# Forces in Gr

Dale
Mentor
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 $t$ 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.
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:
$$A^{\mu}=\frac{dU^\mu}{d\tau}+ {\Gamma^{\mu}}_{\lambda \nu} U^{\lambda} U^{\nu}$$
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.

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Yes, it is. It's just the Minkowski metric in a different coordinate chart (the Rindler chart), at least if I'm remembering that portion of MTW correctly (I don't have it handy right now to check).
Have you checked this ? I still assert it has a non-zero Einstein tensor. Are we talking about the same metric ?

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

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

$${dt}^{2}\,\left( -2\,g\,x-1\right) +{dz}^{2}+{dy}^{2}+{dx}^{2}$$
I think that the Rindler metric is $ds^2=-g^2 x^2 dt^2+dx^2+dy^2+dz^2$

I think that the Rindler metric is $ds^2=-g^2 x^2 dt^2+dx^2+dy^2+dz^2$
Sure, that's flat but $\Gamma^x_{tt}={g}^{2}\,x$ which is not what is required. The metric in my earlier post is not a flat spacetime.
Rereading the posts maybe the important point is that it is flat at x=0 which is obvious from the metric.
More to the point is my #25.

PAllen
2019 Award
Sure, that's flat but $\Gamma^x_{tt}={g}^{2}\,x$ which is not what is required. The metric in my earlier post is not a flat spacetime.
Rereading the posts maybe the important point is that it is flat at x=0 which is obvious from the metric.
More to the point is my #25.
That's interesting (and obviously correct - I double checked the computation because it is surprising to me). That says that Rindler coordinates do not constitute Fermi-Normal coordinates of an accelerated observer in flat spacetime - contrary to some commonly seen claims that this is the case.

As to Mentz's points from #25, the answer is that the metric in the OP is approximate, even for flat space time. It leaves out higher order corrections. The exact, flat spacetime, Fermi-Normal metric of a uniformly accelerating observer is given on page 173 of MTW. The g00 term is:

-(1+g x )^2

and you get -(1+2g x) only by dropping the second order term. In general spacetimes, this is all subsumed in the error term of the metric that was left out of the OP.

I believe (but have not checked the computation) that the issues raised in #25 go away if the exact metric is used.

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Dale
Mentor
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.

Staff Emeritus
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

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

Staff Emeritus
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:
$$A^{\mu}=\frac{dU^\mu}{d\tau}+ {\Gamma^{\mu}}_{\lambda \nu} U^{\lambda} U^{\nu}$$
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
Mentor
. 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.

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
$$\omega_{ab}= \nabla_{[a}u_{b]}+\dot{u}_{[a}u_{b]}$$
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 $(1/2)\epsilon^{abmi}u_b\omega_{mi}$.

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