Light deflection and geodesics

[VladZH]

Technically, there is no such thing as "time curvature". The way curvature is defined is in terms of parallel transport.

• You have two different events (points in spacetime): e_1 and e_2.
• You have two different paths \mathcal{P_1} and \mathcal{P_2} connecting those events (a path being a curve through spacetime).
• You have a vector (direction in spacetime) V^\mu defined at e_1.
• You move along path \mathcal{P_1} from e_1 to e_2, and "parallel transport" V^\mu along the path to get a vector V^\mu_1 defined at point e_2
• You move along path \mathcal{P_2} from e_1 to e_2, and "parallel transport" V^\mu along the path to get a vector V^\mu_2 defined at point e_2
• If V^\mu_1 is different from V^\mu_2, then spacetime is curved.

Note that the paths \mathcal{P_1} and \mathcal{P_2} enclose a 2-dimensional surface in spacetime. So you can't really talk about curvature for a single coordinate, such as t. Curvature necessarily must involve at least 2 coordinates.

OK, let's suppose there is only space, without t. That will be the same way to define curvature?

 

[stevendaryl]

OK, let's suppose there is only space, without t. That will be the same way to define curvature?

Yes, you can talk about purely spatial curvature. It's the same definition: use parallel transport to define curvature.

A particularly simple example is the surface of the Earth, which is a curved 2-D surface. Imagine standing on the equator, at the point of 0^o longitude. Take a spear (representing your vector) and point it parallel to the ground pointing straight north. Now, walk straight north until you get to the North Pole, trying not to twist your spear. Now at the North Pole, your spear is pointing south, along the line 180^o longitude. Now, go back to where you started, at the equator, at 0^o longitude. Instead of going straight north, you go east to the point 90^o east longitude, keeping your spear pointing in the same direction (north). Now you go straight north until you reach the north pole. Your spear will now be pointing south, along the line of 90^o west longitude. So even though you tried to keep your spear pointing in the same direction at all times, the direction it is pointing when you get to the North Pole depends on the path you took. That's what spatial curvature means.

In the case of the bending of starlight by the sun, I haven't done the calculation, so I'm not sure how much of the effect is due to pure spatial curvature.

 

[A.T.]

I'm not sure how much of the effect is due to pure spatial curvature.

I think it's half of the effect for light:
http://mathpages.com/rr/s8-09/8-09.htm

 

[PeterDonis]

I mean absolute (Newtonian) time.

There is no such thing.

can we describe light deflection with GR equations or something?

Certainly. Most GR textbooks treat this problem. A decent online treatment is here:

http://lacosmo.com/DeflectionOfLight/index.html