In Yang-Mills theory the field strength tensor $F_{\mu \nu}$ can be calculated as \begin{equation} F_{\mu\nu} \equiv \frac{i}{g} [D_\mu,D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu -ig[A_\mu,A_\nu]. \end{equation} Where $ \ \ \ D_\mu = \partial_\mu -ig A_\mu $
Is there a physical interpretation for $[D_{\mu},D_{\nu}]$? Every book I read just gives this as the definition of the gauge tensor, because it gives the maxwell tensor in the abelian case and is Lorentz and Gauge invariant... okey, that are some good reasons but I'd like to know if there is physical motivation to use that expression. Why the strength of a gauge field should be related to the non-commutatibity of the covariant derivatives?
Background: I know QFT, some group theory and a little General Relativity. If you start with fiber bundles and stuff, go slow please.
Answer
Classically, the gauge field strength is a curvature of a connection, the same way that the Riemann tensor is. Since $F^a_{\mu\nu}$ lives in the adjoint representation of the gauge group, you can define a 4-index object very analogous to the Riemann tensor:
$$\mathcal{F}^a{}_{b\mu\nu} \equiv F^c_{\mu\nu} f_c{}^a{}_b,$$
where $f_c{}^a{}_b$ are the structure constants defined via
$$[t_c, t_b] = f_c{}^a{}_b \, t_a,$$
modulo factors of $i$ if you care to insert them. In any case, the object $\mathcal{F}^a{}_{b\mu\nu}$ contains information about parallel transport around infinitesimal loops, in the same sense that the Riemann tensor does. But the vector being translated is not a tangent vector; instead it is a vector in gauge space (or "internal" space), whose components are defined via expansion in the generators $t_a$, as in $V = V^a \, t_a$.
So, $\mathcal{F}^a{}_{b\mu\nu}$ describes the change in $V = V^a \, t_a$ as it is parallel-transported (via the covariant derivative $D_\mu \equiv \partial_\mu + A_\mu$, again modulo factors of $i$, etc.) around a small parallelogram in the directions $\partial_\mu, \partial_\nu$.
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