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Homework Help: Doppler Effect aturla frequency

  1. Feb 16, 2005 #1
    In class we were explained that if the source moves toward the stationary observer

    [tex] f' = f_{0} \frac{v}{v-v_{s}} [/tex]
    where f0 is then aturla frequency of the wave, v is hte speed of the wave, and vs is the speed of the source

    and [tex] f' = f_{0} \frac{v_{rel}+v}{v} [/tex]
    where vrel is the speed of the observer

    a)SHow that the result are thesame if the source and observer speed are much less than the wave speed

    Now i can easilky say that [tex] v - v_{s} \neq 0 [/tex] and similarly [tex] v_{rel} + v \neq 0 [/tex] but is that really enough??

    It is worth 5 marks in thsi assignment

    b) FInd the difference between theclassical Doppler shift and the relativitistic Dopper shift up to the second order terms in (v/c)^2

    now i know the relativistic expression is

    [tex] f' = f_{0} \sqrt{\frac{1+\beta}{1-\beta}} [/tex]

    but how do i expand this. I do know how to expand [tex] \frac{1}{1-x} [/tex]
    but in this case i can get v / v-c and c/v-c how would that work??
     
  2. jcsd
  3. Feb 16, 2005 #2
    From your question it seems like you are trying to pove:
    [tex] f' = f_{0} \frac{v}{v-v_{s}} =f_{0} \frac{v_{rel}+v}{v} [/tex]
    [tex] \frac{1}{1-x} = 1 + x \ \ {When\ x\ll1}[/tex]
    so we have
    [tex] f' = f_{0} \frac{v}{v-v_{s}} = f_{0} \frac{1}{1-\frac{v_{s}}{v}} \cong f_{0} ( 1 + \frac{v_{s}}{v} ) = f_{0} \frac{v_{rel}+v}{v} =f_{0} (\frac{v_{rel}}{v}+1)[/tex]
    Since [tex]v_{s} = v_{rel}[/tex]

    hmm I would use the binomial Expansion. I am lazy and did not want to write out the binomial Expansion so here is a link:
    http://hyperphysics.phy-astr.gsu.edu/hbase/alg3.html
     
  4. Feb 16, 2005 #3

    dextercioby

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    Homework Helper

    a)No...U should try to put the expressions in a way in which u can take the relevant limit [itex]v<<v_{s} [/itex]

    b)Use this trick:
    [tex] \frac{1+\beta}{1-\beta}=1+\frac{2\beta}{1-\beta}=1+\frac{1}{something} [/tex]

    and now to take out terms in \beta^{2} outta the square root...

    Daniel.
     
  5. Feb 16, 2005 #4

    HallsofIvy

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    Wouldn't be enough for me! Saying they are not 0 doesn't say anything about the relationship of v- vs and vr- v. What is the relationship between vs and vr?

    What is &beta;? You should know how to expand [itex]\sqrt{1- x^2}[/itex] in a power series. Once you have expanded it in a power series, replace &beta; by its expression in v and c.
     
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