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.

This question Brachistochrone Problem for Inhomogeneous Potential has the obvious extension. Namely the same question, when gravity is treated according to general relativity. To make it specific let's consider the case of Schwarzschild metric. Given to points, outside the event horizon, what is the trajectory (or the worldline) of a test body, which goes from the first point to the second for least time. All considered from the point of view of a distant observer at rest (with respect to the star/back hole). Of course other metrics or the general case are also interesting.

I couldn't find much by searching on the net. There is one article with the tittle suggesting relevance (actually more but this one seems very relevant), but it is over 60 pages of technicalities. So it would be nice to see a shorter answer.

share|improve this question
    
Geodesics? Or are you asking for something more? –  user346 Mar 27 '11 at 15:59
    
I agree with Deepak that the "least time" trajectories are called geodesics, and finding them for a given metric is a standard problem that can usually not be solved analytically. I must just correct a comment: in the Minkowski space, a geodesic actually maximizes the proper time rather than minimizing it! –  Luboš Motl Mar 27 '11 at 16:02
2  
Aren't geodesics extremals for proper time? I am asking for those that make the time in the reference frame of the distant observer minimal. –  MBN Mar 27 '11 at 16:08
1  
As in the standard brachistochrone problem, you want to imagine a stationary frictionless tube that the particle falls through, right? So its path is not a geodesic, but the external force (due to the sides of the tube) is always perpendicular to the direction of motion (as measured by static observers at each point), so that it does no work. Sounds like a well-posed problem, but I bet the solution is messy. –  Ted Bunn Mar 27 '11 at 16:28
    
@ Ted Buun: Yes, that is what I have in mind. I thought that is how the classical problem is formulated. Anyway, the problem will be messy in general, I was hoping that in a specific metric it might be explicitly solvable. I asked about the Schwarzschild's because there many things can be done, but anything else is welcomed too. –  MBN Mar 27 '11 at 16:45
show 5 more comments

2 Answers 2

up vote 9 down vote accepted

Update: I made a number of mistakes in the original version of this post, although I think all the big ideas are right. I tried to fix everything, but I wouldn't be at all surprised if I've made additional errors.

I'm pretty sure that this is a very hard problem! I think I know how to get started, but I doubt I can finish.

I'll work in Schwarzschild coordinates, with $c=1$, Schwarzschild radius $R$, and $(+---)$ signature, so the metric is $$ ds^2=(1-R/r)dt^2-(1-R/r)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta\,d\phi^2). $$ All the action can be taken to lie in the equatorial plane $\theta=\pi/2$.

For a particle traveling on a geodesic in this geometry, the energy-like conserved quantity (i.e., the one arising from time-translation invariance of the metric) is $u_0$ where $u$ is the 4-velocity. I'll call this quantity $E$ (it's really energy per unit mass): $$ E=\left(1-{R\over r}\right)\left(dt\over d\tau\right). $$

I'm going to assume that this quantity is conserved even for our particle whizzing along in the tube. I'm pretty sure this is the correct generalization of the assumption that the constraint forces due to the tube do no work. I think you could prove this by looking at the physics in a local inertial reference frame in which the tube is at rest as the particle whizzes by. In that frame, the above energy conservation law is equivalent to the statement that the particle's speed is constant, which follows from a special-relativistic analysis in that frame, because the tube pushes in a direction perpendicular to the velocity.

Next we use the fact that the four-velocity has unit norm: $$ 1=u_\mu u^\mu=(1-R/r)\dot t^2-(1-R/r)^{-1}\dot r^2-r^2\dot\phi^2, $$ where dots mean $d/d\tau$.

Divide through by $\dot t^2$: $$ \dot t^{-2}=1-{R\over r}-\left(1-{R\over r}\right)^{-1}\left({dr\over dt}\right)^2-r^2\left({d\phi\over dt}\right)^2. $$ Express $\dot t$ in terms of the energy, and rearrange: $$ \left(1-{R\over r}\right)^{-1}\left(dr\over dt\right)^2+r^2\left(d\phi\over dt\right)^2 =1-{R \over r}-\left(1-R/r\over E\right)^2. $$ Say the particle starts from rest at position $r_0$. Then $E=(1-R/r_0)^{1/2}$. So $$ dt=\sqrt{(1-R/r)^{-1}(dr/d\phi)^2+r^2\over 1-R/r-(1-R/r)^2/(1-R/r_0)}\,d\phi. $$ If our initial points are $(r_0,0)$ and $(r_0,\alpha$), then the quantity we want to minimize is $$ t=\int_0^\alpha \sqrt{(1-R/r)^{-1}(dr/d\phi)^2+r^2\over 1-R/r-(1-R/r)^2/(1-R/r_0)}\,d\phi. $$ You can in principle use standard calculus of variations techniques from here to get $r(\phi)$.

That's enough for me! You said in the comments that you'd be happy with just the functional. Are you happy?

share|improve this answer
1  
Yes, I am happy. Thanks a lot. –  MBN Mar 28 '11 at 19:25
    
+1 for nice post. But there's also lot of interesting things going on in the general case... –  Stan Liou Aug 16 '11 at 0:55
add comment

It's worthwhile to extend the thrust of Ted Bunn's approach to more general cases. Suppose you have a general static spacetime $$ds^2 = -e^{2\nu} dt^2 + h_{ij}dx^idx^j,$$ where the Latin indices sum over spatial terms, as usual. The factor of $2$ is purely traditional, meant to simplify various quantities, especially for the spherically symmetric case $h_{ij}dx^idx^j = e^{2\lambda} dr^2 + d\Omega^2$, of which the Schwarzschild spacetime is a particular example. Let's also distinguish "proper" brachistochrones (minimizing proper time $\tau$) from "coordinate" brachistochrones (minimizing coordinate time $t$).

We know that $-d\tau^2 = -e^{2\nu}dt^2 + h_{ij}dx^idx^j,$ and because the spacetime is static, it is also stationary with the Killing field $\partial_t$ that produces a conserved current $\epsilon = -\langle\partial_t,u\rangle = e^{2\nu}(dt/d\tau)$, the specific energy. So for every timelike of such specific energy, substituting the latter into the former gives $$h_{ij}dx^idx^j = -d\tau^2[ 1 - e^{2\nu}(dt/d\tau)^2] = -d\tau^2[1 - \epsilon^2/e^{2\nu}].$$ Aha! The proper brachistochrones of specific energy $\epsilon$ are simply geodesics of the Riemannian manifold with the metric $$dS_\tau^2 = \frac{h_{ij}dx^idx^j}{\epsilon^2/e^{2\nu} - 1}.$$ For the Schwarzschild spacetime, this is $$dS_\tau^2 = \frac{(1-R/r)^{-1}dr^2 + r^2d\Omega^2}{\epsilon^2(1-R/r)^{-1} - 1}.$$

Let's now consider the coordinate brachistochrones. The approach is the same: $$h_{ij}dx^idx^j = dt^2e^{2\nu}[1 - e^{-2\nu}(d\tau/dt)^2] = dt^2e^{2\nu}[1 - e^{2\nu}/\epsilon^2],$$ so that the coordinate brachistochrones are the geodesics of $$dS_t^2 = \frac{h_{ij}dx^idx^j}{e^{2\nu}(1 - e^{2\nu}\epsilon^{-2})}.$$ For the Schwarzschild spacetime, this is: $$dS_t^2 = \frac{(1-R/r)^{-1}dr^2 + r^2d\Omega^2}{(1-R/r)(1-(1-R/r)\epsilon^{-2})},$$ which matches Ted Bunn's result under the assumptions that the trajectory is in the equatorial plane ($d\Omega = d\phi$) and the particle starts out at rest from $r=r_0$ ($\epsilon^2 = (1-R/r_0)$).

An even more general treatment in an arbitrary stationary (not necessarily static!) spacetime is possible, and was done by V. Perlick in J. Math. Phys. 32, 3148 (1991), which treats the problem in a more geometrically abstract form and includes brachistochrones in Gödel's rotating universe.

share|improve this answer
add comment

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.