Light Geodesic Path: Troubleshooting Post Deletion

In summary, you would need to specify the velocity as well as its position to calculate the geodesic.
  • #1
makc
65
0
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:
Physics news on Phys.org
  • #2
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
pervect said:
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). not just g_ij However, you need to specify the velocity as well as its position to calculate the geodesic. exactly, so I said: on step k, we're in x_i, with 'velocity' dx_i (=dt,i=4) ...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). that is, in tangent euclidean coordinates, but x_i I am speaking about, are distant static observer ones

(there must be simple iterative solution)
 
  • #4
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
that is, update à (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 http://physics.csufresno.edu/akira/GR/Theory.html [Broken] 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 by a moderator:
  • #6
Huh? What's à ? 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
pervect said:
Huh? What's à ?

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
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
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?
 

What is a light geodesic path?

A light geodesic path is the shortest path between two points in a spacetime, taken by a beam of light. It is also known as a "null geodesic" or "null ray".

Why would a post be deleted from a light geodesic path?

A post may be deleted from a light geodesic path if it does not follow the shortest path between two points, or if it violates any of the laws of physics.

How can I troubleshoot a post deletion from a light geodesic path?

To troubleshoot a post deletion, you can first check if the post followed the shortest path between two points. You can also check if the post followed the laws of physics, such as the speed of light or the curvature of spacetime. If the post still appears to be deleted, you may need to consult with a physicist or mathematician for further analysis.

Can a post be restored to a light geodesic path after deletion?

It is possible for a post to be restored to a light geodesic path after deletion, but it will depend on the specific circumstances and the severity of the violation. If the post was deleted due to not following the shortest path, it may be possible to amend the post to follow the correct path. If the post violated the laws of physics, it may not be able to be restored.

What are some common mistakes that lead to post deletion from a light geodesic path?

Some common mistakes that may lead to post deletion from a light geodesic path include not considering the curvature of spacetime, not accounting for the speed of light, or not accurately calculating the shortest path between two points. It is important to thoroughly understand the principles of a light geodesic path before attempting to post on it.

Similar threads

  • Special and General Relativity
Replies
4
Views
784
  • Special and General Relativity
Replies
27
Views
4K
  • Special and General Relativity
Replies
1
Views
512
  • Special and General Relativity
Replies
2
Views
1K
  • Special and General Relativity
Replies
20
Views
1K
  • Special and General Relativity
Replies
8
Views
2K
  • Special and General Relativity
Replies
6
Views
2K
  • Special and General Relativity
2
Replies
45
Views
4K
  • Special and General Relativity
Replies
3
Views
1K
  • Special and General Relativity
2
Replies
58
Views
5K
Back
Top