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When we go from the field variables $({\vec E},\vec{B})$ to the potentials $(\phi,{\vec A})$, the number of degrees of freedom describing any electromagnetic field is reduced from $6$ ($3$ components of ${\vec E}$+$3$ components of ${\vec B}$) to $4$ ($\phi$ is a scalar+ $3$ components of ${\vec A}$. A Coulomb gauge condition $\nabla\cdot{\vec A}=0$ can further reduce the number of degrees of freedom by at most one because it amounts to the freedom of choosing only one scalar function. This leaves us with 3 degrees of freedom for a general electromagnetic field. In my way of counting, I end up with $3$ degrees of freedom for a general electromagnetic field. What's wrong with this counting of DOFs?

I asked a related question here. Here I am asking what is wrong with my logic?

Solidification
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1 Answers1

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You are missing one gauge condition on the 4-potential.

The Dirac-Bergmann analysis of the Hamiltonian formulation of the EM Field (metric mostly minus, in vacuum or in matter) shows that for the initial 8 Hamiltonian variables $A_{\mu}, \pi_{\mu}$ one has two 1st class constraints:

$$\pi_0 \approx 0 \quad \tag{1}$$ $$-\partial^i \pi_i \approx 0 \quad \tag{2}$$

which can be fixed by imposing two gauge conditions: $$ A_0 \approx 0 \quad \tag{3}$$ $$-\partial^i \pi_i \approx 0 \quad \tag{4}$$

Now we can count the DoF: $(8-4):2 = 2$ (of course, at each space-time point). If you are not familiar with constrained systems (this is usually not a standard material in US university education), then read the accepted answer here Counting Degrees of Freedom in Field Theories? which uses standard textbook material.

DanielC
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