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Relativistic Doppler Effect

  1. Dec 6, 2015 #1
    [mentor's note - lightly edited to fix the Latex]
    Hi there,

    I was hoping if someone could clear my small misconception for this equation.

    [itex]f' = f \left( \frac{ 1- \beta }{ 1+\beta } \right)^2[/itex]

    I had thought if the numerator is negative and denominator is positive that means the signal of light or what ever would be moving away from the source. But recently I have come across a question which shows the numerator be positive as the source moves away relative to a reference frame.

    Was hoping if someone could tell me a bit more about this topic as it seems to be a bit confusing. I hope that made sense as well haha, if not I can try to explain it better/ give the problem.

    Thanks!
     
    Last edited by a moderator: Dec 6, 2015
  2. jcsd
  3. Dec 6, 2015 #2

    BvU

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    Hello norobo, :welcome:

    1 I don't think there's a square in the expression. Power ##{1\over 2}## is what I find, e.g. here
    2 there's no way ##1-\beta## can be negative
     
  4. Dec 6, 2015 #3
    Hey thanks for the reply!

    Sorry about that it should be 1/2, what I was trying to say is not that it's negative but when the signs are -/+ it means the source is moving apart from the receiver and when the signs are switched the source is going towards the receiver, where f' is the source and f is receiver. Is this true or do I have this backwards? Thanks again!
     
    Last edited: Dec 6, 2015
  5. Dec 6, 2015 #4

    Nugatory

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    As you've written it (and after fixing the square root thing) positive ##\beta## is moving apart. Remembering this is much too much trouble though o_O - it's easier to just remember that when the source is approaching the wave crests get squished closer so the frequency increases and put the smaller or the larger quantity in the numerator and denominator accordingly.
     
  6. Dec 6, 2015 #5
    Ah gotcha, so I want to see if I understand it. If f' is the source and f is the receiver.
    Say a car is approaching past you (at a speed close to light haha) the source being the car, as it approaches the WAVELENGTH is getting smaller hence the frequency increases, so we have [itex]f' = f \left( \frac{ 1+\beta }{ 1-\beta } \right)^{1/2}[/itex] and as it goes away the WAVELENGTH is getting larger so the frequency decreases and we have [itex]f' = f \left( \frac{ 1-\beta }{ 1+\beta } \right)^{1/2}[/itex]
     
  7. Dec 6, 2015 #6

    Ibix

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    You can do what you have done, which is to define ##\beta## as the absolute rate of separation between source and receiver. Then you need your two formulae (which I agree with in this context), one for red-shift and one for blue.

    Alternatively you can give ##\beta## a sign, and make it positive for approaching sources and negative for receding sources (or vice versa). Then you only need one of those formulae - which one depends on your sign convention. I do not know if there is a universally agreed approach to this. You might find it easier to remember that $$f'=f\left( \frac{1\pm\beta}{1\mp\beta}\right)^{1/2}$$and figure out which one is appropriate using Nugatory's menmonic.
     
  8. Dec 6, 2015 #7
    Or, ##f'=f\left( \frac{1+\beta}{1-\beta}\right)^{\pm \frac{1}{2}}##.
     
  9. Dec 7, 2015 #8

    Ibix

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    Indeed. Who says algebra isn't fun?
     
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