Can you give me an intuitive, physical understanding of a "dual field"? For example, the Hodge dual of the gluon field strength matrix $F$ is $\tilde{F}_{\mu \nu}=\epsilon_{\mu \nu \alpha \beta} F^{\alpha \beta}$. What is its physical significance? Why for example does it appear in the strong CP problem where we have the $\theta$-parameter in the Lagrangian $L_\theta = \theta F \tilde{F}$?
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3$\begingroup$ I'm not sure what you mean by "intuitive, physical understanding". What, for example, is your "intuitive, physical understanding" of $F$ itself? Also, whether the dual "appears" in something or not is a matter of how you write it. The usual kinetic term for a gauge field, $F_{\mu\nu}F^{\mu\nu}$ is from some viewpoint more naturally written as $F\wedge\star F$, so the Hodge dual also appears in the kinetic term, while the CP violating term is more naturally written $\theta(F\wedge F)$ without a dual. So the question "why does it appear" is not well-defined. $\endgroup$– ACuriousMind ♦Commented Oct 2, 2015 at 13:55
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$\begingroup$ If you want a more intuitive understanding of form fields themselves (and Hodge duals), the standard "picture" is presented in Misner, Thorne and Wheeler's Gravitation. $\endgroup$– Evan RuleCommented Oct 2, 2015 at 14:01
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$\begingroup$ related: physics.stackexchange.com/q/744511/226902 and physics.stackexchange.com/q/647435/226902 $\endgroup$– QuilloCommented Jan 9, 2023 at 16:14
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I'm going to talk about the Hodge dual in electromagnetism to give you a physical intuition.
In electromagnetism the appearance of the Hodge dual field is a mathematical manifestation of electromagnetic duality. Roughly speaking, if some divine being swapped the electric and magnetic fields, fundamental physics wouldn't change. Of course, this statement is only correct if you believe in magnetic sources which have been hypothesised but never observed.
You can get more of an intuition for the action of the Hodge dual by rewriting the field strength $F$ in terms of more familiar $\mathbf{E}$ and $\mathbf{B}$ fields. Explicitly the definition is
$$F^{0i}=-E^i \quad F^{ij}=\epsilon^{ijk}B_k$$
It is then an easy exercise to check that applying the Hodge star is equivalent to the map
$$(\mathbf{E},\mathbf{B})\to (\mathbf{B},-\mathbf{E})$$
Clearly electromagnetic duality is a symmetry of the QED Lagrangian
$$\mathcal{L}\sim F \tilde F \sim F \wedge \star F$$
Indeed applying the Hodge dual twice gets you back to (minus) the identity, hence the map $F \to \star F$ leaves $\mathcal{L}$ invariant (up to sign).
I'll leave it to you to check that the duality holds at the level of Maxwell's equations (with sources). If you're stuck, take a look at the notes I linked above.
Finally note that the Hodge dual has (in principle) nothing to do with the $\theta$ parameter! You can write down Yang-Mills theory without considering $\theta$ and you'd still get a $\tilde F$ in the Lagrangian. The $\theta$ is in fact a manifestation of the topological structure of solutions to the Yang-Mills equations - a so-called instanton effect. This is an advanced topic, which deserves it's own question!
Note: you can get analogues of electromagnetic duality in Yang-Mills theories, but typically you require supersymmetry for this. In that case, the duality is referred to as S-duality, and does involve changing the $\theta$ parameter in general. If you're feeling brave, head over to Wikipedia and follow some references.
answered Oct 2, 2015 at 14:42