Show that F transforms as an isovector.

[Jimmy Snyder]

Homework Statement

Show that
\delta F^{\mu\nu} = g\epsilon(x) \times F^{\mu\nu}

Homework Equations

F^{\mu\nu} = \partial^{\mu}W^{\nu} - \partial^{\nu}W^{\mu} - gW^{\mu} \times W^{\nu}
\delta W^{\mu} = \partial^{\mu}\epsilon(x) + g\epsilon(x) \times W^{\mu}(x)

The Attempt at a Solution

The equation in part 1 is eqn (3.46) on page 33 of An Informal Introduction to Gauge Field Theories by I. J. R. Aitchison (digitally printed version 2007). The first equation in part 2 is eqn (3.45) on the same page, and the second equation in part 2 is eqn (3.36) on page 32 of the same book. Here is what I get.

\delta F^{\mu\nu} = \delta\partial^{\mu}W^{\nu} - \delta\partial^{\nu}W^{\mu} - \delta(gW^{\mu} \times W^{\nu})
= \partial^{\mu} \delta W^{\nu} - \partial^{\nu}\delta W^{\mu} - g\delta W^{\mu} \times W^{\nu} - gW^{\mu} \times \delta W^{\nu}
= \partial^{\mu}\partial^{\nu}\epsilon + g\partial^{\mu}\epsilon \times W^{\nu} + g\epsilon \times \partial^{\mu}W^{\nu}
+ \partial^{\nu}\partial^{\mu}\epsilon - g\partial^{\nu}\epsilon \times W^{\mu} + g\epsilon \times \partial^{\nu}W^{\mu}
- g\partial^{\mu}\epsilon \times W^{\nu} - g^2\epsilon \times W^{\mu} \times W^{\nu}
- g W^{\mu} \times \partial^{\nu}\epsilon - g^2 W^{\mu} \times \epsilon \times W^{\nu}
Lots cancels out here. Unfortunately, too much does. I get

\delta F^{\mu\nu} = g\epsilon(x) \times (\partial^{\mu}W^{\nu} - \partial^{\nu}W^{\mu})

Where is my mistake?

 

[TSny]

I think you need to be careful with the terms that involve two cross products. Use of parentheses will help. You have two terms: -g²(\epsilonxW^μ)xW^\nu -g²W^μx(\epsilonxW^\nu)

These two terms do not cancel. You can show that the two terms together reduce to -g²\epsilonx(W^μxW^\nu)

See identity #3 in this list http://wwwppd.nrl.navy.mil/nrlformulary/vector_identities.pdf

 

[Jimmy Snyder]

That's it TSny. Thanks for your help.