# Radiation measured from moving star

1. Jul 23, 2013

### asras

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, $\mathcal{O}$, of the star.

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

We now move to the frame of the observer, $\overline{\mathcal{O}}$, who moves with speed $v$ in the positive $x$-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):
$$T^{\overline{\alpha} \overline{\beta} } = \Lambda^{\overline{\alpha}}_{\mu} \Lambda^{\overline{\beta}}_{\nu} T^{\mu \nu}$$
Now if the coordinate of reception is $x$ in the star's frame it must be $R = (1-v)\gamma x$ in the observer's frame (with $\gamma = (1-v^2)^{-1/2}$). Thus we have that
$$T^{\overline{0} \overline{x}} = \frac{L}{4\pi x^2} (1-v)^2 \gamma^2,$$
$$T^{\overline{0} \overline{x}} = \frac{L}{4\pi R^2} (1-v)^4 \gamma^4.$$

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

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

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. Jul 23, 2013

### Bill_K

In the second method, you transformed x, did you forget to transform t as well? N, the "number of photons per second", will change.

3. Jul 23, 2013

### asras

Bill, wouldn't that just be equivalent to
$$N \rightarrow N/\gamma,$$
due to time dilation? (If it were $N \rightarrow N/((1+v)\gamma)$ I would be happy!)

4. Jul 23, 2013

### Bill_K

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.

5. Jul 23, 2013

### asras

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 $\Delta \tau$ the rate of emission is $L = 1/\Delta \tau$. In the frame of the observer the events are separated in time by
$$\Delta t = \Delta \tau \gamma.$$
However the star has moved $v\Delta t$ to the left so the rate of emission in the observers frame becomes
$$L' = \frac{1}{\Delta t + v\Delta t /c } = \frac{1}{\Delta \tau} \frac{1}{(1+v/c) \gamma} = \frac{L}{(1+v/c)\gamma}.$$
(I included the c because I feel it makes it clearer.)

Thanks!