# Counting d.o.f. and gauge fixing $A_{\mu}$ and $\psi$ in $D$-dimensions

Setup:

Let us assume we are in $D$-dimensional Minkowski space-time where $D=d+1$. Consider a free Abelian gauge theory. Then the electromagnetic field will satisfy $$\partial_{\mu}F^{\mu \nu}=0 \tag{1}$$ and this means that $\partial_{\mu} \partial_{\nu} F^{\mu \nu}$ is identically zero.

Question:

1. The gauge field $A_{\mu}$ out of which we construct the field strength comprises $D$ independent components (the indices take a value for each dimension). Then, why people say that the field strength comprises $D-1$ independent components? How is this counting being done?

2. And why these $D-1$ independent components are the off-shell degrees of freedom? In what sense?

3. Finally, what happens with spinors? How many independent components has a spinor in $D$-dimensions and how do we find its off-shell degrees of freedom?

The Maxwell Lagrangian for the gauge field $A_\mu$ is given by,

$$\mathcal L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}.$$

The equations of motion are $\partial_\mu F^{\mu\nu} = 0$, supplemented by the Bianchi identity. Taking as an example the case $D=4$, naively one would think we have $4$ degrees of freedom. However, we know photons have only two degrees of freedom, so it must be reduced somehow.

It turns out the component $A_0$ is not dynamical, since we have no kinetic term for it in the Lagrangian. In other words, $A_0$ is fully determined by the other components through,

$$\nabla^2 A_0 + \nabla \cdot \frac{\partial \vec A}{\partial t}=0$$

which has the solution,

$$A_0 (\vec x) = \int d^3 y \frac{1}{4\pi |\vec x - \vec y|}\frac{\partial \vec A (\vec y)}{\partial t}.$$

Just like in general relativity where we have constraints on what initial data surface to specify, we cannot choose our initial $A_0$. So, in general we've gone from $D \to D-1$ independent components. For the case $D=4$, the degrees of freedom are two due to the additional constraint,

$$\nabla \cdot \vec A = 0$$

known as Coulomb gauge. (One should not think of it per se as a constraint, but rather as exploiting a redundancy in the description of the system due to the gauge symmetry.) This bumps us down to two.

• So in 4 dimensions we have 3 independent components and by analogy in $D$-dimensions we have $D-1$ independent components. Why are they called off-shell though? Also, how did you go from the first equation to the second equation? Jan 9, 2015 at 12:38
• @Marion Don't forget you lose another degree of freedom from gauge symmetry to get two in four dimensions. Jan 9, 2015 at 12:39
• @JamalS, you probably should include that in the answer because you mentioned photons having 2 d.o.f in 4D but you only explain how to get rid of one of them :) Jan 9, 2015 at 12:42
• @PhotonicBoom Updated. Jan 9, 2015 at 13:20