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Transformations of free fields

  1. Mar 6, 2008 #1
    Hi, New here...Can't seem to do latex on here so this post is incomplete until I can work it out.

    This is maybe quite abstract and generic, but here goes. This problem has niggled me for a while and I need some input please.

    I have an action [tex]S=\int d^4 x \sqrt(g(x))\overline\Phi (x)\Omega(x)\Phi(x)[/tex]

    where [tex]\Omega(x)[/tex] is a differential operator and [tex]\Phi(x)[/tex] and [tex]\overline\Phi(x)[/tex] are the free field and it's dual respectively. For example, they may well be scalars [tex]\phi(x)[/tex] and [tex]\overline\phi(x)[/tex] or vectors, or spinnors superfields tensors etc....Any generic spacetime object.

    Say for the moment they are scalars and define a transformation of the fields as follows

    [tex]\delta\phi(x)=\epsilon\phi(x_G)[/tex] and [tex]\delta\overline\phi(x)=-\epsilon\overline\phi(x_{G^{-1}})[/tex]

    where [tex]x \rightarrow x_{G} [/tex] is a finite isometry. Also omega is the operator [tex]g^{\mu\nu}\partial_{\mu}\partial_{nu}[/tex] and is invariant under isometriesThen the change in the action is

    [tex]\delta S=\epsilon\int d^4 x \sqrt(g(x))\overline\Phi (x)\Omega(x)\Phi(x_G)-\epsilon\int d^4 x \sqrt(g(x))\overline\Phi (x_{G^{-1}})\Omega(x)\Phi(x)[/tex]

    which is zero by changing variables in the second integral and using the fact that x goes to xG is an isometry.

    Now, [tex]\delta\phi(x)=\epsilon exp(X^u \partial_{\mu})\phi(x).[/tex]

    where X is a killing vector. I know this for a fact because it's just parameterising the isometry and using the chain rule. The each term in the taylor expansion is also a symmetry. I.e, it is the exponential of the directional derivative, defined as it's taylor expansion. It also happens to be the Lie derivative of a scalar field which pertains to something I will say in a while.

    Now...I have tried to do this for a vector say [tex]A^{\mu}[/tex] and it's dual [tex]\overline A_{\mu}[/tex]

    I am having difficulty deciding what the transformation will be at the moment I have

    [tex]\delta A(x)=\epsilon A^{\mu}(x_G) =\epsilon exp(X\partial)A(x)[/tex] and [tex]\delta \overline A(x)=-\epsilon A_{\mu}(x_{G^-1}}) dx^{\mu}=-\epsilon exp(-X\partial)\overline A[/tex]

    and then the change in the action is

    [tex]\delta S=\epsilon\int d^4 x \sqrt(g(x))\overline A_{\mu}(x) dx^{\mu} e_{\nu}(x)\Omega^{\nu}_{\lambda}(x) dx^{\lambda} e_{\sigma}(x) A^{\sigma}(x_G)-\epsilon\int d^4 x \sqrt(g(x))\overline A_{\mu}(x_{G^-1}}) dx^{\mu} e_{\nu}(x)\Omega^{\nu}_{\lambda}(x) dx^{\lambda} e_{\sigma}(x) A^{\sigma}(x)[/tex]

    Then perform [tex]x \rightarrow x_G[/tex] on the second integral and [tex]dx_G^{\mu} e_{\nu}(x_G)=dx^{\mu} e_{\nu}(x)[/tex] since it is simply the inner product of basis vectors so gives the delta function. Omega is invariant since it is an isometry, and [tex]\sqrt{g} d^4 x[/tex] is also invariant and so the change in the action is zero. So, again, each term in the expansion of [tex]exp(X\partial)[/tex] is a symmetry.

    Now I don't believe this argument for a vector. I think the transformations should include some matrices, and should be the exponential of the Lie derivative of the vector. However, I cant get the argument with matrices to work. I.e. something like this

    [tex]\delta A= U^{-1}\epsilon A(x_G)[/tex] where [tex]U^{-1}[/tex] is the matrix that maps basis vectors at [tex]x_G[/tex] back to x. I was hoping this gives the exponential of the Lie derivative, and it doesn't seem to and also the matrices don't all work out when I do the change of variables in the second integral

    Is my argument for the vector field reasonable? Have I made a conceptual error? Is it simply the exponential of the normal directional derivative. I don't believe it because I would expect the indices to change.

    And how on earth will this work for more arbitrary space time objects [tex]\Phi[/tex]

    Sorry for the long post. I dont think I have made any latex typos
    Last edited: Mar 6, 2008
  2. jcsd
  3. Mar 6, 2008 #2


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    It's [tex] at this site, not [math]. I'm a regular at another vBulletin forum where it's [itex], so it seems to be up to the administrator rather than a vBulletin standard.

    Edit: I see you figured that out. :smile:
  4. Mar 6, 2008 #3
    Hi, yeah thanks worked it out, looked on another thread and clicked on some maths to find the code.

    Any help much appreciated.

    Is my expression for the scalar correct?

    [tex]\delta \phi(x)=\epsilon \phi(x_G)=\epsilon exp (X^{\mu}\partial_{\mu})\phi(x)[/tex]

    I proved this by having the Killing vector field over the manifold, then parameterising the integral curve x goes to xG, taylor expanding and using the chain rule.

    Although I have been reading some riemannian geometry books and this exponential map seems to pertain only to geodesics, and the flows generated by isometries are not geodesics. I don't really understand them though, rather too abstract.
  5. Mar 8, 2008 #4


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    The following might help!

    Cosider a spacetime symmetry whose infinitesimal action on the coordinates is given by

    [tex]x^{a} \rightarrow \bar{x}^{a} = x^{a} + \delta_{f}x^{a}[/tex]


    [tex]\delta_{f}x^{a} = -f^{a}(x)[/tex]

    contains the infinitesimal parameters. Under such symmetry, a multi-component field transforms as

    [tex]\bar{u}_{A}(\bar{x}) = D_{A}{}^{B}(f) u_{B}(x)[/tex]

    where D is a finite dimensional (matrix) representation of the group. To 1st order in the parameters, we write

    [tex]\bar{u}_{A}(x - f) = \left(\delta_{A}{}^{B} + X_{A}{}^{B}(f) \right) u_{B}(x)[/tex]


    [tex]\bar{u}_{A}(x) - f^{a}\partial_{a}u_{A} = u_{A}(x) + X_{A}{}^{B}(f)u_{B}(x)[/tex]

    The X's are set of matrices, appropriate for the field, satisfying the Lie algebra of the symmetry group. For scalar field X = 0, and for covariant vector field it is [itex]\partial_{a}f^{a}[/itex].

    Thus, the infinitesimal change in the form of the field function is

    [tex]\delta_{f}u_{A}(x) \equiv \bar{u}_{A}(x) - u_{A}(x) = f^{a}\partial_{a}u_{A}(x) + X_{A}{}^{B}u_{B}(x)[/tex]

    Except for spinor field, [itex]\delta_{f}[/itex] is nothing but the Lie derivative.


    Last edited: Mar 8, 2008
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