Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $\vec{B}$.

share|improve this question
1  
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 $j^\alpha/|j|$. –  Luboš Motl Jan 10 '13 at 12:46
    
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)? –  Murod Abdukhakimov Jan 10 '13 at 13:37
add comment

2 Answers 2

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 $X$.

share|improve this answer
add comment

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.

share|improve this answer
    
Thanks, Vladimir! I don't know why your answer was downvoted.+1 –  Murod Abdukhakimov Jan 10 '13 at 13:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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