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Homework Help: Relativistic Doppler Shift Equation - Derivation

  1. May 24, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    The full expression for the relativitic Doppler Shift is

    [tex]v' = v \gamma [1- \beta cos \theta][/tex]

    where v' and v are the frequencies of the light source in its own rest frame and and in the laboraty respectivley. [tex] \theta [/tex] is the (laboratory) anglebetween the direction of the photons and the direction in which the source is moving. Show that this is consistent with the equation:

    [tex] v' = v\sqrt{\frac{1 + \beta}{1 - \beta}} [/tex]

    for a source moving driectly away from the observer

    2. Relevant equations

    [tex] v' = v \gamma [1- \beta cos \theta][/tex]

    [tex] v' = v\sqrt{\frac{1 + \beta}{1 - \beta}} [/tex]

    [tex] \gamma = \frac{1}{\sqrt{1 - \beta}} [/tex]

    3. The attempt at a solution

    I have got some way through, but appear to be slightly stuck:

    [tex] v' = v \gamma [1- \beta cos \theta][/tex]

    [tex] v' = v \frac{[1- \beta cos \theta]}{1 - \beta}[/tex]

    for the angle, since moving away, [tex] \theta = 0 [/tex]

    thus, [tex] cos \theta = 1 [/tex]

    [tex] v' = v \frac{[1- \beta (1)]}{1 - \beta}[/tex]

    But I am not sure how to get the square root over the whole of the equation, or to turn the minus to a plus on the top if the fraction.

    Any ideas?

    TFM
     
  2. jcsd
  3. May 24, 2008 #2
    I might be missing something but i think you forgot the root when you subsituted gamma into the expression. Also, since its moving away you would have theta be pi, i believe, whichw ould give you the plus you're looking for
     
  4. May 24, 2008 #3

    TFM

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    Yeah it should really be

    [tex] v' = v\frac{1 - \beta}{\sqrt{1 - \beta}}[/ [/tex]

    Cos of Pi = -1, giving:

    [tex] v' = v \frac{1 + \beta}{\sqrt{1 - \beta}} [/tex]

    So all I need to de know is find out how to get the root over the whole of the fraction, not just the bottom.

    TFM
     
    Last edited: May 24, 2008
  5. May 24, 2008 #4

    TFM

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    Sorry, Latex seems tpo dislike my using ' to represent prime:

    Final equation so far:

    [tex] v` = v\frac{1 + \beta}{\sqrt{1 - beta}} [/tex]

    Any ideas how to get the square root to cover the whole fraction, noit just the bottom half?

    TFM
     
  6. May 24, 2008 #5

    alphysicist

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    Homework Helper

    Hi TFM,

    Your definition of gamma is incorrect. It needs to be:

    [tex]
    \gamma = \frac{1}{\sqrt{1-\beta^2} }
    [/tex]

    Once you use that, you can factor what's inside the square root on the bottom (it has the form of [itex]a^2-b^2[/itex]). After a cancellation, I think you can get the answer. What do you get?
     
  7. May 24, 2008 #6

    TFM

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    I now get:

    [tex] v^| = v\frac{1 + \beta}{\sqrt{(1 + \beta)(1 - \beta)}} [/tex]

    would you now do:

    [tex] v^| = v\frac{(1 + \beta)^{1/2}(1 + \beta)^{1/2}}{(1 + \beta^{1/2})(1 - \beta)^{1/2}} [/tex]

    So you would cancel out - does this look correct?

    TFM
     
  8. May 24, 2008 #7

    alphysicist

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    Homework Helper

    You have an exponent in the wrong place in the denominator, but it looks like it's probably just a typo. (The factor [itex](1+\beta^{1/2})[/itex] needs to be [itex](1+\beta)^{1/2}[/itex]). Other than that, it looks correct, and once you cancel out the identical factors in numerator and denominator you get the desired result.
     
  9. May 24, 2008 #8

    TFM

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    Yeah, the bracket is in the worng place, should have been:

    [tex] v^| = v\frac{(1 + \beta)^{1/2}(1 + \beta)^{1/2}}{(1 + \beta)^{1/2}(1 - \beta)^{1/2}} [/tex]

    So it cancels down to:

    [tex] v^| = v\frac{(1 + \beta)^{1/2}}{(1 - \beta)^{1/2}} [/tex]

    Which is the same as:

    [tex]v^| = v \sqrt{\frac{1 + \beta}{1 - \beta}}[/tex]

    TFM
     
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