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We define the transformation from the Minkowski coordinates $(t,x)$ to the Rindler coordinates $(T,X)$ by $$t=X\sinh T,\\ x=X\cosh T.$$ If an object is moving at fixed Rindler coordinate $X=1/\alpha$, how do we write down the proper time $d\tau$ in terms of $dT$?

Here is my attempt. Since $t=X\sinh T$, we know $dt=\frac{1}{\alpha}\cosh TdT$. Now $dt$ is related to $d\tau$ by the Lorentz factor $\gamma$: $$d\tau=\frac{dt}{\gamma},$$ so we get $$d\tau=\frac{1}{\alpha\gamma}\cosh TdT,$$ a result in which I don't have too much confidence. Is it correct? Is the answer really that simple? Thank you for your patience.

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  • $\begingroup$ For $\gamma$, I'm not sure if there is a moving frame (the object in question?). $\endgroup$
    – Boar
    Commented Nov 13, 2021 at 14:04

2 Answers 2

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Start with the proper-time equation in Minkowski coordinates: $$ d\tau^2=dt^2-dx^2. \tag{1} $$ Write $(t,x)$ in terms of $(T,X)$, using the equations shown at the beginning of the question, to get \begin{align} dt &= (\sinh T)\,dX + (X\cosh T)\,dT \\ dx &= (\cosh T)\,dX + (X\sinh T)\,dT. \tag{2} \end{align} Substitute (2) into (1) to get $$ d\tau^2 = X^2\,dT^2-dX^2. \tag{3} $$ Equation (3) is valid for every worldline. Specializing to the worldline $X=1/\alpha$ gives the final result $$ d\tau^2 = \frac{dT^2}{\alpha^2}, \tag{4} $$ so $d\tau = dT/\alpha$.

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    $\begingroup$ Normally we write the transformation with another parameter, say $t=X\sinh(\kappa T)$ and $x=X\cosh(\kappa T)$, with $\kappa$ chosen so that $X$ and $T$ both have the same units, but that's not required. $\endgroup$
    – Chiral Anomaly
    Commented Nov 13, 2021 at 15:20
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    $\begingroup$ @Steve I'm using $\tau$ to denote proper time, which is defined for timelike worldlines. Your textbook is using $s$ to denote proper length, which is defined for spacelike worldlines. The equations $d\tau^2=dt^2-dx^2$ and $ds^2=-dt^2+dx^2$ both define the same metric (the Minkowski metric), but one is the equation for proper time and the other is the equation for proper length. These two equations are used for two different classes of worldlines (timelike and spacelike), for which their right-hand sides are non-negative. Either equation by itself is enough to implicitly specify the metric. $\endgroup$
    – Chiral Anomaly
    Commented Nov 14, 2021 at 15:51
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    $\begingroup$ @Steve Regardless of the signature convention (mostly-minus or mostly-plus), the equation for proper time in 4d Minkowksi spacetime is $d\tau^2=dt^2-dx^2-dy^2-dz^2$, and the equation for proper length is $ds^2=-dt^2+dx^2+dy^2+dz^2$. Those aren't different conventions, they're different quantities (proper time and proper length). $\endgroup$
    – Chiral Anomaly
    Commented Nov 14, 2021 at 16:11
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    $\begingroup$ @Steve The different conventions arise when deciding how to write the components $g_{ab}$ of the metric tensor itself. The mostly-minus convention uses $g_{ab}=$diag$(1,-1,-1,-1)$, so the proper-time and proper-length equations are $d\tau^2=g_{ab}dx^a dx^b$ and $ds^2=-g_{ab}dx^a dx^b$. The mostly-plus convention uses $g_{ab}=$diag$(-1,1,1,1)$, so the proper-time and proper-length equations are $d\tau^2=-g_{ab}dx^a dx^b$ and $ds^2=g_{ab}dx^a dx^b$. $\endgroup$
    – Chiral Anomaly
    Commented Nov 14, 2021 at 16:11
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    $\begingroup$ @Steve Physicists are sometimes notoriously sloppy with words. I've seen people refer to $ds^2=-dt^2+dx^2+dy^2+dz^2$ as the "mostly-plus convention," but what they really mean is "I'm writing the equation for proper length to implicitly specify the metric, instead of writing the equation for proper time." Since your question was about proper time, I started with the equation for proper time. $\endgroup$
    – Chiral Anomaly
    Commented Nov 14, 2021 at 16:15
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with

$$t=X\,\sinh(T)\\ x=X\,\cosh(T)$$

hence $$\frac{dx}{dt}=v=\frac{\sinh(T)}{\cosh(T)}$$

and $$\gamma=\frac{1}{\sqrt{1-v^2}}=\frac{dt}{d\tau}\quad\Rightarrow\quad\\d\tau=\sqrt{1-v^2}\,dt=\sqrt {1-{\frac { \left( \sinh \left( T \right) \right) ^{2}}{ \left( \cosh \left( T \right) \right) ^{2}}}}X\cosh \left( T \right)\,dT =X\,dT=\frac{1}{\alpha}\,dT$$

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  • $\begingroup$ Thank you, but how did you conclude that $\sqrt{1-v^2}\,dt=X\,dt$? $\endgroup$
    – Boar
    Commented Nov 14, 2021 at 13:50
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    $\begingroup$ @Steve with $~\sqrt {1-{\frac { \left( \sinh \left( T \right) \right) ^{2}}{ \left( \cosh \left( T \right) \right) ^{2}}}}X\cosh \left( T \right) $ $\endgroup$
    – Eli
    Commented Nov 14, 2021 at 14:33
  • $\begingroup$ Thank you very much. I got one last question: why did you use $dt$ instead of $dT$? $\endgroup$
    – Boar
    Commented Nov 14, 2021 at 14:37
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    $\begingroup$ @Steve see new edit $\endgroup$
    – Eli
    Commented Nov 14, 2021 at 15:05

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