This question only considers classical relativistic electrodynamics. I'm also neglecting any kind of radiation reaction force.

The complete dynamics of a system consisting of a finite number of charged particles is given by three equations:

  1. Maxwell's equations $$ \partial_\mu F^{\mu \nu} = J^\nu, \quad \partial_{[\mu} F_{\nu \rho]} = 0 \tag{1} $$ give the dynamics of the EM field in response to the charges. (It's understood that in this case $J^\mu$ represents a sum of charge-weighted world lines for the particles.)
  2. The Lorentz force law $$ (F_i)^\mu = q_i F^\mu_{\ \ \ \nu}(x_i) (U_i)^\nu \tag{2} $$ gives the four-force that the EM field exerts on each particle $i$ given its charge $q_i$, position $x_i$, and four-velocity $U_i$.
  3. The relativistic version of Newton's second law $$(F_i)^\mu = m_i \frac{d^2 (x_i)^\mu}{d \tau^2} \tag {3}$$ connects the dynamical forces to the kinematics of the particle's motion.

Using these three equations, you can evolve both $J$ and $F$ forward in time from given initial conditions.

Now consider the case where $J$ is a continuous relativistic fluid (which is more natural setting for differential equations anyway). In this case eq. (1) is unchanged, and eq. (2) simply generalizes to a Lorentz four-force density field $$f^\mu(x) = F^\mu_{\ \ \ \nu}(x) J^\nu(x). \tag{4}$$ But it's not clear to me what is the continuum equivalent to equation (3) that maps this four-force density field into the actual kinematics of the current field $J$. Most sources that I've come across either simply end the discussion after eq. (4), or immediately jump to specific applications like an ideal fluid or other stat-mech models of particular fluids.

Clearly eqs. (1) and (4) don't contain enough information to fully determine the dynamical evolution of a general $J$ field, because the response to the forces will depend on the mass density profile, which is not specified by the charge density profile $J$. I suspect that if you assume that the fluid is composed entirely of particles with a fixed and known charge-to-mass ratio, then that assumption gives you enough information to determine the dynamics of the $J$ field.

According to these notes, the answer may be that a general relativistic fluid's response to a force density field is given by the equation $$f^\mu = \partial_\nu M^{\nu \mu}, \tag{5}$$ where $M^{\mu \nu}$ is the "material stress-energy tensor". In this case, it seems that in order to fully to determine the current density's dynamics, you need to specify some constituitive relations relating $M^{\mu \nu}$ and $J_\mu$. Is this right?

Q1. What is the general form of the minimal equation(s) that need to be added to eqs. (1) and (4) in order to fully determine the time evolution of the source field - the continuum analogue of eq. (3)? (An example answer could be "Eq. (5) plus a constitutive relation relating $M$ and $J$", although I don't know if that's correct.)

Q2. What is a very simple concrete example of this additional information?


1 Answer 1


For a perfect fluid you need the energy momentum tensor $$ T^{\mu\nu}=(\epsilon+p) u^\mu u^\nu-p g^{\mu\nu} $$ where the $u^\mu$ is the fluid velocity, and relation between $\epsilon$ and $p$ is determined by some constitutive relation (equation of state). The conservation equtaion $$ \nabla_{\mu}T^{\mu\nu}+F^{\mu\nu}J_\mu=0 $$ is the analogue of Euler's equation. Here $J_\mu$ is the charge current that may or may not be related to the fluid velocity. If there is only one component of charged fluid then usually $ J^\mu= \rho u^\mu$.

There can be problems defining $u^\mu$. The frame $u^\mu=(1,0,0,0)$ is usually the the frame in which the fluid is in thermal equilibriumm but it can be frame in which there charge current, or no entropy current, or no enegy current. These do not always coincide.

  • $\begingroup$ I mean this completely respectfully, but this is exactly the kind of answer that prompted me to ask my question, because it seems to bypass the main issue. Every discussion that I've seen jumps straight to some specific example like a perfect fluid, like you did. But what is the general form of the equation of motion that holds for any fluid? $\endgroup$
    – tparker
    Apr 17, 2021 at 19:37

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