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Homework Help: Showing that the real Klein-gordon lagrangian is Lorentz invariant

  1. Nov 10, 2014 #1
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    1. The problem statement, all variables and given/known data
    Hey guys!

    So this question should be simple apparently but I got no idea how to do it. Basically I have the following Lagrangian density

    which should be invariant under Lorentz transformations - that is:


    And so I have to show that under this transformation, the lagrangian density above is Lorentz invariant.

    No idea where to start guys so please help me :D thanks!

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 10, 2014 #2


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  4. Nov 10, 2014 #3
    Thank you once again Shyan.

    The only thing is: in my final answer I have a factor of [itex]\eta_{\mu\sigma}[/itex] floating around, as a result of the product of both Lambdas. What does this mean? how do I get rid of it?

    you can see my final answer here:

    [itex]\mathcal{L}'=\frac{1}{2}(\partial_{\rho}\phi')(\partial^{\sigma}\phi')\Lambda^{\rho}_{\mu}\Lambda^{\mu}_{\sigma} -\frac{m}{2}\phi'^{2}=\frac{1}{2}(\partial_{\rho}\phi')(\partial^{\sigma}\phi')\eta_{\mu\sigma} -\frac{m}{2}\phi'^{2}[/itex].

    Dont know what to do with the metric tensor eta.
  5. Nov 10, 2014 #4


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    You should recheck your contraction of the lambdas. As it stands you have a few indices floating around in the RHS where there are no free indices on the LHS.
  6. Nov 10, 2014 #5
    Okay thank you. Can you please tell me if this equation is true:

    [itex]\partial_{\mu}\phi=\frac{\partial\phi'}{\partial x'^{\rho}}\frac{\partial x'^{\rho}}{\partial x'^{\mu}}=\partial_{\rho}\phi'\Lambda^{\rho}_{\mu}=\partial_{\mu}'\phi'[/itex]
  7. Nov 10, 2014 #6


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    No, that should be as below:
    \partial_\mu \phi=\partial_\mu \phi'=\frac{\partial \phi'}{\partial x^\mu}=\frac{\partial \phi'}{\partial x'^\rho}\frac{\partial x'^\rho}{\partial x^\mu}=\Lambda^\rho_\mu \partial'_\rho \phi'
  8. Nov 11, 2014 #7
    Thanks Shyan but I dont understand how you get [itex]\partial'_{\rho}[/itex] in the final RHS? Where did that prime come from? Essentially I'm asking if its true that [itex]\Lambda_{\mu}^{\rho}\partial_{\rho}=\partial'_{\mu}[/itex]. I think you asked this question in your post that you linked to but no one really said if its right or wrong so I'm just wondering.
  9. Nov 11, 2014 #8
    Cos now I've ended up with the following:


    and I'm not sure if this proves Lorentz invariance?
  10. Nov 11, 2014 #9


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    Do you see that [itex] \partial'_\rho \phi=\frac{\partial \phi}{\partial x'^\rho} [/itex] and that its all just chain rule?
    If by [itex] g^{\rho \sigma}[/itex] you mean the Minkowski metric, then its done because we have: [itex] \partial'_\rho \phi' \partial'_\sigma \phi' g^{\rho \sigma}=\partial'^\sigma \phi' \partial'_\sigma \phi'=\partial'_\rho \phi' \partial'^\rho \phi' [/itex].
  11. Nov 11, 2014 #10
    im just confused about the prime, I'm not sure what the prime in [itex]\partial'_{\rho}\phi[/itex] means hmm. But yea I got the rest...thank you :)
  12. Nov 11, 2014 #11


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    That prime only means we're differentiating w.r.t. primed coordinates. Don't think that if you use different greek letters for super- and sub-scripts, it can represent different coordinate systems. Its a very dangerous idea!!!
    Last edited: Nov 11, 2014
  13. Nov 11, 2014 #12
    Cool thank you man :)
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