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In Maxwell theory we have dual description in terms of dual fields:

$$ \tilde{F}_{\mu\nu} = \partial_\mu \tilde{A}_\nu - \partial_\nu \tilde{A}_\mu = \varepsilon_{\mu\nu\rho\sigma} F^{\rho\sigma} $$ $$ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu $$

How to solve this equation for $\tilde{A}_\mu$ in terms of $A_\mu$?


I have few ideas about solutions: $$ \tilde{A}_\mu \propto \varepsilon_{\mu\nu\rho\sigma}x^\nu \partial^\rho A^\sigma $$

Or something like this: $$ \tilde{A}_\mu = \int^x \varepsilon_{\mu\nu\rho\sigma}dx^\nu \partial^\rho A^\sigma $$

It also looks like that first solution is local, second solution is non-local.


Electro-magnetic duality:

Maxwell equations: \begin{equation} \begin{cases} \partial_\mu F^{\mu\nu} = 0 \\ \partial_\mu \tilde{F}^{\mu\nu} = 0 \end{cases} \end{equation} are invariant under:

$$F^{\mu\nu} \leftrightarrow \tilde{F}^{\mu\nu}$$

So we can formulate theory using dual e-m potential.

Or more formally using path integral:

$$ \begin{equation} \int DA \;e^{-\frac{1}{4e^2} F^{\mu\nu}F_{\mu\nu}} = \int DF D\tilde{A} \;e^{-\frac{1}{4e^2} F^{\mu\nu}F_{\mu\nu} + i\tilde{A}_\mu \varepsilon^{\mu\nu\rho\sigma}\partial_\nu F_{\rho\sigma}} = \int D\tilde{A} \; e^{-\frac{e^2}{4} \tilde{F}^{\mu\nu}\tilde{F}_{\mu\nu}} \end{equation}$$

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3 Answers 3

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Performing a Fourier transform, you find:

$2 \epsilon_{\mu\nu\rho\sigma}p^\rho \tilde{A}^\sigma(p) = p_\mu A_\nu(p) - p_\nu A_\mu(p)~.$

You can view each side of the equation as an operator that maps a vector to an antisymmetric tensor, acting on the vector $\tilde{A}$ on the left and on the vector $A$ on the right. Explicitly, upon defining

$\tilde{M}_{\mu\nu}^{~~~~\sigma}(p) = 2\epsilon_{\mu\nu\rho}^{~~~~~~~\sigma}p^\rho~,\\ M_{\mu\nu}^{~~~~\lambda}(p) = p_\mu \delta_ \nu^\lambda -p_\nu \delta_\mu^\lambda~,$

your equation becomes

$ \tilde{M}_{\mu\nu}^{~~~~\sigma}(p) \tilde{A}_\sigma(p) = M_{\mu\nu}^{~~~~\lambda}(p) A_\lambda(p)~.$

Now your problem becomes that of inverting either one of these two operators, say $\tilde{M}$ for definiteness (they are related to each other by contraction with the epsilon tensor, so it does not make a big difference which one you pick). Unfortunately they are not invertible, because you can easily check that the vector $p^\mu$ is mapped to the vanishing anti-symmetric tensor. That is not so surprising, it is just equivalent to the statement that the field strengths are invariant under gauge tranformations, but still it means that strictly speaking you cannot invert these operators and find $\tilde{A}$ in terms of $A$ (or viceversa). At most you can find a solution up to a gauge transformation, i.e. you can find $\tilde{A}_\mu$ up to a shift of the form $p_\mu C(p)$ where $C(p)$ is an arbitrary function of $p$.

Once you accept that this is the most you can do, you can easily check that the operator

$\tilde{N}_\omega^{~~\mu\nu}(p)\,= \,b \,\frac{1}{p^2}\epsilon_\omega^{~~\mu\nu\tau}p_\tau$

-- with an appropriate choice of the numerical coefficient $b$ that I will not try to fix but can be fixed with a straightforward calculation -- accomplishes the task of almost-inverting $\tilde{M}(p)$, i.e. there exists a $C(p)$ such that

$\tilde{N}_\omega^{~~\mu\nu}(p)\tilde{M}_{\mu\nu}^{~~~~\sigma}(p) \tilde{A}_\sigma(p) = \tilde{A}_\omega(p) + p_\omega C(p)~.$

(To fix the numerical constant $b$ you have to use the identity that relates the contraction of two epsilon tensors to sum of a bunch of products of Kronecher delta's.)

So the best possible answer to your question is --in momentum space--

$\tilde{A}_\omega(p) = \tilde{N}_\omega^{~~\mu\nu}(p) M_{\mu\nu}^{~~~~\lambda}(p) A_\lambda(p) - p_\omega C(p)~.$

Note that this relation is NOT polynomial in momentum, because the operator $\tilde{N}$ contains a $1/p^2$, and therefore is it not local. If you express it in position space it contains something like the inverse of a Laplacian (or D'Alembertian if you are in Lorentzian signature). There does not exist any local relation between a gauge field and its EM dual, basically because you need to invert a derivative to get one from the other.

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  • $\begingroup$ Thank you! Could you give some references? $\endgroup$
    – Nikita
    Apr 23, 2020 at 21:05
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    $\begingroup$ Unfortunately I do not know of a reference where this question is discussed in the language that I used in the answer. Regarding the identity with epsilon tensors you can for instance see it here: demonstrations.wolfram.com/… though you need to pay a little attention to the overall sign if you want to work in Lorentzian signature, because this identity is in Euclidean. $\endgroup$ Apr 24, 2020 at 7:29
  • $\begingroup$ I know such, identity:), thank you. Could you also comment, why first solution from my answer is'not correct? Direct chec confirm, that this is solution of equation. $\endgroup$
    – Nikita
    Apr 24, 2020 at 8:25
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    $\begingroup$ That formula is not covariant under translations, if you conjugate $A$ and $\tilde{A}$ by a unitary that implements a translation $x \to x+a$ the coefficient $x$ that you put on the right-hand side will stay the same. In my approach I implicitly imposed covariance under translations by restricting everything to a sector with definite momentum $p$, and solving within that sector. I could have found more solutions if I had allowed $\tilde{A}(p)$ to be written as a sum (integral) of many $A(p')$ for different $p'$. In fact your formula, written in momentum space, would involve such an integral. $\endgroup$ Apr 24, 2020 at 17:22
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$\newcommand{\d}{\mathrm{d}}$ The dual field is $\tilde{A}$ where $$ \d \tilde{A} = \star\; \d A. \tag{1}$$ $\tilde A$ is non-local in terms of $A$. A closed formula obtaining $\tilde A$ from $A$ is not known (to my knowledge at least). In spirit it should look more like your second relation, although not quite because of the contractions of indices hidden in the exterior derivatives.

One way to come closer to obtaining $\tilde{A}$ from $A$ would be to take equation (1) and integrate it over a manifold $M$ whose boundary is just a single point, $\cal P$. Then \begin{align} \int_M \d \tilde A &= \int_M \star\; \d A \\ \int_{\partial M} \tilde A &= \int_M \star\; \d A \\[5mm] \tilde A (\cal P) &= \int_M \star\; \d A. \end{align} Then you should probably imagine dragging $\cal P$ around to obtain $\tilde A$ everywhere in space, but I don't know how to do it in maths :(

In most cases (not in all cases of course) you don't quite care about what $\tilde A$ is itself because it appears either as $\d\tilde A$, which you know, or coupled like $\tilde A\wedge\d[\text{something}]$ so you can exchange this for $\d\tilde A \wedge[\text{something}]$.

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  • $\begingroup$ Thank you for answer! But if I will formally check "solutions" from my question, I will obtain that they satisfy equations. What is wrong with this? $\endgroup$
    – Nikita
    Apr 22, 2020 at 13:22
  • $\begingroup$ I don't understand what you mean in this comment at all. $\endgroup$ Apr 22, 2020 at 15:18
  • $\begingroup$ Why solutions from answer is not true? $\endgroup$
    – Nikita
    Apr 22, 2020 at 16:31
  • $\begingroup$ The first one stands no chance. That's not how you integrate something. And the second one is close to correct but with the subtlety that I explain in my answer. $\endgroup$ Apr 22, 2020 at 19:41
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Dual in this context is the Hodge dual, as far as I understand.

The dual of a rank-2 tensor in 4d space is also a rank-2 tensor (hence the similarity between $F$ and $\tilde{F}$). The hodge dual of a vector (rank-1 tensor) is rank-3 tensor. You can start with

$\mathbf{A}=A_\mu dx^\mu$

and then find

$\star \mathbf{A} = \tilde{A}_{\nu\zeta\xi}\: dx^\nu dx^\zeta\ dx^\xi=\epsilon_{\mu\nu\zeta\xi}g^{\mu\kappa}A_\kappa dx^\nu dx^\zeta\ dx^\xi$

Where $g^{\mu\kappa}$ is the inverse metric.

I am not sure if it will help you in any way. I suppose if the metric is trivial, the dual $\tilde{A}$ is linearly dependent on $A$. But this is only true with trivial metric. So is this a worthwhile thing to do?

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    $\begingroup$ No. Dual in contex of em duality. You answer is incorrect. $\endgroup$
    – Nikita
    Apr 22, 2020 at 0:11
  • $\begingroup$ @Nikita. How do you define this em duality? Wiki seems to support my point The Faraday tensor's Hodge dual is .... The equation given in Wiki is only valid for trivial metric, but Hodge dual can be defined for any reasonable metric. I don't currently have access to any proper books, but my recollection is that MTW also says something similar $\endgroup$
    – Cryo
    Apr 22, 2020 at 0:25
  • $\begingroup$ I updated question.. $\endgroup$
    – Nikita
    Apr 22, 2020 at 0:50
  • $\begingroup$ @Nikita. I am not an expert on path integrals, perhaps I am missing something there. But I still don't get what you mean by $\tilde{A}$. Your equation 3, ties it to some specific origin (also note that $x^\mu$ is in general not a vector - only works in some coordinate systems), equation 4 seems to tie it to a specific path. Seems like a bad idea. IMHO You will be on firmer ground if you go with differential forms, so then Maxwells Equations are $d\mathbf{F}=\mathbf{0}$ and $d\star\mathbf{F}=\mathbf{J}$ (note the Hodge dual) $\endgroup$
    – Cryo
    Apr 22, 2020 at 1:05
  • $\begingroup$ Having said that, Poincare Lemma, in 4d form, looks like what you are talking about, and it should allow you to go from electromagnetic tensor (2-form) to electromagnetic potential (1-form). Combine it with Hodge dual, and it may make sense, in my humble opinion $\endgroup$
    – Cryo
    Apr 22, 2020 at 1:12

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