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Taylor expansion of the relativistic Doppler effect?

  1. Aug 24, 2016 #1
    [Note from mentor: this thread was originally posted in a non-homework forum, therefore it does not use the homework template.]

    I have been given an equation for the relativistic doppler effect but I'm struggling to see this as a function and then give a first order Taylor expansion. Any help at all would be appreciated as I'm completely stuck.

    fo/fs = √(1 + vrel/c) / √(1-vrel/c)

    Where vrel is the relative speed of the source and the observer with respect to each other, c the speed of light, and vrel > 0 is here presumed to mean that the source and
    the observer move towards each other. Consider the relativistic Doppler effect in the case of (vrel/c) ≪ 1, but (vrel/c) > 0. Write a first-order Taylor expansion for the relativistic Doppler effect, and show that the result is equivalent to either of the two expressions found for sound.

    The expressions for sound are:

    i) The source moves with speed vs towards stationary observer

    fo/fs = 1 / (1-(vs/v))

    ii) The observer moves with speed vo towards the stationary source

    fo/fs = 1 + (vo/v)
     
    Last edited by a moderator: Aug 24, 2016
  2. jcsd
  3. Aug 24, 2016 #2

    stevendaryl

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    This should probably be in the Homework section, right? But just to put this into a more obvious form, let [itex]x = v_{rel}/c[/itex]. Then you want to expand the function

    [itex]F(x) = \frac{\sqrt{1+x}}{\sqrt{1-x}}[/itex]

    in powers of [itex]x[/itex]. Do you know how to do that?
     
  4. Aug 24, 2016 #3
    Sorry this is my first question. Because it's university work I thought it would go here.

    I started by doing that but I got confused with not having what would be a in a normal Taylor expansion question.
     
  5. Aug 24, 2016 #4

    stevendaryl

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    I'm not sure what you mean by a "normal Taylor expansion question". The [itex]F(x)[/itex] that I gave is an ordinary function with an ordinary Taylor expansion.
     
  6. Aug 24, 2016 #5
    Okay I think my Taylor expansion needs some work. But getting confirmation that it's in this form helps. Thanks
     
  7. Aug 25, 2016 #6
    I still have a problem with this. In a Taylor expansion I have a value to expand about. I am not given a value for this. And the question does not ask for a Maclaurin series.
     
  8. Aug 25, 2016 #7

    stevendaryl

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    You're expanding [itex]F(x) = \frac{\sqrt{1+x}}{\sqrt{1-x}}[/itex] about [itex]x=0[/itex]. Do you know how to do that?
     
  9. Aug 25, 2016 #8
    Is this a Maclaurin series?

    I'm not sure because I don't know if I should use binomial expansion or differentiate withTaylor.
     
  10. Aug 25, 2016 #9

    stevendaryl

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    Well, they amount to the same thing. You can use the binomial expansion for a fractional exponent:

    [itex](1+x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha -1)}{1 \cdot 2} x^2 + \frac{\alpha (\alpha -1)(\alpha - 2)}{1 \cdot 2\cdot 3} x^3 + ...[/itex]

    If [itex]\alpha[/itex] is fractional, then the series never terminates, but that doesn't matter.

    But that's the same series you would get if you took derivatives:

    Letting [itex]G(x) = (1+x)^\alpha[/itex],

    [itex]G'(x) = \alpha (1+x)^{\alpha - 1}[/itex]
    [itex]G''(x) = \alpha (\alpha - 1) (1+x)^{\alpha - 2}[/itex]
    [itex]G'''(x) = \alpha(\alpha - 1)(\alpha - 2) (1+x)^{\alpha - 3}[/itex]
    ...

    So you can write: [itex]G(x) = G(0) + x G'(0) + \frac{x^2}{1 \cdot 2} G''(0) + \frac{x^3}{1 \cdot 2 \cdot 3} G'''(0) + ...[/itex]

    You get the same series.

    For this problem, if you're going to use binomial expansions, you have to compute the case [itex]\alpha = +\frac{1}{2}[/itex] for [itex]\sqrt{1+x}[/itex] and compute the case [itex]\alpha = -\frac{1}{2}[/itex] for [itex]\sqrt{1-x}[/itex] (using [itex]-x[/itex]) and then you have to multiply the two series, and group the terms by powers of [itex]x[/itex].
     
  11. Aug 25, 2016 #10
    Thank you, this is clear now and has helped a lot
     
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