Calculating Distance Light Travels Near a Black Hole

[Haorong Wu]

TL;DR Summary

calculate the distance that light travels in a Schwarzschild metric

Suppose that we tangentially send a light from an orbit of radius $h$ to another orbit of radius $l$ near a black hole. I would like to calculate the distance that the light travels.

I start from the Schwarzschild metric, $$ ds^2=-(1-\frac m r) dt^2+\frac 1 {1- \frac m r} dr^2 +r^2 d \theta ^2 + r^2 \sin^2 \theta d \phi^2$$ where $m$ is the Schwarzschild radius.

The null geodesic gives that $$\frac {dr}{d\lambda}=\sqrt{E^2-(1-\frac{m}{r})\frac{L^2}{r^2}}$$ $$\frac{d\phi}{d\lambda}=\frac{L}{r^2}$$ with $E=\hbar \omega$ and $L=hp=hE$ due to the conservation of angular momentum.

So I have the proper length traveled by the photon as \begin{align} s= & \int_h^l \left |g_{ij}dx^idx^j \right |^{1/2}=\int_h^l d\lambda [ \frac 1 {1-m/r} (\frac {dr}{d\lambda}) ^2+r^2(\frac{d\phi}{d\lambda})^2 ]^{1/2}\nonumber \\=&\int_h^l dr [(\frac 1 {1-m/r} (E^2-\frac {L^2}{r^2}+\frac {mL^2}{r^3}) +\frac {L^2}{r^2} )/( E^2-(1-\frac{m}{r})\frac{L^2}{r^2})]^{1/2}. \nonumber \end{align}

I have run the integration in Matlab. Also, I calculate the distance if the light travels in Minkowski space, i.e., $s^{'}=\sqrt{l^2-h^2}$.

These two results are almost identical. Where did I make a mistake?

Thanks!

 

[PeterDonis]

I would like to calculate the distance that the light travels.

The calculation you appear to be doing has nothing do to with spatial distance. Your integral for $s$ is a spacetime interval integral and, if you do it correctly, will give zero since light rays travel on null geodesics.

The spatial distance between the two points you describe will be independent of what kind of object traverses that distance; it is just an integral on the Flamm paraboloid, and will give a somewhat larger answer than the one you give for flat Euclidean space (which is what your $s'$ actually is).

 

[Haorong Wu]

The calculation you appear to be doing has nothing do to with spatial distance. Your integral for $s$ is a spacetime interval integral and, if you do it correctly, will give zero since light rays travel on null geodesics.

The spatial distance between the two points you describe will be independent of what kind of object traverses that distance; it is just an integral on the Flamm paraboloid, and will give a somewhat larger answer than the one you give for flat Euclidean space (which is what your $s'$ actually is).

Thanks, @PeterDonis . But I only integrate the spatial part $g_{rr}dr^2+g_{\phi\phi}d\phi^2$ without the temporal part. Is it still wrong?

I though that when the light travels very near the Schwarzschild radius of a black hole, it will travel a far more distance than in Minkowski space. But in a second thought, its trajectory is bent but the distance does not increase dramatically.

 

[PeterDonis]

I only integrate the spatial part $g_{rr}dr^2+g_{\phi\phi}d\phi^2$

No, you're not. You're not integrating along a spacelike curve. You're integrating along the null curve that is the light ray's worldline. Not including the $dt / d\lambda$ term in the integral just makes it meaningless instead of giving zero as the answer.

I though that when the light travels very near the Schwarzschild radius of a black hole, it will travel a far more distance than in Minkowski space.

To even make this comparison you have to know what points in a surface of constant time in Schwarzschild spacetime correspond with particular points in Euclidean 3-space. You are assuming that the Schwarzschild radial coordinate $r$ is what gives this correspondence. That is one possible assumption, but not the only one. Another would be to equate the radius in Euclidean 3-space to the isotropic radial coordinate instead of the Schwarzschild one. That would give a different answer than the Schwarzschild one. So really there is no unique way to make the comparison.

That said, if you pick a particular basis for comparison, such as the Schwarzschild radial coordinate, the space you need to be integrating in, and whose metric you need to be using, is the Flamm paraboloid. And the curve you need to integrate along will have nothing to do with any constants of geodesic motion, for light rays or anything else. It will just be a particular spacelike curve defined by being purely tangential at the smaller radial coordinate $h$ and ending at the larger radial coordinate $h$. Those specifications should give you a function $\phi(r)$ that allows you to eliminate $\phi$ from the integral and make it an integral over $r$. But it won't be the same as the integral over $r$ you are doing now.

 

[PeterDonis]

its trajectory is bent but the distance does not increase dramatically.

The distance does not increase without bound, no, because the lower radius $h$ has to be above the horizon. I don't know that the increase over the Euclidean distance (given that you are using the radial coordinate $r$ as the basis for comparison, as I said in my previous post) is always small, though.

 

[Haorong Wu]

Thanks, @PeterDonis . I have read two textbooks about GR. But neither of them say anything about the Flamm paraboloid. I would like to study it further. Could you suggest some textbooks on this subject? (Not MTW, please :P )

 

[PeroK]

Thanks, @PeterDonis . I have read two textbooks about GR. But neither of them say anything about the Flamm paraboloid. I would like to study it further. Could you suggest some textbooks on this subject? (Not MTW, please :P )

I think your problems are more fundamental than that.

First, if you measure the spacetime length of a particle trajectory, you will get the proper time experienced by the particle or zero for a massless particle or light ray on a null trajectory.

Second, in order to define a spatial distance, you need a synchronisation convention. Schwarzschild coordinates provide such a convention. That's why the question says "using the Schwarzschild metric". Then you must integrate along the appropriate spacelike curve.

The specifics of the Flamm paraboloid require no more, I suggest, than a google search.

 

[A.T.]

Thanks, @PeterDonis . I have read two textbooks about GR. But neither of them say anything about the Flamm paraboloid. I would like to study it further. Could you suggest some textbooks on this subject? (Not MTW, please :P )

https://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm's_paraboloid

 

[pervect]

Summary:: calculate the distance that light travels in a Schwarzschild metric

Suppose that we tangentially send a light from an orbit of radius h to another orbit of radius l near a black hole. I would like to calculate the distance that the light travels.

I start from the Schwarzschild metric, ds2=−(1−mr)dt2+11−mrdr2+r2dθ2+r2sin2⁡θdϕ2 where m is the Schwarzschild radius.

The null geodesic gives that drdλ=E2−(1−mr)L2r2 dϕdλ=Lr2 with E=ℏω and L=hp=hE due to the conservation of angular momentum.

So I have the proper length traveled by the photon as s=∫hl|gijdxidxj|1/2=∫hldλ[11−m/r(drdλ)2+r2(dϕdλ)2]1/2=∫hldr[(11−m/r(E2−L2r2+mL2r3)+L2r2)/(E2−(1−mr)L2r2)]1/2.

Just to be clear, when you write gij, you restrict i and j from 1 to 3, correct?

That looks right to me - essentially, you seem to be projecting the space-time metric, gμν with μ,ν = 0..3 to a purely spatial metric gij where i,j vary from 1 to 3, though you did not use that exact language.

This is technically a projection operator. We could write $P = g + u \otimes u$, with u being the 4-velocity of a static observer. In the general case this is not always equivalent to omitting time from the line element, but in this particular case it is.

I have not waded through the math in detail, so I don't know if or where you made an algebraic mistake.

 

[PeterDonis]

you seem to be projecting the space-time metric

I don't think it's projecting the metric so much as projecting the worldline of the light ray into a spacelike surface of constant time (which, as you note, requires choosing a simultaneity convention--in this case it is the simultaneity convention of Schwarzschild coordinates). I agree that, in general, that procedure should give the correct answer.

However, for a light ray this projection should be independent of the energy at infinity and the angular momentum; simply specifying that the light ray is purely tangential at the inner radial coordinate $l$ is sufficient to completely specify its worldline. So at the very least, whatever procedure is used to project the light ray's worldline into a spacelike surface of constant time should give a result that is independent of $E$ and $L$. (One possibility might be to see if, given the inner radial coordinate $l$ and the purely tangential initial condition, one could express $E$ and $L$ in terms of $l$.)

 

[Ibix]

(One possibility might be to see if, given the inner radial coordinate $l$ and the purely tangential initial condition, one could express $E$ and $L$ in terms of $l$.)

The ratio $L/E$ is related to the launch angle. See https://www.physicsforums.com/threads/null-geodesics-in-schwarzschild-spacetime.895174/post-5651323.

 

[pervect]

I don't think it's projecting the metric so much as projecting the worldline of the light ray into a spacelike surface of constant time (which, as you note, requires choosing a simultaneity convention--in this case it is the simultaneity convention of Schwarzschild coordinates). I agree that, in general, that procedure should give the correct answer.

I think either approach can work in this case, they achieve the same result.

 

[PeterDonis]

I think either approach can work in this case

I don't think there are two approaches. I was not suggesting an alternate approach to what you described. I was suggesting what I think is a better way of describing the approach you described.