# Lorentz invariance of Klein-Gordon Lagrangian

1. Sep 12, 2014

### ShayanJ

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

2. Sep 12, 2014

### Matterwave

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?

3. Sep 12, 2014

### ShayanJ

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

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

If it is, how should I write $\Lambda^{-1} x$? The direct transformation is $\Lambda_\mu^\nu x^\mu$, what is the inverse transformation in component form?
Or maybe I only should write $\Lambda^{-1}x=x^{'}$ and then using $\Lambda^\nu_\mu\partial^\mu=\partial^{'\nu}$ and $\Lambda^\mu_\nu\partial_\mu=\partial^{'}_{\nu}$, and the invariance is proved?
Or what?

Last edited: Sep 12, 2014
4. Sep 12, 2014

### ChrisVer

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...

5. Sep 12, 2014

### Matterwave

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.

6. Sep 12, 2014

### samalkhaiat

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

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

Sam

7. Sep 12, 2014

### Matterwave

Gosh I hate the notation $x_\mu$... I would much rather write $\partial^\mu=\eta^{\mu\nu}\partial_\nu$ rather than $\partial^\mu=\partial/\partial x_\mu$.

8. Sep 12, 2014

### samalkhaiat

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

$$\partial^{ \mu } \phi ( x ) = \frac{ \partial \bar{ \phi } ( \bar{ x } ) }{ \partial \bar{ x }^{ \sigma } } \partial^{ \mu } \bar{ x }^{ \sigma } = \bar{ \partial }_{ \sigma } \bar{ \phi } ( \bar{ x } ) \Lambda^{ \sigma \mu } . \ \ \ (3)$$

9. Sep 17, 2014

### ShayanJ

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

Last edited: Sep 17, 2014
10. Sep 17, 2014

### dextercioby

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

$$A^{\nu}B_{\nu} = \eta^{\nu\alpha}A_{\alpha}B^{\mu}\eta_{\mu\nu}$$

See where you made the mistake.

11. Sep 17, 2014

### ChrisVer

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:
$A_{\nu} B^{\nu} = \eta_{\nu \mu} A^{\mu} \eta^{\mu \rho} B_{\rho} = \delta^{\rho}_{\mu} A^{\mu}B_{\rho} = A^{\mu} B_{\mu}$