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In deriving the Electromagnetic wave equation in free space we remove all charge sources. The resultant Maxwell vector equations are thus source-free. Using Gaussian units with the speed of light $c=1$, these equations are written as: \begin{align} \nabla \cdot \mathbf{E} &= 0\\ \nabla \cdot \mathbf{B} &= 0\\ \nabla \times \mathbf{E} &= -\partial_t\,\mathbf{B}\\ \nabla \times \mathbf{B} &= \partial_t\,\mathbf{E} \end{align} Also, when introducing the vector potental $\mathbf{A}$ into the mix, it is noticed that since the divergence of $\mathbf{B}$ is zero (i.e. $\nabla \cdot \mathbf{B} = 0$) then $\mathbf{B}$ can be represented in the following way: $$ \mathbf{B}= \nabla \times \mathbf{A} $$

However, in free-space we also have the divergence of $\mathbf{E}$ as zero so why not also have another vector, I will call $\mathbf{G}$ that is gives $\mathbf{E}$ in a similar equation: $$ \mathbf{E}= \nabla \times \mathbf{G} $$

Does such a vector $\mathbf{G}$ have any meaning or usefulness? Does it have a name. Also, if $\,\mathbf{G}$ does not exist, is there a mathematical or physical reason?

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  • $\begingroup$ @Thormund Why would it over determine the system? I suppose it'd under determine the system if you don't cancel the additional gauge redundancies with appropriate gauge "symmetry" identifications. $\endgroup$
    – user87745
    May 14, 2020 at 19:28
  • $\begingroup$ @Javier The point isn't to solve the equations I suppose, the point is to understand different possible gauge structures in which one can formulate a theory of the same physical degrees of freedom, as implied by the tag "duality". $\endgroup$
    – user87745
    May 14, 2020 at 19:34
  • $\begingroup$ see en.wikipedia.org/wiki/Hertz_vector $\endgroup$
    – hyportnex
    May 14, 2020 at 20:23

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Well, I wonder whether it is because introducing two new vector fields rather than one one vector and one scalar, with an arbitrary scalar relationship between them is not really a simplification. $$ \nabla \times \vec{B} = \partial_t (\nabla \times \vec{G})$$ means that $$ \vec{B} = \partial_t \vec{G} + \nabla \psi,$$ where $\psi$ is another arbitrary scalar field.

So now you have replaced two vector fields ($\vec{E}$ and $\vec{B}$) with two vector fields ($\vec{A}$ and $\vec{G}$), which isn't a simplification; or with one vector and one scalar field ($\vec{G}$ and $\psi$, which is no different to the original potential definitions except that it only applies in charge-free regions.

You could of course switch which field is defined as the curl of a vector potential, but one chooses $\vec{B}$ because it applies in all circumstances, rather than the restricted cases that it applies to $\vec{E}$.

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I'll re-write the Maxwell equations so that I can refer to them using equation numbers. \begin{align} \nabla \cdot \mathbf{E} &= 0\tag{1}\\ \nabla \cdot \mathbf{B} &= 0\tag{2}\\ \nabla \times \mathbf{E} &= -\partial_t\,\mathbf{B}\tag{3}\\ \nabla \times \mathbf{B} &= \partial_t\,\mathbf{E}\tag{4} \end{align}


Now, in the vacuum, there are as such two ways in which you can formulate a potential description of the theory.

  • The first formulation is the usual one, where you use Equation $(\text{2})$ to define a vector potential by the relation $$\mathbf{B}=\nabla \times \mathbf{A}\tag{5}$$ and you use Equation $(\text{3})$ to define a scalar potential by the relation $$\mathbf{E}=-\partial_t\mathbf{A}-\nabla\phi\tag{6}$$ However, the physical fields $(\mathbf{E},\mathbf{B})$ remain invariant under the transformations \begin{align} \mathbf{A}&\to\mathbf{A}+\nabla\chi\tag{7a}\\ \phi&\to-\partial_t\chi\tag{7b} \end{align} So, we have to postulate that all potentials related by such transformations are physically identical. In addition, the Maxwell equations read \begin{align} \nabla^2\phi&=-\partial_t(\nabla\cdot\mathbf{A})\tag{7c}\\ \nabla\big(\nabla\cdot\mathbf{A}+\partial_t\phi\big)&=(\nabla^2-\partial_t^2)\mathbf{A}\tag{7d} \end{align}

  • Now, in the vacuum, you can do an equivalent procedure with defining a vector potential via the relation $$\mathbf{E}=\nabla\times\mathbf{G}\tag{8}$$ owing to Equation $(\text{1})$ and then use Equation $(\text{4})$ to define a scalar potential by the relation $$\mathbf{B}=\partial_t\mathbf{G}+\nabla\psi\tag{9}$$ Here, the physical fields $(\mathbf{E},\mathbf{B})$ remain invariant under the transformations \begin{align} \mathbf{G}&\to\mathbf{G}+\nabla\xi\tag{10a}\\ \psi&\to-\partial_t\xi\tag{10b} \end{align} So, we have to postulate that all potentials related by such transformations are physically identical. In addition, the Maxwell equations impose \begin{align} \nabla^2\psi&=-\partial_t(\nabla\cdot\mathbf{G})\tag{10c}\\ \nabla\big(\nabla\cdot\mathbf{G}+\partial_t\psi\big)&=(\nabla^2-\partial_t^2)\mathbf{G}\tag{10d} \end{align}


Notice that we have defined our potentials in such a way that the gauge transformations in both the cases are identical including the signs and they follow the same set of equations (compare Equations $(\text{7})\text{ and }(\text{10})$). This means that a solution of $(\mathbf{A},\phi)$ will also be a solution of $(\mathbf{G},\psi)$. Let's say this solution is some $(\mathbf{V},\upsilon)$. Then we get that from the first formulation we have \begin{align} \mathbf{B}&=\nabla\times\mathbf{V}\equiv\mathbf{F}_1\tag{11a}\\ \mathbf{E}&=-\partial_t\mathbf{V}-\nabla\upsilon\equiv\mathbf{F}_2\tag{11b} \end{align} and from the second formulation we have \begin{align} \mathbf{E}&=\nabla\times\mathbf{V}=\mathbf{F}_1\tag{12a}\\ \mathbf{B}&=\partial_t\mathbf{V}+\nabla\upsilon=-\mathbf{F}_2\tag{12b} \end{align}

Equations $(\text{11})\text{ and }(\text{12})$ tell us that if $(\mathbf{B},\mathbf{E})=(\mathbf{F}_1,\mathbf{F}_2)$ constitutes a solution to the Maxwell equations then so will $(\mathbf{B},\mathbf{E})=(-\mathbf{F}_2,\mathbf{F}_1)$. This is simply an illustration of the famous $(\mathbf{E},\mathbf{B})\leftrightarrow(-\mathbf{B},\mathbf{E})$ duality of the Maxwell equations.

For a nice description of the duality symmetry of the Maxwell equations, see: Symmetry of Maxwell equations for electric-magnetic duality

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    $\begingroup$ Looks interesting. I need some time to dip into your post but unfortunately that will not be for a few days. I'll be back. $\endgroup$
    – K7PEH
    May 16, 2020 at 2:11

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