Number of independent components of the electromagnetic field I'm taking a course on electrodynamics and I'm confused when we start talking about potentials. The electromagnetic field seems to have 6 independent components to me. It's described by six dynamical equations (and 2 non-dynamical constraint equations). The EM field tensor has 6 independent components as well.
However, when describing the field in terms of the EM four-potential, the EM field seems to be completely determined by four independent components. What am I missing here?
 A: The answer is that the electromagnetic field has two degrees of freedom, corresponding to the two polarizations states of electromagnetic waves. There are multiple methods to derive this using different sets of variables (which all agree). It is important to choose a method and set of variables and work with them consistently, because the calculation (although not the answer) is different with different variables and different methods.
If you work with the field variables, then you start with 6 components $E_i$ and $B_i$ for $i=\{x, y, z\}$. Then as you noted, two of Maxwell's equations are constraint equations. When we choose initial data, we must choose initial data consistent with these constraints. Therefore, we are only free to choose 4 of the original 6 components of $E$ and $B$ independently. Since Maxwell's equations are first order in time (in terms of $E$ and $B$, we effectively have 4 "first order" degrees of freedom. In classical mechanics, this is analogous to giving the position and momentum for two particles. Sometimes it is more intuitive to think in terms of a "second order" formulation. In classical mechanics, this is analogous to saying we need to specify the initial position and velocity of two particles. Then there are two second order degrees of freedom, which correspond to the 2 polarizations of electromagnetic waves.
If you work with potential variables, you start with 4 fields, the scalar potential $\phi$ and the three components of the gauge potential $A_i$. Because of the gauge symmetry, you can do the counting of degrees of freedom in different ways, although you will always get that there are two physical degrees of freedom. One method is to use the Dirac procedure. The first step is to perform a Legendre transformation to compute the conjugate momenta for each field and the Hamiltonian. If you do this, you find that the three $A_i$ are dynamical (have a non-zero conjugate momentum), and the scalar potential $\phi$ is a Lagrange multiplier for a first class constraint. Therefore, the phase space is naively 6 dimensional (3 $A_i$ and 3 conjugate momenta). The constraint itself removes one of these degrees of freedom, and the associated gauge symmetry removes another. Thus, after applying the constraints and accounting for gauge symmetry, the phase space has 4 (first-order) degrees of freedom. This can also be thought of as 2 second-order degrees of freedom, which are the two polarization states of electromagnetic waves.
A: I'm not really sure what this question is asking that is different from what the other two asked, but possibly the following is what you want. Consider the following, simpler example. In Newtonian mechanics, the gravitational field is the gradient of the potential. So at a point, there are 3 degrees of freedom, because the field is a vector. But if you want to choose how the field varies in a neighborhood, you can't make the variation be however you like. E.g., you can't give it a nonzero curl. Therefore the 3 d.f. at nearby points are in some sense not independent of each other.
A: This is a poorly defined problem. In a given point the four potential represents four dofs, as it can take any value. The fields can also take any value, which means six more dofs. The gradients of the fields etc. etc.
On the other hand, given the behaviour of a charge in its entire light cone, its field has no degrees of freedom at all.
A: When you get the E and B components from the four potential you take the gradient of phi and time derivative of A and the curl of A. Linear dependence/independence may not be preserved across derivatives. This might be the reason why you get additional linear independences in the EM field.
Example:
Consider the function f(x) = x^2.
Now take a derivative of f w.r.t. x and x^2.
You get 2x and 1 respectively.
Before the derivative you had one linearly independent component, but after the derivative you have two.
