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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{equation} \begin{split} t(\tau) &= \frac{1}{\alpha}\sinh\left(\alpha\tau\right)\,, \\ x(\tau) &= \frac{1}{\alpha}\cosh\left(\alpha\tau\right)\,. \end{split} \end{equation} 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{equation} \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} \end{equation} such that the trajectory gets transformed into \begin{equation} \begin{split} \eta(\tau) &= \frac{\alpha}{a}\tau\,, \\ \xi(\tau) &= \frac{1}{a}\log\frac{a}{\alpha}\,. \end{split} \end{equation} 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?

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The issue you are running into is here

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 $$a^{\mu} \ne \frac{d^{2}x^{\mu}}{d\tau^{2}}=0$$

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  • $\begingroup$ Great. Thank you. Can you please elaborate on why then one prefers to work in the Rindler frame? $\endgroup$
    – Kabouter9
    Commented Oct 31, 2023 at 9:10
  • $\begingroup$ @Kabouter9 The only reason to prefer one frame over another is convenience. Whether that is convenience in describing the problem or in solving it $\endgroup$
    – Dale
    Commented Oct 31, 2023 at 11:56

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