Yang-Mills Stress-Energy Tensor Explained

[ergospherical]

It's given as $T_{\mu \nu} = - \mathrm{tr}(F_{\mu \lambda} {F_{\nu}}^{\lambda} - \frac{1}{4} g_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta})$. Can somebody explain the notation, i.e. what is the meaning here of the trace? (usually I would interpret the trace of a matrix as the number $\mathrm{tr}(a_{\mu \nu}) = {a^{\mu}}_{\mu}$.)

 

[Orodruin]

The trace is over the group indices that are not explicitly written out.

 

[dextercioby]

It would have made more sense to just show "a" as a SU(n) adjoint rep. index on those F's, rather than use Tr which becomes problematic when you consider QCD, that is adding gammas and spinors and their trace(s).

 

[robphy]

It's given as $T_{\mu \nu} = - \mathrm{tr}(F_{\mu \lambda} {F_{\nu}}^{\lambda} - \frac{1}{4} g_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta})$. Can somebody explain the notation, i.e. what is the meaning here of the trace?

For such questions of notation, it would be helpful to give the reference.

 

[samalkhaiat]

It's given as $T_{\mu \nu} = - \mathrm{tr}(F_{\mu \lambda} {F_{\nu}}^{\lambda} - \frac{1}{4} g_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta})$. Can somebody explain the notation, i.e. what is the meaning here of the trace?

Here $F_{\mu\nu}$ is the matrix-valued field tensor: $$F_{\mu\nu} = \mathcal{F}_{\mu\nu}^{a}X_{a} ,$$ where the $X$’s are a set (in fact, any set) of matrices satisfying the Lie algebra of the group $[X_{a},X_{b}] = i C_{ab}{}^{c}X_{c}.$
So $$\mbox{Tr}\left( F_{\mu\nu}F^{\mu\nu}\right) = \mathcal{F}_{\mu\nu}^{a}\mathcal{F}^{\mu\nu b} \ \mbox{Tr} \left( X_{a}X_{b}\right),$$ and $$\mbox{Tr}\left(X_{a}X_{b}\right) \equiv \left(X_{a} X_{b}\right)_{ii}, \ \ i = 1,2, \cdots , p$$ where $p$ is the dimension of the representation. For simple compact Lie groups, we can always choose the $X$’s to be trace-orthonormal $$\mbox{Tr}\left( X_{a}X_{b}\right) = 2C \delta_{ab} ,$$ where $C$ is a constant for each irreducible part of the representation. The matrix notation is useful because it makes gauge-invariance (kind of) obvious: $$\mbox{Tr}\left(gF_{\mu\nu}g^{-1}gF^{\mu\nu}g^{-1}\right) = \mbox{Tr}\left( F_{\mu\nu}F^{\mu\nu}\right).$$