Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let's say that I'm hovering in a rocket at constant spatial coordinates outside a Schwarzschild black hole.

I drop a bulb into the black hole, and it emits some light at a distance of $r_e$ from the center, with a wavelength of $\lambda_{e}$ in the rest frame of the bulb.

What would the wavelength of the light be when it reaches me, at $r_{obs}$ in terms of the radius at which it is emitted, $r_e$?

This is a subquestion from Sean Carroll's Spacetime and Geomtery. Earlier in the chapter, Carroll asserts that any stationary observer $(U^i= 0)$ measures the frequency of a photon following a null geodesic $x^{\mu}(\lambda)$ to be

$$\omega= -g_{\mu\nu}U^\mu\frac{dx^\nu}{d\lambda}$$

I don't understand where this expression comes from. How does one even conceptualize things like wavelength and frequency of light in terms of general relativistic quantities like $U, g_{\mu\nu}, ds^2$, etc?

share|improve this question
    
Into the black hole? Or do you mean closer to the black hole but still outside the event horizon? –  David Z Mar 3 '13 at 4:18
    
Not dropped directly into the singularity, but not necessarily outside the event horizon. It may be the case that the bulb is dropped outside the event horizon, and then emits light from inside it, or that it is dropped from within the event horizon and also emits light there, or that it is dropped from outside the event horizon and emits light outside. Also, we assume that the path that the bulb follows is purely radial, and that the observer coordinates are stationary. –  willie.holdman Mar 3 '13 at 4:38
    
Yes, "into the black hole" means "inside the event horizon." You do know that light can't escape from inside the event horizon, right? –  David Z Mar 3 '13 at 4:42
2  
Yes—so then there should be some mathematical relationship between the radius at which the light is emitted and the wavelength of the observed light that is infinite or undefined for emission points less than 2GM, and well defined for emission points greater than 2GM—I'm just trying to understand how that relationship would be derived. –  willie.holdman Mar 3 '13 at 4:51
    
And sorry! The observer should be outside the black hole— however there are still no constraints on the emission point of the light. –  willie.holdman Mar 3 '13 at 4:55
show 5 more comments

1 Answer

Here are some ideas re your question:

Let us consider the path taken by the torch in the presence of a black hole, and let us assume the observer is outside the horizon. For the sake of simplicity let us assume the light source (the torch) is falling in a rectilinear way.

Some algebra and quite a bit of physics, combining the principle of equivalence and some aspects of special relativity, can show that the geometry of the path of the torch is given by the equation (taking only rectilinear motion into account):

$ds^2=\left(1-\frac{2GM/c^2}{r}\right) c^2dt^2-\frac{dr^2}{\left(1-\frac{2GM/c^2}{r}\right)}$.

Where r the distance of the torch from the centre of the BH and M is the mass of the BH. The coefficients of $dt^2$ and $dr^2$ are the metric “tensor components” of space-time geometry for this particular question. For the light of the torch, however, the path is a geodesic curve – line of shortest path taken by light in a hugely curved space-time, and the above equation becomes $ds^2=0$ hence:

$\left(1-\frac{2GM/c^2}{r}\right) c^2dt^2-\frac{dr^2}{\left(1-\frac{2GM/c^2}{r}\right)}=0$.

The latter equation gives the speed of the light source, the torch, as it falls towards the BH from outside the horizon, and observed by the observer at some very large distance away from the BH

$v(r)=c(1-2GM/c^2r)$.

The frequency shift $z=\frac{f-f_0}{f_0}$ relates to the speed of the light source via the equation

$v(r)=c\frac{z^2+2z}{z^2+2z+2}.$

The last equation gives the way the frequency shift varies as a function of $r$, and how it is affected by the mass, M, of the BH. Here, $f$ is the frequency received by the observer, while $f_0$ is the actual frequency emitted by the light source (the torch.)

share|improve this answer
    
Thanks for this, John! But in this scenario we're actually assuming that the observer is not a very large distance from the Black Hole, but rather is suspended by a rocket at some constant, comoving spatial coordinates ($r_{obs}$) outside the event horizon. $r_{obs}$ is not far enough away from the BH to be taken as flat space. –  willie.holdman Mar 3 '13 at 21:03
1  
Just to nit-pick, $r$ isn't "the distance of the torch from the centre of the BH," it's just a coordinate. It doesn't correspond to radial distances. –  elfmotat Mar 3 '13 at 21:54
    
@danig When I say large distance away from the BH I don't really mean out at anfinity. I mean the observer is not falling with the torch... –  JKL Mar 4 '13 at 19:34
    
@elfmotat How is that? The derivation of this "simple equation" is based on the assumption that $r$ is the radial distance from the gravitating body, which happened to turn into a BH. You need to define what you mean by "just a coordinate", coordinates are usually referred to some reference point otherwise they don't make sense. –  JKL Mar 4 '13 at 19:40
    
You're using Schwarzschild coordinates, –  elfmotat Mar 4 '13 at 19:51
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.