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Doppler Shift for light

  1. Nov 12, 2013 #1
    How do Doppler shifts work for light if, according to special relativity, light is constant velocity for all observers? So if c is unchanged, then surely wavelength and frequency don't change.

    I appreciate that I must be misunderstanding something, because redshift on stars occurs, but I am struggling to explain why.

    Thanks in advance for any help :)
     
  2. jcsd
  3. Nov 12, 2013 #2

    WannabeNewton

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    The frequency of a light wave corresponds to the time-like component of the wave 4-vector of a light wave, and components of 4-vectors are not invariant under non-trivial Lorentz transformations. More generally, energy is a frame dependent quantity. Why should wavelength and frequency of light waves be unchanged between inertial frames just because ##c## is unchanged between inertial frames? They can scale inversely under Lorentz transformations so as to cancel out any change in their ratio.
     
  4. Nov 12, 2013 #3

    Mentz114

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    ## c=\lambda\nu## is the equation relating velocity and wavelength and frequency. So both can change while keeping c constant.

    [edit. got it wrong first time]
     
  5. Nov 12, 2013 #4

    Bill_K

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    One thing which is the same in all rest frames is the phase of the wave. That is, for example, the wave crest always remains a wave crest. A typical wave is exp(i(kx - ωt), where the phase is kx - ωt, or equivalently since ω = ck, the phase is k(x - ct). This quantity must be an invariant.

    Under a Lorentz transformation,

    x' = γ(x - vt)
    t' = γ(t - v/c2 x)

    implying that x - ct just picks up an overall factor:

    x' - ct' = γ(1 - v/c)(x - ct) = √((1 - v/c)(1 + v/c)) (x - ct)

    Invariance requires that the wave vector k picks up the inverse factor:

    k' = √((1 + v/c)(1 - v/c)) k

    which represents the Doppler shift.
     
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