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Proving relativistic blueshift with Lorentz transforms only

  1. Dec 27, 2004 #1

    Curious3141

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    Dear esteemed physicists :)

    I've been trying to satisfy my own curiosity by proving relativistic Doppler using only Lorentz transforms between two inertial frames, one holding a source, the other the receiver.

    I've managed to derive the correct expression [tex] \frac{\nu}{\nu_0} = \sqrt{\frac{c-v}{c+v}}[/tex]

    for the redshift. But the blueshift still eludes me. I don't want to just switch the sign of v in the final expression, I want to be able to set up two inertial frames that are approaching each other and work through the transforms to get there for the blueshift. Preferably, it would be great if the source frame and the receiver frame are approaching (blueshift) then pass each other and then go into redshift as they recede from one another, and everything can be clearly seen from the equations. But I have no idea how to do this.

    Any help please ? This is not homework. Thanks very much.
     
    Last edited: Dec 27, 2004
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  3. Dec 27, 2004 #2

    Andrew Mason

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    I am not sure why you don't want to simply change the sign of v. If you derived the doppler redshift it is simply a matter of changing the sign of v in performing the Lorentz transformation for two pulses of light separated by period T.

    AM
     
  4. Dec 27, 2004 #3

    Curious3141

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    Not quite. The framework for the redshift derivation went like this (just the outline) :

    We have two inertial frames [itex]S[/itex] and [itex]R[/itex]. [itex]S[/itex] holds the source of light [itex]s[/itex] at its origin, [itex]R[/itex] holds the receiver [itex]r[/itex] at its origin. [itex]R[/itex] is moving with constant velocity [itex]v[/itex] along the positive [itex]x[/itex] direction of the [itex]S[/itex] frame.

    From frame [itex]S[/itex], at time [itex]t_0[/itex], [itex]s[/itex] emits a pulse of light, and sees the light being received by [itex]r[/itex] at [itex]t_1[/itex]. [itex]t_1[/itex] and [itex]t_0[/itex] are related in the [itex]S[/itex] frame by the speed of light.

    Those coordinates are all Lorentz transformed to [itex]R[/itex] to get the respective [itex]t'_0[/itex] and [itex]t'_1[/itex] for emission and reception events as seen from [itex]R[/itex].

    The relationship between [itex]t'_1[/itex] and [itex]t_0[/itex] gives the relationship between light reception as seen by [itex]r[/itex] and light emission as seen by [itex]s[/itex]. That relationship comes out to :

    [tex]\frac{t'_1}{t_0} = \sqrt{\frac{c+v}{c-v}}[/tex]

    If successive pulses from [itex]s[/itex] are emitted seperated by period [itex]T[/itex] as seen from [itex]S[/itex], they are received by [itex]r[/itex] with period [itex]T'[/itex] where :

    [tex]\frac{T'}{T} = \sqrt{\frac{c+v}{c-v}}[/tex]

    And hence the ratio of the frequencies is the reciprocal of the above, giving :

    [tex] \frac{\nu}{\nu_0} = \sqrt{\frac{c-v}{c+v}}[/tex]

    That's the formula for the redshift. For the blueshift I tried doing the same, except that [itex]r[/itex] was initially sited an initial distance [itex]d[/itex] on the negative side of the origin of [itex]S[/itex] (by initial, I mean when [itex]t=0[/itex] in the frame [itex]S[/itex]), then letting [itex]R[/itex] move in the positive [itex]x[/itex] direction of [itex]S[/itex]. But I got a mess of an expression, and I can't get the expected one.

    Please help ?
     
  5. Dec 27, 2004 #4

    dextercioby

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    Any decent course on relativistic electromagnetism should have a thorough approach of the both longitudinal and transversal Doppler-Fizeau effect.
    Try looking it up in Jackson's second volume.It should be there.If you don't want to,begin with the fact that em.wave's phase,written in the form:
    [tex] \phi(k,x)=-k^{\mu}x_{\mu} [/tex]
    ,is a Lorentz scalar and use the explicit form for the Lorentz transformation matrix to find the one for the wave 4-vector.

    Daniel.

    PS.U'll get the general formula from which,by chosing appropriate values for the angle [itex] \theta[/itex],u'll derive redshift,blueshift and transversal Doppler-Fizeau effect.
     
  6. Dec 27, 2004 #5

    Andrew Mason

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    Try working out the time between pulses as received at the postions [(x1,t1) and (x2,t2)] of the moving observer, as measured in the source frame. That should be:
    [tex]t_2-t_1 = \frac{T}{1+v/c}[/tex]

    The distance between these two points of reception as measured in the source frame (observer moving toward source) is:
    [tex]x_2-x_1 = -v(t_2-t_1)[/tex]

    Then transform that to the moving observer frame:
    [tex]t'_2 - t'_1 = \gamma ((t_2-t_1) - v(x_2 - x_1)/c^2[/tex]
    [tex]t'_2 - t'_1 = \gamma ((t_2-t_1) - v(-v(t_2 - t_1))/c^2[/tex]

    [tex]t'_2 - t'_1 = \gamma (\frac{T}{1+v/c} + v^2(\frac{T}{1+v/c})/c^2 = \gamma (\frac{T(1+v^2/c^2)}{1+v/c})[/tex]

    This reduces to:
    [tex]T' =T\sqrt{\frac{(1-v/c)}{(1+v/c)}} = T\sqrt{\frac{(c-v)}{(c+v)}} [/tex]

    I can see that in doing the blueshift a -2v/c term appears and it takes a little algebra to make it disappear. Perhaps this is where you had a problem.

    AM
     
  7. Dec 27, 2004 #6

    pervect

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    I'm not sure I see the difficulty. Imagine an object sitting at the origin of a coordinate system, emitting light waves. Imagine an incoming particle, which arrives at the origin at t=0. For t<0, there will be a blueshift. At t>0, there will be a redshift. Any form of analysis which gives the redshift should easily be able to get the blueshift simply by considering t<0 rather than t>0
     
  8. Dec 27, 2004 #7

    Curious3141

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    Thank you very much, you have helped me "see the light" (sorry, bad pun) :D

    I now "get" exactly what I'm supposed to do for the Longitudinal doppler both red and blueshifts. My mistake was in not being careful with the direction of change of the x-coordinates in the blueshift. All resolved now.

    BTW, I'm afraid there's a mistake in your derivation :

    If [itex]x_2 - x_1 = -v(t_2 - t_1)[/itex] then the correct Lorentz transform for observer moving toward source would have to be :

    [tex]t'_2 - t'_1 = \gamma ((t_2-t_1) + v(x_2 - x_1)/c^2)[/tex]

    Note the change of sign in the RHS.

    Then everything works out OK.

    The error in your derivation becomes apparent because this expression :

    [tex]\gamma (\frac{T(1+v^2/c^2)}{1+v/c})[/tex]

    cannot be factored without bringing in complex factors. The correct expression in its place should be :

    [tex]\gamma (\frac{T(1-v^2/c^2)}{1+v/c})[/tex]

    with the change of sign.

    But that's a minor error. The rest is fantastic, really helped to clarify my thinking. Thanks a googolplex !! ;)
     
    Last edited: Dec 27, 2004
  9. Dec 27, 2004 #8

    Andrew Mason

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    Quite right. I should have been more careful. The expression should have been:

    [tex]t'_2 - t'_1 = T' = \gamma (\frac{T(1-v^2/c^2)}{1+v/c})[/tex]

    which becomes:
    [tex]t'_2 - t'_1 = T' = \gamma (1-v/c)T[/tex]


    Glad to be of help.

    AM
     
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