speeds between the light source, a point on the wheel, and the camera. Since the paper argues that no Doppler exists for*any*movement velocity V, it would also apply to each point on the wheel whether its relative movement was rotational in nature or not. Therefore, proving Doppler effects of a sliding wheel is equivalent to proving Doppler effects on a rotating one.]Here we see the wheel receding from the light source and moving towards the camera.

$$f_0$$ = frequency emitted by light source

$$f_{wheel}$$ = frequency emitted by wheel

$$f_{s'}$$ = frequency perceived by the camera

$$f_{wheel} = \frac{f_0\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}cos\phi_{1}$$

This represents the red-shifting of $$f_{0}$$ due to the relative movement between the light source and the wheel.$$f_{wheel}$$ = frequency emitted by wheel

$$f_{s'}$$ = frequency perceived by the camera

$$f_{wheel} = \frac{f_0\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}cos\phi_{1}$$

$$f_{s'} = \frac{f_{wheel}(1+\frac{v}{c}cos\phi_{1})}{\sqrt{1-\frac{v^2}{c^2}}}$$

This represents the blue-shifting of $$f_{wheel}$$ due to the relative movement between the wheel and the camera.Combining them cancels out the Doppler effect such that $$f_{s'}=f_0$$. I'm sure you agree with all of this, Tach, because I'm basically just parroting your paper. (I'm doing this mostly because I've always needed an excuse to learn TEX )

Anyway, your mistake is in neglecting the optical nature of a colored, matte wheel. Unlike a mirror, the wheel would only reflect light of 600THz in the frame of the wheel, not light of 600THz in the frame of the light source. In this case:

$$f_{wheel}$$=600THz so

$$\frac{f_0\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}cos\phi_{1}$$=600THz

$$\frac{f_0\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}cos\phi_{1}$$=600THz

Solving for $$f_{0}$$...

$$f_0 = \frac{f_{wheel}(1+\frac{v}{c}cos\phi_{1})}{\sqrt{1-\frac{v^2}{c^2}}}$$

Arbitrarily picking

$$\frac{v}{c} = .5$$ and $$\phi_1 = \frac{\pi}{4}$$

We get$$f_0 = \frac{600THz(1+.3535)}{.75} = $$1082THz

Since we have already established that $$f_{s'}=f_0$$, we conclude that$$f_{s'} = $$1082THz

Q.E.D.

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