I Yang-Mills Stress-Energy Tensor Explained

Click For Summary
The Yang-Mills stress-energy tensor is defined using the equation T_{\mu \nu} = - tr(F_{\mu \lambda} F_{\nu}^{\lambda} - (1/4) g_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta}). The trace in this context refers to the trace over the group indices of the matrix-valued field tensor F_{\mu\nu}, which is expressed as F_{\mu\nu} = F_{\mu\nu}^{a}X_{a}, where X’s are matrices from the Lie algebra. The trace operation helps in maintaining gauge invariance, as shown by the property tr(gF_{\mu\nu}g^{-1}gF^{\mu\nu}g^{-1}) = tr(F_{\mu\nu}F^{\mu\nu}). The notation can be complex, especially when incorporating additional elements like gamma matrices and spinors in quantum chromodynamics (QCD). Understanding these notations and their implications is crucial for grasping the underlying physics of Yang-Mills theories.
ergospherical
Science Advisor
Homework Helper
Education Advisor
Insights Author
Messages
1,100
Reaction score
1,387
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}##.)
 
Physics news on Phys.org
The trace is over the group indices that are not explicitly written out.
 
  • Informative
Likes ergospherical
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).
 
  • Like
Likes vanhees71 and ergospherical
ergospherical said:
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.
 
  • Like
Likes vanhees71 and ergospherical
ergospherical said:
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).
 
  • Like
  • Informative
Likes ergospherical, vanhees71 and dextercioby
Being material observers, we do not expand with the universe. Our ruler for measuring its increasing size does not expand either - its scale does not change. If I identify the ruler with a metric, then from my perspective it should be invariant both spatially and temporally. If it expanded with the universe, then its size measured with this ruler would be constant. Why then do we use a metric with the spatial scale expanding with the universe and constant temporal scale to measure the...

Similar threads

Replies
38
Views
711
  • · Replies 16 ·
Replies
16
Views
2K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 7 ·
Replies
7
Views
474
Replies
2
Views
404