- #26

Dale

Mentor

- 29,746

- 6,082

This is correct. The reason is because it is not directly the Christoffel symbol which is equal to the fictitious force. If you look at the equation for the four-acceleration you see:Thinking it over some more, I'm not sure this makes sense globally. The Christoffel symbol is indeed as given here, but it only corresponds to the observed "acceleration due to gravity" along one particular observer's worldline--the observer whose proper time is equal to the coordinate time [itex]t[/itex] in this chart. Other observers at rest in this chart (the other Rindler observers) have different proper accelerations and so will observe a different "acceleration due to gravity", but the Christoffel symbols in the global chart will be the same.

[tex]A^{\mu}=\frac{dU^\mu}{d\tau}+

{\Gamma^{\mu}}_{\lambda \nu} U^{\lambda} U^{\nu}[/tex]

Where U is the four-velocity (unit tangent vector) as a function of the proper time, τ. I haven't worked it for Rindler yet, but when you contract with [itex]U^{\lambda} U^{\nu}[/itex] you should get the correct expression.

In general, when you expand the Christoffel symbol terms, you can consider any of those to be fictitious forces (divided by mass). You could also consider them to be coordinate accelerations. There is no general way to distinguish the two other than what side of Newton's second law equation you write them on, it is simply a matter of preference and whim.

Last edited: