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Derivation of Doppler Shift from Frequency to Wavelenght

  1. Aug 30, 2009 #1

    IBY

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

    Derive the doppler shift equation from the equation in terms of frequency to one in terms of wavelenght. Clues: frequency=c/lamda, use Taylor's expansion, velocity of source is much smaller than velocity of wave.

    c-velocity of wave
    v-velocity of source
    f-frequency
    lambda-wavelenght

    2. Relevant equations

    Doppler shift in terms of frequency
    [tex]f'=f(1+\frac{v}{c}cos \theta)[/tex]
    Relationship between wavelenght and frequency
    [tex]\lambda=\frac{c}{f}[/tex]
    Taylor's expansion
    [tex]\sum_{n=0}^2 \frac{f'^n(f(1+\frac{v}{c}cos \theta)}{n!}x^n[/tex]

    3. The attempt at a solution
    I used the wavelenght-frequency relationship equation in order to substitue for f in the doppler frequency equation to get:

    [tex]\frac{c}{\lambda'}=\frac{c}{\lambda}(1+\frac{v}{c}cos \theta)[/tex]

    [tex]\frac{c}{\lambda'}=\frac{c}{\lambda}+\frac{c}{\lambda}\frac{v}{c}cos \theta[/tex]

    [tex]\frac{c}{\lambda'}=\frac{c}{\lambda}+\frac{v}{\lambda}cos \theta[/tex]

    [tex]\lambda'=\frac{c}{\frac{c}{\lambda}+\frac{v}{\lambda}cos \theta}[/tex]

    [tex]\lambda'=\lambda+\frac{\lambda c}{v cos\theta}[/tex]

    I used Taylor's expansion to the second degree, and I got:

    [tex]\lambda'=\frac{\lambda c}{2v}x^2+{\lambda+\frac{\lambda c}{v}[/tex]

    I small angle approximate, so x^2=sin^2 (theta), use trig id to get sin^2 (theta)=1-cos^2 (theta)

    [tex]\lambda'=\frac{\lambda c}{2v}-\frac{\lambda c}{2v}cos^2 \theta+\lambda+\frac{\lambda c}{v}[/tex]

    [tex]\lambda'=\frac{3\lambda c}{2v}-\frac{\lambda c}{2v}cos^2 \theta+\lambda [/tex]

    And now, I am dead in the water.
     
    Last edited: Aug 30, 2009
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  3. Aug 30, 2009 #2

    gabbagabbahey

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    You might want to double check this step:wink:
     
  4. Aug 30, 2009 #3

    IBY

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    @gabbagabbahey
    I don't see it. If one divides by fraction, aren't you supposed to multiply with the bottom part reciprocated?
     
  5. Aug 30, 2009 #4

    gabbagabbahey

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    [tex]\frac{A}{A+B}\neq1+\frac{A}{B}[/tex]

    You can't divide up the denominator in this manner
     
  6. Aug 30, 2009 #5

    IBY

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    If so, then I use Taylor's expansion on this?

    [tex]\lambda'=\frac{\lambda c}{c+vcos \theta}[/tex]
     
  7. Aug 30, 2009 #6

    gabbagabbahey

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    I would divide both the denominator and numerator by [itex]c[/itex] first:

    [tex]\lambda'=\frac{\lambda}{1+\frac{v}{c}\cos\theta}=\lambda\left(1+\frac{v}{c}\cos\theta\right)^{-1}[/tex]

    The reason you want to do this is because you know [itex]v\ll c[/itex], so [itex]\frac{v}{c}\cos\theta[/itex] will be a small number and you are left trying to Taylor expand [itex](1+\text{small number})^{-1}[/itex] which you can do easily.
     
  8. Aug 30, 2009 #7

    IBY

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    Okay, one last fact checking (hopefully) before I go away for good. :smile:

    So there is:

    [tex]\lambda(1+\frac{v}{c}cos \theta)^{-1}[/tex]

    I derive it:

    [tex]\lambda \frac{v}{c}sin \theta(1+\frac{v}{c}cos \theta)^{-2}x[/tex]

    I do it again, use the multiplication rule:

    [tex](\lambda (\frac{v}{c})^{2}sin^2 \theta(1+\frac{v}{c}cos \theta)^{-3}+\frac{\lambda \frac{v}{c}cos \theta(1+\frac{v}{c}cos \theta)^{-2}}{2})x^2[/tex]

    Small number, so sin of theta is zero and cos of theta is 1, x=sin so including all of the equation above, I am left with:

    [tex](\frac{\lambda \frac{v}{c}(1+\frac{v}{c})^{-2}}{2})sin^2 \theta+\lambda(1+\frac{v}{c})^{-1}[/tex]

    Trig id sin^2 theta is 1-cos^2 theta

    [tex](\frac{\lambda \frac{v}{c}(1+\frac{v}{c})^{-2}}{2})(1-cos^2 \theta)+\lambda(1+\frac{v}{c})^{-1}[/tex]

    For some reason, I feel like I went into a dead end.
     
  9. Aug 30, 2009 #8

    gabbagabbahey

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    Huh?! With respect to what variable are you taking the derivative and why? Also, what is [itex]x[/itex] in the above equation?:confused:
     
  10. Aug 30, 2009 #9

    IBY

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    Hhhmm
    Evidently, I made the mistaken assumption that the "theta" was the variable. Probably, I am making this more complicated than it is supposed to be.
     
  11. Aug 30, 2009 #10

    gabbagabbahey

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    Well, if you have some function [itex]f(x)[/itex] and [itex]x[/itex] is small, then [itex]f(x)\approx f(0)+f'(0)x[/itex] right?

    You know that the quantity [itex]\frac{v}{c}\cos\theta[/itex] is small, so why not call that quantity [itex]x[/itex]?

    When you do that, you have [itex]\lambda'=\lambda(1+x)^{-1}[/itex] right? So define the function [itex]f(x)=(1+x)^{-1}[/itex] and expand about small 'x'...make sense?

    The reason you don't want to use [itex]\theta[/itex] as your variable is because you don't know that [itex]\theta[/itex] is small, so you would need to keep all terms in the Taylor expansion, which doesn't help you at all.
     
  12. Aug 30, 2009 #11

    IBY

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    Okay, I think I got it. After I expanded it, I got:

    [tex]\lambda'=\lambda(1+x)^{-1}-\lambda(1+x)^{-2}+\frac{2\lambda(1+x)^{-3}}{2}[/tex]

    x is small, so it is around 0

    [tex]\lambda'=\lambda-\lambda x+\lambda x^2[/tex]

    x is v cos (theta)/lambda

    [tex]\lambda'=\lambda-\lambda\frac{v}{c}cos \theta+\lambda(\frac{v}{c}cos \theta)^2[/tex]

    Well, it turns out I only need the first order expansion in order to get a parallel equation with wavelenght:

    [tex]\lambda'=\lambda(1-\frac{v}{c}cos \theta)[/tex]

    So, I think I got it! Thanks for the help. :smile:
     
    Last edited: Aug 30, 2009
  13. Aug 30, 2009 #12

    gabbagabbahey

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    Looks good to me!:approve:
     
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