# Magnitude Proper Accelaration Changes Under Coordinate Transformation

Rindler transformation are studied in the book of Carroll in Section 9.5. He starts from the Minkowski metric and a trajectory with uniform acceleration given by $$$$\begin{split} t(\tau) &= \frac{1}{\alpha}\sinh\left(\alpha\tau\right)\,, \\ x(\tau) &= \frac{1}{\alpha}\cosh\left(\alpha\tau\right)\,. \end{split}$$$$ Notice that the magnitude of the accelaration ($$a^{\mu} = d^{2}x^{\mu}/d\tau^{2}$$) is given by $$\sqrt{a_{\mu}a^{\mu}} = \alpha\,,$$ which is uniform. Next, he considers a coordinate transformation $$$$\begin{split} t &= \frac{1}{a}e^{a\xi}\sinh\left(a\eta\right)\,, \\ x &= \frac{1}{a}e^{a\xi}\cosh\left(a\eta\right)\,, \end{split}$$$$ such that the trajectory gets transformed into $$$$\begin{split} \eta(\tau) &= \frac{\alpha}{a}\tau\,, \\ \xi(\tau) &= \frac{1}{a}\log\frac{a}{\alpha}\,. \end{split}$$$$ By calculating the magnitude of the acceleration in the transformed frame, you obtain zero. However, as far as I know, this quantity is a scalar, meaning that it stays invariant under a coordinate transformation. How is this possible?

Notice that the magnitude of the accelaration ($$a^{\mu} = d^{2}x^{\mu}/d\tau^{2}$$) is given by $$\sqrt{a_{\mu}a^{\mu}} = \alpha\,,$$ which is uniform
In your expression $$a^{\mu} = \frac{d^{2}x^{\mu}}{d\tau^{2}}$$ you have used the ordinary derivative $$d$$. You need to use the covariant derivative along the curve, $$D$$. So $$a^{\mu} = \frac{D^{2}x^{\mu}}{D\tau^{2}}$$ You "got away" with it in the inertial frame because in the inertial frame the covariant derivative $$D$$ is the same as the ordinary derivative $$d$$ because all of the Christoffel symbols are zero. However, in the Rindler frame the Christoffel symbols are non-zero so the two derivatives differ. Meaning that in the Rindler frame $$a^{\mu} = \frac{D^{2}x^{\mu}}{D\tau^{2}}=\alpha$$ while $$a^{\mu} \ne \frac{d^{2}x^{\mu}}{d\tau^{2}}=0$$