1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Showing that the real Klein-gordon lagrangian is Lorentz invariant

  1. Nov 10, 2014 #1
    • Note that the attempt part of the homework template is mandatory
    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
    [itex]\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)(\partial^{\mu}\phi)-\frac{m}{2}\phi^{2}[/itex]

    which should be invariant under Lorentz transformations - that is:

    [itex]\phi^{'}(x^{'})=\phi(x)[/itex]

    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

    ShayanJ

    User Avatar
    Gold Member

  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

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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

    ShayanJ

    User Avatar
    Gold Member

    No, that should be as below:
    [itex]
    \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'
    [/itex]
     
  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:

    [itex]\mathcal{L}'=\frac{1}{2}(\partial_{\rho}\phi')(\partial_{\sigma}\phi')g^{\rho\sigma}-\frac{m}{2}\phi'^{2}[/itex]

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

    ShayanJ

    User Avatar
    Gold Member

    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

    ShayanJ

    User Avatar
    Gold Member

    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 :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Showing that the real Klein-gordon lagrangian is Lorentz invariant
Loading...