# What is the energy-momentum tensor of a current density?

A charge $q$ in an electric potential $\phi$ acquires potential energy $q\phi$.

Likewise, a charge density $\rho(x)$ aquires potential energy density $\rho(x)\phi(x)$.

A moving charge $q$ in an electromagnetic 4-potential $A$ acquires'' an additional momentum $qA$ so that the canonical momentum equals $qA$ plus the kinetic momentum.

Q2. Is there a physically meaningful notion of the energy-momentum tensor $T(x)$ which a 4-current density $j(x)$ acquires'' in a 4-potential $A(x)$?

Q3. Is there a tensor field $T(x)$ sush that

1. $T(x)$ depends only on $j(x)$, the electromagnetic 4-potential $A(x)$, and maybe their derivatives;

2. $T(x)=0$ identically, once $j(x)=0$ identically;

3. $T(x)+T_{field}(x)$ is conserved, where $T_{field}(x)$ is the (Belinfante-Rosenfeld) energy-momentum tensor of the field itself.

One expects the same tensor field $T(x)$ for Q2 and Q3.

An immediate idea could be to apply the Noether theorem. But the theorem is not applicable here because the Lagrangian explicitly depends on $x$ through the term $A_\mu(x)j^\mu(x)$, hence is NOT translational invariant. Notice that the field $j(x)$ is given, it is not dynamical.

A guess for the required tensor could be $T^\mu_\nu=A_\nu j^\mu-\delta^\mu_\nu A_\lambda j^\lambda$. But this does not work because the divergence $\partial_\mu T^\mu_\nu=F_{\mu\nu}j^\mu-A_\mu\partial_\nu j^\mu$ contains an extra term in addition to the Lorentz 4-force, which contradicts to point 3 above.

Setting formally $T(x)=-T_{field}(x)$ does not work because it contradicts to point 2.

In the literature known to me (e.g., Landau-Lifshitz, sections 32-33 and 94 of volume 2) such tensor $T(x)$ is never constructed. Instead, a particular type of particles producing the current $j(x)$ is chosen and the energy-momentum tensor of the particles is added to $T_{field}(x)$. This results in a conserved tensor. But the result depends on the particular type of the particles, not just on $j(x)$, which contradicts to point 1.

I would be very grateful to you for any insight, optimally including an explicit YES/NO answer for either Q2 or Q3.

EDIT Qestion Q1 has been removed (Q1:Is the quantity $\rho(x)\phi(x)$ a part of some Lorentz covariant tensor field $T(x)$? The answer is trivially yes: just take $T(x)=A_\nu (x)j^\mu(x)$.)

• What makes you think that there is anything missing? The standard form (in terms of the fields) already satisfies the correct equation $\partial_\mu T^{\mu\nu}=F^{\nu\alpha}j_\alpha$. This is well known in electrostatics: The integral over $\rho\phi$ is the same as the integral over $E^2$. – Thomas Mar 25 '17 at 20:10
• Thank you for your comment. Now trying to edit the question to make this clearer. – Mikhail Skopenkov Mar 27 '17 at 11:29
• Are you asking whether there is a suitable stress tensor $T_j$ that depends on the current so that the equation $\partial_\mu T^{\mu\nu} =F^{\alpha\beta}j_\alpha$ can be written as $\partial_\mu (T^{\mu\nu}+T_j^{\mu\nu})=0$? – Thomas Mar 27 '17 at 16:09
• @Thomas: yes, exactly. – Mikhail Skopenkov Mar 27 '17 at 17:42
• This is sometimes discussed in connection with relativistic MHD, see, for example, equ.(2.55) in Anile, "Relativistic Fluids and Magnetofluids". – Thomas Mar 27 '17 at 18:32

Let me try to summarize what we learned: You asked whether we can write the standard equation of energy conservation in EM $$\partial_\mu T^{\mu\nu}= F^{\nu\alpha}j_\alpha ,$$ which includes the work done by the currents as a source term, in such a way that total energy conservation is manifest, $\partial_\mu (T^{\mu\nu}+T^{\mu\nu}_m)=0$, where $T_m$ corresponds to the matter fields (currents).

Now, obviously, this requires some equation of motion for the charges (the Lorentz-force equation). We could start from QED. Then the stress tensor is $$T^{\mu\nu}=\bar\psi D^{(\mu}\gamma^{\nu)}\psi +\frac{1}{4}F^{\mu\alpha}F^{\alpha}_\nu$$ which satisfies everything, but is not written in terms of the current.

So we have to be more macroscopic. The next idea would be to use kinetic theory. We write down a distribution function $f$ which satisfies the Vlasov equation. Then the current is $$j_\mu(x,t) =\int d\Gamma\, qv_\mu f(x,p,t)$$ and the stress tensor is $$T_{\mu\nu}(x,t) = \int d\Gamma\, v_\mu p_\nu f(x,p,t) + {\rm fields}$$ which is correct, but again $T_{\mu\nu}$ is not directly written in terms of the current.

I think the closest one can come is MHD. The ideal MHD matter stress tensor is $$T^{\mu\nu} = (\epsilon+P)u^\mu u^\nu + Pg^{\mu\nu}$$ and the current is $$j^\mu = qu^\mu + {\rm dissipative}$$ As explained, for example, in Anile chapter 2 this satisfies conservation of total stress-energy (with some extra terms needed in medium, or if dissipative terms are included).

• The idea to use QED lagrangian is amazing. Still it does not give an answer. Probably, it is the time to accept the most relevant reply although no answer to the question has been obtained. Do you think the question is appropriate for reposting at physicsoverflow.org ? – Mikhail Skopenkov Apr 2 '17 at 12:38

Q2. Is there a physically meaningful notion of the energy-momentum tensor T(x) which a 4-current density j(x) acquires'' in a 4-potential A(x)?

Since you have in mind a non-dynamical, external current, to me it would make more sense to speak of the e-m tensor acquired by the field (the only dynamical entity at this point) because of the presence of an external current, and not vice-versa. The physical interpretation of the e-m tensor is then the usual energy-momentum density/flux.

edit:i made a mistake and removed the guilty paragraph

• Thank you very much for your answer. I agree with you completely that it is better to call $T(x)$ the e-m tensor acquired by the field. But it seems that $T^\mu_\nu=\pm A_\lambda j^\lambda\delta^\mu_\nu$ does not satisfy property 3. The divergence $\partial_\mu T^\mu_\nu=\pm F_{\nu\lambda}j^\lambda\pm(\partial_\lambda A_\nu)j^\lambda\pm A_\lambda(\partial_\nu j^\lambda)$ contains extra terms in addition to the Lorentz 4-force, contradicting to point 3. Maybe I am missing something obvious. It is unclear how the lagrangian can be translational invariant. E.g., take $j(x)=(x^0,-x^1,0,0)$. – Mikhail Skopenkov Mar 27 '17 at 18:04
• @MikhailSkopenkov You are right: while the lagrangian is scalar, translations are not a symmetry here (so we don't get any conserved $T^{\mu\nu}$ ). I made a naive mistake deriving the current, and i deleted that part of the answer. What one could try to do is take the energy-momentum tensor of the free field, and look at what its divergence looks like in the case with external current. One can try to use eq.s of motion to put it in a form dependent by the current $j$, and that would probably be (the divergence of ) the tensor you look for. – tbt Apr 6 '17 at 20:56
• thank you for your comment. Still cannot find the tensor, who thought the question is that hard... – Mikhail Skopenkov Apr 10 '17 at 19:57

Q2. Is there a physically meaningful notion of the energy-momentum tensor $T(x)$ of a 4-current density $j(x)$?

The energy-momentum tensor of matter describes momentum and energy of matter and their fluxes in space.

Momentum and energy of matter usually depend on many things like density, temperature, chemical composition, not on charge and current density alone.

Energy and momentum of matter depend on mass distribution and velocities, but mass distribution, in general, is not deducible from charge and current density.

• Thank you very much for your answer. You write about energy-momentum energy of matter depending on many things besides current desnsity. My question is if the contribution of the current density in the total energy-momentum energy can be separated in a sense. Now trying to edit the question to make this clearer. – Mikhail Skopenkov Mar 27 '17 at 11:27