# Difference between these 4-vector derivatives?

1. Oct 21, 2014

Hey everyone,

So I've come across something in my notes where it says that these two Lagrangian densities are equal:

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

$\mathcal{L}_{2}=-\phi^{\dagger}\Box\phi - m^{2}\phi^{\dagger}\phi$

where $\Box = \partial^{\mu} \partial_{\mu}=\frac{1}{c^{2}}\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}$

How does this come about? I know that the second term in each density is equal of course, but the first term in each...how are they equal? can someone explain please?

Thank you!

2. Oct 21, 2014

### ShayanJ

The first one has the square of the first derivative of the field where the second one has has the second derivative with power one multiplied by another field variable. This suggests there have been an integration by parts and so by equal, the note means the action integrals for this two Lagrangian densities are equal. Of course provided that the fields vanish at infinity.

3. Oct 21, 2014

Hey, erm I'm sorry I dont really understand lol....:S

4. Oct 21, 2014

### ShayanJ

For getting the equation of motion from the Lagrangian, at first the action integral should be written.
$S=\int \mathcal L \ d^4 x$
Now for your case, we have:
$S=\int \partial_\mu \phi^\dagger \partial^\mu \phi \ d^4 x - \int m^2 \phi^\dagger \phi \ d^4 x$
Integrating the first integral by parts, gives:
$S=\phi^\dagger \partial^\mu \phi |_{(0,-\vec{\infty})}^{(\infty,\vec{\infty})}-\int \phi^\dagger \partial_\mu \partial^\mu \phi -\int m^2 \phi^\dagger \phi \ d^4 x$
But the first term is zero, so we get your second Lagrangian density.

The actions coming from the two Lagrangians are equal and so give the same equations of motion. Your Lagrangian densities are equal in this sense!

5. Oct 22, 2014

Omg thank you so much that's so simple! The only thing is, how do we know that the action integral is from -infty to +infty? and the fields are defined to vanish in these limits right?

6. Oct 22, 2014

### ShayanJ

The action integral is $S=\int_{t_1}^{t_2}L dt$(The time limits in the integral of my previous post was wrong). Varying this w.r.t. whatever dynamical variables, gives us the equations governing the evolution of those variables from $t_1$ to $t_2$. But for continuous systems, we define Lagrangian density($\mathcal L$) by $L=\int \mathcal L d^3 x$. To make the procedure manifestly relativistic, people abandon Lagrangian and work with Lagrangian density. Anyway, just like when you want to find total energy from energy density, you should integrate over all of space. The space integrations in the action integral and its limits come from here.

7. Oct 22, 2014