Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equivalence of differential operator terms in action

  1. May 7, 2013 #1
    Hi guys, I'm sure I'm being very stupid here but I'm reading through notes which contain various actions for fields, most of which are very similar, however there is some discrepancies with the way differential operators are shown acting on the fields and I can't for the life of me work out which are equal to which so any advice would be great.

    The three forms I can see are, for a real field [itex]\phi[/itex]

    [itex]
    1) (∂^\mu \phi)(∂_\mu \phi)
    [/itex]

    [itex]
    2) ∂^\mu ∂_\mu |\phi|^2
    [/itex]

    [itex]
    3) \phi (∂^\mu ∂_\mu) \phi
    [/itex]

    These seem to often be used in very similar places but I can't really see how they relate? This is likely painfully simple and I'm just overthinking it but even just pointing in the right direction would be great, thanks!
     
  2. jcsd
  3. May 8, 2013 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Perhaps it is clearer if we just write [itex]\partial_\mu \phi[/itex] as the gradient [itex]\nabla\phi[/itex]. In that case, the three expressions you have written are:
    1) [itex](\nabla\phi) \cdot (\nabla\phi)[/itex]
    2) [itex]\nabla^2 |\phi|^2[/itex]
    3) [itex](\nabla^2\phi) \phi[/itex]
    where
    [tex]\nabla \equiv \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right)[/tex]
    and
    [tex]\nabla^2 \equiv \nabla \cdot \nabla = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}[/tex]

    It may also help you if you write out these expressions for a simple example, like [itex]\phi(x, y, z) = xyz[/itex].
     
  4. May 9, 2013 #3
    I could be wrong, but I think (2) as you've written it doesn't describe a charged scalar field. I think you may have meant [itex]|\partial \phi|^{2}=(\partial^{\mu}\phi^{*})(\partial_{\mu}\phi)[/itex]. Anyways, (1) and (3) are equivalent as long as you remember that these quantities appear in an integral, so that [itex]\int d^{4}x (\partial^{\mu}\phi)(\partial_{\mu}\phi)=-\int d^{4}x\, \phi \partial^{\mu}\partial_{\mu}\phi[/itex]. This means that the action for a real scalar field can be written as [itex]S=\frac{1}{2}\int d^{4}x\, [(\partial \phi)^{2}-m^{2}\phi^{2}]=-\frac{1}{2}\int d^{4}x \, \phi(\partial^{2}+m^{2})\phi[/itex]. (Note [itex]\partial^{2}\equiv\partial^{\mu}\partial_{\mu}[/itex])
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equivalence of differential operator terms in action
Loading...