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Usually the Rindler coordinate transform is written as $$T = \frac{1}{a} \sinh(at) \tag{1}$$ $$X = \frac{1}{a} \cosh(at) \tag{2}$$ where $a$ is the uniform acceleration. The transforms lead to the hyperbolic coordinates as known. In particularly the transform (1) for time coordinate is based on the special relativistic Lorenz transform as following $$ dt = dT \gamma=\frac{dT}{(1-\frac{v^2}{c^2})^{1/2}} $$ substitution of $v=a T$ into this (allowing $c=1$) $$ dt =\frac{dT}{(1-(a T)^2)^{1/2}} \tag{3}$$ and integration yields $$ t =\int \frac{dT}{(1-(a T)^2)^{1/2}}=\frac{1}{a} a\sin(aT) $$ Inverting the equation we get $$T = \frac{1}{a} \sin(a t)$$ So instead of hyperbolic sine (1) we end up with the harmonic sine. The same harmonic coordinate transforms (as per Wikipedia) were used by Sommerfeld (1910), von Laue (1911) and Pauli (1921), but later Lemaître, Einstein & Rosen used the hyperbolic functions. In order to get (1) the literature uses (3) but with $inverse$ sign $$ dt =\frac{dT}{(1+(a T)^2)^{1/2}} \tag{4}$$ So my question is: how is it possible to come up with hyperbolic functions in Rindler coordinates instead of harmonic functions?

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  • $\begingroup$ Why $v=aT$? $v= dX/dT = \tanh at$ instead, producing (4) after some manipulation... $\endgroup$ – Valter Moretti Jul 31 '19 at 13:56
  • $\begingroup$ "substitution of v=aT..." - Note that, in the context of SR, uniform coordinate acceleration is impossible since that would lead to $v \gt c$. The Rindler coordinates are applicable to observers with uniform proper acceleration. $\endgroup$ – Alfred Centauri Jul 31 '19 at 14:14
  • $\begingroup$ @Valter. Thank you for your tip! Seems to be elegant way to obtain sinh() $\endgroup$ – Eddward Jul 31 '19 at 14:31
  • $\begingroup$ @Alfred, thank you! $\endgroup$ – Eddward Jul 31 '19 at 14:33
  • $\begingroup$ @Eddward I missed the physical point. You assumed that $v=aT$ since you interpreted "uniform acceleration" in classical sense as Alfred Centauri noticed. It is impossible in relativistic context. Uniform acceleration must be interpreted as uniform proper-time acceleration. In fact $a$ is here the constant Lorentzian norm of the four-acceleration. All (hyperbolic) formulas you wrote are consistent with this idea. $\endgroup$ – Valter Moretti Jul 31 '19 at 16:35
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The assumption that $v=aT$ is wrong. One has to use the proper acceleration $a$ that remains constant. If $a_c=\frac{dv}{dt}$ is the coordinate acceleration, t is coordinate time, then the proper acceleration is given by $$a= \frac{a_c}{(1-\frac{v^2}{c^2})^{3/2}}$$ So $$\frac{dv}{dt}= a(1-\frac{v^2}{c^2})^{3/2}$$ Resolving this in respect to $v(t)$ leads to $$v=\frac{at}{\sqrt{1+(\frac{at}{c})^2}}$$ Now $v(t)$ can be substituted into the Lorez transform for time coordinate leading to the hyperbolic sine.

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