Hi, New here...Can't seem to do latex on here so this post is incomplete until I can work it out.(adsbygoogle = window.adsbygoogle || []).push({});

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

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