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I Invariance of an equation

  1. May 5, 2017 #1
    Hey. When talking about invariance of a function f under some transformation T we mean that T(f)=f. But what is meant by invariance of an equation f=0? As far as I can see it makes sense to call an equation invariant when the transformed equation T(f)=T(0) is equivalent to the original equation f=0, or maybe just if T(f)=T(0) implies f=0.

    To be specific, I am asking because the source free Yang-Mills equation DμFμν=0 is said to be invariant under gauge transformations and i am wondering what exactly is meant by this. When preforming the gauge transformation U we obtain UDμFμνU=0 which is equivalent to the original equation but not the same as the original equation.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. May 5, 2017 #2

    jedishrfu

    Staff: Mentor

  4. May 5, 2017 #3

    fresh_42

    Staff: Mentor

    Not necessarily. It can also mean ##f \circ T = f##, that is a transformation of coordinates leaves the relationship intact. One should always consider the context and how it is meant. ##T \circ f = f## should better be noted as ##f## is an eigenvector of the transformation operator ##T##.
    This would only mean, that ##T## is injective. It's not called invariant.
    Have you heard of Noether's theorem?
     
  5. May 5, 2017 #4

    jedishrfu

    Staff: Mentor

    and the punch line: Noether heard of it.
     
  6. May 5, 2017 #5

    fresh_42

    Staff: Mentor

    Noether heard of it, but it took Little (aka F. Klein) to be published! :cool:
     
  7. May 5, 2017 #6
    The Lagrange function generating these Yang-Mills equations are gauge invariant L=L', but as mentioned the Yang-Mills equations themselves transform according to the adjoint representation (DμFμν)'=UDμFμνU which does not seem invariant to me, still people say that the are invariant under gauge transformations. My question is, in what sense are they invariant?

    Yes I have heard about Nöethers theorem I do not see how it relates. For every continuous symmetry of the action there exists a continuity equation.
     
    Last edited: May 5, 2017
  8. May 5, 2017 #7

    fresh_42

    Staff: Mentor

    Noether (without "ö") defined as an invariant of a Lie group for her theorem the existence of a relation
    $$P(x,u,\frac{\partial u}{\partial x},\frac{\partial^2 u}{\partial x^2},\ldots) = P(y,v,\frac{\partial v}{\partial y},\frac{\partial^2 v}{\partial y^2},\ldots)$$
    which shows, that it is a statement about the variables, coordinates.
     
  9. May 5, 2017 #8
    Thank you for your answer. If i understand your post the statement of invariance is a statement of the form of an equation independent of variables. But then is not every equation involving only the gauge field Aμ and coordinates invariant under gauge transformations since the gauge transformation of the equation makes Aμ→Aμ', and so the transformed equation has the same form as the original with Aμ' in place of Aμ as variable?
    Please go into details.
     
    Last edited: May 5, 2017
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