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In a continuous medium the Lorentz force density is known to be written in the form:

$f_\alpha = F_{\alpha \beta} J^\beta$,

where $F_{\alpha \beta}$ is an electromagnetic field tensor, and $J^\beta$ is a charge-current density.

Whould it be correct saying that the action of this force on charge-current density reads as follows:

$\frac{dJ^\alpha}{dt} = f^\alpha = F^{\alpha \beta}J_\beta$ ?

It seems reasonable because in this case the charge-current density 4-vector undergoes Lorentz transformation, i.e. it is "accelerated" along direction of $\vec{E}$ proportionally to the magnitude of $\vec{E}$, and "rotated" around direction of $\vec{B}$ to the angle proportional to the magnitude of $\vec{B}$.

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    $\begingroup$ Nope, the force isn't the derivative of the charge-current vector with respect to time. The force is the derivative of the momentum (which may be extended to 4-momentum) with respect to time! Momentum and current of charge are totally different things. When the charge is densitized, the 4-momentum must also be densitized to the stress-energy tensor. So the right equation will contain a multiple of $T_{\alpha\beta}$ on that place, perhaps contracted with the vector $j^\alpha/|j|$. $\endgroup$
    – Luboš Motl
    Commented Jan 10, 2013 at 12:46
  • $\begingroup$ Thanks, Luboš. My question is related to the problem from particle physics theory. As far as I know, there are no particles for which the change of momentum is not proportional to the change of charge-current. Does it mean that the formula in question can be used for particles (maybe with some dimensional multiplier)? $\endgroup$
    – Murod Abdukhakimov
    Commented Jan 10, 2013 at 13:37

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As as been pointed out, four-force (density) has to be the derivative of four-momentum (density) with respect to proper time.

Now, in a charged fluid, what is the relationship between four-momentum-density and four-current-density? Well, if the fluid has a constant charge-to-mass ratio, then they ought to be proportional. The four-momentum-density ought to be $\rho_m u$ for some velocity field $u$ and density field $\rho_m$ (m for mass), while the four-current-density is $j = \rho_c u$. If the fluid has a constant charge-to-mass ratio, then $\rho_c = \rho_m X$ for some constant factor $X$.

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For a cold (no pressure) charged gas the electromagnetic filed must contain a self-field contribution too in order to describe the density variations due to repulsion of charges. Plasma equations contain it all.

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  • $\begingroup$ Thanks, Vladimir! I don't know why your answer was downvoted.+1 $\endgroup$
    – Murod Abdukhakimov
    Commented Jan 10, 2013 at 13:17

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