# Yang-Mills Field Strength Tensor

1. Dec 23, 2006

### neevor

I was wondering why for
$$F_{\mu \nu} = [D_{\mu},D_{\nu}] = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}+[A_{\mu},A_{\nu}] [\tex] the term [tex] A_{\mu}\partial_{\nu} - A_{\nu}\partial_{\mu} [\tex] vanishes. Last edited: Dec 23, 2006 2. Dec 23, 2006 ### cristo Staff Emeritus Since no summation is implied, one can simply swap the indices \mu and \nu in the second term of the expression, and hence the result. Note, to display latex use the [ tex ] and [ / tex ] tags (without spaces inside the square brackets) in place of \begin{displaymath}. 3. Dec 23, 2006 ### neevor Really? because A_{\mu} is not equal to A_{\nu} in general. So simply swapping the two indecies would just give, A_{\nu}\partial_{\mu} - A_{\mu}\partial_{\nu} leaving me with the same problem. 4. Dec 23, 2006 ### cristo Staff Emeritus Oh yes, sorry. I just glance at it and typed before I thought really! I'm blaming it on the fact that it's late. With regard to the question, I'm not too sure.. is there anything special about A? Sorry I can't be of more help! 5. Dec 23, 2006 ### George Jones Staff Emeritus The term [tex] A_{\mu}\partial_{\nu} - A_{\nu}\partial_{\mu}$$

does not vanish.

There is a similar term, but of opposite sign, that comes from the product rule in terms like $\partial_\mu A_\nu$.

$$\partial_\mu \left(A_\nu \psi \right) = \left( \partial_\mu A_\nu \right) \psi + A_\nu \partial_\mu \psi$$

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