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Deriving the gravitational red shift?

  1. Jan 18, 2015 #1
    I am trying to derive the gravitational red shift effect but I think I am going about it all wrong. Specifically, I want to derive the change in frequency/ wavelength when a photon moves away from the surface of a star mass M and radius R.

    So I tried to use relativistic mass of the photon and I got something along the lines of:
    [itex]\Delta E= h\Delta f = GMm_{rel}\left(\frac{1}{R}-\frac{1}{r}\right)[/itex]

    And then substituting in [itex]m_{rel}=\frac{hf}{c^2}[/itex]

    Would give [itex]\Delta f =\frac{GMf}{c^2}\left(\frac{1}{R}-\frac{1}{r}\right)[/itex]

    But then I realised that f would be changing as the photon leaves the surface, so I thought that maybe I have to integrate?

    Considering the photon rising a small distance [itex]\delta r [/itex] and rearranging would give

    [itex]\int\frac{1}{f} df =\frac{GM}{c^2}\int\frac{dr}{r(r+dr)}[/itex]

    So it is a big mess right now! Would appreciate if someone could tell me where I am going wrong...
     
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  3. Jan 18, 2015 #2

    stevendaryl

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    It depends on what, exactly, you're trying to do. If you are trying to compute the frequency shift of a light signal traveling height [itex]h[/itex] radially outward from the surface of the Earth, your first calculation gives a good approximation:

    [itex]\frac{\delta f}{f} = \frac{GM}{c^2}(\frac{1}{R} - \frac{1}{R+h})[/itex]

    To get a more accurate calculation, you have to go beyond this energy loss approach. That approach is using Newtonian gravitational energy. To get a more accurate result, you have to use General Relativity.
     
  4. Jan 18, 2015 #3
    Thank you for your reply! But what about the f on the left hand side? It would not remain constant, and that is why I wanted to integrate. Because for large changes in height ( say if you wanted to find the change in frequency for the photon escaping the gravitational field of the star completely, then surely you would have to integrate, even if this is just an approximation using Newton's laws. Because f would not be constant?
     
  5. Jan 18, 2015 #4

    stevendaryl

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    Well, the integral approach can be done, although as I said, I don't think it's worth doing, because you should really be doing GR.

    The differential equation for the pseudo-Newtonian approach would be this:

    [itex]df = f MG/c^2 (\frac{1}{r+dr} - \frac{1}{r}) = - f MG/(c^2 r^2) dr[/itex]

    (This uses the fact that \frac{1}{r+dr} = \frac{1}{r} - \frac{dr}{r^2} + ...)

    So if you get all the [itex]f[/itex]s to one side, you have:

    [itex]df/f = - MG/c^2 \frac{dr}{r^2}[/itex]

    Integrating both sides gives:

    [itex]log(f/f_0) = MG/c^2 (1/r - 1/r_0)[/itex]

    or

    [itex]f = f_0 e^{MG/c^2 (1/r - 1/r_0)}[/itex]

    But using [itex]e^x = 1 + x + x^2/2 + ...[/itex], we can write this as:

    [itex]f = f_0 (1 + \delta U/c^2 + 1/2 (\delta U)^2/c^4 + ...)[/itex]

    where [itex]\delta U = MG/r - MG/r_0[/itex]

    But the higher-order terms are not to be trusted, since we only used a Newtonian approximation for energy. So I don't think that the more accurate integration result is any better than the simple result.
     
  6. Jan 18, 2015 #5
    This has been so useful! Thank you! I just have two final questions.

    Firstly, is the f in the approximation the initial frequency, or the final frequency after the photon has moved away from the star?[

    And secondly, with ragrds to the expanded approximation, you said that
    I am not sure why this would be the case. Do the higher order terms not just give small corrections to the first few terms since the higher order terms are being divided by such large powers of c? I would think that these higher order terms would be unnecessary because this is just an approximation, but I am not sure why they shouldn't be trusted. And also, surely this integration method should approximate the original approximation method of
    [itex]\frac{\Delta f}{f}=\frac{GM}{c^2}\left(\frac{1}{R}-\frac{1}{R+h}\right)[/itex], however in the integration method if I just take the first two terms, I get

    [itex]f-f_0=\Delta f=\frac{GM}{c^2}\left(\frac{1}{r}-\frac{1}{r_0}\right)[/itex], so the factor of 1/f has disappeared on the left! And the 1/(old distance) and 1/(new distance) have swapped around?

    EDIT: I see that the swapping around is just because I now have a negative change in frequency delta f , although I am still unsure about where the 1/f went!
     
  7. Jan 18, 2015 #6

    stevendaryl

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    [itex]\delta f[/itex] should definitely be proportional to [itex]f[/itex]. Otherwise, the units don't even match (the left side, [itex]\delta f[/itex] has dimensions of frequency, while the right side is dimensionless).

    As for the sign, it should be negative: The frequency decreases as the light signal rises in the gravitational field. So it should be:

    [itex]\delta f = - \frac{GM}{c^2} (\frac{1}{r_0} - \frac{1}{r}) = - \frac{GM}{c^2} (\frac{1}{R} - \frac{1}{R+h})[/itex]
     
  8. Jan 18, 2015 #7
    Oh! I see where I went wrong! When expanding out the bracket I didn't multiply the second term by f_0! That makes sense! And it also answers my second question I suppose- in the first approximation, the f on the denominator is the original frequency of the photon as it was at the surface of the star!

    Thank you so much!
     
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