In the paper Dynamics of a Charged Particle the author claims after equation (10):

However, this equation is mathematically inconsistent because both $\dot v^\mu$ and $\dot F^\mu$ are spacelike fourvectors, i.e. are perpendicular to the velocity $v^\mu$, while $\ddot v^\mu$ is not.

I don't believe this is correct since the time component of the four-acceleration is zero in the proper frame. Differentiating this wrt the proper time will again give a zero time component for the proper four-jerk giving another space-like four vector.


The answer to the question "Is the four-jerk time-like or space-like?" is addressed by a paper by Russo and Townshend ("Relativistic kinematics and stationary motions", 9 October 2009, Journal of Physics A: Mathematical and Theoretical, Volume 42, Number 44 - https://doi.org/10.1088/1751-8113/42/44/445402 - preprint at https://arxiv.org/abs/0902.4243 ).

In view of these observations, it seems remarkable that the relativistic generalization of jerk, snap, etc. has attracted almost no attention in more than a century since the foundation of special relativity. It might be supposed that this is because there is little new to relativistic kinematics once one has defined the D-acceleration A, in a D-dimensional Minkowski spacetime, as the proper-time derivative of the D-velocity U: $$A =\frac{dU}{d\tau}=\gamma \frac{dU}{dt},\qquad \gamma= \frac{1}{\sqrt{1 − v^2}} . \qquad(1.2) $$ In particular, it is natural to suppose that one should define the relativistic jerk as $J = \frac{dA}{d\tau}$. However, J is not necessarily spacelike. This was pointed out in our previous paper [3] and it led us to define the relativistic jerk as $$\Sigma = J − A^2U , \qquad J = dA/d\tau.\qquad (1.3)$$ Observe that $U\cdot\Sigma ≡ 0$, which implies that $\Sigma$ is spacelike if non-zero.

$\qquad$ Following the posting in the archives of the original version of this paper, it was brought to our attention that relativistic jerk arises naturally in the context of the Lorentz-Dirac equation,...

  • $\begingroup$ +1 Interesting, but it's a pity the paper is about branes, which makes the paper difficult to understand for me and probably the OP as well. $\endgroup$ – Larry Harson Apr 6 '17 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.