1
$\begingroup$

Please bear with me, and don't get upset if i have lack in knowledge about spacetime.

Brachistochrone: Given two points A and B in a vertical plane, what is the curve traced out by a point acted on only by gravity, which starts at A and reaches B in the shortest time.

So from my understanding light moves in a 'straight' way (it moves in all ways but the outer paths cancel out?). And it always takes the fastest route. And when light 'bends' around a massive object it actually is still moving in a straight line on a geodesic, which is the fastest route.

So is it fair to say that a brachistochrone is a straight line in the curved space, as its the shortest time path from a to b under a constant gravitational force?

$\endgroup$
4
$\begingroup$

No. Note that in the brachistochrone problem, gravity is not the only force acting on the object - you also have the normal force from the surface the object is sliding on.

Spacetime geodesics are the trajectories of freely-falling objects. On the surface of the Earth, they correspond to objects falling straight downward with acceleration $g=9.8$ m/s$^2$, or following a parabolic arc as per the usual projectile problem (more accurately, they follow elliptical or hyperbolic orbits as per the solution to the Newtonian gravity problem). Any trajectory which deviates from this is not a spacetime geodesic.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But light does not 'fall' straight down when bending around a massive object. Still it moves on a geodesic. Or is light just a special case because it moves so fast? And what about a ball you throw horizontaly and then it starts falling while moving forward, isn't that a Spacetime geodesic either ? Sorry but i just want but I want to fully understand it. And thanks for your help! $\endgroup$ – Alex bries Dec 6 '19 at 23:47
  • $\begingroup$ @Alexbries No, you’re correct. That was silly of me, please see my edit. $\endgroup$ – J. Murray Dec 6 '19 at 23:50
  • $\begingroup$ thanks a lot Murray. $\endgroup$ – Alex bries Dec 6 '19 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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