Reconciling Lorentz Four-Force Density w/ Spacelike Four-Current (##c = 1##)

[SiennaTheGr8]

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.

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.

 

[Dale]

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?

No, it is not a problem. Just go the other direction. Maxwell’s equations are specified in terms of the four-current density. So start with the densities as the primary definition. Then from that it is straightforward to derive the four-force on a moving charge.

Is that equality general, or does it hold only for timelike jν? If the equality is general, how would one show that it holds even when the "continuous charge distribution" model doesn't hold?

I don’t think it matters. You can always take a spacelike four current density and write it as the sun of two timelike ones. Maxwell’s equations are linear, so you can go back and forth as needed.

Why is the special-case j→=ρ0γu→ substitution made to derive Poynting's theorem?

I have not seen that done before. It is not necessary. I would look for a better derivation that doesn’t do that.

 

[SiennaTheGr8]

I don’t think it matters. You can always take a spacelike four current density and write it as the sun of two timelike ones. Maxwell’s equations are linear, so you can go back and forth as needed.

Thanks for the response.

So, I'm aware that in a "composite" scenario, you could treat the various "types" of charged particles separately, and assign a timelike $j^{\mu}_{\textrm{part}}$ to each of them, where the "total" (potentially spacelike) $j^{\mu}$ is equal to their sum. (I think Rindler covers this much fairly well in Relativity: Speical, General, and Cosmological, 2nd ed.)

I think what I'm missing is the "step" between that "sum-of-timelike-densities" idea and the $f^{\mu} = F^{\mu}_{\; \; \nu} j^{\nu}$ business. Could you do a $f^{\mu}_{\textrm{part}} = ( F^{\mu}_{\; \; \nu} j^{\nu} )_{\textrm{part}}$ thing for each particle-type, where again you'd have a "total" $f^{\mu}$ equal to their sum? Is that what you're getting at, or am I still missing the point?

 

[Dale]

I think what I'm missing is the "step" between that "sum-of-timelike-densities" idea and the $f^{\mu} = F^{\mu}_{\; \; \nu} j^{\nu}$ business. Could you do a $f^{\mu}_{\textrm{part}} = ( F^{\mu}_{\; \; \nu} j^{\nu} )_{\textrm{part}}$ thing for each particle-type, where again you'd have a "total" $f^{\mu}$ equal to their sum? Is that what you're getting at, or am I still missing the point?

Yes, that is what I meant. You don’t even have to be right on each particle type’s actual $j^\nu$, as long as they add up to the right amount.

 

[vanhees71]

What you are considering are convection currents only. On a microscopic level this is indeed ok, but in the macroscopic description of electrodynamics in the usual linear-response approximation you also deal with conduction currents. In the (local) rest-frame of the medium you have $\vec{j}_{\text{cond}}=\sigma \vec{E}$, $j_{\text{cond}}^0=0$. If the medium moves with a four-velocity $u^{\mu}(x)$ field, you get the manifestly covariant version of Ohm's Law,
$$j_{\text{cond}}^{\mu}=\sigma F^{\mu \nu} u_{\nu},$$
which is always space-like and of course the invariant charge density, $\rho_{0,\text{cond}} = u_{\mu} j_{\text{cond}}^{\mu}=0$ (as it must be, because it's a scalar, i.e., if it's 0 in any frame (here the local rest frame) it's 0 in any frame). If $\rho_0 \neq 0$, then you have both a conduction and a convection current,
$$j_{\text{conv}}^{\mu} = \rho_0 u^{\mu},$$
and the total current is
$$j^{\mu}=j_{\text{cond}}^{\mu} +j_{\text{conv}}^{\mu}.$$
Of course this is again consistent with the scalar charge-density of the fluid element at rest since $u_{\mu} u^{\mu}=1$ and $u_{\mu} j_{\text{cond}}^{\mu}=0$.

On a more microscopic level of description of, say, a piece of metal, you describe this as a bound lattice of positively charged ions with one four-velocity $u_{\text{lat}}^{\mu}$ and a (nearly perfect) gas of conduction electrons with another four-velocity $u_{\text{cond}}^{\mu}$. For an example using this simple classical model to understand DC currents in wires relativistically, see

https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf

 

[Dale]

convection currents

That is a good word for it!

the manifestly covariant version of Ohm's Law,
jcondμ=σFμνuν,
which is always space-like

Amazing. I have not seen this before, which in itself surprises me. Or maybe I saw and forgot because I didn’t use it.

 

[SiennaTheGr8]

What you are considering are convection currents only. On a microscopic level this is indeed ok, but in the macroscopic description of electrodynamics in the usual linear-response approximation you also deal with conduction currents. In the (local) rest-frame of the medium you have $\vec{j}_{\text{cond}}=\sigma \vec{E}$, $j_{\text{cond}}^0=0$.

$\sigma$ being the surface charge density (in the rest frame)?

 

[SiennaTheGr8]

Yes, that is what I meant. You don’t even have to be right on each particle type’s actual $j^\nu$, as long as they add up to the right amount.

And then the "total" four-force density would be what's equal to the four-divergence of the EM stress–energy tensor?

 

[vanhees71]

Unfortunately, in modern introductory textbooks the manifestly covariant in-medium electrodynamics is not treated. I've found it some years ago in the brillant first textbooks on SR and GR, written by von Laue in 1912 and the early 1920ies, respectively. There's also a short chapter in a newer German Theoretical Physics textbook (Bartelmann et al, Theoretische Physik, 4 Vols.).

The theory goes back to Minowski. It has been posthumously published with Born as the editor. Of course, there are also some loose ends. Particularly the pretty deep question about how to interpret the energy-momentum-stress tensor and its split in "matter" and "field pieces" is even discussed today. The problem is that (a) you can shift pieces from one part to the other and (b) there's the problem of "pseudo-gauge invariance" inherent to the definition of the local densities of the conserved quantities of "Noether symmetries" and (c) the question of whether you consider "physical" or "canonical" momenta.

 

[vanhees71]

$\sigma$ being the surface charge density (in the rest frame)?

No, $\sigma$ is electric conductivity, i.e., a "transport coefficient" in linear-response theory. As such it's defined in the local rest frame of the medium and thus a scalar, as all transport coefficients.

It's indeed a bit unfortunate that I didn't mention this, because indeed $\sigma$ is often used for surface-charge densities in the standard E&M notation.

 

[vanhees71]

And then the "total" four-force density would be what's equal to the four-divergence of the EM stress–energy tensor?

It depends a bit on the level you describe your medium. In plasma physics you usually have a two-fluid (electrons + ions) or a three-fluid (electrons + ions + neutrals) model, which can often simplified to a single-fluid equation, which doesn't follow the ions and electrons (+neutrals) separately but only the flow of electric charge (usual magneto hydrodynamics).

The hydrodynamical equations of motion are equivalent to the (local) conservation laws for energy, momentum, and electric charge. In addition you need an equation of state to close the system. The equation of motion thus reads
$$\partial_{\mu} T_{\text{mech}}^{\mu \nu}=-\partial_{\mu} T_{\text{em}}^{\mu \nu} = F^{\mu \nu} j_{\nu}.$$