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Homework Help: Show that F transforms as an isovector.

  1. Jul 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that
    [tex]\delta F^{\mu\nu} = g\epsilon(x) \times F^{\mu\nu}[/tex]

    2. Relevant equations
    [tex]F^{\mu\nu} = \partial^{\mu}W^{\nu} - \partial^{\nu}W^{\mu} - gW^{\mu} \times W^{\nu}[/tex]
    [tex]\delta W^{\mu} = \partial^{\mu}\epsilon(x) + g\epsilon(x) \times W^{\mu}(x)[/tex]

    3. 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.

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

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

    Where is my mistake?
  2. jcsd
  3. Jul 10, 2012 #2


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    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([itex]\epsilon[/itex]xWμ)xW[itex]\nu[/itex] -g2Wμx([itex]\epsilon[/itex]xW[itex]\nu[/itex])

    These two terms do not cancel. You can show that the two terms together reduce to -g2[itex]\epsilon[/itex]x(WμxW[itex]\nu[/itex])

    See identity #3 in this list http://wwwppd.nrl.navy.mil/nrlformulary/vector_identities.pdf [Broken]
    Last edited by a moderator: May 6, 2017
  4. Jul 10, 2012 #3
    That's it TSny. Thanks for your help.
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