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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?

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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 $c\ne1$).

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  • $\begingroup$ 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? $\endgroup$ – AccidentalFourierTransform Aug 11 at 22:25

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