So, a few weeks back, Ben posed the challenge of writing these equations in terms of differential forms. Well, the answer is not hard to find, in fact I have a couple of texts here which, like, "blurt it out". Being an idiot, that is not good enough for me, so I have been thinking about this, off and on, ever since. Oh, and not only am I an idiot, I am also a non-physicist, so if I goof on either count just shout.

Suppose I call the electric field as \mathbf{E} = E_x \text{i} + E_y \text{j} + E_z \text{k}, and likewise for the magnetic field (replace \mathbf{E} by \mathbf{B}.

Now we know that, for any p-form, there is a one-to-one correspondence with some field. So, for reasons which will become clear (I hope), I will let p=2. Moreover, we physicists (!) know that the EM field is a single field on Minkowski 4-spacetime.

So, suppose I write, say, \mathbf{E} = E_{\kappa} dx^{\mu}\wedge dx^{\nu} to be succinct, where I am assuming summation over. Likewise for \mathbf{B}

I now note that, by the axioms of the wedge product, that dx^{\mu} \wedge dx^{\nu} = 0 when \mu = \nu. I further recall that, by the same axioms, dx^{\mu} \wedge dx^{\nu} = - dx^{\nu} \wedge dx^{\mu}. This strongly implies antisymmetry, of course.

Now, I know the components E_{\kappa},\;B_{\kappa} are tensorial (the proof is messy, but not hard, I think), so I will define the antisymmetric (0, 2) tensor F_{\mu \nu} =\begin{pmatrix}0&-E_x &-E_y& -E_z\\E_x&0& B_x & -B_y\\E_y & -B_z & 0 &B_x\\E_z & B_y & B_x &0 \end{pmatrix}.

Then I will say that there is a 2-form F = \frac{1}{2}F_{\mu \nu} dx^\mu \wedge dx^{\nu}, by virtue of the aforementioned axioms, now easily seen from the matrix above. (The factor of one half simply means I am free to choose the upper right or lower left matrices, but notice that, whichever I choose, although the \mathbf{E}-field components are consistently signed, those of the \mathbf{B}-field are not).

So I have two tricks up my sleeve, but first this; I am assuming that the magnetic field is static, whereas the electric field is not (I hope this is correct). So, at some stage I am going to have to factor in time. Anyway, for now, the obvious trick is to make use of the antisymmetry dx^{\mu} \wedge dx ^{\nu} = -dx^{\nu} \wedge dx^{\mu}, which, factoring in time (and setting c = 1 I am going to write explicitly as

F = (E_x\; dx \wedge dt + E_y \;dy \wedge dt+ E_z \;dz \wedge dt)+ (B_x \;dy \wedge dz + B_y \;dz \wedge dx + B_z \;dx \wedge dy).

Looking at my matrix, and remembering the axioms, I easily see this makes it as nice as it could be. You do see this don't you? It's a little hairy, but it just requires keeping track of signs under inversion.

Fuck, I am exhausted! More another time......

Fuck, I am exhausted! More another time......

Eck, I am fuxhausted just from reading all this ! No more anytime ........

Agreed... his mathematical knowledge, is my undoing

OK, for the math illiterati (which includeth mine humble self) I present 2 exhibits. One on visualising differential forms (good for newbies to tensor calculus), another more symbolic discursion which, among other interesting shit, does a derivation of Maxwell's stuff, from 2-forms.
So, for comparison, or to keep tally on how QH is doing, read thusly:
http://homepage.mac.com/sigfpe/Mathematics/forms.pdf
http://www.math.purdue.edu/~dvb/preprints/diffforms.pdf

You can write Maxwell's equations in terms of evolving forms on 3-d space, which is also interesting. For example, E and H will be 1-forms, and D and B will be 2-forms. The material constitutive laws will be expressed in terms of Hodge operators.

The material constitutive laws will be expressed in terms of Hodge operators.
The Hodge, or * operator.
Both those refs talk about the * operator, or the Hodge dual.
Is it a coincidence that it looks like a complex conjugate symbol? It's kind of a "reflection operator"?

A closer look at that ref from purdue.edu shows, I think, pretty much exactly what you say about Maxwell's formalism. They use 3-forms, or show the equivalence between the diff. forms with Hodge duals. Something like that.

Yes, it is something like dual. It is actually the Poincare dual mapped into the primal side using the inner product. They use not only 3-forms, but 0- (scalar potential), 1- (electric field and magnetizing field), and 2-forms (magnetic induction and displacement field) too.

OK, let me just try and guide your intuition here. Note this is not the true definition of the Hodge operator, but it might help you to see what's going on.

Suppose I have a bog-standard vector space, let's say 3-dimensional. All this means is there are 3 basis vectors (these are the vectors for which any other vector can be expressed in terms of). I now define a "choice space" by, say, from 3 choose 1 at a time. These choices I will collectively call 1-forms, x, y and z, and the space of all these choices a 1-form space.

I now define the "remainder space" by, for x chosen in choice space, I will have yz in remainder space. The Hodge operator is simply the map that takes elements in choice space to the corresponding elements in remainder space, as *(x) = yz etc. So the same is true for the "choose 2" space: *(xy) = z etc etc for any p = 1, 2, ...., n

These spaces are amusingly named \Lambda^p(V_n) and \Lambda^{ n-p}(V_n). respectively, so the Hodge operator is \ast: \Lambda^p(V_n) \to \Lambda^{n-p}(V_n) where, for \omega \in \Lambda^p(V_n) you will have \ast\omega = \gamma \in \Lambda^{n-p}(V_n). This latter equality defines the Hodge dual.

I warn you again, this is only a rough and ready guide.

In three dimensional Euclidean space actually 1- and 2-forms can be identified with vector fields and then the Hodge operator is just the identity map.

F = (E_x\; dx \wedge dt + E_y \;dy \wedge dt+ E_z \;dz \wedge dt)+ (B_x \;dy \wedge dz + B_y \;dz \wedge dx + B_z \;dx \wedge dy). Anyhoo, to continue with my challenge, I quote myself for mine own reference.

Looking in some crappy text here I find that the non-existence of magnetic monopoles is expressed as \nabla \cdot \text{B} = \text{div} \text{B} = 0.

Now the divergence operator in the language of differential forms is dB = \sum\frac{\partial B_{\mu}}{\partial x^{\mu}}, hence d B = \text{div} B.

Now looking at my quote, and noticing that the "E part" plays no role here, I may as well write this as \nabla \cdot B} = 0 \Rightarrow dF= 0.

Now, again looking in my book, I find the Law of Faraday as \nabla \times \text{E} = -\frac{\partial \text{B}}{\partial t} which I will rewrite as \nabla \times \text{E} + \frac{\partial \text{B}}{\partial t}=0, where I again setting c = 1.

I will also rewrite my self-quote above as

F = (E_x dx + E_y dy + E_z dz )\wedge dt+ (B_x dy \wedge dz + B_y dz \wedge dx + B_z dx \wedge dy).

Now recalling that the exterior derivative operator sends a p-form to a p+1-form I will have

dF = (E_x dx \wedge dy + E_y dy \wedge dz + E_z dz \wedge dx )\wedge dt+ (B_x dy \wedge
dz + B_y dz \wedge dx + B_z dx \wedge dy)\wedge dt.

But this is just \nabla \times \text{E} + \frac{\partial \text{B}}{\partial t}, since the exterior derivative of a 1-form (i.e. the stuff inside my first parenthesis) is the curl. This follows in a straightforward way from the equivalence between a 1-form and a vector field that temur mentioned (actually he called it an identity, which I don't like so well). quote: "The proof is left as an exercise for the reader" unquote

Then if the latter vanishes, then so does the former: d F = 0 So non-existence of monopoles, and Law of Faraday can be simply expressed thusly. How very nice (of course I knew this already, but it took me a bit of work to see why).

So now we can see why this must be true. Recall from my diff. forms thread the Lemma of Poincaré: d^2 = 0, that is, for some p-form \omega,\; d(d \omega) = 0. It seems the converse is true, though the proof is fiendish: if there is some p-form for which d \gamma = 0, then there must be some p-1 form, \alpha such that d \alpha = \gamma, whereby d \gamma =d(d \alpha) = 0, which is Poincaré.

So, by the above, define the 1-form A such that the 2-form F = dA. This is called the vector potential. Now look in our book. We find, in terms of fields, for example, that curl A = B \Rightarrow dA = B \Rightarrow dB = d(d A) = 0 and

grad \text{A}_t = \frac{\partial \text{A}}{\partial t} + \text{E} , and noticing that A_t is a 0-form, this implies that grad \text{A}_t = d A_t is a 1-form, then d E= d(dA_t) =0. It suffices to recall that F = E + B, and that dE + dB = d(E + B) = dF, and we're done.

What about the other two equalities. Ah, that will be a different bed-time story, children.

Gosh I am a long-winded bastard! Who likes the sound of their own voice, hmmm?

Then if the latter vanishes, then so does the former: d F = 0 So non-existence of monopoles, and Law of Faraday can be simply expressed thusly. How very nice (of course I knew this already, but it took me a bit of work to see why).

So now we can see why this must be true. Recall from my diff. forms thread the Lemma of Poincaré: d^2 = 0, that is, for some p-form \omega,\; d(d \omega) = 0. It seems the converse is true, though the proof is fiendish: if there is some p-form for which d \gamma = 0, then there must be some p-1 form, \alpha such that d \alpha = \gamma, whereby d \gamma =d(d \alpha) = 0, which is Poincaré.

So, by the above, define the 1-form A such that the 2-form F = dA. This is called the vector potential. Now look in our book. We find, in terms of fields, for example, that curl A = B \Rightarrow dA = B \Rightarrow dB = d(d A) = 0 and

grad \text{A}_t = \frac{\partial \text{A}}{\partial t} + \text{E} , and noticing that A_t is a 0-form, this implies that grad \text{A}_t = d A_t is a 1-form, then d E= d(dA_t) =0. It suffices to recall that F = E + B, and that dE + dB = d(E + B) = dF, and we're done.


Beautiful, isn't it?

So you have the equations of motion (dF = 0), and the statment of gauge invariance (d(F+dA) = dF + d(dA) = 0).

Now all you need is an action. I may be off by a numerical factor, but you can show that the action is given as

\mathcal{S} = \int F \wedge F.

In my opinion, this is the most beautiful equation in all of physics :)

Yeah, it is beautiful, but it's better than that. Let's argue backwards.

Suppose the F-field, the 2-form F, is given by a section of the principal bundle on Minkowski spacetime, a manifold recall, whose structure group G is a Lie group. Now define, ad hoc, so to speak, the 1-form A, the vector potential. We will assume, again on the fly, that this is a connection on our bundle.

Then I can show, with some difficulty I grant you, what Alpha presented (where's he gone, by the way - I was rather fond of the bastard) that, by definition of the covariant derivative, D F = d A is the curvature of A. Notice that curvature is a Lie algebra-valued (\mathfrak{G}-valued) 2-form by definition.

So, all is well, since we know that F is a 2-form. Alpha also showed us that, by definition,

DF= dF + A \wedge F - F \wedge A = dF +[A,F] = 0, the Bianchi identity.

But, if we already know that dF = 0 then it must be that [A,F] =0, then this requires that our structure group has an algebra with all brackets zero. And if we further require our group to be unitary, then we're stuffed unless we choose U(1), where of course all Lie brackets vanish.

Thus have we derived the necessary (but not sufficient) structure group for "ordinary" electromagnetism. In spite of my crappy logic, this is so much fun!

haha

I wish I could have fun doing something like that.

haha

I wish I could have fun doing something like that.

haha

I wish I could have fun doing something like that.

My problem is that I don't think I enjoy solving physics/math problems enough. I'm fine with reading about physics and even watching the professor's solve problems, making them seem so easy, but when it comes down to it, I'm kind of lazy and question my abilities as well.

I did not say the map between 1-forms and vectors is the identity (the spaces are different), but when you represent 1- and 2-forms as vectors, then the Hodge operator (now sending vector to vector) becomes identity. On second thought it is more accurate to say that it is possible to represent 1- and 2- forms as vectors such that the Hodge operator is the identity map.

So, once again, looking in the only text available to me, I find the remaining Maxwell equations to be

\nabla \cdot \text{E} = 4\pi \rho and \nabla \times \text{B}- \frac{\partial \text{E}}{\partial t} = 4 \pi \text{J}; these are Maxwell's derivations of the Laws of Coulomb and Ampere, respectively.

First I note the almost perfect symmetry with the other 2 equations of Maxwell, which suggests that it may be permissible to interchange the E and B fields.

However, I see a problem, so I need to learn a bit of physics; specifically, I need to know what \rho and J are.

So here it is. I can do the above interchange using the Hodge operator, no problem, but somewhat lengthy to describe. But now...

according to my limited understanding, \rho is electric charge density, J is electric current density, and it appears that these are rather simply related;

J=\rhov, the v being velocity. So I need to find a way of combining these guys into a p-form. But first, what is p? Well, I anticipate, that I will need to take the exterior derivative of \ast F which give me a 2 + 1 = 3-form, so I will want to express J and
\rho combined as a 3-form.

This is tough for me, I cannot get this to work in the way that I know it must. So, the other thing I read about is the "continuity condition"; \frac{\partial \rho}{\partial t} + \rho \text{v} =0, so when \text{J} = \rho \text{v} this apparently implies conservation of charge, which sort of makes sense.

But I have been mucking around, off and on, for a couple of days now, and I cannot use this knowledge to derive a "J_{\rho}" 3-form that, together with the derivative of the Hodge dual of F, will give be the equalities I know I must have.

I came within a hair's breadth a couple of times, but seemed to have the "dt" on the wrong term, or wrongly signed, each time.

Any hint (no more, please) would be appreciated.

I think J is a 2-form, not 3-form, and te continuity equation is \frac{\partial\rho}{\partial t}+\mathrm{div}\,J=0. The continuity equation \frac{\partial\rho}{\partial t}+\rho\mathbf{v}=0 you gave does not seem to make sense because the partial derivative of rho is not a vector, while rho times velocity is.

Whoops-a-daisy, you are right, I goofed, The continuity equation is indeed as you said, so back to the drawing board.

So, once again, looking in the only text available to me, I find the remaining Maxwell equations to be

\nabla \cdot \text{E} = 4\pi \rho and \nabla \times \text{B}- \frac{\partial \text{E}}{\partial t} = 4 \pi \text{J}; these are Maxwell's derivations of the Laws of Coulomb and Ampere, respectively.

First I note the almost perfect symmetry with the other 2 equations of Maxwell, which suggests that it may be permissible to interchange the E and B fields. ...Your abilities are amazing. It has been a long time since I struggled to do boundary value equations etc. with even the simple notation you give here (I do not even know what your earlier "wedge product" is so of course that form is "pearls before me")

I only post to mention that back in the darkages, I was once required to re-write Maxwell's equations so that they are perfectly symmetric in E & B. This of course requires introducing a symbol for the magnetic monopole charge density and for someone like you, making them symmetric is trivial I am sure.

What made it interesting is that when this is done, and the speed of light preserved, charge and mass on the electron, also (as I recall, perhaps wrongly) it forces the monopole to be very heavy. Perhaps only if one assume it has the same charge to mass ratio as the electron, for "full symmetry." I forget the details.

I also forgot the details but I think what you are saying about the mass of a monopole is implied after you quantize the field. For the classical EM field I don't see how one can derive a relation between the masses of the electron and magnetic monopole. I also remember the electric charge quantization comes for free when there is magnetic monopole.

Billy, I don't deserve your flattery, as I am about to demonstrate. Sad bastard that I am, I spent my lunch hour in the library, I realized this:

I was getting myself in all sorts of tangles with signs and time derivatives because of my failure to realize 3 elementary physical facts;

the electric current density J, a vector field, is not the same as the electric current itself I, which is a 4-vector. I was confused because I looked in a text which used J for both ;

I had forgotten we were in Minkoswki spacetime, where the metric has signature {-1, 1, 1, 1};

I hadn't twigged that the symmetry of the Maxwell equations is no mere mathematical accident, rather they were deliberately cast that way by him to express the experimental facts that, just as a moving electric field induces a magnetic field, so a moving magnetic field induces an electric field.

So, I freely confess that what follows is not entirely my own work!

So anyway, let's return to the EM field strength, whose components are tensorial and may be written as the totally antisymmetric tensor F_{\mu \nu}.

I now define the tensor dual F^{\mu \nu} by F^{\mu \nu} = g^{\mu \alpha}g^{\nu \beta}F_{\alpha \beta}. I am going to be anal and write this down

F^{\mu \nu} = \begin{pmatrix}0&E_x&E_y&E_z\\
\\
-E_x&0&B_z&B_y\\
\\
-E_y&-B_z&0&B_x\\
\\
-E_z&B_y&-B_x&0
\end{pmatrix}

If anyone is sufficiently bored, they can compare this with my earlier matrix, and see that the "E's" have changed sign, but the "B's" haven't. This motivates the following;

I define the Hodge dual 2-form as \ast F = \frac{1}{2}F^{\mu \nu}dx_{\mu} \wedge dx_{\nu} and write this explicitly as

\ast F = -(B_x dx + B_y dy + B_z dz) \wedge dt +(E_x dy \wedge dz + E_y dz \wedge dx + E_z dx \wedge dy)

This works because \ast (dx \wedge dy) = -dz \wedge dt etc. and dx \wedge dy = -dy \wedge dx.

Now to the part that was causing me grief. Let me say that the current density j has components j_x, j_y, j_z. So I now make the current 4-vector J \equiv (\rho,
j_x, j_y, j_z) and then make the current 1-form

J = -\rho dt +j_x dx + j_y dy +j_x dz . The Hodge dual to this will be the current 4 - 1 = 3-form

\ast J = \rho dx \wedge dy \wedge dz -(j_x dy \wedge dz + j_y dx \wedge dz + j_z dx \wedge dy)\wedge dt, where I justify my signs in the same way as above.

If you don't agree, look at this;

d \ast J = (\frac{\partial \rho}{\partial t} + \sum \frac{j_{\mu}}{\partial x^{\mu}}) = (\frac{\partial \rho}{\partial t}+\text{div j})dt \wedge dx \wedge..... = 0 (continuity condition) because of the way I arranged my signs. So there!

Anyway, calculating through, I get d \ast F = \ast J, and, with the same sort of hand waving as for the other Maxwell equations, where \text{div}E = 4\pi \rho (Coulomb) and \text{curl}B -\frac{\partial E}{\partial t} = 4 \pi J (Ampere) I get that

d \ast F = 4 \pi \ast J is implied by both the Laws of Coulomb and of Ampere, which I confess is not quite what I was expecting. I wish I could say it was fun, but it wasn't!

Billy, I don't deserve your flattery,...I think you do. You can even recognize your own mistakes! I was not able to follow fully your posts, much less notice any mistakes in them. I often fail to see my own mistakes.

Besides, I need to say something nice to someone ever now and then - at least once for every 25 times I tell someone their post is "nonsense."

I am sort of a self appointed "chief corrector" of extreme nonsense. I figure that is my duty so that you, Temur, BenTheMan, Pete,* James R, etc. do not need to waste your more valuable time doing these simple tasks. I am often adequate for this, and as I am in retirement, I have time to do this job.
--------------
*Pete sometimes helps, but he is too polite to do this job with my blunt force. Unlike me, he is not lazy and will do some complex calculations etc to show what is correct. I just point out what is nonsensical.