What would a relativistic set of 'suvat' equations look like? [duplicate]

Question

In newtonian dynamics 'suvat' equations are relations between the initial and final speeds $u$ and $v$ of an object which covers a displacement $s$ during time $t$ under constant acceleration $a$, and they are essentially \begin{align} v &= u + at\\ s &= ut + \frac{1}{2}at^2, \end{align} and so on. Is there a corresponding set of equations for relativistic dynamics? I imagine that, for instance, the two equations above would have to factor in length contraction or time dilation, so is there a valid equivalent of 'suvat' equations?

Answer 1

Then SUVAT equations all follow from the differential equation for the acceleration:

$$ \frac{d^2x}{dt^2} = a $$

where for the SUVAT equations the acceleration $a$ is assumed constant. Integrating once gives us:

$$ \frac{dx}{dt} = v = at + u $$

where $u$ is the constant of integration and this is of course just the initial velocity. Integrating again gives:

$$ x = \tfrac{1}{2}at^2 + ut + x_0 $$

where this time the constant of integration $x_0$ is the initial value of $x$, though we usually leave that out. These are the two SUVAT equations you cite.

The problem with trying to do the same in special relativity is that different observers will disagree about the acceleration. Suppose I am floating stationary in space watching you blasting away in your rocket. By acceleration do we mean the acceleration I measure or the acceleration you measure? This matters because the acceleration I measure, $a'$, is related to the acceleration you measure, $a$, by:

$$ a' = \frac{a}{\gamma^3} $$

where $\gamma$ is the Lorentz factor:

$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$

So if we require the acceleration to be constant, as in the SUVAT equations, does that mean $a$ is constant or $a'$ is constant? In fact $a'$ can't be constant because that would mean I observe you to reach and exceed the speed of light in a finite time, which we know is impossible.

Generally speaking we'd say the acceleration you measure is constant. This makes intuitive sense because if your rocket motor produces a thrust of $1g$ then that means you on the rocket feel a $1g$ acceleration. And as long as you run your rocket motor at a constant rate you'll feel that constant $1g$ acceleration.

If we make the assumption that your acceleration is constant our equivalent of the equation of motion is:

$$ \frac{d^2x'}{dt'^2} = \frac{a}{\left(1 - v^2/c^2\right)^{3/2}} $$

where $x'$ is the distance I measure and $t'$ is the time I measure. For a gory looking equation this turns out to give fairly simple equations for your distance and velocity as measured by me. These are given in Phil Gibbs' article on the relativitic rocket, along with a lot more information. I refer you to taht article, but for completeness I'll give the velocity and distance equations:

$$ v = \frac{at}{\sqrt{1 + (at/c)^2}} $$

$$ s = \frac{c^2}{a} \left(\sqrt{1 + (at/c)^2} − 1\right) $$

Note these assume the initial velocity, $u$, is zero so we don't get the $u$ and $ut$ terms in the equations.

Answer 2

The relativistic generalization of "uniform acceleration" is a timelike curve of constant curvature in Minkowski spacetime.

One should use really use rapidities [the analogue of the angle].

$v=u+at$ is generalized to $\theta=\theta_0+\frac{\rho}{c} t$ (where $\rho$ is the constant proper-acceleration), which can be expressed in terms of velocity by using $v=c\tanh\theta$. So, we have $$v=c\tanh\theta=c\tanh(\theta_0+\frac{\rho}{c} t)=c\frac{(v_0/c)+\tanh(\frac{\rho}{c} t)}{1+(v_0/c)\tanh(\frac{\rho}{c} t)}.$$

One would have to integrate that out to get the position function. From my notes [and my rushed attempt to restore the c's], I get $$y-y_0=\frac{c^2}{\rho}\left( \sqrt{1+\left(\frac{\rho}{c} t+\gamma_0 \frac{v_0}{c}\right)^2}-\gamma_0 \right),$$ where $\gamma_0=\frac{1}{\sqrt{1-(v_0/c)^2}}=\cosh\theta_0$.