Field momentum of Klein-Gordon Lagrangian

Question

Given the Lagrangian $L$ of the field $\phi$ the field momentum $\Pi$ reads:

$$L_{KG}=-\frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}m^2\phi^2$$

$$\Pi=\frac{\partial L}{\partial(\partial_\mu\phi)}=\partial_\mu\phi$$

I dont see how the derivative above gives this result. How do we perform this derivative? The wrong way I thought is doing it like this:

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

PS: The Minkowski signature convention is $(-,+,+,+)$.

Answer 1

First there is an issue with your definition. The canonically conjugate momentum is

$$\pi=\dfrac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}$$

In fact notice that in your equation the LHS carries no indices and the RHS carries one which should indicate something is actually wrong.

So you must differentiate $\mathcal{L}$ with respect to $\dot{\phi}=\partial_0\phi$.

Why is so? Well, this is a straightforward generalization of the canonically conjugate momentum from classical mechanics, where the momentum conjugate to $q$ is

$$p = \dfrac{\partial L}{\partial \dot{q}}$$

Now how do we compute this for the KG lagrangian? Well the Lagrangian is

$$\mathcal{L}[\phi,\partial_\mu\phi]=\frac{1}{2}\partial_\mu \phi \partial^\mu\phi-\frac{1}{2}m^2\phi^2=\frac{1}{2}\eta^{\mu\nu}\partial_\mu \phi\partial_\nu \phi-\frac{1}{2}m^2\phi^2$$

Hence it is a function of $\phi$ and $\partial_\mu \phi$ for $\mu=0,1,2,3$.

You should when differentiating regard $\phi,\partial_0\phi,\partial_1\phi,\partial_2\phi,\partial_3\phi$ as five different and independent coordinates!

So we compute

$$\frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}=\frac{1}{2}\eta^{\mu\nu}\frac{\partial}{\partial(\partial_0\phi)}(\partial_\mu \phi \partial_\nu\phi)$$

Where the last term vanishes because $\phi^2$ doesn't depend on $\partial_0\phi$.

Next we have

$$\frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}=\frac{1}{2}\eta^{\mu\nu}\left[\frac{\partial (\partial_\mu\phi)}{\partial(\partial_0\phi)}\partial_\nu\phi + \frac{\partial(\partial_\nu\phi)}{\partial(\partial_0\phi)}\partial_\mu \phi\right] =\frac{1}{2}\eta^{\mu\nu}\left[\delta_{\mu 0}\partial_\nu\phi+\delta_{\nu 0}\partial_\mu \phi\right]$$

Using the Kronecker deltas this is

$$\frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}=\frac{1}{2}\eta^{0\nu}\partial_\nu\phi+\frac{1}{2}\eta^{\mu 0}\partial_\mu \phi.$$

Finally if one works in the $(+,-,-,-)$ signature, $\eta^{\mu 0}= \delta_{\mu 0}$ and hence you get

$$\pi = \partial_0 \phi = \dot{\phi}.$$

Answer 2

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

Where: $\phi=\phi_i(t,x,y,z)$ and $\partial_\mu\phi_i\equiv\frac{\partial\phi_i}{\partial x^\mu}$

$$\frac{\partial L}{\partial(\partial_\mu\phi)}=\partial^\mu\phi\quad\text{?}$$ you can "expand" the Lagrangian like this: (signature $+,-,-,-$)

\begin{align*} & {L}=\frac{1}{2}\left[\partial_0\phi\partial_0\phi- \partial_1\phi\partial_1\phi-\partial_2\phi\partial_2\phi- \partial_3\phi\partial_3\phi\right]-\frac{1}{2}\,m^2\phi^2\\\\ &\text{so}:\\\\ &\frac{\partial{L}}{\partial(\partial_0\phi)}=\partial_0\phi=\partial^0\phi\\ &\frac{\partial{L}}{\partial(\partial_1\phi)}=-\partial_1\phi=\partial^1\phi\\\\ &\text{and so on}\\\\ &\text{and}\\\\ &\frac{\partial{L}}{\partial\phi}=-m^2\,\phi\\\\ &\text{and hence the EL formula obtain:}\\\\ &\partial_\mu\partial^\mu\,\phi+m^2\phi=0\quad \surd \end{align*}