# Form of Maxwell's equations

Do the Maxwell equations in the usual 3-vector form have the same form in any Lorentz frame? For example, the one that says ##\nabla \cdot \vec B = 0## will be valid in another, primed Lorentz frame? That is ##\nabla' \cdot \vec {B'} = 0##?

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Dale
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Do the Maxwell equations in the usual 3-vector form have the same form in any Lorentz frame? For example, the one that says ##\nabla \cdot \vec B = 0## will be valid in another, primed Lorentz frame? That is ##\nabla' \cdot \vec {B'} = 0##?
Yes. It is a tedious exercise, but you can work through it and prove it.

Orodruin
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Yes. It is a tedious exercise, but you can work through it and prove it.
How is it tedious? Knowing the covariant form of Maxwell’s equations I find it pretty straight forward ...

Dale
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How is it tedious? Knowing the covariant form of Maxwell’s equations I find it pretty straight forward ...
It is tedious to me because using three-vector equations is always tedious when you know the covariant equations.

PeterDonis
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using three-vector equations is always tedious when you know the covariant equations

IIRC, MTW give some examples where the covariant (4-D tensor) equations can be used to derive the 3-vector equations more easily than the 3-vector view itself can, and one of the examples has to do with transforming EM fields.

Orodruin
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It is tedious to me because using three-vector equations is always tedious when you know the covariant equations.
The easiest way of doing it is first showing that the full covariant form implies the three-vector form. This in itself is just a couple of lines.

pervect
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The easiest way of doing it is first showing that the full covariant form implies the three-vector form. This in itself is just a couple of lines.

It seemed messy to me as well, but I suppose if you look up the dual of the Faraday tensor (here denoted as G) in terms of the E and B field components from wiki https://en.wikipedia.org/wiki/Electromagnetic_tensor or some other more reliable textbook, you can get ##\nabla \cdot B = 0## from ##\partial_a G^{a0} = 0##

Doing all the equations still strikes me as a pain, though.

Orodruin
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Doing all the equations still strikes me as a pain, though.
It really isn’t. I would do it but I am on my phone and writing TeX on the phone is a pain.

Orodruin
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It really isn’t. I would do it but I am on my phone and writing TeX on the phone is a pain.
So here goes (up to arbitrary unit-dependent constants). The main thing to remember is that (using Greek letters for indices going from 0 to 3 and Latin letters for those going from 1 to 3):
$$E^i = F^{i0} \quad \mbox{and} \quad B^i = -\epsilon_{ijk}F^{jk}/2 \quad \mbox{or} \quad F^{ij} = -\epsilon_{ijk} B^k.$$
From ##\partial_\mu F^{\mu\nu} = J^\nu## follows that
$$\rho = J^0 = \partial_0 F^{00} + \partial_i F^{i0} = 0 + \partial_i E^i = \nabla \cdot \vec E$$
and that
$$\vec J = \vec e_i J^i = \vec e_i \partial_\mu F^{\mu i} = \vec e_i [\partial_0 F^{0i} + \partial_j F^{ji}] = - \partial_t \vec E - \vec e_i \epsilon_{jik} \partial_j B^k = - \partial_t \vec E + \nabla \times \vec B.$$
The remaining Maxwell equations follow in exactly the same way but for the dual field tensor with no source term. I honestly don't think that this is "a pain". It is just about splitting the sum in ##\partial_\mu F^{\mu\nu}## into the temporal and spatial parts.

Last edited:
SiennaTheGr8 and weirdoguy
Dale
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$$B^i = -\epsilon_{ijk}F^{ij}$$
See, I find even just writing down this theee vector right here tedious. I mean, I always worry about making a mistake while writing the components of ##\epsilon_{ijk}##.

SiennaTheGr8
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See, I find even just writing down this theee vector right here tedious. I mean, I always worry about making a mistake while writing the components of ##\epsilon_{ijk}##.
But much less tedious than actually doing it in the three-vector formalism ...

Dale
It's actually possible to obtain the electric and magnetic fields from four-vector relations alone—no rank-2 tensors, no Ricci-calculus notation. Haven't seen anyone else do it this way, but it works, and I find it helpful. Your mileage may vary.

The "trick," if you want to call it that, is that the four-tensor expression ##(\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}) U_{\nu}## is "like" the "cross product" between ##\mathbf U## and the "curl" of ##\mathbf A##. I use scare quotes because there is no such thing as a cross product or curl of four-vectors (unless one tweaks some definitions, which some people do), but the analogy is perfect if you use the "bac - cab" expansion of the vector triple product.

Here's what I mean. With Feynman notation, we have the following three-vector identity:

## \mathbf a \times (\nabla \times \mathbf c) = \nabla_{\mathbf c} (\mathbf a \cdot \mathbf c) - (\mathbf a \cdot \nabla) \mathbf c ##.

The analogy I'm drawing is with the right side of that expression:

## (\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}) U_{\nu} = \vec \partial_{\mathbf A} ( \mathbf U \cdot \mathbf A ) - ( \mathbf U \cdot \vec \partial ) \mathbf A ##,

where ##\vec \partial = (\partial^t, - \nabla) ## is the four-del operator (sorry for mixing arrow notation in there—emboldening ##\partial## doesn't work).

So the Lorentz four-force can be notated like this (##c = 1##):

##\mathbf F = q \left( \vec \partial_{\mathbf A} ( \mathbf U \cdot \mathbf A ) - ( \mathbf U \cdot \vec \partial ) \mathbf A \right) ##,

where ##\mathbf U = (U^t, \mathbf{u} )## is the four-velocity of the test charge and ##\mathbf A = (A^t, \mathbf{a})## is the four-potential. (And the Feynman notation here is extraneous superfluous [edited], since the particle's four-velocity has no dependence on four-position anyway.) If you write that in "1 + 3" component form, after some work (including the three-vector "bac - cab" rule) you end up with:

##\mathbf F = q \left( \mathbf{u} \cdot (- \nabla A^t - \partial^t \mathbf{a}), \, U^t (-\nabla A^t - \partial^t \mathbf{a}) + \mathbf{u} \times (\nabla \times \mathbf{a}) \right)##.

Then define the three-vector fields ##\mathbf e## and ##\mathbf b## to simplify that:

##\mathbf F = q \left( \mathbf{u} \cdot \mathbf{e}, \, U^t \mathbf{e} + \mathbf{u} \times \mathbf{b} \right)##.

Orodruin
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That’s just replacing F by dA, which is a rank 2 tensor.

Yes, but both ##\vec \partial (\mathbf U \cdot \mathbf A)## and ##(\mathbf U \cdot \vec \partial) \mathbf A## are four-vectors. I'm saying that you can start with the Lorentz four-force written in terms of them, without bringing the Faraday tensor into the picture at all.

My point is really that the Lorentz four-force is "like"

##\mathbf F = q \left( \mathbf U \times ( \vec \partial \times \mathbf A ) \right)##,

except that those individual operations involving ##\times## aren't defined (you've got to use the "bac - cab" rule).

Here, covariant electrodynamics (Lorenz gauge) with four-vector notation alone:

##\Box \mathbf A = \mathbf J##

##\mathbf F = q \left( \vec \partial (\mathbf U \cdot \mathbf A) - (\mathbf U \cdot \vec \partial) \mathbf A \right) ##.

Is all this entirely uninteresting? I don't know, I kind of like it. The d'Alembertian is like the Laplacian, and the second equation is like a vector triple product involving a curl. Ricci calculus is obviously incredibly powerful, but I'm just a hobbyist, and vector notation is far easier on my eyes and pea-brain.

Orodruin
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The d'Alembertian is like the Laplacian, and the second equation is like a vector triple product involving a curl.
Well, yes, it has to be. It is how the curl and the Laplace operator generalise to a 4D Lorentzian spacetime (i.e., the exterior derivative of a one-form and just the trace when the gradient is taken twice).
and vector notation is far easier on my eyes and pea-brain.
I don't see what is wrong with ##F = dA## and ##d*F = *J## from that perspective - if removing indices is what you are after.

I don't see what is wrong with ##F = dA## and ##d*F = *J## from that perspective - if removing indices is what you are after.

Sure, the exterior algebra stuff is clearly superior... if you're comfortable with it.

Dale