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Finally, we can invert the push-forward to switch the integration variables. Most importantly, $d^3 \breve{x} \to \gamma d^3 x$, so:

$q = \frac{1}{c^2} \int d^3 x \, \gamma u_\mu J^\mu$

As originally stated.

It would seem therefore that the important part is the time at which you evaluate your integral. If you are evaluating at fixed lab time, which is like the algebraic derivation, I gave originally, then you need the Lorentz factor. If, on the other hand, you are evaluating at fixed proper time, like the differential forms approach above, then the Lorentz factor is not needed.

Consider an integral of the left-hand-side over $\Omega$, the 3d sub-space of the space-time that is perpendicular the the four-velocity of the particle at some time in its history. At that time, therefore $\Omega$ is aligned with the rest-frame of the particle ($\tilde{S}$).

Consider the integral of the left-hand-side over $\Omega$, the 3d sub-space of the space-time that is perpendicular the the four-velocity of the particle at some time in its history. At that time, therefore $\Omega$ is aligned with the rest-frame of the particle ($\tilde{S}$).

Consider an integral of the left-hand-side of over $\Omega$, the 3d sub-space of the space-time that is perpendicular the the four-velocity of the particle at some time in its history. At that time, therefore $\Omega$ is aligned with the rest-frame of the particle ($\tilde{S}$).

Where I have used the integral form of the Gauss' law.

Where $J^\mu$ is the usual current density. The integral over $\Omega$ may still be written with no problem since differential forms are frame-independent:

So:

Consider an integral of the left-hand-side over $\Omega$, the 3d sub-space of the space-time that is perpendicular the the four-velocity of the particle at some time in its history. At that time, therefore $\Omega$ is aligned with the rest-frame of the particle ($\tilde{S}$).

Where I have used the integral form of the Gauss' law, and $\mathbf{\hat{n}}$ is the outward pointing normal vector.

Where $J^\mu$ is the usual current density. The integral over $\Omega$ may be now cosidered in terms of the current density:

So (with some more work):