- #26

DrGreg

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A few comments on thisFor a particle undergoing constant proper acceleration ##\alpha## in a straight line, we have, in inertial coordinates in units where ##c = 1##,$$\begin{align*}

t &= \frac{1}{\alpha} \sinh \alpha \tau \\

x &= \frac{1}{\alpha} \cosh \alpha \tau \\

\gamma &= dt/d\tau = \cosh \alpha \tau \\

\text{celerity} &= dx/d\tau = \sinh \alpha \tau \\

v &= \frac{dx}{dt} = \frac{dx/d\tau}{dt/d\tau} = \tanh \alpha \tau \\

p &= \gamma m v = m \sinh \alpha \tau \\

f &= \frac{dp}{dt} = \frac{dp/d\tau}{dt/d\tau} = \frac{ m \alpha \cosh \alpha \tau}{ \cosh \alpha \tau} = m \alpha

\end{align*}$$assuming ##m## is constant. (That assumption would be wrong if the train is burning fuel carried by the train.)

- The constancy of the proper acceleration was hardly used. My proof still applies if you delete the first two equations for ##x## and ##t##, replace ##\alpha \tau## by ##\varphi## (rapidity) throughout, and observe that ##d\varphi/d\tau## is the proper acceleration ##\alpha##, an invariant. What this shows is that for a particle of constant mass accelerating in a straight line, all inertial frames agree on the magnitude of the 3-force acting on the particle (even if the force varies over time).
- It seems to be "just coincidence" that this is so; there doesn't seem to be any obvious reason for it, it's just something that drops out of the maths.
- The result may well be false if the mass isn't constant or if the motion isn't along a straight line.
- I can't claim credit for this result; I've seen it before (but it was so long ago that I can't remember if my method of proof was used).