GR: Gravitational Forces Represented by Christoffel Symbols

[Dale]

Sure, that's flat but \Gamma^x_{tt}={g}^{2}\,x which is not what is required.

But for a stationary worldline in Rindler coordinates
{\Gamma^{x}}_{\lambda \nu} U^{\lambda} U^{\nu} = \frac{1}{x}
which is what is required.

 

[pervect]

Have you checked this ? I still assert it has a non-zero Einstein tensor. Are we talking about the same metric ?

ds^2={dt}^{2}\,\left( -2\,g\,x-1\right) +{dz}^{2}+{dy}^{2}+{dx}^{2}

I did check, and it did have a nonzero stress-energy tensor,

There are alternate formulations of the Rindler metric, but the one I usually use is

ds^2={dt}^{2}\,\left(-g^2\,x^2 -2\,g\,x-1\right) +{dz}^{2}+{dy}^{2}+{dx}^{2}

 

[pervect]

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:
A^{\mu}=\frac{dU^\mu}{d\tau}+<br /> {\Gamma^{\mu}}_{\lambda \nu} U^{\lambda} U^{\nu}
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 U^{\lambda} U^{\nu} 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.

I think this hit the nail on the head. Thanks - it serves as a much better definition of four acceleration.

On a somewhat related note, if we consider \nabla_a u^b, we know that multiplying by u^a gives us the 4-acceleration, which is what motivates the above result. What would we need to do to get the information about the rate of rotation an observer moving along the worldline would experience? I'm guessing that it's also buried in this expression.

 

[Dale]

. What would we need to do to get the information about the rate of rotation an observer moving along the worldline would experience? I'm guessing that it's also buried in this expression.

I don't know. I would guess that you would need to parallel transport the rank three angular momentum tensor along the worldline. If you expand it in terms of the Christoffel symbols then you could interpret those terms as being due to the fictitious gravitational forces in the particular coordinate system.

 

[Mentz114]

I think this hit the nail on the head. Thanks - it serves as a much better definition of four acceleration.

On a somewhat related note, if we consider \nabla_a u^b, we know that multiplying by u^a gives us the 4-acceleration, which is what motivates the above result. What would we need to do to get the information about the rate of rotation an observer moving along the worldline would experience? I'm guessing that it's also buried in this expression.

You are right, the rotation is given by the rank-2 antisymmetric tensor
<br /> \omega_{ab}= \nabla_{[a}u_{b]}+\dot{u}_{[a}u_{b]}<br />
where \dot{u}_{a} = \nabla_b u_a u^b. The axis of rotation and the angular velocity can be found from the vorticity vector (1/2)\epsilon^{abmi}u_b\omega_{mi}.