The box notation and Lagrangians in field theory

[aqualone]

I have some questions about scalar field Lagrangians, using the box notation defined as \Box \equiv \frac{\partial^2}{\partial t^2} - \nabla^2. It's a basic, perhaps silly issue, but somehow I've managed to sweep it under the rug for a long time.

So, usually, the Lagrangian of a free scalar field is given as \mathcal{L} = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu} \phi - \frac{1}{2}m^2 \phi^2 and then using the Euler-Lagrange equation gives the equation of motion, (\Box + m^2)\phi = 0. Which is all nice and makes sense. But sometimes I see \mathcal{L} = -\frac{1}{2}\phi(\Box+m^2)\phi, and the only justification I've heard is that it is related to the above Lagrangian by integration by parts. Which is clear too, but I have no idea how you get (\Box + m^2)\phi = 0. (Naively) applying the Euler-Lagrange equation gives zero for \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} and then using \frac{\partial\mathcal{L}}{\partial \phi} = 0 gives (\frac{1}{2}\Box + m^2)\phi = 0, which is not quite right.

Is it that the \Box notation has some other meaning? That is, in \phi\Box\phi is there some sort of differentiation being done on the \phi on the left, or is it just what it appears to be?

Thanks in advance; this is really confusing me.

 

[vanhees71]

The point is that you have to take the variation of the action, which doesn't change (applying appropriate boundary conditions) from the one to the other form. Since in the variation in the sense of Hamilton's principle, you have $\delta x=0$, you simply get using the 2nd form of the Lagrangian
$$A[\phi]=-\frac{1}{2} \int \mathrm{d}^4 x \phi(x) (\Box+m^2) \phi(x)$$
for the variation
$$\delta A[\phi]=-\frac{1}{2} \int \mathrm{d}^4 x [\delta \phi(x) (\Box+m^2) \phi(x)+ \phi(x) (\Box+m^2) \delta \phi(x)].$$
Integrating by parts twice, assuming that boundary integrals are 0, gives
$$\delta A[\phi]=-\int \mathrm{d}^4 x \delta \phi(x) (\Box+m^2) \phi(x).$$
Since this should be $0$ for any variations $\delta \phi$, you again get the free Klein-Gordon equation
$$(\Box+m^2) \phi(x)=0.$$

 

[aqualone]

Thanks for the reply; it makes much more sense now. I haven't had a chance to work through the details yet, but I gather that the essential idea is that the integration by parts argument relates to a step in the derivation of the Euler-Lagrange equation. So, a naive application of \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}= \frac{\partial\mathcal{L}}{\partial\phi} to \mathcal{L} = -\frac{1}{2}\phi(\Box + m^2)\phi does indeed give the wrong answer, because the form of the E-L equation has changed, but we get the right equation of motion if we start from the principle of least action.

In interacting theories, sometimes the free part of the Lagrangian is written as -\frac{1}{2}\phi(\Box + m^2)\phi. So, to get the equations of motion in those cases, you just make use of \mathcal{L} = -\frac{1}{2}\phi(\Box + m^2)\phi \implies (\Box+m^2)\phi = 0 and then use the standard E-L equation for the other terms? What's the point of writing -\frac{1}{2}\phi(\Box + m^2)\phi, anyway? I know in some path integral calculations there is a practical reason to, but cases where it seems to make little difference, is it just a matter of style?

 

[haushofer]

I have some questions about scalar field Lagrangians, using the box notation defined as \Box \equiv \frac{\partial^2}{\partial t^2} - \nabla^2. It's a basic, perhaps silly issue, but somehow I've managed to sweep it under the rug for a long time.

So, usually, the Lagrangian of a free scalar field is given as \mathcal{L} = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu} \phi - \frac{1}{2}m^2 \phi^2 and then using the Euler-Lagrange equation gives the equation of motion, (\Box + m^2)\phi = 0. Which is all nice and makes sense. But sometimes I see \mathcal{L} = -\frac{1}{2}\phi(\Box+m^2)\phi, and the only justification I've heard is that it is related to the above Lagrangian by integration by parts. Which is clear too, but I have no idea how you get (\Box + m^2)\phi = 0. (Naively) applying the Euler-Lagrange equation gives zero for \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} and then using \frac{\partial\mathcal{L}}{\partial \phi} = 0 gives (\frac{1}{2}\Box + m^2)\phi = 0, which is not quite right.

Is it that the \Box notation has some other meaning? That is, in \phi\Box\phi is there some sort of differentiation being done on the \phi on the left, or is it just what it appears to be?

Thanks in advance; this is really confusing me.

If you have second order derivatives in your lagrangian L, you should use the euler-lagrange equations with three terms. You're missing the derivative of L with respect to the second order derivative of phi. See e.g. Ortin's Gravity and Strings, eqn 2.4.

 

[dextercioby]

[...] What's the point of writing -\frac{1}{2}\phi(\Box + m^2)\phi, anyway? I know in some path integral calculations there is a practical reason to, but cases where it seems to make little difference, is it just a matter of style?

That's precisely the reason, the Lagrangian density enters the path integral, thus it makes sense to use the box notation because it emphasizes the inverse of the propagator in the coordinate representation.

 

[vanhees71]

Thanks for the reply; it makes much more sense now. I haven't had a chance to work through the details yet, but I gather that the essential idea is that the integration by parts argument relates to a step in the derivation of the Euler-Lagrange equation. So, a naive application of \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}= \frac{\partial\mathcal{L}}{\partial\phi} to \mathcal{L} = -\frac{1}{2}\phi(\Box + m^2)\phi does indeed give the wrong answer, because the form of the E-L equation has changed, but we get the right equation of motion if we start from the principle of least action.

$\newcommand{\Lag}{\mathcal{L}}$
Sure, using the Lagrangian in the form with 2nd derivatives needs an extension of the formula for the equations of motion. So let $\mathcal{L}=\mathcal{L}(\phi,\partial_{\mu} \phi, \partial_{\mu} \partial_{\nu} \phi)$. Then you get
$$
\begin{split}
\delta A&=\int \mathrm{d}^4 x \left [\delta \phi \frac{\partial \Lag}{\partial \phi} + \partial_{\mu} \delta \phi \frac{\partial \Lag}{\partial (\partial_{\mu} \phi)} + \partial_{\mu} \partial_{\nu} \frac{\partial \Lag}{\partial(\partial_{\mu} \partial_{\nu} \phi)} \right ] \\
&= \int \mathrm{d}^4 x \delta \phi \left [\frac{\partial \Lag}{\partial \phi} - \partial_{\mu}\frac{\partial \Lag}{\partial(\partial_{\mu} \phi)} + \partial_{\mu} \partial_{\nu} \frac{\partial \Lag}{\partial(\partial_{\mu} \partial_{\nu} \phi)} \right],
\end{split}$$
where I again integrated by parts and assumed that the boundary terms can be neglected.

The action is stationary if the bracket vanishes everywhere, and that provides the more general Euler-Lagrange equations.