Anonymous at Fri, 28 Mar 2025 04:51:55 UTC No. 16630540

Enumerate the list of first steps you tried

Anonymous at Fri, 28 Mar 2025 05:46:23 UTC No. 16630576

>>16630159
The null vector is distributed, not localized.
>>16630540
It's probably a homework question and OP is an H1B.

Anonymous at Fri, 28 Mar 2025 05:49:05 UTC No. 16630578

>>16630540
>>16630576
Retarded pseuds indistinguishable from bots

Anonymous at Fri, 28 Mar 2025 10:33:46 UTC No. 16630749

>>16630159
It can actually!
Just use this EM tensor for a point-particle with trajectory [math]x(\tau)[/math] and velocity [math]u^a[/math]:
[eqn]
T^{a b}(x)=m \int u^a u^b \delta^{(4)}(x-x(\tau)) d \tau
[/eqn]

Anonymous at Fri, 28 Mar 2025 10:52:54 UTC No. 16630757

>>16630159
If you mean the general geodesic equation of general relativity i.e. the Christoffel connection, then I suspect empty space is not enough to determine it amongst other possible connections but I could be retarded.

Anonymous at Fri, 28 Mar 2025 13:22:09 UTC No. 16630851

>>16630159
>the field equation in empty space
You mean “Ricci tensor = 0”? Because the geodesic equation is just a geometric property of the manifold. It holds regardless of what the manifold is. Whereas “Ricci tensor = 0” just specifies a class of manifolds, Einstein manifolds. You’re asking to derive a more fundamental result from a less fundamental one.

Anonymous at Fri, 28 Mar 2025 14:08:00 UTC No. 16630870

I want to elaborate on my previous post a little bit. The geodesic equation is
[eqn]\ddot{x} = -\Gamma \dot{x}^2[/eqn]
This is just a fancy relativistic Newton's 2nd law. The Christoffel symbols play the role of the gravitational potential. In the non-relativistic limit and limit of uniform field, this reduces to ma=mg.

Let's turn it down a notch and just consider your regular-old non-relativistic scalar potential in 1D. The equation of motion is
[eqn]m\ddot{x}=\frac{\partial V}{\partial x}[/eqn]
This equation holds regardless of the particular form of the potential V because it's just the Euler-Lagrange equation for x. The particular form of V is specified by the Poisson's equation
[eqn]\nabla V = J[/eqn]
plus the boundary conditions. You solve for V first, then plug it back into EL equations to solve for x. Both are independent of one another. So OP is asking to derive the equation's of motion from GR's analogue of Laplace's equation. Obviously, you can't do that. In action principle terms, these are two separate field equations, one for [math]\delta S/\delta x = 0[/math] and the other for [math]\delta S/\delta g = 0[/math].