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Ryder's SU(2) example

  1. Nov 23, 2015 #1

    ShayanJ

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    In section 3.5 of his textbook Quantum Field Theory, Ryder discusses an example of a non-Abelian gauge theory. He considers a 3D internal space and rotations in this space.

    At first he shows that the fields in this internal space transform like ##\delta \vec \phi=-\vec \Lambda \times \vec \phi ## under a rotation ## \vec \Lambda ## in the internal space. Then he shows that ## \delta(\partial_\mu \vec \phi)=-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi##.

    He then introduces the covariant derivative ## D_\mu \vec \phi=\partial_\mu \vec \phi+g\vec W_\mu \times \vec \phi ## and demands that ## \delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi) ## which gives him the transformation rule for ## \vec W_\mu## which is ## \delta \vec W_\mu=-\vec \Lambda \times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda ##.

    Now here is where my confusion begins. He writes ##\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi) ## and then using the above expressions, he verifies that ## \delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi) ##.
    But it seems to me that without using ##\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi) ##, I still should be able to get the same conclusion with just replacing the transformed quantities in the expression for the covariant derivative, so I write:
    ## (D_\mu \vec \phi)_{transformed}=\partial_\mu \vec \phi-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi+g\left( \vec W_\mu-\vec\Lambda\times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda \right) \times \left( \vec\phi-\vec\Lambda\times \vec \phi \right)##.

    But when I do the calculations, the result is far from what Ryder gets(which should already be obvious from my expression).

    What's wrong with this method?
    Thanks
     
  2. jcsd
  3. Nov 23, 2015 #2

    nrqed

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    Nothing is wrong with your method if you keep in mind that we are supposed to work in first order of the transformation parameter ##\Lambda##. If you drop all your terms with two powers of ##\Lambda## and then look at the difference between your expression and ## D_\mu \phi ##(because you want the variation ## \delta D_\mu \phi ##) then you should get his expression.
     
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