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

+ 3 like - 0 dislike
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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 $T(x)=A_\nu (x)j^\mu(x)$.)


This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Mikhail Skopenkov

asked Mar 25 in Theoretical Physics by Mikhail Skopenkov (15 points) [ revision history ]
edited Apr 5 by Dilaton
Most voted comments show all comments
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 $E^2$.

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Thomas
Thank you for your comment. Now trying to edit the question to make this clearer.

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Mikhail Skopenkov
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 $\partial_\mu (T^{\mu\nu}+T_j^{\mu\nu})=0$?

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Thomas
@Thomas: yes, exactly.

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Mikhail Skopenkov
This is sometimes discussed in connection with relativistic MHD, see, for example, equ.(2.55) in Anile, "Relativistic Fluids and Magnetofluids".

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Thomas
Most recent comments show all comments
Yes, but I think this means that you should view $T_1$ (the usual stress tensor of an ideal fluid) as the part that represents $j$ (because $j=q u$).

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Thomas
Oh cool! Why did not I think about it! But... There is a technical difficulty: $u$ must satisfy $u^\mu u_\mu=c^2$ (the identity is crucial for the conservation of e-m). Hence $j^\mu j_\mu=c^2/q^2$, but the left-hand side is even not necessarily positive. Anyway, what we are getting could hardly be called ``the e-m tensor acquired by a 4-current''. Did not expect the question to be that complicated!

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Mikhail Skopenkov

4 Answers

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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).

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Thomas
answered Mar 29 by tmchaefer (40 points) [ no revision ]
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 ?

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Mikhail Skopenkov
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One particular example of a stress-energy tensor which formally fulfils what you ask is the stress-energy tensor of charged dust. Dust has a stress-energy tensor written as
$$T^{\mu\nu}_\mathrm{d} = \rho u^\mu u^\nu $$

where $\rho$ is the mass density and $u^\mu$ its four-velocity. The current is then related to the quantities showing up in here as

$$j^\mu = \frac{q}{m}\rho u^\mu\,, \; u^\nu = \frac{j^\nu}{\sqrt{-g_{\alpha \beta} j^\alpha j^\beta}}\,,$$

where $q$ and $m$ are the charge and mass of the particles.

This means that the dust stress-energy tensor can be written in terms of the current (and the metric) as

$$T^{\mu\nu}_\mathrm{d} = \frac{m}{q} \frac{j^\mu j^\nu}{\sqrt{- g^{\alpha \beta} j_\alpha j_\beta}}. $$

When you compute the equations of motion $T^{\mu\nu}_{\;\;;\nu} =...$, you use the current conservation $(j^\mu)_{;\mu} = 0$ (equivalently $(\rho u^\mu)_{;\mu} = 0$) and the fact that the particles of the dust are subject to the Lorentz force to obtain

$$T^{\mu\nu}_\mathrm{d\;;\nu} = \rho a^\mu = F^{\mu \kappa}j_\kappa$$

But it is obvious that what you are looking for is something slightly different: a way to obtain the full set of Maxwell's equations with source terms from a single stress-energy tensor without having to worry about the pesky and dirty physics of the specific particles in the current. Here, however, we are considering yet another physical model of particles sourcing the fields with a free parameter in the form of specific charge $q/m$. I.e., different values of $q/m$ lead to theories of particles with different response to the Lorentz force.

I believe that what you are looking for does not exist for very physical reasons. Either your current is dynamical and will thus inevitably have its own "dirty" specific dynamic, or it will be a fixed external field which breaks the translational symmetry and the applicability of the stress-energy formalism.

answered Apr 6 by Void (1,505 points) [ revision history ]
edited Apr 6 by Void

@Void Thanks much for your answer. Now joining your belief that the required tensor does not exist. Let me disagree that the proposed tensor $T^{\mu\nu}=\frac{m}{q}\frac{j^\mu j^\nu}{\sqrt{-j_\alpha j^\alpha}}$ formally answers the question. It does not seem to satisfy point 3 for general $j$. First, the expression under the square root is not even necessarily positive (as mentioned in the comments to the question). Second, the equation $\rho a^\mu=F^{\mu k}j_k$ required for point 3 is yet another very strong restriction on $j$ (after $a^\mu$ is expressed though $j$). Thus Q3 remains open (but probably not interesting physically).  

@michail.skopenkov $-j^{\alpha} j_\alpha$ will always be positive as long as 1) it is positive initially, and 2) the dynamics follow the equation of motion $\rho a^\mu = F^{\mu\kappa}j_\kappa$.

I agree that the dynamics I present here are only a restriction to a subspace of possible dynamics of the charged matter. However, the notion that a stress-energy tensor characterizing any possible dynamics of charged matter even may exist is simply unfounded.

Consider it from the point of the Lagrangian formalism: what even is $j^\mu$? Without a matter field, it has no meaning because $j^\mu=0$. Is it a field which is varied in the action? Etc. etc. If you give a precise mathematical statement of Q3 including a precise description of what $j^\mu$ is, mathematically, in your formalism, it will probably be possible to immediately prove that the answer is either "no", or "yes, but it is non-unique".

@Void Agreeing with you that Q3 is more mathematical than physical question. For me it seems to be not easy at all. Let me give a precise statement (unfortunately, I have no rights to edit the question itself):
Let $j:R^4\to R^4$ and $A:R^4\to R^4$ be arbitrary $C^\infty$ functions related by the source equation $\square A^\mu=j^\mu$ (here we use the Lorentz gauge for simplicity). Does there exist a tensor-valued functional $T(x)=T(j(x),A(x))$ satisfying points 1-3 in Q3 above?
An argument for the existence was that for a charged particle rather than a medium an analogous energy-momentum vector does exist, independently of the dynamics (see the beginning of the question).

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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.

Before answering Q3 let's focus on this:

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 $Aμ(x)j^μ(x)$, hence is NOT translational invariant. Notice that the field $j(x)$ is given, it is not dynamical.

Be careful: Translational invariance of the lagrangian/action does not require the lagrangian density $\mathcal L$ to be independent of x. In fact, the term $A^{\mu}(x)j_{\mu}(x)$ does not spoil the translational invariance of the lagrangian. Both $A$ and $j$ are 4-vectors and their contraction is a scalar under Poincarè transformations: $A'{\mu}(x') j^{' \mu }(x') = A_{\mu}(x)j^{\mu}(x)$.

We can now apply Noether's theorem and answer Q3. We're dealing with a non-dynamical source so the lagrangian density is $\mathcal L = -\frac{1}{4} F^{\mu\nu}F_{\mu \nu} -A_{\mu}j^{\mu} $

and the canonical energy-momentum tensor is $\tilde T ^{\mu}_{ \nu}= F^{\mu \alpha} \partial_{\nu}A_{\alpha} + \mathcal L \delta^{\mu}_{\nu}= F^{\mu \alpha} \partial_{\nu}A_{\alpha} -\frac{1}{4} F\cdot F \delta^{\mu}_{\nu} - A\cdot j \delta^{\mu}_{\nu} = T^{\mu}_{\nu \, (free)} - A\cdot j \delta^{\mu}_{\nu} $

The canonical tensor is not symmetric, but the term $ - A\cdot j \delta^{\mu}_{\nu}$ relevant to your question is, and it is maintained after symmetrisation. It satifies the requirements Q3 1/2/3.

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user tbt
answered Mar 27 by tbt (0 points) [ no revision ]
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 $j(x)=(x^0,-x^1,0,0)$.

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Mikhail Skopenkov
+ 0 like - 1 dislike

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.

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Ján Lalinský
answered Mar 26 by Ján Lalinský (10 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2017-04-05 15:37 (UTC), posted by SE-user Mikhail Skopenkov

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