Acceleration in Rindler coordinates

Question

Can you please point out the basic flaw in the following reasoning?

I use Minkowski $x^\mu$ and Rindler coordinates $\xi^\mu$

$$ x^\mu = (t,x) $$

$$ \xi^\mu = (\eta, \rho) $$

$$ x^\mu(\xi) = \rho \, (\sinh\eta, \cosh\eta) $$

$$ (x^1)^2 - (x^0)^2) = \rho^2; \qquad \frac{x^0}{x^1} = \tanh\eta $$

$$ ds^2 = -dt^2 + dx^2 = -\rho^2 \, d\eta^2 + d\rho^2 $$

and the world-line, 2-velocity and 2–acceleration

$$ x^\mu(\tau) = a^{-1} \; (\sinh a\tau, \cosh a\tau) $$

$$ \dot{x}^\mu(\tau) = (\cosh a\tau, \sinh a\tau) $$

$$ \ddot{x}^\mu(\tau) = a \,(\sinh a\tau, \cosh a\tau) $$

with

$$ \ddot{x}_\mu \ddot{x}^\mu = a^2 $$

Fine.

Transforming this world-line to Rindler coordinates results in

$$ \xi^\mu(\tau) = (a\tau, a^{-1}) $$

$$ \dot{\xi}^\mu(\tau) = (a, 0) $$

so — as expected — this world-line „sits“ at $ \xi^1(\tau) = \text{const.} $

However

$$ \ddot{\xi}^\mu(\tau) = 0 \quad \implies \quad \ddot{\xi}_\mu \ddot{\xi}^\mu = 0 $$

Where did the acceleration disappear to?

Answer 1

Ok, I must have been blind.

In flat space but curved coordinates one has to calculate the acceleration using

$$ a^\mu = \ddot{\xi}^\mu + \Gamma^\mu_{\kappa\lambda} \dot{\xi}^\kappa \dot{\xi}^\lambda $$

which works out nicely.

Thanks for the hint!