# Light geodesic path

1. Jan 4, 2005

### makc

can't delete post ?

w/e.

i'm trying to follow the path numerically.
in ase of light geodesic, given gij, xi, and very small deltas dxi on step k, what would be dxi on step k+1?

Last edited: Jan 4, 2005
2. Jan 4, 2005

### pervect

Staff Emeritus
You'll need to clarify a bit. It sounds like you are asking for the null geodesics (the paths that light follow) given the metric of space-time (g_ij). However, you need to specify the velocity as well as its position to calculate the geodesic. In this case we know the magnitude of the velocity (because it's a null geodesic) but not its direction.

Furthermore, if you know the coordinates at some particular time, the velocity alone will entirely deterimine the coordinates at some time t+dt for small enough dt - the acceleration will have a very small effect, on the order of dt^2. Thus there isn't any way to answer your question as posed, if I'm understanding it correctly, because while geodesic deviation will cause light beams to "curve", that curvature is equivalent ot an acceleration, and it won't affect the velocity of light (which after all is always equal to 'c' locally).

3. Jan 4, 2005

### makc

(there must be simple iterative solution)

4. Jan 4, 2005

### pervect

Staff Emeritus
If you are looking as strong gravitational fields, you need general relativity, and you'd need an approach

as discussed here on physics forum

to calculate the path of a geodesic.

I'm not clear if that's actually the problem you are attempting to solve, though, your notes on gij suggest to me that there may be some miscommunication going on.

5. Jan 6, 2005

### makc

that is, update Ã (4x4x4x4 elements), calculating 3 g derivatives for each, then raise index, and then... solving full-fledged geodesic ecuation, on EVERY STEP? come on, you are kidding me, don't you? look, for example, Eq5 here for Schwarzschild solution looks far easier thing to do; I'd use it, but the idea is to use arbitrary g_ij. more thoughts, anyone?

Last edited: Jan 6, 2005
6. Jan 6, 2005

### pervect

Staff Emeritus
Huh? What's Ã ? Calculate 3 g derivitaves (what's being differentiated) "for each" (each what?)

It does look like you are interested in geodesic motion, at least.

For the Schwarzschild metric, it turns out that there are two conserved quantities, something that looks like energy, and something that looks like angular momentum. These exist because of the symmetries (Killing vectors) in the Schwarzschild metric. If you have a totally arbitrary g_ij where the g_ij are functions of time, you won't have ANY Killing vectors, and no conserved quantities. If the g_ij are constant (not functions of time) you should have a timelike Killing vector which will give you one conserved quantity. Helpful, but you'll only be able to reduce the problem to a two dimensonal one (not counting time), with the two conserved quantities in the Schwarzschild metric, you can reduce the problem to a 1-d problem (not counting time).

7. Jan 6, 2005

### makc

Sorry, in my code page that was
+--
|
|
that is, capital gamma.

I still think one can get away with g_ij.

Consider ardbitrary 2D space with g_ij=0 for i!=j. For null geodesics, we have:
0 = g_x(x, t) * dx^2 - g_t(x, t) * dt^2.

Then, we just set delta_t = 0.00...01, starting point x, t, and direction
a=+1 or -1. Geodesic then done like this:

delta_x = a * sqrt( g_t(x, t)/g_x(x, t) ) * delta_t
x_new = x + delta_x
t_new = t + delta_t

now, that can't be MUCH harder just because we add a couple of dimensions, can it?

8. Jan 6, 2005

### pervect

Staff Emeritus
You probably won't have g_ij diagonal at any point in space time for the arbitrary case you are describing. But you could make it diagonal (or even better yet, a unit diagonal) with a linear change of variables. With such a change in variables, goedesics become straight lines in the transformed coordinate system, but only for a small region of space-time. You'd have to basically do a matrix inversion to find the required linear change of variables at a specific point, and as you travel along the geodesic you compute, you'd have to periodically recompute the transformation matrix to continue extending the line when you've moved far enough that g_ij gets too far away from being diagonal.

9. Jan 6, 2005

### makc

diagonality is not important, it just shows how simple things could be.
for 2D ex-e above, arbitrary:
ds^2=0=g_xx*dx^2-g_tt*dt^2+2*g_xt*dxdt
once g_ij and dt are known, we only have to find dx, which is as simple as ax^2+bx+c=0.
the key thing here is a number of free variables (which is 1 - delta_x). in 2D, light has no option to turn. in 3D, we already have light cone. so, the thing that escapes me, the thind I am missing here, the thing I am asking about - relation between dx, dy, and dz (back to 4D). what makes photon follow one path and not another? the reason should be the same reason which makes general geodesic equation you referred to. what is it?