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Radiation measured from moving star

  1. Jul 23, 2013 #1
    My problem is how to calculate the radiation measured from a moving star. I figure there's two ways to do it, both of which I go through below, but they do not produce the same answer!

    For the first part everything is in the rest frame, [itex] \mathcal{O} [/itex], of the star.

    The star emits [itex] N [/itex] photons of frequency [itex] \nu [/itex] each second. The total energy radiated per second is thus [itex] L := Nh\nu [/itex]. It can be shown the stress-energy tensor in this case has entries for events [itex] (t,x,0,0) [/itex]:
    [tex] T^{00} = T^{0x} = T^{x0} = T^{xx} = \frac{L}{4\pi x^2}, [/tex]
    and all other entries zero.

    We now move to the frame of the observer, [itex] \overline{\mathcal{O}} [/itex], who moves with speed [itex] v [/itex] in the positive [itex] x [/itex]-direction and who is also located on the x-axis. The components of the stress-energy tensor in this frame are given by (using the usual rules for tensor transformation):
    [tex] T^{\overline{\alpha} \overline{\beta} } = \Lambda^{\overline{\alpha}}_{\mu} \Lambda^{\overline{\beta}}_{\nu} T^{\mu \nu} [/tex]
    Now if the coordinate of reception is [itex] x [/itex] in the star's frame it must be [itex] R = (1-v)\gamma x [/itex] in the observer's frame (with [itex] \gamma = (1-v^2)^{-1/2} [/itex]). Thus we have that
    [tex] T^{\overline{0} \overline{x}} = \frac{L}{4\pi x^2} (1-v)^2 \gamma^2, [/tex]
    [tex] T^{\overline{0} \overline{x}} = \frac{L}{4\pi R^2} (1-v)^4 \gamma^4. [/tex]

    So this gives the radiation the observer measures. However if we approach the problem using doppler-shift of the photons we get [itex] \nu' = (1-v)\gamma \nu [/itex], [itex] R = (1-v)\gamma x [/itex], and we have

    [tex] T^{\overline{0} \overline{x}}= \frac{Nh\nu}{4\pi R^2} (1-v)^3 \gamma^3. [/tex]


    So which is correct? I'm fairly sure the first method is correct, but I can't figure out exactly what I'm missing in the second calculation.
     
  2. jcsd
  3. Jul 23, 2013 #2

    Bill_K

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    In the second method, you transformed x, did you forget to transform t as well? N, the "number of photons per second", will change.
     
  4. Jul 23, 2013 #3
    Bill, wouldn't that just be equivalent to
    [tex] N \rightarrow N/\gamma, [/tex]
    due to time dilation? (If it were [itex] N \rightarrow N/((1+v)\gamma) [/itex] I would be happy!)
     
  5. Jul 23, 2013 #4

    Bill_K

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    It's not just time dilation. Like you say it's γ(1 + v/c), the same as the Doppler shift factor. It's the number of photons you're intercepting per second. One way to see the need for the (1 + v/c) factor is that N will be different depending on whether the star is moving toward you or away from you, which changes the sign of v.
     
  6. Jul 23, 2013 #5
    Ah, I see. If we imagine two emission events (lets just say that one photon is emitted at each event) separated in the rest frame of the star by [itex] \Delta \tau [/itex] the rate of emission is [itex] L = 1/\Delta \tau [/itex]. In the frame of the observer the events are separated in time by
    [tex] \Delta t = \Delta \tau \gamma. [/tex]
    However the star has moved [itex] v\Delta t [/itex] to the left so the rate of emission in the observers frame becomes
    [tex] L' = \frac{1}{\Delta t + v\Delta t /c } = \frac{1}{\Delta \tau} \frac{1}{(1+v/c) \gamma} = \frac{L}{(1+v/c)\gamma}. [/tex]
    (I included the c because I feel it makes it clearer.)

    Thanks!
     
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