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Variation of Lagrange Density under field transformation

  1. Oct 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey guys!

    So I have a Lagrangian with two coupled fields like so:

    [itex] \mathcal{L} = \frac{1}{2}(\partial_{\mu}\phi_{1})(\partial^{\mu}\phi_{1}) +\frac{1}{2}(\partial_{\mu}\phi_{2})(\partial^{\mu}\phi_{2})-\frac{m_{1}^{2}}{2}(\phi_{1}\phi_{1})-\frac{m_{2}^{2}}{2}(\phi_{2}\phi_{2})-g(\phi_{1}\phi_{2})^2 [/itex]

    where g is a coupling constant.

    So I have to find the variation of this lagrangian under the transformation

    [itex]\delta\phi_{1}=\epsilon\phi_{2}, \delta\phi_{2}=-\epsilon\phi_{1}[/itex]

    2. Relevant equations


    3. The attempt at a solution
    I dont know what to do - do I just plug these into the Lagrangian? and if I do, how do I compute the new [itex]\partial_{\mu}\phi[/itex]?

    Thanks a lot guys!
     
  2. jcsd
  3. Oct 15, 2014 #2

    ShayanJ

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    Gold Member

    You should replace [itex] \phi_i [/itex] with [itex] \phi_i+\delta \phi_i [/itex] in the Lagrangian.
     
  4. Oct 15, 2014 #3
    Okay so i plugged them in but I'm still a bit stuck.

    For example consider this term:
    [itex]\frac{1}{2}(\partial_{\mu}(\phi_{1}+\epsilon\phi_{2}))[/itex]

    How do I compute this? (yes I am a noob with this stuff) Would it be equal to [itex]\frac{1}{2}(\partial_{\mu}\phi_{1}+\epsilon\partial_{\mu}\phi_{2})[/itex]?

    Also do I omit terms of order [itex]\epsilon^{2}[/itex]?

    Thanks...
     
  5. Oct 15, 2014 #4

    ShayanJ

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    Gold Member

    Yes.
    Yes.
     
  6. Oct 15, 2014 #5
    So i went through it all and here's where I am:

    [itex]\begin{split}\mathcal{L}=\frac{1}{2}[(\partial_{\mu}\phi_{1})(\partial^{\mu}\phi_{1})] +\frac{1}{2}[(\partial_{\mu}\phi_{2})(\partial^{\mu}\phi_{2})] \\
    -\frac{m_{1}^{2}}{2}(\phi_{1}^{2}+2\epsilon\phi_{1}\phi_{2}) -\frac{m_{2}^{2}}{2}(\phi_{2}^{2}-2\epsilon\phi_{1}\phi_{2}) -g(\phi_{1}\phi_{2}-\epsilon\phi_{1}^{2}+\epsilon\phi_{2}^{2})
    \end{split}[/itex]

    Is this looking okay...?

    Thanks so much...
     
    Last edited: Oct 15, 2014
  7. Oct 16, 2014 #6

    ShayanJ

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    Gold Member

    I get the following:
    [itex]
    \mathcal{L}'=\frac{1}{2} \partial_{\mu} \phi_1 \partial^{\mu} \phi_1+\frac{1}{2} \partial_{\mu} \phi_2 \partial^{\mu} \phi_2-\frac{1}{2}m_1^2(\phi_1+2\epsilon \phi_2)\phi_1-\frac{1}{2}m_2^2(\phi_2-2\epsilon \phi_1)\phi_2-g[\phi_1\phi_2-2\epsilon(\phi_1^2-\phi_2^2)]\phi_1\phi_2
    [/itex]
     
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