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I am trying to learn a bit about differential forms. I saw a question and answer noting that the homogeneous Maxwell equations can be written as $dF=0$. However, as noted there, depending on the topology of the space, it might be possible for $dF=0$ without $F = dA$. In such a case, the answer by Hyperon noted that one typically still postulates $F=dA$ in order to recover the equations of motion via an action principle.

Suppose I don't postulate $F=dA$, but I keep $dF=0$. The manifold the fields live on will be such that $dF=0$ does not imply $F = dA$. Please note that for this question about solutions to Maxwell's equations, I just want to consider a classical system.

Can I completely summarize the effect of this choice as the following? The solutions to generalized equations "$dF=0$ and $d \star F = J$" can always be written as a solutions to the usual "$F=dA$ and $d \star F = J$" plus a solution to the generalized vacuum equations "$dF=0$ and $d \star F = 0$". That is, I suspect the weakening of $F=dA$ down to $dF=0$ just increases the number of vacuum solutions which can then be added to solutions of the usual non-vacuum Maxwell equations, but I'm unsure how to rule out the potential existence of more exotic solutions to the general equations that cannot be decomposed in this way.


I hope the question is clear. Just to be 100% clear, here is an equivalent reformulation of the question.

A solution to "$dF=0$ and $d \star F = J$" will be called $F_{g, n}$, for generalized, non-homogeneous.

A solution to "$dF=0$ and $d \star F = 0$" will be called $F_{g, h}$, for generalized, homogeneous.

A solution to "$F=dA$ and $d \star F = J$" will be called $F_{u, n}$, for usual, non-homogeneous.

A solution to "$F=dA$ and $d \star F = 0$" will be called $F_{u, h}$, for usual, homogeneous.

Can I always decompose an $F_{g,n}$ into a sum of a $F_{g,h}$ and a $F_{u,n}$?

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2 Answers 2

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It is wrong to postulate $F = dA$, indeed this equation is not true in general.

I think you are confusing a few concepts. It is true that there are certain occasions when $dF = 0 \Leftrightarrow F = dA$ for some globally defined one-form $A$. But then in these situations I don't see why - using your notation - you need the difference between $F_{g,...}$ and $F_{u,...}$ since they are exactly the same as each other. What you write is then trivially true, just solving a differential equation as a sum of homogeneous and non-homogeneous terms.

In more general topologies, $F = dA$ has no solution, so there are no solutions $F_{u,n}$ at all. So then you cannot split $F_{g,n}$ into $F_{g,h}$ and $F_{u,n}$ because you are missing the non-homogeneous terms.


In the remainder of this answer, I will give you some more details on a more complete story when one encounters non-trivial topologies.

From here-on, let $M$ be a manifold with open covering $\{U_{\alpha}\}$ of $M$ by open sets and $\Omega^k(S)$ be the set of $k$-forms defined on a subset $S \subset M$ so that $\Omega^0(S)$ is the set of functions defined on $S$ (often called $C(S)$). Denote the restriction of $M$ to $U_\alpha$ by $r_{\alpha}$.

One starts with the Maxwell equation $dF = 0$, where $F$ is a two-form defined on manifold $M$. The Poincaré lemma tells us that, among other things, given a covering $\{U_\alpha\}$ of $M$ by open sets, there is a family $\{A_\alpha\}$ of locally defined one-forms $A_\alpha \in \Omega^1(U_\alpha)$ with $r_\alpha(F) = dA_\alpha$ on $U_\alpha$.

Note that each $A_{\alpha}$ is only defined up to a gauge transform $A_\alpha \mapsto A_{\alpha} + \Omega_\alpha$ where $\Omega_\alpha$ is a closed one-form, and we are already local so that $\Omega_{\alpha} = d\Lambda_{\alpha}$ for some function $\Lambda_{\alpha} \in \Omega^0(U_{\alpha})$.

There is a rich mathematical theory here. By considering compatibility on overlaps $U_\alpha \cap U_\beta$ (restrictions should commute), one sees that there is a corresponding family $\{\lambda_{\alpha\beta}\}$ of functions defined on $U_\alpha \cap U_\beta$ such that $A_{\alpha}-A_{\beta} = d\lambda_{\alpha\beta}$ on the restriction $U_{\alpha \beta}$. Note that $\lambda_{\alpha\beta} = -\lambda_{\beta\alpha}$. Repeating this compatibility argument on triple intersections easily leads to $d(\lambda_{\alpha \beta} + \lambda_{\beta\gamma} + \lambda_{\gamma \alpha}) = 0$ and hence to $\lambda_{\alpha\beta} + \lambda_{\beta \gamma} + \lambda_{\gamma \alpha} = c_{\alpha\beta\gamma}$ where $\{c_{\alpha\beta\gamma}\}$ are constants. It turns out, and this is a familiar theme with topology in general (and especially Cech cohomology-related things) that quartic overlaps and higher add no new constraints.

Thus we have the equivalence: $$\begin{aligned}dF = 0 \qquad \longleftrightarrow \qquad &\text{families } \{A_{\alpha}\} \text{ modulo local gauge equivalence }A_{\alpha} \sim A_{\alpha} + d\Lambda_\alpha,\\ &\text{transition functions }\{\lambda_{\alpha\beta}\},\\ &\text{`cocycles' } \{c_{\alpha\beta \gamma}\} \end{aligned}$$

One might worry that the right hand side of the above is dependent on the specific local covering $\{U_\alpha\}$. However, it turns out that it is not, thanks to the `modulo gauge equivalence' part. The right hand side forms what is called a principal bundle with respect to the group $U(1)$ or $\mathbb{R}$. In the quantum theory, a subtle new compatibility condition emerges: the cocycles must be integral (up to $2\pi$ factors).

In typical `undergraduate' spaces such as $\mathbb{R}^n$, which are homotopic (read topologically equivalent) to a point, Poincaré's lemma applies globally and so we only need a single $A_{\alpha}$, usually called $A$, and no transition functions or cocycles. But then they often implicitly use spaces such as $\mathbb{R}^n \backslash \{0\}$ and $\mathbb{R}^n - \{\text{line}\}$, which have non-trivial topologies.

Now let's assume we are solving $dF = 0$ and $d\star F = J$, first on $M = \mathbb{R}^n$. Then we can write $F = dA$ for some one-form $A \in \Omega^1(M)$. The second equation is now $d \star d A = J$, which is essentially a Laplacian. Of course, you should solve this in the usual way, as the sum of a homogeneous and non-homogeneous terms. It is impossible to say more without restricting $J$ further and in general this is just a set of coupled differential equations, in practice solved numerically outside of a few textbook cases.

Now let us solve the Maxwell equations on more general $M$. Now we can write $F = dA_{\alpha}$ on each $U_{\alpha}$. Similar to the above we get a Laplacian $d \star d A_{\alpha} = J$ restricted to $U_{\alpha}$. We solve these subject to the gluing conditions of the $A_{\alpha}$ as discussed above. Again it is impossible to say more without further knowledge of $M$ or $J$.

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  • $\begingroup$ +1 Thank you for your answer. I agree that in the special case of a space where $dF = 0 \Leftrightarrow F = dA$, then what I wrote is trivially true. When you write that $F=dA$ for a globally-defined $A$ may have no solutions, could you give an example? When I think of spaces like $\mathbb{S}^2 \times \mathbb{R}$, I think I can still write down nontrivial globally-defined one-forms $A$ and hence can find a set of nonvanishing $F = dA$. I might be misunderstanding. $\endgroup$
    – user196574
    Commented May 6 at 4:18
  • $\begingroup$ Do you know of any example in physics where $F$ isn't $dA$ - that is, EM fields $\mathbf E,\mathbf B$ cannot be globally expressed as $\mathbf E = -\partial_t \mathbf A /c - \nabla \varphi$, $\mathbf B = \nabla \times \mathbf A$? Even in tokamak, whose vacuum chamber space is a torus, we can just ignore this topological feature of the vacuum chamber, and treat EM fields there as living in simple euclidean space, because walls only stop the plasma from penetrating them, not EM fields. $\endgroup$ Commented May 7 at 21:49
  • $\begingroup$ @JánLalinský If I find time tonight I will update my answer to address both these comments. $\endgroup$ Commented May 8 at 10:30
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The answer is yes, you can always decompose $F_{gn}$ as a sum of $F_{gh}$ and $F_{un}$ by linearity.

Depending on the topology of the domain, $dF=0$ and $F=dA$ are actually equivalent. The former means that $F$ is closed and the latter that $F$ is exact. By definition, you have the equivalence between the two iff the second Betti number is zero. For example, this is the case for a convex space, so it applies for usual EM defined on the entire space time $\mathbb R^4$. An example where it is not the case is to consider $\mathbb R^4$ minus a worldline. There are some fields $F$ that are not exact but closed.

For example consider $F$ to be the field of a magnetic monopole traveling on the excluded worldline. Note that since most domain that you consider is locally convex, you can construct $A$ locally, but you’ll encounter obstructions when trying to define it globally.

More explicitly say that you are in an inertial frame and the worldline is $x=y=z=0$ (the monopole is stationary). The field of a unit charge magnetic monopole is given in spherical coordinates: $$ F = \frac1{4\pi}\sin\theta d\theta\wedge d\phi $$ However, if $F = dA$, then the integral over any closed surface gives zero. This would a fortiori hold for any sphere about the worldline (like the sphere $t=0,x^2+y^2+z^2=1$), which is not the case here, you'd rather get $1$.

For the variational principle, the usual approach is to vary $A$ and you’ll directly get Maxwell’s equation. If you want a variational purely in terms of $F$ so that it is closed, the usual approach is to enforce the constraint by Lagrange multipliers.

Hope this helps

Answer to comment

The simple trick is to enforce the eight Maxwell's equations by 8 corresponding Lagrange multipliers which you can regroup 2 auxiliary vector fields. Maxwell's equations are: $$ \begin{align} \partial_\rho F_{\mu\nu}^\star &= 0 & \partial_\mu F^{\mu\nu} &= J^\nu \end{align} $$ and you can get them from the Lagrangian density: $$ \mathcal L = -\frac14 F^{\mu\nu}F_{\mu\nu}-A_\mu J^\mu $$ by varying $A$ or from: $$ \mathcal L = F_{\mu\nu}\epsilon^{\mu\nu\rho\sigma}(\partial_\rho B_\sigma-\partial_\sigma B_\rho)+F^{\mu\nu}(\partial_\mu C_\nu-\partial_\mu C_\nu)+C_\mu J^\mu $$ by varying $F$ and the two auxiliary fields $B,C$. Note that it's not much of an improvement at this stage, but it can be useful as an intermediate step for Hubbard-Stratonovitch transformations.

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  • $\begingroup$ +1 Thank you for the answer. I'm currently trying to square yours and SvenForkbeard's answers. I hadn't realized there was a variational principle purely in terms of $F$, could you write a little more on that or point me to a reference? I can also ask that as a separate question if you'd prefer. $\endgroup$
    – user196574
    Commented May 6 at 4:03

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