# Redshift of two kinds

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.

But this is nowhere near close to the answer, which should be ##\omega_{\infty} = \omega(1 - 2M/r)## so where did I go wrong?
Is this the text book answer? I cannot fault your arguments or calculations so maybe you are "not wrong".

George Jones
Staff Emeritus
Gold Member
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)##

According to whom? I think that, except for a possible sign error, your answer is correct.

The photon wave 4-vector at emission as represented in ##O''## is ##k = \omega e_{\hat{t}} - \omega e_{\hat{r}}##.

Since the photon is moving radially outwards, shouldn't this be ##k = \omega e_{\hat{t}} + \omega e_{\hat{r}}##?

Using simplistic methods, I got the same result as you, except for a sign difference. See

According to whom? I think that, except for a possible sign error, your answer is correct.

Is this the text book answer? I cannot fault your arguments or calculations so maybe you are "not wrong".

Thanks for the replies! If I read the passage correctly then yes this is what the text says *should* be the answer. More specifically, this is exercise 8.4(a) on page 356 of Padmanabhan's GR text (which I know George has). It says that the ratio of the received wavelength to the emitted wavelength should satisfy ##\frac{\lambda_{\infty}}{\lambda} \propto (1 - 2M/r)^{-1}## whereas I have ##\frac{\lambda_{\infty}}{\lambda} = (1 - \sqrt{2M/r})^{-1} = (1 - 2M/r)^{-1}(1 + \sqrt{2M/r}) ##.

Since the photon is moving radially outwards, shouldn't this be ##k = \omega e_{\hat{t}} + \omega e_{\hat{r}}##?

Using simplistic methods, I got the same result as you, except for a sign difference. See

I had ##e_{\hat{r}}## pointing in the direction of motion of the radially infalling observer so wouldn't the outgoing light ray be pointing in the ##-e_{\hat{r}}## direction? I got the same sign when I did it the second way (last paragraph of post #1) wherein I didn't use a local Lorentz frame so I don't know if I'm making some kind of algebraic mistake.

Also if ##\frac{\lambda_{\infty}}{\lambda} = 1 + \sqrt{2M/r} ## as in the link then wouldn't we fail to get ##\frac{\lambda_{\infty}}{\lambda} \rightarrow \infty ## as ##r \rightarrow 2M##? Whereas if ##\frac{\lambda_{\infty}}{\lambda} = (1 - \sqrt{2M/r})^{-1}## then ##\frac{\lambda_{\infty}}{\lambda} \rightarrow \infty## as ##r \rightarrow 2M##. Shouldn't we have infinite redshift of the light ray that is sent out to the distant static observer right when the radially infalling observer reaches the event horizon?

Thanks again for the replies!

George Jones
Staff Emeritus
Gold Member
Thanks for the replies! If I read the passage correctly then yes this is what the text says *should* be the answer. More specifically, this is exercise 8.4(a) on page 356 of Padmanabhan's GR text (which I know George has). It says that the ratio of the received wavelength to the emitted wavelength should satisfy ##\frac{\lambda_{\infty}}{\lambda} \propto (1 - 2M/r)^{-1}##

Yes, this is the way I read it.

I had ##e_{\hat{r}}## pointing in the direction of motion of the radially infalling observer so wouldn't the outgoing light ray be pointing in the ##-e_{\hat{r}}## direction?

Okay. I thought the notation ##e_{\hat{r}}## indicated a unit vector in the ##\partial_r## direction.

I got the same sign when I did it the second way (last paragraph of post #1) wherein I didn't use a local Lorentz frame so I don't know if I'm making some kind of algebraic mistake.

Also if ##\frac{\lambda_{\infty}}{\lambda} = 1 + \sqrt{2M/r} ## as in the link then wouldn't we fail to get ##\frac{\lambda_{\infty}}{\lambda} \rightarrow \infty ## as ##r \rightarrow 2M##? Whereas if ##\frac{\lambda_{\infty}}{\lambda} = (1 - \sqrt{2M/r})^{-1}## then ##\frac{\lambda_{\infty}}{\lambda} \rightarrow \infty## as ##r \rightarrow 2M##. Shouldn't we have infinite redshift of the light ray that is sent out to the distant static observer right when the radially infalling observer reaches the event horizon?

I think so. I now realize that we calculated different things. My calculation shows that an observer falling freely from rest at infinity doesn't experience infinite blueshift at the event horizon.

I probably won't have another chance to look at this until tomorrow morning west coast time (parent-teacher meeting, daughter's soccer, etc.)

George Jones
Staff Emeritus
Gold Member
Some last comments before I leave work.

First, I found a book that gives both our results,

Also, I noticed something interesting when our two results are combined.

Suppose an observer at infinity emits wavelength ##\lambda_1## light in a radially inwards direction. This light is received at ##r## with wavelength ##\lambda_2## by an observer, Fred, who falls freely from infinity. Fred immediately emits light of the same wavelength (##\lambda_2##) in a radially outwards direction. This light is observed with wavelength ##\lambda_3## at infinity.

What is the relationship between ##\lambda_1##, ##\lambda_3##, and ##r##?

Yes, this is the way I read it.
Cool, so now that just leaves the question of how Padmanabhan gets that result, in light of the above.

Okay. I thought the notation ##e_{\hat{r}}## indicated a unit vector in the ##\partial_r## direction.
Ah ok, sorry about the confusion. I should have used some other notation.

I probably won't have another chance to look at this until tomorrow morning west coast time (parent-teacher meeting, daughter's soccer, etc.)

Have fun!

Some last comments before I leave work.

First, I found a book that gives both our results,
Yey! And what book is that, if you don't mind me asking?

Suppose an observer at infinity emits wavelength ##\lambda_1## light in a radially inwards direction. This light is received at ##r## with wavelength ##\lambda_2## by an observer, Fred, who falls freely from infinity. Fred immediately emits light of the same wavelength (##\lambda_2##) in a radially outwards direction. This light is observed with wavelength ##\lambda_3## at infinity.

What is the relationship between ##\lambda_1##, ##\lambda_3##, and ##r##?

Well using what you had, ##\frac{\lambda_2}{\lambda_1} = 1 + \sqrt{2M/r}##, and the above result that ##\frac{\lambda_3}{\lambda_2} = (1 - \sqrt{2M/r})^{-1}## we would get ##\frac{\lambda_3}{\lambda_1} = \frac{\lambda_3}{\lambda_2}\frac{\lambda_2}{\lambda_1} = \frac{1 + \sqrt{2M/r}}{1 - \sqrt{2M/r}}## hence ##\lambda_3 > \lambda_1##; does this agree with what you had?

More specifically, this is exercise 8.4(a) on page 356 of Padmanabhan's GR text (which I know George has). It says that the ratio of the received wavelength to the emitted wavelength should satisfy ##\frac{\lambda_{\infty}}{\lambda} \propto (1 - 2M/r)^{-1}## whereas I have ##\frac{\lambda_{\infty}}{\lambda} = (1 - \sqrt{2M/r})^{-1} = (1 - 2M/r)^{-1}(1 + \sqrt{2M/r}) ##.
Suspiciously, Padmanabhan's answer uses the 'proportional to' symbol rather than the equality symbol, implying some factor is missing and it would seem that factor is your ##(1 + \sqrt{2M/r}) ##. I have checked out both George's and your solutions and they appear to be correct (including signs) for the slightly different contexts they are used in.

Suspiciously, Padmanabhan's answer uses the 'proportional to' symbol rather than the equality symbol, implying some factor is missing and it would seem that factor is your ##(1 + \sqrt{2M/r}) ##

That's what I wanted to think but usually the proportionality symbol is written for constants whereas here the ##(1 + \sqrt{2M/r}) ## factor is variable with the point of emission ##r##.

That's what I wanted to think but usually the proportionality symbol is written for constants whereas here the ##(1 + \sqrt{2M/r}) ## factor is variable with the point of emission ##r##.
I agree that the missing factor is *usually* a constant. I was being generous to the author. :tongue:

George Jones
Staff Emeritus
Gold Member
Padmanabhan should have started his question with "(a) In the asymptotic limit as ##r \rightarrow 2M_+##, compute ..."

In this asymptotic limit, your result and Padmanabhan's result agree. For example, write ##2M/r = 1 -\epsilon##.

I was motivated to try this by Straumann's statement: "we obtain the asymptotic expression

$$1 + z \simeq \frac{4ME}{r - 2M}.$$

Ah ok I see, thank you so much for the help George! So does that mean that the exponential relationship ##\lambda_{\text{rec}}/\lambda_{\text{em}} \propto e^{t_{\text{rec}}/4M}## is really only valid when the infalling observer is very close to the event horizon (part b of the exercise)?

George Jones
Staff Emeritus