# Acceleration in Rindler coordinates

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?

• Use the covariant derivative. Aug 17, 2020 at 8:43
• Must have been blind - thx
– TomS
Aug 17, 2020 at 9:01

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