Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It seems unnatural to me that it is so often worthwhile to replace physical objects with their Hodge duals. For instance, if the magnetic field is properly thought of as a 2-form and the electric field as a 1-form, then why do they show up in Ampere's and Gauss' as laws as their duals, i.e.

$$ \int_{\partial M} \star \mathcal B^2 = \int \int_M \left( 4\pi j^2 + \frac{\partial \star \mathcal E^1}{\partial t} \right)$$

$$ \int \int_{\partial U} \star \mathcal E^1 = 4 \pi Q_{\mathrm{enc}} $$

Similarly, angular momentum is considered nearly everywhere as a pseudo-vector instead of as a 2-form. Do these laws have formulations that do not use Hodge duals? Is this just for the sake of simplicity, since tensors are less familiar to the physics community than vectors?

share|improve this question
    
I meant that tensors are less familiar than vectors, not differential forms. So people are more comfortable thinking of the magnetic field as its associated pseudo-vector than as an antisymmetric tensor. –  ZachMcDargh Mar 22 at 17:47
    
OP: "Do these laws have formulations that do not use Hodge duals?" - Have you checked Wikipedia? Gauss' laws and Ampere's laws may be written in terms of the $E$ and $B$ fields using vector calculus. –  JamalS Mar 22 at 17:48
    
@JamalS Yes, but treating the magnetic field as a vector is implicitly using its dual. I am looking for something that expresses Ampere's law treating $B$ as a 2-form. –  ZachMcDargh Mar 22 at 17:55
    
I have a good picture of what $\vec B$ and $\vec E$ mean physically and what effect they have onto classical charged particles. I do not have a good picture of a two-form, yet. I'd like to, but I think I will have to wait until I did GR to get this. –  queueoverflow Mar 22 at 18:28
    
You might enjoy pages 136-138 of supermath.info/ma430.pdf there I investigate a toy 5-dimensional E and M where the difference between the electric and magnetic field are more pronounced. –  James S. Cook Mar 23 at 1:43

1 Answer 1

up vote 8 down vote accepted

In the language of differential forms in spacetime, the field strength $2$-form $F = E\wedge\mathrm{d}\sigma + B$ gives Gauss's law for magnetism and Faraday induction: $$\mathrm{d}F = 0\text{.}$$ Meanwhile, the electromagnetic excitation $2$-form $H = -\mathcal{H}\wedge\mathrm{d}\sigma + \mathcal{D}$ provides a natural formulation of Gauss's law and Ampère's circuital law: $$\mathrm{d}H = J\text{,}$$ where $J$ is the electric current 3-form.

For instance, if the magnetic field is properly thought of as a 2-form and the electric field as a 1-form, then why do they show up in Ampere's and Gauss' as laws as their duals, i.e. ...

Because it's a qualitatively different role: $F$ being a closed form is a necessary property to ensure conservation of magnetic flux and that the existence of a potential $1$-form $A$ for which $F = \mathrm{d}A$. But $H$, instead of conservation of magnetic flux, expresses the conservation of charge, with $H$ acting as a "potential" for the electric current $J$.

Of course, if you know that $H\propto\star F$, then you can eliminate the $(\mathcal{D},\mathcal{H})$ excitation fields put everything in terms of $(E,B)$ only. Or the reverse, if you wished. This naturally introduces at least an implicit Hodge dual into the equations, as you have above. But doing so obscures the fundamentally metric-free character of Maxwell's equations: the only place the metric appears is in the Hodge dual. So instead, one can think of the Hodge dual as providing a simple constitutive relation for free space, with vacuum having its own meaningful $\mathbf{D}$ and $\mathbf{H}$ fields.

In that kind of presentation, the appearance of the Hodge dual is natural and necessary to turn electromagnetism into a fully predictive theory--the metric must make an appearance eventually, but Maxwell's equations themselves are metric-free!

There are other possible relations between $H$ and $F$ independent of Maxwell's equations per se, leading to alternative theories of electromagnetism, such as Born-Infeld theory and Heisenberg-Euler vacuum polarization, etc. Generally, the requirements of the relation being local and linear gives $36$ independent components, which $15$ are dissipative and don't contribute to Lagrangian ("skewon") and $1$ that contributes to Lagrangian but doesn't affect light propagation or electromagnetic stress energy (a ghostlike "axion").

For the differential form presentation of electromagnetism that emphasizes the logically independent roles of $F$ and $H$, a good place to start is Hehl and Obukhov's arXiv:physics/0005084, since it works exclusively in $1+3$ decomposition and hence much more clearly corresponds to the more usual presentation of electromagnetism in terms of $(\mathbf{E},\mathbf{B})$ and $(\mathbf{D},\mathbf{H})$. They have also the book on this: Foundations of Classical Electrodynamics, though it's more demanding.

Additionally, MTW's Gravitation has many nice illustrations of what would be $F$ and $H$, although in MTW's presentation they correspond to the "Faraday tensor" and the "Maxwell tensor", respectively, and differ by a conversion factor.

share|improve this answer
    
Thank you for this fantastic answer! It seems from what you've said here that the answer to the title question is that the Hodge dual is not completely necessary, but it is often convenient, and can be a useful way to incorporate additional information, e.g. about the metric. –  ZachMcDargh Mar 23 at 0:07
    
@ZachMcDargh: Yes and no. It's not needed for Maxwell's equations, but the info it incorporates is necessary for EM to be fully predictive--$H\propto\star F$ of standard EM relates $(\mathbf{E},\mathbf{B})$ with $(\mathbf{D},\mathbf{H})$, which is necessary, and as an important side-effect, connects the light cone with the causal structure of spacetime. In H&O's book, they investigate a general $H_{ab}\propto\kappa_{ab}{}^{cd}F_{cd}$ and some of the assumptions on $\kappa$ for which the light cone even exists, and for which which metric could be derived from it, up to some conformal factor. –  Stan Liou Mar 23 at 6:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.