Acceleration and relativity

Question

Usually in physics books this equation used $a=\frac {dv}{dt}$ to calculate the relativistic acceleration. It is true that speed $v=\beta c$ don't have relativistic (Lorentz) factor, But time have a relativistic factor $t=\tau \gamma$, what about acceleration!?

Answer 1

Assuming I've understood your comment to twistor59's answer correctly: in SR (this changes in GR) acceleration by itself doesn't cause any relativistic effects. If you compare two frames that are instaneously at rest relative to each other there will be no relativistic effects no matter how fast one of the frames is accelerating.

Accelerating objects do experience relativistic effects, but only because over any finite timespan their velocity is different to your reference frame.

Chapter 6 of Gravitation by Misner, Thorne and Wheeler explains how to calculate the effects of acceleration, or John Baez's article on the Relativistic Rocket gives the headlines.

Answer 2

Just restricting to motion in one dimension:

If the $x'$ frame is moving wrt to the $x$ frame with a speed $v$ in the positive x direction then a speed $u'$ measured in the $x'$ frame transforms as $$u=\frac{u'+v}{1+\frac{vu'}{c^2}}$$, so $$du = \frac{(1-\frac{v^2}{c^2})du'}{(1+\frac{vu'}{c^2})^2}$$ $$=\frac{du'}{\gamma^2(1+\frac{vu'}{c^2})^2}$$ Similarly $$t=\gamma(1+\frac{vx'}{c^2})t'$$ so $$dt=\gamma(1+\frac{vu'}{c^2})dt'$$ So the acceleration transformation is $$a=\frac{du}{dt}=\frac{1}{\gamma^3(1+\frac{vu'}{c^2})^3}\frac{du}{dt}=\frac{1}{\gamma^3(1+\frac{vu'}{c^2})^3}a'$$