Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    samalkhaiat

    User Avatar
    Science Advisor

    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]

    where

    [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]

    or,

    [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.

    regards

    sam
     
    Last edited: Mar 8, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Transformations of free fields
  1. Free Field Vacuum (Replies: 7)

Loading...