# I Counting degrees of freedom in field theory

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1. Jan 19, 2017

### Frank Castle

I'm having a bit of trouble with counting the number of physical ("propagating") degrees of freedom (dof) in field theories. In particular I've been looking at general relativity (GR) and classical electromagnetism (EM).

Starting with EM:

Naively, given the 4-potential $A^{\mu}$ has four components one would think there are 4 dof in EM. However, it is known that the photon only has 2 dof, its polarisation states (how is this known? Is it from empirical data, or can it be shown mathematically? Does it follow from Maxwell's equations, i.e. one can only satisfy Maxwell's equations in a consistent manner if the $\mathbf{E}$ and $\mathbf{B}$ fields are both transverse to the direction of propagation (and mutually orthogonal)?! In the case of monochromatic light, say one oscillates horizontally and the other vertically, then an individual photon has two possible polarisation states.)

To solve this discrepancy, we first notice, from studying the Lagrangian for EM (in vacuum) $$\mathcal{L}_{EM}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ that $A^{0}$ has no kinetic term and hence is not dynamical, meaning that if we are given some initial data $A^{i}$ and $\dot{A}^{i}$ at time $t_{0}$, then $A^{0}$ is fully determined by the equation of motion (eom) $\nabla\cdot\mathbf{E}=0$. Hence $A^{0}$ is not independent: we don't get to specify it arbitrarily on the initial time slice. As such, this reduces the number of dof to 3.

Up to this point I think I understand the argument, however, I hit a bit of a stumbling block when gauge invariance is taken into account.....
.... Next noting that $\mathcal{L}_{EM}$ is invariant under a gauge transformation of $A^{\mu}$, i.e. $A^{\mu}\rightarrow A^{\mu}+\partial^{\mu}\Lambda(x)$. Hence, we can choose $\Lambda(x)$ such that $A^{\mu}$ satisfies $$\partial_{\mu}A^{\mu}=0$$
Is the point here that this equation places a constraint on the remaining 3 fields $A^{i}$, such that, given the knowledge of $A^{0}$ (already determined by the initial data and eom $\nabla\cdot\mathbf{E}=0$), one of the fields $A^{i}$ is fully determined by the other two (and A^{0}). Hence, the gauge symmetry permitting a arbitrary choice of $\Lambda(x)$ enables one to remove another dof, thus leaving 2 physical dof remaining?!

Next, considering GR:

Again, starting with the metric tensor $g_{\mu\nu}(x)$, given that it is symmetric one would naively think that the theory has 10 physical dof. However, if one takes into account the Bianchi identities: $$\nabla_{\mu}G^{\mu\nu}=\nabla_{0}G^{0\nu}+\nabla_{i}G^{i\nu}=0$$ we see that these can be used to remove 4 dof (is the argument here that, given the knowledge of $G^{0\nu}$, one can fully determine the components $G^{i\nu}$ by noting that $\nabla_{i}G^{i\nu}=-\nabla_{0}G^{0\nu}$, or vice verse?!). This leaves us with $10-4=6$ dof.
Secondly, we note that GR is diffeomorphism invariant, i.e. the Einstein-Hilbert action $$S_{EH}=\frac{1}{8\pi G}\int d^{4}x\sqrt{-g}\,\mathcal{R}$$ is invariant under diffeomorphisms: $$x^{\mu}\rightarrow x^{\mu}+\xi^{\mu}$$ (where $\zeta^{\mu}$ is a vector field generating the diffeomorphism).

This is where I'm slightly unsure again....
.... I know that under a diffeomorphism the metric transforms (to first order) as $$g'_{\mu\nu}(x')=\frac{\partial x^{\alpha}}{\partial x'^{\mu}}\frac{\partial x^{\beta}}{\partial x'^{\nu}}g_{\alpha\beta}(x)=g_{\mu\nu}(x)-g_{\mu\beta}(x)\partial_{\nu}\xi^{\beta}-g_{\alpha\nu}(x)\partial_{\mu}\xi^{\alpha}$$

Is the point that we are free to choose the 4 components $\xi^{\mu}$ such that $g_{\mu\nu}(x)$ satisfies a particular constraint (say $\nabla^{\mu}g_{\mu\nu}(x)=\nabla^{0}g_{0\nu}(x)=\nabla^{i}g_{i\nu}(x)=0$), thus removing another 4 dof?!

Hence, we find that GR has $10-4-4=2$ physical dof, which are the polarisation states of the graviton.

Hopefully the questions I have are clear in the text above; I've tried to highlight each one by putting them in italics. The basic summary is how does one use gauge (or diffeomorphism) invariance to remove dof, and are they unphysical dof precisely because one can remove them by making a gauge transformation?

Last edited: Jan 19, 2017