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Doppler Effect

  1. Aug 16, 2005 #1
    Why is it that the extent of the doppler effect on sound depends on whether, for example, you are moving towards the source or the source is moving towards you? Why does this not happen for light?
     
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  3. Aug 16, 2005 #2

    russ_watters

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    It does happen for light (how do you think that cop knew how fast you were driving?). But since the speed of light is a lot faster than the speed of sound, you have to be moving a lot faster (or have sensitive equipment) to notice it.
     
  4. Aug 16, 2005 #3
    The book I'm reading seems to tell otherwise but maybe I'm interpreting it wrong:

    Nigel Calder's "Einstein's Universe"
     
    Last edited: Aug 16, 2005
  5. Aug 17, 2005 #4

    russ_watters

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    I don't think that first paragraph is correct. A quick google shows that the doppler shift equation for sound doesn't differentiate who is really moving.

    There is a difference, in that velocities don't add in Einstein's relativity in the same way as in Newton's. But that doesn't appear to be what he means.

    Anyone else have any insight....?

    edit: LINK
     
    Last edited: Aug 17, 2005
  6. Aug 17, 2005 #5

    rbj

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    not really insight, but just a vote: i don't think that "the extent of the doppler shift depends on whether the source of the sound is moving towards the listener or the listener is moving towards the source of sound" is "ecause sound waves travel throught a medium - the air". the doppler shift is because of how the actual oscillation of whatever source is observed at a distance from the POV of the speed of the propagation of the resulting wave. relative doppler has the added effect that the observed frequency of oscillation would also be different than from only a classical POV.

    whatever.
     
  7. Aug 17, 2005 #6

    Chronos

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    Redshift related Doppler shift discrepancies are still of interest. Lorentz invariance remains under the magnifying glass.
     
  8. Aug 17, 2005 #7
    the doppler effect accure all the time waves exist. light and sound are waves so it will happen for them
     
  9. Aug 17, 2005 #8

    Meir Achuz

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    "The precise reckoning of the doppler effect was a matter of great importance to Einstein, and he found that light did not behave in exactly the same way as sound. Because sound waves travel throught a medium - the air - the extent of the doppler shift depends on whether the source of the sound is moving towards the listener or the listener is moving towards the source of sound.
    [...] In Einstein's democratic universe, that cannot make any difference: all that matters is he relative speed of the start and the onlooker."

    These two quotes state the correct situation.
    The details of the difference depend on the different derivations
    (and can be seen in the formula for each case), but the basic difference is that there is a medium for air, and not for light.
     
  10. Aug 17, 2005 #9

    rbj

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    if the author means "velocity" (as a vector) instead of speed, then i agree. but the doppler effect on light coming from a source moving toward an observer will be different than the doppler effect from the same source moving away from the observer at the same speed. red shifting is different than blue shifting.

    the effect that speed has on the rate of oscillation creating the light wave as observed from the observer is independent of direction. it's, [tex] \sqrt{1-v^2/c^2} [/tex], a function of [tex] |v|^2 [/tex], the magnitude of the velocity vector.

    we agree on that.
     
    Last edited: Aug 17, 2005
  11. Aug 17, 2005 #10
    I don't know if this makes any difference or not, but the author goes on to talk about the discrepancy of redshifts and blueshifts with respect to energy. In the end he's using all this to describe Einstein's line of thought when coming up with [itex]E=mc^2[/itex].
     
  12. Aug 17, 2005 #11

    rbj

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    i dunno what Einstein's line of thought was to get [itex] E = m c^2 [/itex], but the way it was done in my sophmore physics book was, after time dilation, length contraction, and relativistic mass (the Lorentz transformations, IIRC) are figgered out, the question was asked: in a known force feild, how much energy does it take to accelerated a body of rest mass [itex] m_0 [/itex] to a velocity of [itex] v [/itex] considering that the mass is increasing with increasing velocity and force is

    [tex] F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt} [/tex]

    and you get an answer for

    kinetic energy: [tex] T = \left( \frac{m_0}{\sqrt{1-v^2/c^2}} - m_0 \right) c^2 [/tex]

    or

    [tex] T = m c^2 - m_0 c^2 = E - E_0 [/tex]

    where [itex] E = m c^2 [/itex] is interpreted as the "total energy" and [itex] E_0 = m_0 c^2 [/itex] is interpreted as the "rest energy". the difference beint "kinetic energy".

    dunno how others learned it.
     
  13. Aug 18, 2005 #12

    Meir Achuz

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    I will have to give the relativistic formula for the Doppler shift:
    w'=w gamma[1+(v/c) cosA], where v is the speed of the star and A is the angle between the star's velocity and the line from the star to you, all in your rest system. The formula is the same whether you or the star is moving, but the light is always observed by you in your rest system.
     
    Last edited: Aug 18, 2005
  14. Aug 18, 2005 #13

    rbj

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    i dunno what "gamma" is (Gamma function??) but the [itex] 1 + |v|/c \cos(A) [/itex] does not contradict what i thought i was saying. [itex] |v| \cos(0) [/itex] is the opposite sign as [itex] |v| \cos( \pi ) [/itex] which means that red shifting is different than blue shifting. and it should not matter who is moving, since it is relative. neither the observer nor the star have any absolute claim on being the unique stationary position.

    edit: i know what [tex] \gamma = \left( 1 - v^2/c^2 \right)^{-\frac{1}{2}} [/tex] is. just didn't recognize the term at first.
     
    Last edited: Aug 19, 2005
  15. Sep 4, 2010 #14
    The Doppler effect of sound does depend on who is moving. If I were to move backwards at a speed a little greater than the speed of sound from a speaker I would out run the sound and never hear it. But if I moved the speaker back and the same speed i would eventually hear the sound.
     
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