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Lorentz invariance of Klein-Gordon Lagrangian

  1. Sep 12, 2014 #1

    ShayanJ

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    I wanna prove the invariance of the Klein-Gordon Lagrangian [itex] \mathcal{L}=\frac 1 2 \partial^\mu \phi \partial_\mu \phi-\frac 1 2 m^2 \phi^2 [/itex] under a general Lorentz transformation [itex] \Lambda^\alpha_\beta [/itex] but I don't know what should I do. I don't know how to handle it. How should I do it?
    Thanks
     
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  3. Sep 12, 2014 #2

    Matterwave

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    Well, I think the easiest way to do it would be to apply a Lorentz transformation and see that the Lagrangian doesn't change. Have you tried doing that?
     
  4. Sep 12, 2014 #3

    ShayanJ

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    Yeah, I tried but I don't know how should I apply it. Is the following correct?

    [itex]
    \mathcal{L}^{'}=\frac 1 2 \Lambda^\nu_\mu \partial^\mu[\phi^{'}(\Lambda^{-1} x)] \Lambda^\mu_\nu \partial_\mu [\phi^{'}(\Lambda^{-1} x)]-\frac 1 2 m^2 [\phi^{'}(\Lambda^{-1} x)]^2
    [/itex]

    If it is, how should I write [itex] \Lambda^{-1} x [/itex]? The direct transformation is [itex] \Lambda_\mu^\nu x^\mu [/itex], what is the inverse transformation in component form?
    Or maybe I only should write [itex] \Lambda^{-1}x=x^{'} [/itex] and then using [itex] \Lambda^\nu_\mu\partial^\mu=\partial^{'\nu} [/itex] and [itex] \Lambda^\mu_\nu\partial_\mu=\partial^{'}_{\nu} [/itex], and the invariance is proved?
    Or what?
     
    Last edited: Sep 12, 2014
  5. Sep 12, 2014 #4

    ChrisVer

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    what is the definition of a scalar field transformation then? or try to expand it in taylor around x [I guess then you can reach a form of the same lagrangian + a total derivative term]
    In any case there are some problems in your definition... you should avoid writting two summed indices the same...[in your case a mu pair should change... then you'll get the metric which will give you again the double partial derivative...
     
  6. Sep 12, 2014 #5

    Matterwave

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    If you just replace it with ##\partial^{\nu'}## and call it done, then you might as well have just said the Lagrangian is Lorentz invariant by inspection.

    Maybe one can try this hint:

    $$A^{\mu'}A_{\mu'}=\Lambda^{\mu'}_{~~\nu}A^\nu \Lambda^{\rho}_{~~\mu'} A_\rho=\Lambda^{\rho}_{~~\mu'}\Lambda^{\mu'}_{~~\nu}A^\nu A_\rho=\delta^{~~\rho}_{\nu}A^\nu A_\rho=A^\rho A_\rho$$

    See how you can apply this to your calculation.
     
  7. Sep 12, 2014 #6

    samalkhaiat

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    Start from the “transformation” of the scalar field
    [tex]\phi ( x ) = \bar{ \phi } ( \bar{ x } ) . \ \ \ (1)[/tex]
    So, [itex]\phi^{ 2 } ( x ) = \bar{ \phi }^{ 2 } ( \bar{ x } )[/itex] and this takes care of the second term.

    Now, differentiate (1) with respect to [itex]x^{ \mu }[/itex] and use the chain rule:
    [tex]\partial_{ \mu } \phi ( x ) = \frac{ \partial \bar{ \phi } ( \bar{ x } ) }{ \partial \bar{ x }^{ \rho } } \frac{ \partial \bar{ x }^{ \rho } }{ \partial x^{ \mu } } = \bar{ \partial }_{ \rho } \bar{ \phi } ( \bar{ x } ) \ \Lambda^{ \rho }{}_{ \mu } . \ \ \ (2)[/tex]
    Next, differentiate (1) with respect to [itex]x_{ \mu }[/itex]:
    [tex]\partial^{ \mu } \phi ( x ) = \frac{ \partial \bar{ \phi } ( \bar{ x } ) }{ \partial \bar{ x }^{ \sigma } } \frac{ \partial \bar{ x }^{ \sigma } }{ \partial x_{ \mu } } = \bar{ \partial }_{ \sigma } \bar{ \phi } ( \bar{ x } ) \ \Lambda^{ \sigma \mu } . \ \ \ (3)[/tex]
    Multiply (2) and (3) and use the defining property of the Lorentz transformation
    [tex]\Lambda^{ \rho }{}_{ \mu } \ \Lambda^{ \sigma \mu } = \Lambda^{ \rho }{}_{ \mu } \ \Lambda^{ \sigma }{}_{ \nu } \ \eta^{ \mu \nu } = \eta^{ \rho \sigma } .[/tex]

    Sam
     
  8. Sep 12, 2014 #7

    Matterwave

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    Gosh I hate the notation ##x_\mu##... I would much rather write ##\partial^\mu=\eta^{\mu\nu}\partial_\nu## rather than ##\partial^\mu=\partial/\partial x_\mu##.
     
  9. Sep 12, 2014 #8

    samalkhaiat

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    Yes, hating it is a good practise. Ok, the equation below is for your taste :)

    [tex]\partial^{ \mu } \phi ( x ) = \frac{ \partial \bar{ \phi } ( \bar{ x } ) }{ \partial \bar{ x }^{ \sigma } } \partial^{ \mu } \bar{ x }^{ \sigma } = \bar{ \partial }_{ \sigma } \bar{ \phi } ( \bar{ x } ) \Lambda^{ \sigma \mu } . \ \ \ (3)[/tex]
     
  10. Sep 17, 2014 #9

    ShayanJ

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    Just one more thing.
    Consider the quantity [itex] A^\mu B_\mu [/itex]. Is it the same as [itex] A_\mu B^\mu [/itex]?
    This is how I tried to prove it:
    [itex] A^\nu B_\nu=\eta_{\nu\mu}\eta^{\nu\mu}A_\mu B^\mu[/itex].
    The problem is, [itex] \eta_{\nu\mu}\eta^{\nu\mu} [/itex] is the twice contracted product of Minkowski metrics and is equal to [itex] \pm 2 [/itex]. But regardless of the fact that maybe [itex] A^\mu B_\mu=A_\nu B^\nu [/itex] isn't true, a coefficient of two doesn't seem to be true!
     
    Last edited: Sep 17, 2014
  11. Sep 17, 2014 #10

    dextercioby

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    No, you haven't understood the Einstein summation convention properly. Compare what you wrote with what's right:

    [tex] A^{\nu}B_{\nu} = \eta^{\nu\alpha}A_{\alpha}B^{\mu}\eta_{\mu\nu} [/tex]

    See where you made the mistake.
     
  12. Sep 17, 2014 #11

    ChrisVer

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    avoid having summed indices more than once- that's a rule. Otherwise summation doesn't make much sense.
    what you wrote would have to be:
    [itex]A_{\nu} B^{\nu} = \eta_{\nu \mu} A^{\mu} \eta^{\mu \rho} B_{\rho} = \delta^{\rho}_{\mu} A^{\mu}B_{\rho} = A^{\mu} B_{\mu} [/itex]
     
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