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I Adjoint transformation of gluon

  1. Jan 8, 2017 #1

    CAF123

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    It is commonly written in the literature that due to it transforming in the adjoint representation of the gauge group, a gauge field is lie algebra valued and may be decomposed as ##A_{\mu} = A_{\mu}^a T^a##. For SU(3) the adjoint representation is 8 dimensional so objects transforming under the adjoint representation are 8x1 real Cartesian vectors and 3x3 traceless hermitean matrices via the lie group adjoint map. The latter motivates writing ##A_{\mu}## in terms of generators, ##A_{\mu} = A_{\mu}^a T^a##.

    My first question is, this equation is said to be valid independent of the representation of ##T^a## - but how can this be true? In some representation other than the fundamental representation, the ##T^a## will not be 3x3 hermitean traceless matrices and thus will not contain 8 real parameters needed for transformation under the adjoint rep. But we know the gluon field transforms under the adjoint representation so does this line of reasoning not constrain the ##T^a## to be the Gell mann matrices?

    Consider the following small computation:
    $$A_{\mu}^a \rightarrow A_{\mu}^a D_b^{\,\,a} \Rightarrow A_{\mu}^a t^a \rightarrow A_{\mu}^b D_b^{\,\,a}t^a$$ Now, since ##Ut_bU^{-1} = D_b^{\,\,a}t^a## we have ##A_{\mu}^a t^a \rightarrow A_{\mu}^b (U t^b U^{-1}) = U A_{\mu}^b t^b U^{-1}##.
    The transformation law for the ##A_{\mu}^a## is in fact ##A_{\mu} \rightarrow UA_{\mu}U^{-1} - i/g (\partial_{\mu} U) U^{-1}##.
    1) What is the error that amounts to these two formulae not being reconciled?
    2) The latter equation doesn't seem to express the fact that the gluon field transforms in the adjoint representation. I was thinking under SU(3) colour, since this is a global transformation, U will be independent of spacetime so the derivative term goes to zero but is there a more general argument?

    Thanks!
     
  2. jcsd
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