0

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!?

Qmechanic
  • 201,751
Neo
  • 1,025
  • 11
  • 27

2 Answers2

2

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.

John Rennie
  • 355,118
  • Is this actually by assumption in SR? When you treat an accelerating particle, you one basically says if one goes to the particle (instantanous) rest-frame that frame is like a normal rest frame. So if one says that an accelerated frame is nothing but a sequence of non-accelerated frames, nothing should happen. (Of course one could also apply this reasoning to a moving frame, saying it is just a sequence of stationary frames,... but here SR says no) – lalala Apr 23 '21 at 15:28
  • @lalala have a look at What is time dilation really? as this explains the physical origin of it. – John Rennie Apr 23 '21 at 15:32
  • Thanks a lot, nice writeup. Yes, the (standard) treatment of splitting the curve into infinitesimal straight lines is what I meant (and assuming for every ds you have an inertial frame, allowing you to use ds^2=dt^2-dr^2). I think this is an additional (if we think axiomatically) assumption in SR (which might be a good idea). It is a good idea if you want to calculate the twin-paradox time delay, but maybe not so good for Ehrenfest paradox . – lalala Apr 23 '21 at 15:57
1

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 $$a=\frac{du}{dt}=\frac{1}{\gamma^3(1+\frac{vu'}{c^2})^3}\frac{du}{dt}=\frac{1}{\gamma^3(1+\frac{vu'}{c^2})^3}a'$$

twistor59
  • 16,746
  • This is acceleration transformation from one frame work to another, i am asking about relativistic factor for acceleration by itself, – Neo Nov 30 '12 at 09:37
  • Well the relation $t=\tau \gamma$ relates the time in a frame for which a particle is at rest (i.e proper time) and the time in another frame which is moving relative to it. So I guess the analogous result for acceleration would be that, for an instant in which the particle was at rest in the primed frame, then the accelerations are related just by $\frac{1}{\gamma^3}$ – twistor59 Nov 30 '12 at 10:00