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

1. Nov 10, 2014

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

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.

2. Relevant equations

3. The attempt at a solution

2. Nov 10, 2014

### ShayanJ

3. Nov 10, 2014

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.

4. Nov 10, 2014

### Orodruin

Staff Emeritus
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. Nov 10, 2014

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

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

6. Nov 10, 2014

### ShayanJ

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

7. Nov 11, 2014

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.

8. Nov 11, 2014

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?

9. Nov 11, 2014

### ShayanJ

Do you see that $\partial'_\rho \phi=\frac{\partial \phi}{\partial x'^\rho}$ and that its all just chain rule?
If by $g^{\rho \sigma}$ you mean the Minkowski metric, then its done because we have: $\partial'_\rho \phi' \partial'_\sigma \phi' g^{\rho \sigma}=\partial'^\sigma \phi' \partial'_\sigma \phi'=\partial'_\rho \phi' \partial'^\rho \phi'$.

10. Nov 11, 2014

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 :)

11. Nov 11, 2014

### ShayanJ

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
12. Nov 11, 2014