2

For point-like particles, the term "degree of freedom" seems rather clear: It's the number of independent coordinate functions $q_i(t)$ that we need to specify to completely describe the system at all points in time. This number coincides with the dimension of the configuration-space. So for $N$ pointlike particles, there are $3N$ degrees of freedom.

Now for the electric and magnetic field, I need (at any point in time) 6 components for each of uncountably infinite positions $\vec{x}$. Obviously it is not meaningful to talk about infinite degrees of freedom, so my first guess would be

  1. to call the degrees of freedom of the electromagnetic field the number of independent functions of $t$ and $\vec{x}$ that one needs to fully specify the system (without using the maxwell equations).

Is that the right way to think about it? If so, why can we reduce the DoF from 6 to 4, making use of the potentials $\vec{A}, \Phi$? To do so, we already used Maxwell's equations. And reducing the number even more by choosing appropriate gauges (like the temporal gauge, $\Phi = 0$ also employs the gauge freedom that Maxwell's equations have given us.

So to make is short, I ask for a rigorous definition of the terminology "degree of freedom" along with the information which equations are allowed to play a role here, and which don't. Especially, introductory texts to QED mention that the field actually only has two degrees of freedom, which I can see for one classical solution with wave vector $\vec{k}$, but not in general.

Qmechanic
  • 201,751
Quantumwhisp
  • 6,733
  • 2
  • 19
  • 50

1 Answers1

2

Dynamically, more natural notion is phase space, which is the space where the evolution is first order in time. Intuitively, it's all the information you need to make the system evolve. Configuration space then arises when phase space has additional structure. Typically, when it is the cotangent bundle of a manifold, configuration space is naturally identified as the manifold in question. In a lot of cases, it makes sense to view configuration space as having half the dimensions of phase space. Intuitively, this works if you can define a Lagrangian/Hamiltonian, or more abstractly a symplectic structure (the fact that your phase space has even dimensions is already a good indicator). You cannot define configuration space without considering the dynamics as well.

For electromagnetism in vacuum, there are two degrees of freedom ($2$ polarisation states). And there are many ways to see this.

You can get this directly from Maxwell's equations. You start with $6$ components of the EM fields. You get rid of $2$ due to the "static" Maxwell equations: $$\nabla \cdot E = 0\\ \nabla \cdot B = 0$$ so you have $4$ components. Since the equations are of first order in time derivative, this means that $4$ actually corresponds to the dimension of phase space, so halving it, you obtain the $2$ dof's.

From the potentials, you can get the same results. The idea is that Maxwell's equations can be derived from the Lagrangian density: $$ \mathcal L = \frac{1}{2}(E^2-B^2) \\ = \frac{1}{2}\left((\nabla V+\partial_t A)^2-(\nabla\times A)^2\right) \\ $$ This time, it looks as if there are $4$ dof's. However, the time derivative of $V$ does not appear, so it is not a dynamical variable, but rather enforces a static constraint (a Lagrange multiplier). So you only have three dof's. Finally, adding gauge invariance, you lose an additional dof, so you obtain $2$ as well. Note that you can similarly get the same result by specifying the gauge (Lorenz, Coulomb...).

If you are familiar with general relativity, you can similarly prove that the Einstein's equations actually leave $2$ dof's even if you seem to have $10$ dof's for the metric tensor (with the added subtlety that there are more constraint equations and more gauge transformations).

Hope this helps.

hft
  • 19,536
LPZ
  • 11,449
  • 1
  • 4
  • 24
  • If there are sources (currents, charges) present, do I have more degrees of freedom? (This https://physics.stackexchange.com/a/408253/102243 answer states that there are 2 degrees of freedom in general, not only for the free theory). – Quantumwhisp Nov 08 '22 at 09:28
  • It depends. If the sources are independent of the field the they simply add the linear term $$\mathcal L_{source} = -j\cdot A+\rho V$$ so you still have $2$ dof's. More generally, they could depend on the field but would need to preserve gauge invariance and not involve higher time derivatives up to a gauge transformation. – LPZ Nov 08 '22 at 12:09