- #1

WannabeNewton

Science Advisor

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## Main Question or Discussion Point

**Redshift of two "kinds"**

Hello! Consider the following problem: A space ship on a circular orbit at radius ##r## in a Schwarzschild metric emits a photon with the rest frame frequency ##\omega## at an angle ##\alpha## outward from the tangential direction of the motion, in the plane of the orbit. What is the frequency of the photon as seen by a stationary observer at a large distance?

Call the spaceship frame ##O## and imagine that at the emission event there is a stationary frame ##O'##. At that event, ##O## has axes ##e_{\hat{x}}, e_{\hat{y}}## in the plane of the orbit with ##e_{\hat{x}}## pointing along the tangential direction of the orbit at that event. The wave 4-vector at this event of the emitted photon as represented in ##O## is just ##k = \omega_0 e_{\hat{t}} + \omega_0\cos\alpha e_{\hat{x}} + \omega_0 \sin\alpha e_{\hat{y}}##. Furthermore, the speed of ##O## relative to ##O'## is given by ##v = \sqrt{\frac{M/r}{1 - 2M/r}}## (this can easily be derived by using the fact that ##(u^{\mu})_{O}(u_{\mu})_{O'} = -\frac{1}{\sqrt{1 - v^2}}##). Thus, using an inverse Lorentz boost in the ##e_{\hat{x}}## direction to go from ##O## to ##O'## at this event, we have the frequency ##\omega'## as measured in ##O'##: ##\omega' = k^{0'} = \Lambda^{0'}{}{}_{\nu}k^{\nu} = \omega\gamma(1 + v\cos\alpha)##. Therefore ##\omega_{\infty} = (1 - 2M/r)^{1/2}\omega' = \omega\gamma(1 + v\cos\alpha)(1 - 2M/r)^{1/2}## which is what my text lists as the correct answer.

Now I know this may seem like a roundabout way of solving this problem but I did it this way for a particular reason. Consider now the scenario wherein an observer with frame ##O''## falls radially inwards from rest at infinity. At some ##r## this observer emits a photon, with frequency ##\omega## in ##O''##, radially outwards which is received by a static observer at infinity. Consider again a static frame ##O'## at the emission event. At this event ##O''## has a radial axis ##e_{\hat{r}}## in the direction of motion. The photon wave 4-vector at emission as represented in ##O''## is ##k = \omega e_{\hat{t}} - \omega e_{\hat{r}}##. The speed of ##O''## relative to ##O'## is ##v = \sqrt{2M/r}## so performing an inverse Lorentz boost in the ##e_{\hat{r}}## direction to go from ##O''## to ##O'## we have ##\omega' = \gamma \omega (1 - v)## hence ##\omega_{\infty} = (1 - 2M/r)^{1/2}\omega' = \gamma \omega (1 - v)(1 - 2M/r)^{1/2} = \omega(1 - v) = \omega(1 - \sqrt{2M/r})## since ##\gamma = (1 - 2M/r)^{-1/2}##.

But this is nowhere near close to the answer, which should be ##\omega_{\infty} = \omega(1 - 2M/r)## so where did I go wrong? Why did this method work for the case of the circular orbit but not for the case of the radial free fall from rest at infinity?

By the way I should note that I did the radial case in the direct way and still didn't get the right answer. By "direct way" I mean simply evaluating ##\omega = -k_{\mu}(u^{\mu})_{O''}## in the Schwarzschild basis, using the fact that ##u_{O''} = (1 - 2M/r)^{-1}\partial_{t} - \sqrt{2M/r}\partial_{r}##. For the wave 4-vector ##k = k^t \partial_t + k^r \partial_r## we have that ##k^{r}\rightarrow 1## as ##r \rightarrow \infty## which fixes the conserved energy ##e = k^r = 1## hence ##k^t = (1 - 2M/r)^{-1}##. Therefore ##\omega = (1 - 2M/r)^{-1}(1 + \sqrt{2M/r}) = (1 - \sqrt{2M/r})^{-1}## hence ##\omega_{\infty} = 1 = \omega(1 - \sqrt{2M/r} )## which is the same result as above so I still don't get where I'm going wrong.

Thanks in advance!