Are Maxwell's equations linearly dependent?

[cianfa72]

TL;DR

About the linearly independence of Maxwell's
PDE equations

HI,
consider the 4 Maxwell's equations in microscopic/vacuum formulation as for example described here Maxwell's equations (in the following one assumes charge density $\rho$ and current density $J$ as assigned -- i.e. they are not unknowns but are given as functions of space and time coordinates).

Two of the equations are scalar (divergence based equations) while the other two give rise to 6 equations in 6 unknowns (curl based equations).

Therefore it seems there are 8 equations in 6 unknowns ($E$ and $B$ field components).

Are the above partial differential equations (PDEs) actually linearly dependent ? Thanks.

 

[anuttarasammyak]

Maxwell equations with EM potential is
$$\square A^\mu - \partial^\mu(\partial_\nu A^\nu)=-\mu_0 j^\mu$$
where $\mu,\nu$=0,1,2,3. Four equations for Four variables. Does this meet your point ?

 

[cianfa72]

Four equations for Four variables. Does this meet your point ?

Ok, from the solution of the above four PDEs one then get the EM fields.

BTW is the EM potential $A^{\mu}$ solution given in a specific gauge ?

 

[cianfa72]

Coming back to the original question, the 4 Maxwell equations give conditions on divergence and curl of fields $E$ and $B$. Perhaps the point is that to uniquely define each field such two conditions are actually necessary and sufficient.

 

[Meir Achuz]

We can use the fact that $\bf B$ has zero divergence to introduce a vector potential $\bf A$, in terms of which $\bf B$ is given by

$$\bf B=\nabla\times A.$$



Now, the $\nabla\times{\bf E}$ equation can be written as

$$\nabla\times\left({\bf E}+\frac{1}{c}\partial_t{\bf A}\right)=0.$$

The vanishing of this curl, permits us to relate $\bf E$ to the vector potential $\bf A$ and
a scalar potential $\phi$ by
$${\bf E}=-\nabla\phi-\frac{1}{c}\partial_t{\bf A}.$$

So $\bf E$ and $\bf B$ are determined by 4 variables.

 

[cianfa72]

What about the other two equations one for the divergence of $E$ and the other for the curl of $B$ ?

 

[Meir Achuz]

They led to the two equations I wrote for B and E.

 

[cianfa72]

Ok, I see. You started from $\nabla \cdot B = 0$ that enable us to introduce the vector potential $A$ such that $B=\nabla\times A$.

Then exchancing $\nabla$ operator with derivative operator $\partial / \partial_t$ you get from $\nabla \times E = - \partial B /\partial_t$

$$\nabla\times\left({\bf E}+\frac{1}{c}\partial_t{\bf A}\right)=0.$$

Vanishing of this curl enable us to write it as the gradient of a smooth function $\phi$ (the scalar potential).

The vanishing of this curl, permits us to relate $\bf E$ to the vector potential $\bf A$ and
a scalar potential $\phi$ by
$${\bf E}=-\nabla\phi-\frac{1}{c}\partial_t{\bf A}.$$

What about the last Maxwell equation for the divergence of $E$, namely $\nabla \cdot E = \rho$ ?

 

[Andy Resnick]

Just a quick note- this whole discussion seems limited to medium-free electromagnetics. If matter is present, 2 additional equations (constitutive relations) are required to, for example, connect E with D and B with H.

 

[Meir Achuz]

D and H do not enter the two homogeneous Maxwell's equations.

 

[Andy Resnick]

D and H do not enter the two homogeneous Maxwell's equations.

To be sure, I'm somewhat confused by the OP's relating 'microscopic' and 'vacuum', since they seem to refer to very different physical situations (Wiki notwithstanding).

If I understand your comment in the context of the OP's use of "microscopic/vacuum formulation", then I could argue that the constitutive relations D = ε₀E and B = μ₀H were snuck in without explanation.

My go-to reference for this stuff is Post's "Formal structure of electromagnetics"- for example, he states that the decomposition of the second order d'Alembertian wave equation into four first-order Maxwell field equations is not unique.

 

[Meir Achuz]

Are you confusing microscopic with macroscopic?
There is no D or H for microscopic.
$\epsilon_0$ and $\mu_0$ are just put there to confuse things.

 

[cianfa72]

Coming back to post #8 I believe the point is as follows: potential vector $A$ and scalar potential $\phi$ are defined so that the homogenous Maxwell's equations are automatically satisfied (it is actually an iff condition).

Then the expressions obtained for $E$ and $B$ using those potentials are replaced into non-homogenous Maxwell's equations (i.e. equations involving the sources $\rho$ and $J$).

By solving those 4 equations for the 3 unknowns components of vector potential + scalar potential, one is actually solving the complete system of 4 Maxwell's equations.

 

[cianfa72]

Are you confusing microscopic with macroscopic?
There is no D or H for microscopic.
$\epsilon_0$ and $\mu_0$ are just put there to confuse things.

Yes, the microscopic Maxwell's equations (as in the OP) do not involve D or H.