# Showing that the real Klein-gordon lagrangian is Lorentz invariant

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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
$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)(\partial^{\mu}\phi)-\frac{m}{2}\phi^{2}$

which should be invariant under Lorentz transformations - that is:

$\phi^{'}(x^{'})=\phi(x)$

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

## The Attempt at a Solution

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ShayanJ
Gold Member
Thank you once again Shyan.

The only thing is: in my final answer I have a factor of $\eta_{\mu\sigma}$ 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:

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

Dont know what to do with the metric tensor eta.

Orodruin
Staff Emeritus
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.

Okay thank you. Can you please tell me if this equation is true:

$\partial_{\mu}\phi=\frac{\partial\phi'}{\partial x'^{\rho}}\frac{\partial x'^{\rho}}{\partial x'^{\mu}}=\partial_{\rho}\phi'\Lambda^{\rho}_{\mu}=\partial_{\mu}'\phi'$

ShayanJ
Gold Member
No, that should be as below:
$\partial_\mu \phi=\partial_\mu \phi'=\frac{\partial \phi'}{\partial x^\mu}=\frac{\partial \phi'}{\partial x'^\rho}\frac{\partial x'^\rho}{\partial x^\mu}=\Lambda^\rho_\mu \partial'_\rho \phi'$

Thanks Shyan but I dont understand how you get $\partial'_{\rho}$ in the final RHS? Where did that prime come from? Essentially I'm asking if its true that $\Lambda_{\mu}^{\rho}\partial_{\rho}=\partial'_{\mu}$. 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.

Cos now I've ended up with the following:

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

and I'm not sure if this proves Lorentz invariance?

ShayanJ
Gold Member
Thanks Shyan but I dont understand how you get $\partial'_{\rho}$ in the final RHS? Where did that prime come from? Essentially I'm asking if its true that $\Lambda_{\mu}^{\rho}\partial_{\rho}=\partial'_{\mu}$. 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 $\partial'_\rho \phi=\frac{\partial \phi}{\partial x'^\rho}$ and that its all just chain rule?
Cos now I've ended up with the following:

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

and I'm not sure if this proves Lorentz invariance?
If by $g^{\rho \sigma}$ you mean the Minkowski metric, then its done because we have: $\partial'_\rho \phi' \partial'_\sigma \phi' g^{\rho \sigma}=\partial'^\sigma \phi' \partial'_\sigma \phi'=\partial'_\rho \phi' \partial'^\rho \phi'$.

im just confused about the prime, I'm not sure what the prime in $\partial'_{\rho}\phi$ means hmm. But yea I got the rest...thank you :)

ShayanJ
Gold Member
im just confused about the prime, I'm not sure what the prime in $\partial'_{\rho}\phi$ 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!!!

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Cool thank you man :)