# Rindler metric and proper acceleration.

1. Apr 9, 2013

### Scherejg

1. The problem statement, all variables and given/known data
Show that observers whose world-lines are the curves along which only
ξ varies undergo constant proper acceleration with invariant magnitude
ημν(duμ/dτ )(duν/dτ ) = 1/χ2.

2. Relevant equations

ds2 = −a2χ2dξ2 + dχ2 + dy2 + dz2
xμ = {ξ, χ, y, z}
t = χsinh(aξ)
x =χcosh(aξ)

3. The attempt at a solution

The trajectory of this particle in the rest frame would be:
xκ={χsinh(aξ),χcosh(aξ),0,0} as a function of ξ.

dxκ/ dξ = {aχcosh(aξ), aχsinh(aξ),0,0} Velocity of xκ.

Since I want proper time, I would expand dxκ/ dξ into (dxκ/ dξ)*(dξ/dτ)
So I need to define dξ/dτ. For this I would use the metric:
letting ds^2= -dτ^2
dτ/dξ= aχ
γ= dξ/dτ = 1/aχ
The acceleration is then: γ^2{a^2χsinh(aξ),a^2χcosh(aξ),0,0} = d2xk/dτ^2
ημν(duμ/dτ )(duν/dτ ) = γ^4(-1*a^4χ^2sinh^2(aξ) + 1*a^4χ^2cosh^2(aξ))
Which cancels out to: 1/χ^2. I just wanted to make sure that although the math works out, my reasoning is okay.
Where I took ds^2= -dτ^2 my first thought was that dτ^2= (a^2)(χ^2)(dξ)^2 - (V(ξ))^2(dL)^2. I realized that dropping the velocity term would allow me to get the right answer.. so it must = 0. Since the velocity itself is not actually zero I would be inclined to guess that dividing the equation by (dξ)^2 would leave (dL)^2/(dξ)^2 and that must equal zero?

2. Apr 10, 2013

### TSny

Since only ξ changes, that means dχ = dy = dz = 0. I think that's why the "velocity term" is zero. That is,

ds2 = −a2χ22 + dχ2+ dy2 + dz2 = −a2χ22.

So, dτ2 = a2χ22