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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 $(-,+,+,+)$.

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    $\begingroup$ @Cham: Not if he's using the sign convention of $(-,+,+,+)$. Which is, of course, the one true sign convention that all right-thinking physicists agree upon. ;-) $\endgroup$ – Michael Seifert Mar 29 at 14:58
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    $\begingroup$ @MichaelSeifert, the sign is hidden inside the metric. The lagrangian with an explicit sign in front of the kinetic term is a "phantom field". The Lagrangian above doesn't give the KG equation. See the answers below! $\endgroup$ – Cham Mar 29 at 16:02
  • $\begingroup$ @Cham: My point is that under the "relativist's" sign convention $(-,+,+,+)$, the quantity $- \frac{1}{2} \partial^\mu \phi \partial_\mu \phi = \dot{\phi}^2 - (\vec{\nabla} \phi)^2$. This is exactly the same as you get if you write the kinetic term as $+\frac{1}{2} \partial^\mu \phi \partial_\mu \phi$ under the "particle physics" sign convention $(+,-,-,-)$, and so the equations of motion are equivalent. $\endgroup$ – Michael Seifert Mar 29 at 16:36
  • $\begingroup$ @MichaelSeifert, there isn't a single "relativist's" sign convention. There are many! In the US, there's the East coast sign convention, and the West coast sign convention. There's also the particles relativists convention, and some cosmologists sign convention. Even for the cosmologists convention, you may find several point of views. See the discussion about the sign conventions in MTW's "telephone" big black book (Misner, Thorne, Wheeler). $\endgroup$ – Cham Mar 29 at 17:30
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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

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

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  • $\begingroup$ The derivative could be simplified a lot, since $\partial_a \phi \, \partial^a \phi = (\partial_0 \phi)^2 - (\partial_1 \phi)^2 - (\partial_2 \phi)^2 - (\partial_3 \phi)^2$, in cartesian coordinates. $\endgroup$ – Cham Mar 29 at 12:34
  • $\begingroup$ Thanks a for the clean derivation $\endgroup$ – Kiji Mar 29 at 15:05
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$$L=\frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi)-\frac{1}{2}\,m^2\,\phi^2$$

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*}

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