Finding the number of independent degree of freedom of the Electromagnetic field We know that there are only two independent degree of freedom, every point, in an Electromagnetic field. There seems to be two inequivalent ways to arrive at this conclusion.

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*We just declare that the electromagentic field is described by a guage field $A_\mu$. This means that the physical degree of freedom are actually the equivalence classes under the equlivalence relation $A_\mu \sim A_\mu +\partial_\mu\alpha$. To extract out the physical content, we can define the tensor $F_{\mu \nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. It has six independent components: $E_i=F_{0i}$ and $B_i=\epsilon_{ijk}F^{jk}$ and clearly depend only on the equivalence classes. Using the guage freedom we can eliminate, say, the $A_0$ component. However, there is still a residual guage freedom: $A_\mu \sim A_\mu +\partial_\mu\lambda$ where the function $\lambda$ is independent of time. Now, we note that $A_i$ (the space components) can be written as $A_i=A_i^T +A_i^L$ where $A_i^T= A_i-\partial_i(\Box^{-1}\partial_jA^j)$ and $A_i^L=\partial_i(\Box^{-1}\partial_jA^j)$ with $\Box^{-1}$ is the inverse of the Laplace operator. The component $A_i^L$ is pure divergence and lies in the equivalence class of $A_\mu=0$ under the residual guage equivalence. This shows that $A_{\mu}$ actually has two degree of freedom. In this argument we never used any dynamics (that is Lagrangian/Hamiltonian).



*As I understand, there is another way to get the same result. We start with the Maxwell action $S=-\frac{1}{4}\int d^4 x F_{\mu \nu}F^{\mu \nu}$. As usual, we notice that the canonically conjugate momentum of the $A_0$ is identically zero. So, using Hamilton's equation of motion we find that $\dot{A_0}=0$ and the constraint $\partial_iE_i=0$. The constriant elimniates one degree of freedom and the guage symmetry generated by the constraint eliminates one more. So again we get two degree of freedom.

It seems strange to me that in the first case we got the two degree of freedom just by using the guage principle, while in the second case we do seem to make use of a dynamical principle. If someone can suggest a way to relate these two approaches, it will be very helpful.
Edit: I think I found a flaw in my argument in the case-I. After fixing $A_0 =0$ only time indepndent guage transformations are allowed, as I mentioned. This means that in the expression of $A^L_i = \partial_i(\Box^{-1}\partial_j A^j)$, the function $\lambda=\Box^{-1}\partial_j A^j$ should be time independent. This cannot happen unless $\partial_iE_i=0$ which is basically the EOM for the $A_0$, using the standard Maxwell Lagrangian. I think this is how the EOM enters. This also clarifies my other related doubt, that, why the EM guage field is said to have three off shell degree of freedom. The guage freedom can only eliminate one d.o.f if we do not impose EOM.
 A: The crucial point here is the first sentence of your first case: You can't "just declare" that $A_\mu$ is a gauge field! You can declare that it is a vector field, but saying that it is "gauge" is already a statement about dynamics.
In terms of formal mechanics, we have to do something like Lagrangian or Hamiltonian mechanics, with generalized fields/coordinates, an action formalism and so on and so forth. This notion of formal mechanics doesn't have a notion of "a gauge field" - a field is just a space(time)-dependent dynamical variable, and apart from its target space (what the field is "valued in") the only physical properties such a field has arise from the action and the laws of mechanics.
So, yes, sure, if you already know that $A_\mu$ is a gauge field and that hence equivalence classes $A \sim A+\mathrm{d}\lambda$ should be viewed as physically equivalent, you can figure out the d.o.f. associated with this field. The point is, you can't know that without deriving that statement from your dynamics, in this case by looking at the action $S[A] = \int F\wedge {\star F}$ with $F= \mathrm{d}A$ and observing that this has gauge symmetry - either the Hamiltonian way as in your second case or the Lagrangian way by observing that the solutions to the equations of motion are underdetermined and the set of all valid solutions to a given initial condition are related by the gauge transformations $A\mapsto A+\mathrm{d}\lambda$.
That this must be so - that being a "gauge field" must involve the dynamics - can also be seen from another angle: The gauge symmetry, after all, has to be a local symmetry of the action (that's what having a symmetry means!), and so it is impossible to talk about a physical field being "gauge" without specifying its dynamics via an action - a theory that is not invariant under the potential gauge transformations of the field is not a gauge theory, and it would be inconsistent to try to declare the field as a "gauge field" in such a theory whose dynamics are not invariant under the gauge transformations.
