# Meaning of time derivative of the Lorentz factor $\gamma$?

This question about the Lorentz factor $$\gamma$$ in special relativity. I know what $$\gamma$$ means and how to drive. I'm wondering if I have time derivative of $$\gamma$$, what dose it mean conceptually?

For a test particle, it's essentially the power being delivered to a test particle by a force, because $$E=m\gamma$$ (or $$E=m\gamma c^2$$ in units where $$c\ne1$$).
• One typically sets $c=1$ to save time, but here it seems it had the opposite effect :-) If you had written $E=\gamma mc^2$ from the beginning, you could have avoided the parenthetical clarification altogether. Anyway, I guess the more precise statement that $\dot\gamma$ is the power per unit rest-energy, right? Does that make it more transparent? Also, IIRC, it is the zero component of the four-force, right? – AccidentalFourierTransform Aug 11 at 22:25