# Derivation of hyperbolic motion in Special Relativity

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

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.

• 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 $x_0=c^2/\alpha$. Commented Aug 28, 2013 at 5:28
• @Pulsar Woops. Thanks for the careful read! Edit made. Commented Aug 28, 2013 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? Commented Jan 13, 2016 at 2:39
• @MaddeAnerson This may be helpful: physics.stackexchange.com/questions/66839/… Commented Jan 13, 2016 at 3:19
• @joshphysics Thank you! So the reason for $a = \gamma^3 \alpha$ is that $\frac{d^2 x}{d \tau^2} = \gamma^2 \frac {d^2x}{dt^2} = \gamma^2\frac {d^2}{dt^2}\gamma (x' + vt)= \gamma^3\frac{d^2x'}{dt^2}$? Commented Jan 13, 2016 at 12:34

Would it not be simple enough to decalre that y^3 d^2(x')/dt^2 = y^3 dx/dt dx(vt) to which can be simplified to equal y( y (dx/dt) + y(dx(vt)) = y^2(dx/dt) + y^2 (dx(vt)) where singling out y as a single for a= y(v)^3a there is a y^2 plus you can also solve for y here witht he singular for a+/-y where one negative y in brackets encumbers the positive y where in the equation of B intercepts y there exists a negative y to which is squared to make a positive y for a such as a + y = (-y^2) + a is the same as a + b = b+ a but is in y-gamma for relations such that any representation of a positive y is a negatitive y^2 and should be limiteed in the fundamaentals of calculas to be a rule for y-gamma so y ^3 is superceding that the negative cannot be squared becasue it cannot be put in brackets due to format of the equation and as such has to be reduced to two simpler forms for y^2 in its own assimilation not equating for B intercepts y and that is y^3 dx/dt (d(vt)) where the dx/dt must be flipped and is part of the integreation of the equation itself this format will give a positive y to whould be negative but is not expressed duie to the nature of the equation has to be reduced forst to y ( y(dx/dt)) (y(d(vt)) where it is now expressable as two postive y take away one negative y being the outside singluar y making the whoel equation negative is now eliminated by the positivea and now the equation can be reduced even farther to y(dx/dt) y (d(vt)) =

ydx/ydy transpose y(dv(dt))

where the cross multiplication is not a solution tot he equation but a factor to the construction of a new equation and for B int y would equal ydx/ydt + ydy/ydv where velocity squares a negative y into one positive for y(dx)/y(dy)