1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic Doppler Shift Equation - Derivation

  1. May 24, 2008 #1

    TFM

    User Avatar

    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

    User Avatar

    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

    User Avatar

    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

    User Avatar
    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

    User Avatar

    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

    User Avatar
    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

    User Avatar

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativistic Doppler Shift Equation - Derivation
  1. Doppler shift (Replies: 2)

Loading...