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SiennaTheGr8

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- TL;DR Summary
- If I'm not mistaken, the relation between four-velocity and four-current density is necessary to derive the Lorentz four-force density from the Lorentz four-force. But that relation doesn't hold for spacelike four-current densities. Is this a problem?

(##c = 1##)

The general definition of the four-current density is ##j^{\mu} = (\rho, \vec j)##, where ##\rho## is the charge density and ##\vec j## is the three-current density. This vector may be timelike, lightlike, or spacelike, because both positive and negative charges may be involved with various trajectories.

My understanding is that when the "continuous charge distribution" model applies, there's the additional simple relationship ##j^{\mu} = \rho_0 u^{\mu}##, where ##\rho_0## is the proper charge density of an element of the distribution and ##u^{\mu} = \gamma ( 1, \vec u ) ## is its four-velocity. In this case, the four-current density is necessarily timelike (or zero).

I'm struggling to reconcile the general definition with the definition of the Lorentz four-force density ##f^{\mu}##, which is obtained by differentiating the Lorentz four-force ##F^{\mu} = q F^{\mu}_{\; \; \nu} u^{\nu}## with respect to proper volume ##V_0## (##q## being charge and ##F^{\mu}_{\; \; \nu}## being the Faraday tensor):

## f^{\mu} = \frac{d F^{\mu}}{d V_0} = \rho_0 F^{\mu}_{\; \; \nu} u^{\nu} ##

(because ##\rho_0 = \frac{d q}{d V_0}##).

So far so good, but from there I ~always see the special-case ##j^{\nu} = \rho_0 u^{\nu}## substitution made, giving ## f^{\mu} = F^{\mu}_{\; \; \nu} j^{\nu} ##. Sometimes there's an acknowledgment that this is "for a continuous charge distribution," but what's bothering me is that this seems like a pretty important result, and I see it equated to the divergence of the electromagnetic stress–energy tensor as though the equality were a perfectly general one.

I suppose an alternative way to pose my questions is in terms of the three-current density ##\vec j## and Poynting's theorem. Why is the special-case ## \vec j = \rho_0 \gamma \vec u ## substitution made to derive Poynting's theorem? Does Poynting's theorem

I feel like I'm missing something obvious here.

The general definition of the four-current density is ##j^{\mu} = (\rho, \vec j)##, where ##\rho## is the charge density and ##\vec j## is the three-current density. This vector may be timelike, lightlike, or spacelike, because both positive and negative charges may be involved with various trajectories.

My understanding is that when the "continuous charge distribution" model applies, there's the additional simple relationship ##j^{\mu} = \rho_0 u^{\mu}##, where ##\rho_0## is the proper charge density of an element of the distribution and ##u^{\mu} = \gamma ( 1, \vec u ) ## is its four-velocity. In this case, the four-current density is necessarily timelike (or zero).

I'm struggling to reconcile the general definition with the definition of the Lorentz four-force density ##f^{\mu}##, which is obtained by differentiating the Lorentz four-force ##F^{\mu} = q F^{\mu}_{\; \; \nu} u^{\nu}## with respect to proper volume ##V_0## (##q## being charge and ##F^{\mu}_{\; \; \nu}## being the Faraday tensor):

## f^{\mu} = \frac{d F^{\mu}}{d V_0} = \rho_0 F^{\mu}_{\; \; \nu} u^{\nu} ##

(because ##\rho_0 = \frac{d q}{d V_0}##).

So far so good, but from there I ~always see the special-case ##j^{\nu} = \rho_0 u^{\nu}## substitution made, giving ## f^{\mu} = F^{\mu}_{\; \; \nu} j^{\nu} ##. Sometimes there's an acknowledgment that this is "for a continuous charge distribution," but what's bothering me is that this seems like a pretty important result, and I see it equated to the divergence of the electromagnetic stress–energy tensor as though the equality were a perfectly general one.

*Is*that equality general, or does it hold only for timelike ## j ^{\nu} ##? If the equality is general, how would one show that it holds even when the "continuous charge distribution" model doesn't hold? (And if it isn't general, then why do I never see what the formulation would be in the non-continuous case?)I suppose an alternative way to pose my questions is in terms of the three-current density ##\vec j## and Poynting's theorem. Why is the special-case ## \vec j = \rho_0 \gamma \vec u ## substitution made to derive Poynting's theorem? Does Poynting's theorem

*only*hold for a continuous charge distribution, and if not, how would you show that it holds generally?I feel like I'm missing something obvious here.

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