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In my language, a differential two-form on $\mathbb{R}^4$ (viewed as a differentiable manifold with coordinates $t,x,y,z$) is a differentiable choice at each point of an alternating bilinear function from the tangent space at that point to $\mathbb{R}$, or equivalently something of the form $a_{xy} dx \wedge dy + a_{yz} dy \wedge dz + a_{zx} dz \wedge dx + a_{tx} dt \wedge dx + a_{ty} dt \wedge dy + a_{tz} dt \wedge dy$. A one-form just gives a cotangent vector. The most natural operation on a two-form is to take two tangent vectors at a given point and to apply the function to them to get a number. Another natural operation is to integrate this over an oriented surface in $\mathbb{R}^4$. As well, divergence is a natural operation on two-forms, whereas curl is a natural operation on one-forms, even though physicists often take the curl or the divergence of the same type of quantity, which they call a "vector field" (which at least seems bad from a mathematician's perspective).

We also have a metric on $\mathbb{R}^4$ (known by the name Lorentz), meaning a symmetric bilinear function on the tangent space at each point (that varies differentiably). Since $\mathbb{R}^4$ is a linear and homogenous space, we can actually think of that form as a form on space (manifold) as a whole. This means that there is a natural identification between tangent and cotangent vectors at each point, implying an identification between two-forms on the $T_p(\mathbb{R}^4)$ and elements of $T_p(\mathbb{R}^4) \otimes_{\mathbb{R}} T_p(\mathbb{R}^4)$ at each point $p$, and this identification extends differentiably to differential two-forms. Well, okay, it isn't really "natural" in the purely categorical sense, but because we have specified a metric, it kind of is.

This means that Lorentz transformations, ones which preserve the metric, commute with the identification between two-forms and elements of $T_p(\mathbb{R}^4) \otimes_{\mathbb{R}} T_p(\mathbb{R}^4)$ (equivalently, bilinear functions on the cotangent space). That is, if we have a Lorentz transformation sending $p$ to $q$, then we have a commutative diagram where the vertical arrows are the isomorphisms between the tangent and cotangent spaces (or more specifically their second tensor powers) at $p$ and $q$ respectively, and the horizontal arrows are the morphism and comorphism induced on the cotangent spaces of the Lorentz transformation (which is a diffeomorphism).

I think it also means something along the lines of the idea that a Lorentz transformation preserves the "form" of the equation, meaning that if you apply the Lorentz transformation as a change of coordinates, the coordinate-independent object (say the bilinear function) has another set of coordinates, but because we are using a Lorentz transformation, the coordinates are the same.

Diffeomorphisms which are not Lorentz transformations still send tangent and cotangent vectors at points to the same things at other points in a functorial way (specifically, a functor from the category of pointed differentiable manifolds to the category of vector spaces), but they don't necessarily satisfy this nice commutativity.

So could someone please explain electromagnetism in a simple way in this language? Also, how does it relate to the classical picture? E.g. does it look something like $E_x dy \wedge dz + \cdots + B_y dt \wedge dy + B_z dt \wedge dz$ or whatever? (This is wrong, but it's the kind of answer I'm looking for.) Once we have, say, a two-tensor, what natural operations (in the mathematical sense) can we do to it to get the basic physical quantities? E.g., if we have a two-form, what two vectors (or vector fields) do we plug into it? (I mean, that's what a two-form is made for - it's a bilinear function that you can plug two tangent vectors into! That and you can also integrate it and take its divergence.) How do Maxwell's equations work in this context?

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closed as too localized by Ryan Budney, Qiaochu Yuan, Willie Wong, Steve Huntsman, S. Carnahan Feb 14 '11 at 5:19

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David, there are quite a few textbooks on this topic. For example, Misner, Thorne and Wheeler is a standard reference but there's lots of upper-level undergrad / 1st year grad SR/GR type texts that cover this material. I'm not seeing why this is on mathoverflow. –  Ryan Budney Feb 11 '11 at 17:13
How is divergence a natural operation on two-forms before you specified the metric or volume form? –  Willie Wong Feb 11 '11 at 19:17
By divergence I mean exterior differentiation of two-forms. –  David Corwin Feb 11 '11 at 20:35
See my comment below. –  David Corwin Feb 13 '11 at 6:52
This question is not at an appropriate level for MathOverflow. The standard answer is that the classical photon field is a connection on a principal $U(1)$-bundle over spacetime, and its curvature 2-form is the field strength. Maxwell's equations in a vacuum are just $dF=d*F=0$. If you choose a preferred time direction, you get a splitting into $E$ and $B$ fields. If you want a longer answer, you should ask (a less wordy version of) this question at physics.stackexchange.com or math.stackexchange.com –  S. Carnahan Feb 14 '11 at 5:28

2 Answers 2

The current approach to this subject is to regard electromagnetism as a special cace (the abelian case) of a gauge field (aka a connection). I have written a book called "The Geometrization of Physics" that explains this (see in particular page xi of the introduction). It is freely available here:


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I looked at this, and it doesn't explain the answer in the form I'm looking for. On p.68, it gives the standard description of the Maxwell tensor in physics textbooks. I think this question is relevant because I've talked to many math students who find the physicists' tensor notation to be very confusing and who only learn tensors in the context of tensor products. –  David Corwin Feb 13 '11 at 6:52

This is only a rather partial (in all senses of the word) answer. The elctro-magnetic field is nothing but a closed $2$-form $\omega$ over $\mathbb R^4$. It cannot be split in a natural way into an electric and a magnetic parts, but once we choose space-time coordinates, then we may define $E$ and $B$ by $$\omega=\sum_{j=1}^3E_jdx_j\times dt+\sum_{\epsilon(i,j,k)=1}B_idx_j\times dx_k.$$ The fact that $\omega$ is closed gives $\partial_tB+{\rm curl}E=0$ and ${\rm div}B=0$.

The rest of Maxwell's system comes from a variational principle $$\delta{\mathcal L}[\omega]=0,\qquad{\mathcal L}[\omega]:=\int\int L(B,E)dxdt.$$ Special relativity tells us that the integrand $L$ must be invariant under Lorentz transformations. This means that there exists a function $L_0:\mathbb R^2\rightarrow\mathbb R$ such that $$L(B,E)=L_0(E\cdot B,\frac12(|E|^2-|B|^2)).$$

If $L_0=\frac12(|E|^2-|B|^2)$, then you get the usual linear Maxwell's equations. But other choices have been made, in order to resolve the paradox of the infinite energy of a single particle. One of them, $$L_0=-\sqrt{1+|B|^2-|E|^2-(E\cdot B)^2}$$ gives the Born-Infeld model. This is related to models in string and brane theory.

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This is exactly what I wanted! I still don't see why this is so irrelevant to MathOverflow, since I've met multiple math students who were looking for an answer like this one. –  David Corwin Feb 15 '11 at 4:12

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