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My question pertains to the equation of hyperbolic motion in special relativity: $$x^{2} - c^{2}t^{2} = c^{4}/\alpha^{2}.$$ As far as I am aware, this equation is the key to calculating coordinate time for accelerating frame as they approach the speed of light. I cannot find a source for the derivation of this equation. I found some information at What is the general relativistic calculation of travel time to Proxima Centauri? but it is pretty unclear as to the mathematical processes concerning rapidity.

The differentiation of this function: $$\phi(u) = \phi(v) + \phi(u'),$$ into
$$\frac{\mathrm{d}}{\mathrm{d}t}\phi(u) = \frac{\mathrm{d}}{\mathrm{d}t'}\phi(u')\frac{\mathrm{d}t'}{\mathrm{d}t}$$
and the subsequent rewriting into
$$\frac{\mathrm{d}}{\mathrm{d}t}\phi(u) = \frac{1}{c}\gamma^{2}(u)\frac{\mathrm{d}u}{\mathrm{d}t},$$
$$\gamma^{3}(u')\frac{\mathrm{d}u'}{\mathrm{d}t'} = \gamma^{3}(u)\frac{\mathrm{d}u}{\mathrm{d}t}.$$
None of the process is particularly clear. Of course, the question above could be entirely wrong and there could be a much simpler derivation.

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up vote 4 down vote accepted

Hyperbolic motion in a single spatial dimension occurs when an object has constant proper acceleration $\alpha$. Proper acceleration is related to acceleration $a$ as measured in an inertial frame by \begin{align} \alpha = \gamma(v)^3 a \end{align} where \begin{align} \gamma(v) = \left(1-\frac{v^2}{c^2}\right)^{-1/2} \end{align} and $v$ is the velocity of the object as measured in the inertial frame. For a given proper acceleration $\alpha(t)$, the relation between proper acceleration and acceleration as measured by the inertial observer can be regard as a differential equation for the velocity $v(t)$; \begin{align} \alpha(t) = \gamma(v(t))^3\dot v(t) \end{align} If $\alpha$ is constant, then this differential equation can be solved by direct integration. Let's assume the initial condition is $v(0) = 0$, then we get \begin{align} \int_0^{v(t)} dv\,\gamma(v)^3 = \int_0^tdt \,\alpha \end{align} which gives \begin{align} v(t)\gamma(v(t)) = \alpha t \end{align} and solving for $v(t)$ gives \begin{align} v(t) = \frac{c t \,\alpha}{\sqrt{c^2 + t^2\alpha^2}} \end{align} Now, noting that the velocity is the time derivative of position; $v(t) = \dot x(t)$, we regard this as a differential equation for $x(t)$; \begin{align} \dot x(t) = \frac{c t \,\alpha}{\sqrt{c^2 + t^2\alpha^2}} \end{align} Again, we solve this by integration, this time we just integrate both sides with respect to time and use the initial condition $x(0) = x_0$; \begin{align} \int_0^tdt'\,\dot x(t') = \int_0^tdt'\frac{c t' \,\alpha}{\sqrt{c^2 + t'^2\alpha^2}} \end{align} and we get \begin{align} x(t) = x_0+\frac{c}{\alpha}\left(\sqrt{c^2+t^2\alpha^2}-c\right) \end{align} If we set $x_0 = c^2/\alpha$, square both sides, and rearrange this gives \begin{align} x(t)^2 -c^2t^2 = \frac{c^4}{\alpha^2} \end{align} as desired.

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It's almost correct, but the result of the integral is $x(t) = c/\alpha\left(\sqrt{c^2 + t^2\alpha^2}-c\right)$, which does satisfy $x(0)=0$. And this can be rewritten into the form $(x+x_0)^2-c^2t^2 = c^4/\alpha^2$, with $x_0=c^2/\alpha$. – Pulsar Aug 28 '13 at 5:28
@Pulsar Woops. Thanks for the careful read! Edit made. – joshphysics Aug 28 '13 at 6:17
Excuse me, but I can't for me life understand how $\alpha = \gamma^3 a$. Can you explain where you get that relation from? Also, is this true for any worldline $\alpha = const.$; that is, they follow a hyperbola? – Madde Anerson Jan 13 at 2:39
@MaddeAnerson This may be helpful:… – joshphysics Jan 13 at 3:19
@joshphysics Thank you! So the reason for $a = \gamma^3 \alpha$ is that $\frac{d^2 x}{d \tau^2} = \gamma^2 \frac {d^2x}{dt^2} = \gamma^2\frac {d^2}{dt^2}\gamma (x' + vt)= \gamma^3\frac{d^2x'}{dt^2}$? – Madde Anerson Jan 13 at 12:34

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