0

For free electromagnetic fields, it is possible to choose a gauge such that the scalar potential $\phi(t,{\bf x})=0$ and the vector potential ${\bf A}(t,{\bf x})$ satisfies the Coulomb gauge condition $\nabla\cdot{\bf A}(t,{\bf x})=0$. If the Fourier amplitudes of the spatial Fourier transform of ${\bf A}(t,{\bf x})$ are denoted by $\tilde{\bf A}_{\bf k}$, the latter condition is equivalent to $${\bf k}\cdot\tilde{\bf A}_{\bf k}=0,~~ \text{for each} ~{\bf k}.\tag{1}$$ Therefore, the $\tilde{\bf A}_{\bf k}$ vector is confined to a plane transverse to ${\bf k}$, and therefore has two independent components. This explains why there are two independent degrees of freedom! But $(1)$ is true for each ${\bf k}$ and there are infinitely many ${\bf k}$ in the Fourier decomposition of ${\bf A}(t,{\bf x})$.

Why doesn't that make the number of degrees of freedom from two to infinitely many?

Qmechanic
  • 201,751
SRS
  • 26,333
  • That boils down to conventions on how to count DOF in field theory. Possible duplicate: https://physics.stackexchange.com/q/185092/2451 – Qmechanic Apr 12 '20 at 16:11
  • @Qmechanic To me, it seems that we have two DOFs if we count for $\tilde{A}_{\bf k}$, but infinitely many if counted for $A(t, {\bf x})$. Both can't be right! I know 2 is the right number but what's wrong/incomplete about my point? – SRS Apr 12 '20 at 16:14

1 Answers1

3

One counts degrees of freedom per space point or at each (spatial) momentum. (This is the only sane definition.) Then there is no discrepancy.