The metric components in a two-dimensional spacetime are given in terms of the coordinates $(t, x)$ by$$ds^2 = -\cosh x\,dt^2 + dx^2.$$Consider a particle that is "held in position" at $x = 1$. What is the acceleration of this particle, i.e., if the particle has unit mass, how much force must be exerted to hold it in place?
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Parametrize the particle's worldline w.r.t. $t$: $$~x^\mu (t) = (t,1)$$
Its four-velocity is: $$u^\mu =\frac{d x^\mu}{d \tau}$$
To evaluate this we use the fact that:
$$d \tau^2 = coshx~dt^2-dx^2$$
Also use $x=1$ and $dx=0$:
$$d \tau^2 = cosh(1) dt^2$$
Therefore:
$$u^\mu = \frac{d x^\mu}{d \tau} = \frac{d t}{d \tau} \frac{d x^\mu}{d t} = \frac{1}{\sqrt{cosh(1)}} (1,0)$$
Its four-acceleration is given by:
$$a^\mu = \frac{du^\mu}{d \tau}+ \Gamma^{\mu}_{\alpha \beta} u^\alpha u^\beta$$
The first term we can immediately see is zero. Since the x-component of velocity is zero we can also discard a number of the Christoffel terms so that we are left with:
$$a^\mu = \Gamma^{\mu}_{00} u^0 u^0$$
If you work out the Christoffel symbols (I'll leave that to you) to get an expression for $a^\mu$, you can find the proper acceleration experienced by the particle by taking the magnitude of the four-acceleration:
$$|a| = \sqrt{g_{\mu \nu} a^\mu a^\nu}$$
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