General Relativity Effective Potentials

[Auburnman]

I am having some trouble interpreting different effective potentials

The first potential is V = (L^2)/(2r^2) - (r^2)/(2R^2) - (L^2)/(R^2)
The second potential is V = (-1/2) + (L^2)/(2r^2) -(L^2)/(R^2)

What I am having a hard time identifying do these potentials have stable orbits?
And are particles attracted to r = 0? And can particles reach r = 0?

 

[xepma]

Remember that the force follows from the potential through

F = -\nabla V

or in your 1D case F = -\partial_r V.

The requirement for a stable orbit is that the force is zero, F = 0

Now apply this to your potentials.

 

[Auburnman]

...dude this is general relativity get out of here with your Forces lol, that's Newtonian physics your talking about

 

[Mentz114]

Try writing the Lagrangian for a particle and solving the equations of motion. The general Lagrangian is just L =( kinetic energy - potential energy) so no problem there.

[I see now that this is what xepma has already done]

 

[xepma]

...dude this is general relativity get out of here with your Forces lol, that's Newtonian physics your talking about

That's a very nice attitude you got there. But my description still applies.

The difference between relativistic and Newtonian gravition would only result in a different effective potential. The principle of a stable orbit does not change.

 

[George Jones]

The difference between relativistic and Newtonian gravition would only result in a different effective potential. The principle of a stable orbit does not change.

I agree, but, for a stable circular orbit, I think that another condition has to added to dV/dr = 0, i.e., whether locally the potential is a "hill" or a "valley".