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

Discussion in 'Formal debates' started by James R, Nov 10, 2011.

  1. AlphaNumeric Fully ionized Moderator

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    Tach PM'd me to have a go at me for complaining about it. I'll reply here, in general terms, to illustrate what I was referring to.

    If the wheel is suspended on an axis and just spinning then I might agree. However the setup involves a wheel rolling on the ground (the no slip condition is explicitly mentioned somewhere in all of this). In that case you have both the rotational motion and the lateral motion. Hence my post above, where the front of the wheel basically acts like a normal mirror moving forwards at speed v towards the observer and emitter, which obviously then see a shift.

    If you lifted the wheel just off the ground and just let it spin at the same rate than obviously the motion of the bit of the wheel at the 'front' would now have no forward component, it would be at right angles to the direction of the photon bouncing back to the observer with the torch and thus no shift occurs.

    This is why I said it was poorly written (which Tach also complained about), because the picture doesn't mention anything to do with the motion of the wheel on the ground, yet there's been considerable discussion about it.

    It's easy to compute the motion of a particular piece of the wheel, it'll be the sum of it's rotational motion about the central axis and the motion of the axis, which you can express in terms of the angle the piece in question is making with some particular direction. When the piece lines up with the ground, so it's at the front or back of the wheel, wrt its motion, then it'll only have motion parallel to the ground due to the overall motion of the axis but that is non-zero. The only part of the wheel which is stationary at a given instant is the bit on the ground, that's what the no slip condition is!

    So, if the wheel is indeed rolling on the ground the claim of Tach's is false, trivially and obviously so. This obviously generalises to instances where the wheel's axis is moving in some arbitrary direction, so the only way Tach's claim could even possibly be true is if the axis, the emitter and the receiver are all at rest wrt one another.

    Can someone tell me if I'm just misremembering something? Did someone else mention the no slip condition has something to do with all of this? If it was Tach then his claim is immediately falsified.

    /edit

    Having just scrolled up in the thread I see the no slip conditon is mentioned in JamesR's openning post so if the debate involves a rolling wheel Tach is wrong. If he weren't wrong then speed cameras wouldn't work. I'd also point out he didn't actually address any of my previous post in his PM, he just complained about me posting it and how no one else had misunderstandings. It would seem my understanding was fine, as the rolling of the wheel is entirely relevant. Tach, if you plan to reply to me via PM or in the debate thread then I'd appreciate you addressing the explicit points I've raised rather than just saying what amounts to "Stop complaining and asking for people to clarify things!".
     
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  3. przyk squishy Valued Senior Member

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    I don't think so. If that's what he's thinking about, why does he have a complicated formula for the Doppler shift depending on an angle?

    I don't think that's really the problem here. The thing is, while Tach may not be much good at communicating what he's doing, others are. In James R's post in the debate thread he explains exactly which problem he is considering:
    If Tach was really thinking of something else, why didn't he take issue with James R right there for solving the "wrong" problem? Instead Tach passes right over this and complains he thinks James R is using the wrong formula. I can only take from this that Tach really believes that if you fire a light pulse at a wheel with a mirror-like rim rolling toward you, it'll come back to you with the same frequency as when you fired it.
     
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  5. RJBeery Natural Philosopher Valued Senior Member

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    Tach's math is correct, his framing is wrong. For a given mirror angle there's basically a single point along the mirror's trajectory that the camera can see the light source. He's chosen the arrangement where the mirror's velocity is parallel to its orientation. This is a special case, which does not allow us to make any general claims.

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    This is not true when the mirror's angle and velocity differ.

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  7. RJBeery Natural Philosopher Valued Senior Member

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    Tach also made the claim that it doesn't matter if it's a wheel or a mirror. This is also false but for a different reason. In the following picture the Doppler shift is strictly due to the relative motion between the apple and the camera (which, in this case, there is none).

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    Now, if there is movement relative to the light source you can see that the wavelengths between the apple and the camera do not change! Therefore, Doppler between the apple and the camera is purely a function of the relative motion between the apple and the camera. The relative motion of the light source is irrelevant, as long as the apple's "color" is within the blue-shifted or red-shifted incident light.

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  8. RJBeery Natural Philosopher Valued Senior Member

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    Lastly, here is Tach's apparent source of information. I don't know if he's the author or not, and we can discuss all day about the intricacies of an intentionally obfuscated thought experiment, but in the end results are what matter. To this, I also include a real-world counter-example to Tach's claims.
     
  9. RJBeery Natural Philosopher Valued Senior Member

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    ...at the risk of beating a dead horse, I'd like to preempt any possible retort Tach may be conjuring. First of all, the fact that the wheel itself is spinning is completely irrelevant, as Tach correctly pointed out in his paper; however, there's actually another presumption, namely that the wheel is the only object in relative motion to the light source and the camera. If we relax this presumption we can see that Doppler effects do propagate through the spinning mirror when the direction of its velocity is parallel to its plane orientation.

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  10. Aqueous Id flat Earth skeptic Valued Senior Member

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    Hey Pete.

    Tach's paper shows that he is considering the surface of the mirrored tread.
     
  11. Aqueous Id flat Earth skeptic Valued Senior Member

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    Hey PJBeery.

    Are you saying that they should be framing this as transverse velocity? Normally, wouldn't the longitudinal case be assumed, since that is the optimal for signal return with doppler?
     
  12. AlphaNumeric Fully ionized Moderator

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    Right, so Tach sent me another PM which failed to address what I asked for clarification on (absolutely no mention of the no slip condition) and just whined about how I'm wrong to criticise him.

    JamesR's challenge involved the no slip condition. This implies the wheel is rolling along the ground. Naturally I imagine an emitter and an observer on the ground too, so they are stationary wrt the ground and the wheel's centre is moving wrt to the ground and them. In the scenario I mentioned previously this leads to a clear Doppler shift, as the front of the wheel acts precisely like a flat mirror.

    Tach complained I didn't do any mathematics, as if the set up I just mentioned wasn't clear and obvious enough. He also says the wheel and the observer and emitter aren't moving relative to one another. So why is the no slip condition mentioned explicitly in JamesR's challenge? Why mention some relation with the ground if you're then going to stipulate that the ground is irrelevant and everyone's at relative rest to one another? It seems to me Tach's document is not directly addressing JamesR's challenge but a slightly different one.

    Since Tach complained I didn't do any mathematics let's go through the trivial stuff then. The wheel has radius R and rotates at \(\dot{\theta} = \omega\) angular velocity about the central axis. Therefore in the axis rest frame the piece at angle \(\theta\) from the vertical (ie 0 is pointing up, like a clock) is moving at speed \(\mathbf{v} = \omega R (\cos \theta,-\sin \theta) \equiv v(\cos \theta,-\sin \theta)\).

    The no slip condition says that the motion of the bottom of the wheel is 0 relative to the ground. But it's rotational motion at \(\theta = \pi\) is \(\mathbf{v} = -v(1,0)\) so the axis of the wheel is moving in the +ve direction with speed v, so in the ground's frame \(\mathbf{v} = v(1+\cos \theta,-\sin\theta)\).

    Now an observer with a torch up ahead of the rolling wheel. Clearly the bit which reflects the light to the observer is the front part, \(\theta = \frac{\pi}{2}\). The observer is moving at speed \(\mathbf{w}\) in the ground frame and while I could compute the relativistic difference I'm not going to bother. Suffice to say that what the Doppler shift cares about is the component of this different in the direction of the line of sight, which is the x direction. Tach uses \(\phi\) to describe this, I'll just do the dot product. The proportion is therefore just the x component of the relative velocity. Thus the second part of \(\mathbf{v}\) isn't important, only the first part, when computing whether there's a Doppler shift. Clearly this occurs unless \(\mathbf{w}\) and \(\mathbf{v}\) have the same x component, ie no relative motion. This means that if the no slip condition is ignored and you've just got a wheel spinning in space, at relative rest wrt to the guy with a torch, then there's no shift. For a wheel rolling under the no slip condition towards someone standing on the ground then it's exactly like the example I previously mentioned, a normal flat mirror approaching the guy with a torch at speed v, except now we can say \(v = \omega R\) due to it's origin from the spinning wheel which isn't slipping.

    Tach, you've now sent me 2 PMs, neither one of which mentions anything about the no slip condition. If you weren't discussing that then you shouldn't have put your document forward as a counter to JamesR's statement which explicitly mentions it. If you think your claim is true under such conditions, you're demonstrably false. Either way just coming back with "Pay attention! Don't criticise me!" does nothing more than make it seem like you struggle to engage in discussion. I'm asking for clarification as it seems you're talking past what JamesR challenged you to discuss. Your abrasive manner does you no favours but reinforce the things I said to you in PM.
     
  13. Pete It's not rocket surgery Moderator

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    Because the angle between the mirror-source vector and the velocity vector can be anything.
    As long as the mirror surface is parallel to the velocity vector, there is no doppler shift.

    There is ambiguity in the definition of the rim.
    James is clearly thinking of the outer cylindrical surface.
    But I believe that Tach is thinking of the side of a wheel between the spokes and the tread, like the rim of a bicycle wheel.
    This is a surface that is parallel to the direction of motion, and Tach's objections to James's post suddenly make a lot more sense.
     
  14. Pete It's not rocket surgery Moderator

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    ...having another look...

    Oh, dear.
    From Tach's document:

    So, you are correct... he is considering the surface of the mirrored tread.

    But he's used v for a spinning wheel rather than a rolling wheel (as AlphaNumeric alluded to), so he is only consider a velocity vector parallel to the surface.

    So, yeah... worse than I thought. Not a misunderstanding, just wrong.
     
  15. przyk squishy Valued Senior Member

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    This is exactly the point I am making: the incident and reflected frequencies are identical in that case. So if that was what Tach was thinking about, why would he present a complicated formula for the Doppler shift with an angular dependence

    This too is the point I was making here: I have a hard time believing Tach couldn't find this clear. So if he was really considering something different, why didn't he say "Hey, that's not the problem I'm considering!" at this point?

    I had exactly the same problem with Tach near the end of the other thread: I clearly spelled out which problem I was solving more than once, explaining exactly how I was defining the difference between the upper and lower parts of the rolling wheel. Tach repeatedly ignored these and skipped to pointing out "errors" in my calculations. Yet it's clear from the "errors" he pointed out that he was really applying different definitions of the "upper" and "lower" halves of the wheel. So Tach clearly wanted me to solve a different problem than the one I'd solved (one that would get him the result he wanted), but for some reason never wanted to say so openly.
     
  16. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    Then I guess we need to clarify: are we actually talking about a rolling wheel, or are we talking about a spinning wheel instead?
     
  17. RJBeery Natural Philosopher Valued Senior Member

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    This whole business is silly. I would ask Tach to answer the following two questions:

    1) Does a mirror moving toward a light source reflect blue-shifted frequencies, and does a mirror moving away from a light source reflect red-shifted frequencies?
    2) Does the rolling wheel you are discussing have the mirror moving toward/away from the light source?

    #1 is obvious, and #2 depends on the setup. Either way, a mirror reflects light equal in frequency to the incident light (as measured from the mirror's frame), and coming up with a convoluted scenario where the mirror happens not to be moving toward or away from the light source is not pointing out something profound.
     
  18. Neddy Bate Valued Senior Member

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    In reading Tach's latest post in the debate thread, it turns out he was talking about a specific case. He has a light source and a camera, (both fixed to the road frame), some distance apart. The mirrored wheel rolls along the distance between the light source and the camera. There is enough of an angle that light can still reflect off the mirror and be received by the camera. So the light that is received by the camera is the same frequency as the light that was emitted by the light source. That's all it was. I want my money back.
     
  19. OnlyMe Valued Senior Member

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    That is Tach's interpretation, but it is also one he applied to the original discussion. The original discussion was more accurately described by the proof James gave.

    It seemed to me that both Pete and przyk were definning the hypothetical as it was argued by James. I don't think Tach ever made an attempt to understand the problem.., everyone else was discussing.
     
  20. AlphaNumeric Fully ionized Moderator

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    przyk, Pete and Cpt, that's precisely the issues I've been having and Tach has sent me multiple PMs whining about how dense I'm being because I asked for clarification.

    In the most recent PM he's stated he says that in the scenario I outlined the axle indeed moves sideways but the front of the wheel, on the circumference, doesn't produce a shift, even through it clearly has the same x component velocity as the axle or else the wheel would be changing circumference.

    His statement is, and I quote, "What enters the calculations of the Doppler effect is the projection of the tangential velocity onto the incident and, respectively reflected rays as measured in the frame comoving with the point of incidence.".

    Indeed, if you are working in a frame where the motion of the mirrored bit is orthogonal to the normal vector defining the reflection then there'll be no Doppler shift from that frame's point of view. Someone holding a mirror will rightly say that any photon bouncing off it will have its frequency unchanged and provided the emitter and observers are also stationary from said person's point of view then Tach is right. What he hasn't realised is that the frequencies the person holding the mirror sees before and after might not be the same frequency the emitter saw when he emitted the photons in his rest

    Tach's claims only work if you ask "Does the mirror see the same frequency before and after reflection?" rather than "Do the emitter and observer agree?". To get into a frame where the mirror isn't moving at all in the direction normal to its surface (which is the x direction in my set up) you might have to do a Lorentz boost, which means the frequency of the photons change.

    Tach, consider my set up. The torch emits photons in the -ve x direction at some frequency. You then boost into a frame where the front of the wheel is only moving in the y direction (ie the axle rest frame). This changes the frequency of the photons by some manner. The reflection occurs and we use your result that the frequencies in said frame don't change. The front of the wheel acts just like a bog standard flat stationary mirror parallel to the y direction. Fine. Then you boost back. Except now the photons are moving in the opposite direction and the boost affects them differently and you get a different frequency. And before you whine about the lack of maths let's go through it.

    We use coordinates (t,x) and (t',x'). (t,x) are the coordinates in the ground (and emitter and observer) rest frame. The photon starts with energy E (with E=hf = f in usual units but E suffices) and it moving in right to left so \(p^{\mu} = (E,p) = (E,-E)\) via \(p^{\mu}p_{\mu} = 0\) for photons. A Lorentz boost can be done via the following matrix

    \(\Lambda(v) = \gamma(v) \left( \begin{array}{cc} 1 & -v \\ -v & 1 \end{array} \right)\)

    We boost by \(\Lambda(-v)\) to get into the axle rest frame, which gives us \((p')^{\mu} = \gamma E(1-v)(1,-1)\). Using Tach's result (which no one denies), there's no motion of the mirrored surface in the relevant direction so no change in frequency. The photon's momentum changes sign though due to it now moving in the opposite direction, so \((q')^{\mu} = \gamma E(1-v)(1,+1)\).

    We now boost back by using \(\Lambda(v)\) and we get \(q^{\mu} = \gamma^{2}E(1-v)^{2}(1,-1)\). Using the expression for \(\gamma[tex] we get [tex]q^{\mu} = \frac{1-v}{1+v}(E,-E) = \frac{1-v}{1+v}p^{\mu}\)

    Oh look, a double Doppler shift effect, due to the motion of the mirror wrt to the emitter and detector by precisely what we expected.

    Thus while we can say that the tangential motion due to the spinning of the wheel's circumference doesn't give a Doppler shift if the emitter and observers are at rest wrt to the axle it isn't a universal truth that no shift occurs, as this pretty much simplest possible example shows. Quite why Tach thought I was talking about the axle being a reflector when I explicitly mentioned vector addition to get the motion of the circumference wrt to the ground I don't know.

    He doesn't deny the no slip condition, he just believes it irrelevant which is not only demonstrably false but demonstrated false.
     
  21. OnlyMe Valued Senior Member

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    This is from Tach's opening statement in the Debate thread,
    Bold emphasis mine... It seems from this that, the wheel was rolling not spinning.
     
  22. Emil Valued Senior Member

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    I do not understand why using "mirror".
    Experiment is different for "mirror" and "no mirror"?
    In my opinion is the same.
    The difference between "mirror" and "no mirror" is reflective surface roughness.
     
  23. RJBeery Natural Philosopher Valued Senior Member

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    No, actually there is a difference. A mirror would produce double the Doppler shift of a colored item. This is because the colored item becomes an effective "filter" of any color-shifting of the incident light, wiping out half of the reflected effect.
     

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