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?
Answer
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:
$$|a| = \sqrt{g_{\mu \nu} a^\mu a^\nu}$$
No comments:
Post a Comment