Conjugate momenta in field theory

In classical mechanics for particles, the Euler-Lagrange equations are given by:

$$\frac{d}{dt} \frac{\partial L}{\partial \dot q} = \frac{\partial L}{\partial q}$$

and the momentum conjugate to q is defined as:

$$p = \frac{\partial L}{\partial \dot q}$$.

In particular, the conjugate momentum shows up on the L.H.S. of the Euler-Lagrange equations. However, in field theory, the Euler-Lagrange equations are given by:

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

and the conjugate momenta are defined as:

$$\pi = \frac{\partial L}{\partial \dot \phi}$$.

In this case, the conjugate momentum is different from the term that appears in the derivative on the L.H.S. of the Euler-Lagrange equation.

What causes this difference between particle theory and field theory?

• Fields are functions of spacetime, while the $q$'s are a function of time alone. – knzhou Sep 23 '19 at 4:59
• I understand that aspect, but I'm mainly curious why the useful definition is $\pi = \frac{\partial L}{\partial \dot \phi}$ rather than $\pi^\mu = \frac{\partial L}{\partial (\partial_\mu \phi)}$. The later definition seems to be a closer in analogy. – Ricky P Sep 23 '19 at 5:49
• You can define $\pi^\mu$ if you want. However, $\pi$ is also used as the canonical momentum in the Hamiltonian formulation. Hamiltonians generate evolution in time, regardless of how many spatial dimensions there are. Since the formulation singles out time, it turns out you only want $\pi$ and not the full $\pi^\mu$. – knzhou Sep 23 '19 at 5:54