Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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.)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook