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

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

Page 9 of 16
  1. Tach

    Tach Banned Banned

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    This incorrect idea was advanced in the beginning by pete. He has given up on it.

    Pauli, Bateman (and many others) have proven your claim false. It has been false since 1909. Did you not read that?
     
    Last edited: Nov 20, 2011
    Tach, Nov 20, 2011
    #161

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  3. RJBeery

    RJBeery Natural Philosopher Valued Senior Member

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    Actually, what OnlyMe says is right but he doesn't understand that you're discussing ambient light sources from all directions.
     
    RJBeery, Nov 20, 2011
    #162

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  5. Tach

    Tach Banned Banned

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    Actually, I am not discussing ambient light. You need to consider the fact that the camera has a film, that collects rays from many directions. This is how pictures are formed.So, the light rays bouncing of the rim definitely do not converge in one point. THIS is his mistake.
     
    Tach, Nov 20, 2011
    #163

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  7. Pete

    Pete It's not rocket surgery Registered Senior Member

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    Wrong then, wrong now.
    Read what I wrote again, and think about the angle I described.
    The angle between the mirror surface and the observer velocity is clearly not zero except at a single point on the wheel.

    Not in the frame of the axle, true.

    In it's own frame, the microfacet obviously does not move at all.
    As you said in your PM, it does have acceleration... but that is irrelevant to this discussion, and it's perpendicular to the plane of the facet anyway.

    But that's beside the point anyway... you previously said that "in ALL frames, the microfacets composing the wheel rim move parallel with themselves."

    We agree that there is at least one frame in which the microfacets move parallel with themselves.
    Do we agree that there are frames (the ground frame for example), in which the microfacets do not all move parallel with themselves?

    You have no right to speak for me Tach, and you clearly have a very shaky grasp on what I'm arguing.
     
    Pete, Nov 20, 2011
    #164
  8. Tach

    Tach Banned Banned

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    Yes.

    Yes.

    Nope, zero angles transform in zero angles. I have just showed you that.
     
    Tach, Nov 20, 2011
    #165
  9. RJBeery

    RJBeery Natural Philosopher Valued Senior Member

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    I hadn't thought there would continue to be discussion on this particular topic, otherwise I would've waited to challenge Tach to the other debate.
     
    RJBeery, Nov 20, 2011
    #166
  10. Trippy

    Trippy ALEA IACTA EST Staff Member

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    Actually....
    First off:
    Secondly, consider. Your post in the debate:

    So, given that its the velocity of the wheel under consideration what we actually have is this:

    \(f_{reflected}=\sqrt{\frac{1+v/c}{1-v/c}} f_S\)

    for wheel approaching the gun and

    \(f_{reflected}=\sqrt{\frac{1-v/c}{1+v/c}} f_S\)

    for the wheel receding from the gun.

    So, if we set \(k=\frac{1-\frac{v}{c}}{1+\frac{v}{c}}\)

    So then we have:
    \(f_{reflected}=\sqrt{\frac{1}{k}} f_S\)

    for wheel approaching the gun and

    \(f_{reflected}=\sqrt{k} f_S\)

    for the wheel receding from the gun.

    So therefore the amount of doppler shift, as viewed from the perspective of an observer on the wheel is equal, but inverse, or if you consider the log of the changes, equal and opposite.

    Again, I repeat, it's not me making the goofs here.

    Wow. Congratulations on evading the point.

    You found a typo. Feel better that you found something you can actually comment on?

    Maybe you should see about a rebate.
     
    Trippy, Nov 20, 2011
    #167
  11. Tach

    Tach Banned Banned

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    How quickly you forget.

     
    Last edited: Nov 20, 2011
    Tach, Nov 20, 2011
    #168
  12. Trippy

    Trippy ALEA IACTA EST Staff Member

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    This +1
     
    Trippy, Nov 20, 2011
    #169
  13. Tach

    Tach Banned Banned

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    Really? So they have equal magnitudes and opposite signs because you were working all along in logarithmic space, right? This is why you are "summing" the Doppler effects?
     
    Tach, Nov 20, 2011
    #170
  14. Pete

    Pete It's not rocket surgery Registered Senior Member

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    I already explained why the approach you used isn't appropriate.

    And we're back to this simple question you did not answer before:
    Consider the leading surface of a rolling wheel. The vertical microfacet obviously has a horizontal velocity component, correct?

    I see I missed this from your previous post:
    Not by the relativistic aberration formula, that's for sure.
    Wrong, as I've already explained.
    In my diagram, \(\phi_v\) is the angle between the mirror surface and the observer velocity, not the mirror velocity.
    You said earlier that you've honestly considered my argument, but you're not showing that you've even looked properly at the diagram.
     
    Pete, Nov 20, 2011
    #171
  15. Pete

    Pete It's not rocket surgery Registered Senior Member

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    Exactly.

    What gives you the impression that I've "given up" on that statement, particularly since I've reiterated it in recent posts.
     
    Pete, Nov 20, 2011
    #172
  16. Tach

    Tach Banned Banned

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    No, you didn't, you simply denied that angles can be transformed through the aberration formula.




    But you are applying the formulas in the microfacet frame (you just admitted that earlier), so the Doppler formulas must take into consideration the microfacet instantaneous velocity , right? Your "observer" is attached to each microfacet.

    Actually, I did. The problem is that you are assuming a non-zero angle between the microfacet instantaneous velocity and its direction of motion (which is absurd, they both coincide with the tangent vector to the rim). This is precisely the error that I am trying to get you to admit. Well, at least you admitted that this is the case in the axle frame.
     
    Last edited: Nov 20, 2011
    Tach, Nov 20, 2011
    #173
  17. Trippy

    Trippy ALEA IACTA EST Staff Member

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    Allow me to reiterate:

    I'm sorry, it's not my fault I just pwned you at your own game.

    You asked me to demonstrate my point, and I have. Now I'm man enough to admit that I might have messed up the language I used in my initial post, mea culpa, I could probably have worded it better, I'm going to blame the two screaming children I have had in the background (among other things). Believe it or not, you, and this discussion, do not rate highly on my list of priorities.

    I would have expected that anyone with at least an intermediate level of prose literacy to have been able to infer that what I was trying to communicate was precisely what I just demonstrated, that the reason there is no doppler shifting in the instance you have demonstrated it is because the red shifting and blue shifting cancel - the factor of the shift is the equal, but inverse.

    Instead, you've got your head jammed so far up your own arse and you're so focused on proving everybody else wrong, that you'll grab on to what amounts to an irrelevant error - and it is irrelevant, to the point being made. They cancel, that stands proven. The cancel multiplicatively, rather than by addition (unless you consider the log) mea culpa, I even subsequently aknowledged that I might have erred in that regard, but it doesn't actually invalidate my point - that the red shift and the blue shift cancel.
     
    Trippy, Nov 20, 2011
    #174
  18. Tach

    Tach Banned Banned

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    Good, so you are subscribing to my position now. Thank you for your support.
    In the case of mirror self-parallel translation, the Doppler effects cancel because in logarithmic space, the two shifts are equal magnitude and opposite sign.
    In the case of rolling motion, the magnitudes are equal and of opposite signs in logariithmic space, so the Doppler effect is just as null.
     
    Last edited: Nov 20, 2011
    Tach, Nov 20, 2011
    #175
  19. Trippy

    Trippy ALEA IACTA EST Staff Member

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    Nonsense!!

    I'm saying that same thing I have said right from the beginning - that there might be special cases where you are correct, but it is far from being the general rule.

    What you just did is fraudulent. It's intellectual dishonesty, especially when you were replying to a post that contained a clause that directly contradicts your assertion.
     
    Trippy, Nov 20, 2011
    #176
  20. Tach

    Tach Banned Banned

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    So, a wheel rolling between a source and a receiver does not exhibit Doppler effect of equal magnitude and opposite signs in logarithmic space?
     
    Tach, Nov 20, 2011
    #177
  21. Aqueous Id

    Aqueous Id flat Earth skeptic Valued Senior Member

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    6,152
    ...
     
    Aqueous Id, Nov 20, 2011
    #178
  22. Pete

    Pete It's not rocket surgery Registered Senior Member

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    I explained that the aberration formula applies to the apparent direction of light rays. You're claiming without support that it applies to arbitrary angles.
    And I've given you a specific counterexample that you seem to be avoiding:

    Consider the leading surface of a rolling wheel. The vertical microfacet obviously has a horizontal velocity component, correct?

    Be definition, the microfacet frame is the frame in which the microfacet instantaneous velocity is zero.

    And I was referring to the observer of the reflected light, not an observer on the microfacet. Sorry for the confusion, but that should really be clear from the diagram, and it's spelled out in the attached post:
    But I see that the "observer" label has been left off the diagram, so maybe that explains it.

    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.


    Why would I admit to such nonsense?

    - In the microfacet frame of my diagram, the microfacet instantaneous velocity is zero, so it's meaningless to speak of its direction of motion.
    - In the ground reference frame, it's obvious that the direction of motion of the microfacets is in general not tangent to the rim.

    Once again, you might like to answer this simple question:
    Consider the leading surface of a rolling wheel. The vertical microfacet obviously has a horizontal velocity component, correct?

    Hardly an "admission", since that's never been in dispute.
     
    Pete, Nov 20, 2011
    #179
  23. Aqueous Id

    Aqueous Id flat Earth skeptic Valued Senior Member

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    6,152
    Tach is in the amplitude domain when he adds the potential from the arriving light waves. It is linear and it adds. Signals out of phase will add in pairs under his ideal conditions. The net effect is a new waveform for each pair, at phase zero. The sum of all of these pairs, in the amplitude domain, is zero.

    Tach's approach is, to me, simple and elegant in this regard and arguable under the ideal conditions of the perfect reflector.

    There still seems to be a reluctance to give credit where credit is due. And this seems to overshadow the actual information he conveyed at the outset of his paper.
     
    Aqueous Id, Nov 20, 2011
    #180
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