Showing that the real Klein-gordon lagrangian is Lorentz invariant

  • Thread starter Dixanadu
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  • #1
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Homework Statement


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!

Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #3
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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.
 
  • #4
Orodruin
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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.
 
  • #5
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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]
 
  • #6
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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]
 
  • #7
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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.
 
  • #8
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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?
 
  • #9
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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.
Do you see that [itex] \partial'_\rho \phi=\frac{\partial \phi}{\partial x'^\rho} [/itex] and that its all just chain rule?
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?
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].
 
  • #10
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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 :)
 
  • #11
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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 :)
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:
  • #12
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Cool thank you man :)
 

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