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In special and general relativity, the magnetic field is defined as $$B^\mu = F^{*\mu\nu}u_\nu, \label{tag1}\tag{1}$$ where $F^{*\mu\nu} = \frac12 \varepsilon^{\mu\nu\rho\sigma}F_{\sigma\rho}$, and $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is the standard Maxwell tensor, $A_\mu$ is the electromagnetic potential, and $u_\mu $ are the covariant components of the 4-velocity associated with a ceratin frame of reference, which is chosen rather arbitrary.

Can we give any physical meaning to the 4-divergence $$\partial_\mu B^\mu \qquad \text{or} \qquad \nabla_\mu B^\mu$$ of $B^\mu$?

At least it does not look that it is zero because $$\partial_\mu B^\mu = u_\nu\partial_\mu F^{*\mu\nu} + F^{*\mu\nu}\partial_\mu u_\nu = F^{*\mu\nu}\partial_\mu u_\nu \neq 0.$$

EDIT 1: Note that \eqref{tag1} defines a proper 4-vector because $B^\mu$ transform as contravariant components of a 4-vector under general coordinate transformation $x^\mu \to x^{\mu'}$. It is, of course, frame-dependent as it explicitly depends on the definition of the 4-velocity, which is, to my understanding, totally fine in electrodynamics. Also, note that \eqref{tag1} is the standard definition of the magnetic field in the GRMHD (general relativistic magnetohydrodynamics) literature.

EDIT 2: I am thinking of $u^\mu$ as the 4-velocity of a fluid, which implies that the frame associated with it (fluid particles rest frame) is non-inertial in general. Is that the problem of definition \eqref{tag1}? Does \eqref{tag1} make sense only for inertial frames?

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    $\begingroup$ What is $u_\nu$? is it irrotational? $\endgroup$ – AccidentalFourierTransform Jul 24 at 15:09
  • $\begingroup$ Maybe the covariant form of Maxwell's equations will help. en.wikipedia.org/wiki/… $\endgroup$ – puppetsock Jul 24 at 16:29
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    $\begingroup$ Is $u^\mu = \frac{d x^\mu}{d\tau}$, where $x^\mu(\tau)$ is the world line of a particle? In that case you get an expression $B^\mu(x(\tau))$, a function of $\tau$ only. Or is $u_\mu$ a velocity field? For example, because you bring the example of hydrodynamics, the velocity field of a fluid? i mean the latter case is just an assortment of continuous particles so in principle one would still get $B^\mu$ as a function of a parameter counting fluid molecules and not of spacetime. $\endgroup$ – Lorenz Mayer Jul 24 at 21:23
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There is no such thing as a 4-magnetic field. What you've constructed is a 4-tuple of numbers $(0,B_x,B_y,B_z)$. This object doesn't transform as a 4-vector. Therefore it doesn't make sense to apply a 4-gradient to it.

To make the conceptual issue more clear, here's a simpler example. I'm going to define a special-relativistic variable $t$ called the scalar-time. It's defined like this:

$$t=x^\nu u_\nu.$$

Here $x$ is the displacement in spacetime from a particular reference event chosen as the origin, and $u$ is the arbitrarily chosen velocity vector of an observer. This definition is notated so as to look like a scalar, but it's not. It's the time coordinate of the Minkowski coordinates associated with a particular observer.

Of course you can say that $t$ is really frame-independent, but that's kind of silly. The definition explicitly refers $t$ to a particular frame. Analogously, the fact that the Aztecs believed human sacrifice to be a good thing is simply a fact, not a belief -- but that doesn't mean that it's a fact that human sacrifice is good.

AccidentalFourierTransform says in a comment:

In particular, if u is the velocity field of some fluid, then Bμ as defined in the OP is the "rest" magnetic field, the magnetic field measured by a comoving observer, or something like that. It is a well defined object, and it is meaningful to ask how it varies with position.

You're now saying that we might have some physical body that is a natural thing to consider as being at rest. In that case, we have a "scalar-time" $t$ which is naturally preferred. This is what happens in cosmology, for example: we're usually interested in the proper time of an observer moving with the Hubble flow. But that doesn't mean that a time coordinate is coordinate-independent, it just means that there is a preferred time coordinate. Nor does it mean that a time coordinate is a relativistic scalar; it has nontrivial transformation properties under a change of coordinates.

We can form an expression like $\nabla t$, but it won't transform as a covector, because $t$ doesn't transform as a scalar. This is in fact exactly what Hoyle did with the "C-field" in his steady-state cosmology. The gradient of his preferred time coordinate was the preferred velocity vector. For an observer with this preferred velocity vector, hydrogen atoms would pop into existence at rest.

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    $\begingroup$ His definition of $B^\mu$ does transform like a $4$-vector. $\endgroup$ – Lorenz Mayer Jul 24 at 16:15
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    $\begingroup$ This is wrong. The object in the OP is manifestly a 4-vector. You cannot pick a specific frame of reference, $u=(1,0,0,0)$, evaluate a tensor, and claim that the result depends on the frame of reference: it was you who introduced the apparent dependence by evaluating it at $u=(1,0,0,0)$. $\endgroup$ – AccidentalFourierTransform Jul 24 at 17:25
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    $\begingroup$ I believe the tensor in the OP is something like the interior product of the (dual) electromagnetic two-form and a vector $u$; this is a well-defined (pseudo)tensor (more like a three form, but anyway). In particular, if $u$ is the velocity field of some fluid, then $B^\mu$ as defined in the OP is the "rest" magnetic field, the magnetic field measured by a comoving observer, or something like that. It is a well defined object, and it is meaningful to ask how it varies with position. $\endgroup$ – AccidentalFourierTransform Jul 24 at 17:30
  • $\begingroup$ @AccidentalFourierTransform: I understand your point, but I don't think your way of describing this is the best way. I've added some more material to my answer to try to explain more. $\endgroup$ – Ben Crowell Jul 24 at 17:57
  • $\begingroup$ Isn't this scalar $t$ you cook up actually a scalar? It's the time coordinate value in a particular frame of reference. As such, it takes the same value no matter what chart you're in, because its definition involves a prescription to calculate it in a particular chart - the situation is very similar to the reasoning often given for why rest mass is a scalar (that is, its measured in the rest frame so can't be frame dependent). $\endgroup$ – jacob1729 Jul 24 at 22:26
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Your confusion arises from the fact that you're thinking of $u_\mu$ as a non-constant vector. In the context of "the magnetic field measured by an inertial observer in a flat spacetime", $u^\mu$ is a constant vector field on Minkowski spacetime, everywhere parallel to the observer's four-velocity; it can be thought of as the vector $(\partial/\partial t)^\mu$ for the inertial observer's time coordinate $t$, which is specified everywhere in spacetime. So $\partial_\mu u_\nu = 0$ in this context, and $\partial_\mu B^\mu = 0$ as well.

This statement has a generalization to curved spacetime as well. Suppose that $u^\mu$ are the tangents to a timelike geodesic congruence; in other words, they are the tangents to the worldlines of a family of freely falling observers through spacetime. Suppose further that this congruence is hypersurface-orthogonal; this corresponds roughly to the idea that there is some notion of "time coordinate" in spacetime that the observers can all agree on. (An example would be the "cosmic time" in an FRW cosmology.) In this case, it can be shown (see Wald's General Relativity, §9.2) that we must have $\nabla_{[\mu} u_{\nu]} = 0$, and so the quantity $F^{*\mu \nu} \nabla_{\mu} u_\nu$ still vanishes.

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  • $\begingroup$ that's right, I am thinking of $u^\mu$ as a non-constant vector. Specifically, as a velocity of a fluid. Is that the problem of definition (1) from OP? $\endgroup$ – peshenator Jul 24 at 22:22

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