0

I'm trying to do actual Schwarzschild Metric calculations. In looking at this video Schwarzschild Proper Distance at 1:20 he shows the calculation for moving directly outward on a radius from the black hole. This equation has the $M$ figure in it so it has significant curvature the closer we are to the mass of the black hole. But then at 3:00 he shows the calculation for moving along the sphere (with constant $r$) and there is no use of $M$ in the equation. This would indicate to me that there is no curvature due to mass when moving along the sphere, regardless of how close I am to the black hole. Is this a correct interpretation? It's very hard to envision the first curvature without the second.

Qmechanic
  • 201,751
foolishmuse
  • 4,551
  • In that equation for the circumference you mention at minute 3 he set dt=0, so there is no motion, that describes a spacelike path. – Yukterez Jan 05 '23 at 16:51
  • @Yukterez H did the same thing in the first equation, dt=0. – foolishmuse Jan 05 '23 at 16:53
  • At 1:20 he does the spacelike radial distance, that is in contrast to the circumference not euclidean in that coordinates. gtt and grr are not euclidean, but gθθ and gφφ are, in the Schwarzschild Droste coordinates he uses we have 2πr for the circumference – Yukterez Jan 05 '23 at 16:56
  • @Yukterez would that make a difference to my basic question? Is there an equation for the circumference that does include M? – foolishmuse Jan 05 '23 at 17:53
  • Sure, if you want the lorentzcontracted circumference in the frame of someone orbiting in the transverse direction you need the M to determine his circular velocity. In the coordinates you use the local clocks and rulers are stationary though, so the circumference for them is just the euclidean one, but radially they have to accelerate in order to stay stationary. – Yukterez Jan 05 '23 at 18:47
  • @Yukterez That shows me his velocity, but I'm looking for curvature of spacetime. Does it show me that as well? Or do I derive curvature from his velocity? – foolishmuse Jan 05 '23 at 18:53
  • 1
    For what direction curves into which for the specified local observer in your coordinates you need the Riemann curvature tensor, see here at Output 4 (in that link the indices {1,2,3,4} are {t,r,θ,φ} so they start with 1, not 0). That depends on the motion of the local observers, therefore its components are different in different coordinates for the same metric. – Yukterez Jan 05 '23 at 19:01
  • Indeed, the Schwarzschild space is curved (length contracted) only in the radial direction. – safesphere Jan 07 '23 at 02:20

1 Answers1

2

It's due to the way the radial coordinate $r$ is defined. It's natural for beginners to assume it's a radial distance, but it isn't. It is defined as the circumference of a circle centred on the black hole divided by $2\pi$. So if you are in a circular orbit round any black hole the circumference of your orbit is always $2\pi r$ regardless of the mass of the black hole.

The actual radial distance has to be calculated by integrating $ds$ along a line of constant $t$, $\theta$ and $\phi$. For more on this see Exact meaning of radial coordinate of the Schwarzschild metric.

John Rennie
  • 355,118
  • Going back to my original question, this indicates that there is no curvature in spacetime when travelling in a circle around a black hole, or perhaps that we just don't measure the curvature. I can see how we would say that around a BH there is a constant curvature throughout the circle. But it seems counterintuitive to say that this constant curvature is the same BH with much larger mass. How can we compare the curvature between BH? – foolishmuse Jan 06 '23 at 16:34
  • @foolishmuse yes, the surface on which you are travelling is just like a sphere in flat space. But if you let down a tape measure to measure the radial distance from the centre of the sphere you would get a result larger than the circumference C divided by 2. This difference between the measured distance and C/2 is what depends on the mass of the black hole. – John Rennie Jan 06 '23 at 16:49
  • Now I'm going to refer back to your answer to my earlier question here: https://physics.stackexchange.com/questions/738753/how-does-loop-quantum-gravity-deal-with-spaghettification-near-a-black-hole?noredirect=1#comment1653936_738753 Your answer to my current question would indicate that space is curved only in the vertical direction, not horizontally. So now looking at the earlier question, it would indicate that LQG particulate space is also stretched in the vertical direction. This might not be spagettification, but certainly is stretching vertically. Your thoughts on this idea? – foolishmuse Jan 06 '23 at 18:48