I Parabolic Motion from Hyperbolic Motion

[SiennaTheGr8]

I thought this was nice. In what follows, $c=1$.

For a particle undergoing constant proper acceleration $\alpha$ in the positive $x$-direction, an inertial observer can use:

$x (\tau) = x_0 + \dfrac{\gamma(\tau) - \gamma_0}{\alpha}$,

where the Lorentz factor is:

$\gamma (\tau) = \cosh \phi (\tau)$

and the rapidity is:

$\phi (\tau) = \phi_0 + \alpha \tau$

($\tau$ is the particle's proper time, naught subscripts indicate values at $\tau = 0$).

When $\phi(\tau) \ll 1$, the Taylor series for the $\cosh$ function gives $\cosh \phi(\tau) \approx 1 + \frac{1}{2}(\phi_0 + \alpha \tau)^2$:

$x (\tau) \approx x_0 + \dfrac{ \left[ 1 + \frac{1}{2} (\phi_0 + \alpha \tau)^2 \right] - \gamma_0}{\alpha} \\[20pt] \qquad = x_0 + \dfrac{ 1 + \frac{1}{2} \phi^2_0 - \gamma_0}{\alpha} + \phi_0 \tau + \dfrac{1}{2} \alpha \tau^2 .$

If also $\phi_0 \ll 1$, then we can use $\tau \approx t$ (assume $t_0=0$), $\phi_0 \approx v_0$, $\phi^2_0 \approx 0$, $\gamma_0 \approx 1$, and $\alpha \approx a$:

$x (\tau) \approx x_0 + v_0 t + \dfrac{1}{2} a t^2$.

Usually this reduction is done from a simplified version of the $x(\tau)$ function ($\phi_0 = 0$), so that the $v_0 t$ term isn't included.

 

[Ibix]

$x (\tau) \approx x_0 + \dfrac{ \left[ 1 + \frac{1}{2} (\phi_0 + \alpha \tau)^2 \right] - \gamma_0}{\alpha} \\[20pt] \qquad = x_0 + \dfrac{ 1 + \frac{1}{2} \phi^2_0 - \gamma_0}{\alpha} + \phi_0 \tau + \dfrac{1}{2} \alpha \tau^2 .$

If also $\phi_0 \ll 1$, then we can use $\tau \approx t$ (assume $t_0=0$), $\phi_0 \approx v_0$, $\phi^2_0 \approx 0$, $\gamma_0 \approx 1$, and $\alpha \approx a$:

$x (\tau) \approx x_0 + v_0 t + \dfrac{1}{2} a t^2$.

Nice. Could you use $\gamma_0\approx 1+\phi_0^2/2$ to get directly from your penultimate expression to the final one without the extra approximations?

 

[SiennaTheGr8]

Yes, that's good. So that would give (for $\phi(\tau) \ll 1$ and $\phi_0 \ll 1$):

$x(\tau) \approx x_0 + \phi_0 \tau + \dfrac{1}{2} \alpha \tau^2$,

though you'd still need further substitutions to "convert" to coordinate time/velocity/acceleration if you want to recover the Newtonian approximation.