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

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

  1. James R Just this guy, you know? Staff Member

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    This thread is for discussion of the Formal Debate between Tach and James R on the following topic:

    That that the light reflected off the circular rim of a mirror-like wheel shows zero Doppler effect between the source and the receiver in the case of the wheel rolling without slipping

    Links:

    [thread=110880]Proposal thread[/thread]
    [thread=110904]Debate thread[/thread]
     
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  3. James R Just this guy, you know? Staff Member

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    Tach:

    Recall that you agreed to conduct this debate according to the Standard rules. Hopefully you have read them.

    Rule 1:

    1. The debater for the affirmative side of the debate will create the "Debate" thread and post the first post, setting out his or her main arguments.

    You have set out no arguments to support your case, so there is nothing for me to rebut so far.

    I'm happy for you to edit your first post if you wish.

    Please also note points 8 and 9 of the Standard Rules.
     
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  5. Tach Banned Banned

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    OK, I have added the argument, have a go at it.
     
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  7. Pete It's not rocket surgery Moderator

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    Interesting. I think that considering a rolling wheel is too complex - why not stick with the simple case of a plan mirror?

    Using a rolling wheel strictly makes it necessary to specify the motion of the source and receiver relative to the ground. Are you assuming they are at rest relative to the ground?

    I also note "the circular rim of a mirror like wheel" doesn't define a particular surface.
    Are you considering all surfaces?
    Or only the cylindrical outer surface on which the wheel rolls (assuming a cylinder-like wheel)?
     
  8. Tach Banned Banned

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    Because it makes things more challenging.

    Yes.
     
  9. James R Just this guy, you know? Staff Member

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    Tach:

    Please review rule 9 of the Standard Rules, which you agreed would govern this debate. Thanks.
     
  10. prometheus viva voce! Moderator

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    I'm still slightly suspicious that Tach is not posting any arguments at all, and is referring exclusively to an external document over which he has control. Surely it would be fairer to reproduce the document in the debate thread? There can't be that much work involved - there are only 4 equations of any substance that would require texing.
     
  11. Aqueous Id flat Earth skeptic Valued Senior Member

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    I would offer that those few equations can be reduced to one basic concept that is either true or false.
     
  12. Pete It's not rocket surgery Moderator

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    Whoops, my bad. I shouldn't have asked the question.
    My apologies to you both.
     
    Last edited: Nov 10, 2011
  13. RJBeery Natural Philosopher Valued Senior Member

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    Are we allowed to discuss the debate topic in this thread during the debate? Or are we limited to evaluating the debaters' arguments?
     
  14. RJBeery Natural Philosopher Valued Senior Member

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    It's like Name That Tune: "I can dismantle the supporting paper in 1 post!"

    Please Register or Log in to view the hidden image!

     
  15. Aqueous Id flat Earth skeptic Valued Senior Member

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    when does it start? I'm interested to see where this goes.
     
  16. OnlyMe Valued Senior Member

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    James added a 4 day window for responses, instead of the normal 2 day window, citing time constraints on his end. Both James and Tach have up to four days to respond to the other. Add to that, each gets only two debate posts and a summary post.., if I read the rules correctly.
     
  17. AlphaNumeric Fully ionized Moderator

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    The pictures in the document are pretty poorly done, they don't set out the situation in a clear manner. Plus the picture in question involves a plane mirror moving in a specific direction, rather than a wheel.

    Perhaps I'm reading it wrong but the simplest counter example to Tach's claim is to consider a mirrored wheel moving at speed v in the x direction, rolling along the ground. The observer is also the light emitter and it up ahead of the wheel. From their point of view the bit of the wheel closest to them is reflecting light from their torch right back to them, it's perpendicular to the direction the wheel is moving in. It's also not stationary in the x direction, the only bit of the wheel where that is true is the bit touching the ground. As such the front bit of the wheel is basically a mirror moving forwards at instantaneous speed equal to the wheel's speed v. Thus we've reduced that bit of the wheel to being just like a bog standard reflector and thus it will produce a Doppler effect. If it didn't speed camera's wouldn't work. Thus Tach's claim cannot be universally true, regardless of whether more convoluted angle setups manage to have zero shift.

    Am I not understanding the scenario? The description people talk about seems quite different from the document, which (to be honest) makes little sense to me due to it being poorly written and extremely unclear in what it's trying to do in terms of physical layout. Is Tach the author or is he just getting it from somewhere?
     
  18. Aqueous Id flat Earth skeptic Valued Senior Member

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    Tach's position, as I read it, is to analyze the surface as a sum of infinitessimals, wherein each point above axis is approaching the detector, and has a twin at the same angle below axis that is regressing. No matter which point you pick there will be a pair in which one is approaching and one one is receding so the net shift is zero for each pair. Integrate over the entire surface, all the pairs cancel, and the answer is zero.
     
  19. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    It seems like there are a lot of semi-classical assumptions being made as to how a Relativistic wheel would reflect light in order to apply the Doppler shift formula in the first place. But if we take these assumptions as given, I would personally side with James R's arguments, because what matters isn't the instantaneous velocity of any given segment of the wheel, but rather the instantaneous position of the reflecting surface as it reflects each pulse. I think the most straightforward thing to do is to consider the source frame, place the camera at the same position and velocity as the source, and consider where the reflecting pulses are coming from at given times and thereby calculate the resultant frequency. My bet is you'll get the same result as James did by switching to and from the wheel frame; the difference is you don't need to take the Doppler formula as given, because you're deriving everything from more fundamental principles.
     
  20. OnlyMe Valued Senior Member

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    That makes sense. Mostly because it represents some attempt at describing the problem as one of physics, rather than just mathematics.

    Mathematics is a very powerful tool, but in and of itself it is not physics. Physics is always limited to a description of experience, observation and the world. While mathematics can add to that description, it is not limited in the same way. It is almost always a predictive tool rather than descriptive. Which means sometimes it describes physics and sometimes it is no more than an exercise in mathematical logic. And the logic of math does not always agree with the world or physics.
     
  21. Aqueous Id flat Earth skeptic Valued Senior Member

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    I think this will ultimately boil down to a single concept: has the problem been correctly modeled? Each person is making certain assumptions. If the assumptions don't match, then you are essentially pitting two dissimilar models, so then you have to sort out which one (if either) is correct.
     
  22. OnlyMe Valued Senior Member

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    Doppler radar is used to measure wind speeds, by frequency shift. If wind speed results in a Doppler shift a solid object will.

    So it really boils down to are we talking physics or ....?
     
  23. Pete It's not rocket surgery Moderator

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    I have it.

    Tach is thinking only about the side of the wheel, not the outer tread.

    Since that surface that is parallel to its motion relative to the light source and receiver, there is no doppler shift

    That's why he has the velocity vector aligned with the mirror in his diagram.

    It could be that this is all he was thinking right from the beginning... "There is no doppler shift of a moving mirror [with direction of motion parallel to its surface.]

    Once again, miscommunication starts fights.
     

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