I have the following Lagrangian density $\mathcal{L}$ where

$$ \mathcal{L}=\frac{1}{2}\left(c[\partial_{t}\phi(x,t)]^{2}-\frac{1}{l}[\partial_{x}\phi(x,t)]^{2}+\frac{1}{\omega_{J}^{2}l}[\partial_{x}\partial_{t}\phi(x,t)]^{2}+\gamma[\partial_{x}\phi(x,t)]^{4}\right)\tag{1} $$

where $c,l,\omega_{J},\gamma$ are constants. Defining the usual conjugate momenta $\pi$ such that $$ \pi=\frac{\partial\mathcal{L}}{\partial(\partial_{t}\phi(x,t))}.\tag{2} $$

How should I evaluate the third term $[\partial_{x}\partial_{t}\phi(x,t)]^{2}$ where there is also an $x$-derivative?

Edit: I found a solution to this. It seems that I cannot use the regular convention for defining conjugate momenta. Rather I have to define it such that $$ \pi=\frac{\delta\mathcal{L}}{\delta[\partial_{t}\phi]}=\frac{\partial\mathcal{L}}{\partial[\partial_{t}\phi]}-\partial_{x}\frac{\partial\mathcal{L}}{\partial[\partial_{x}\partial_{t}\phi]} = c\partial_{t}\phi-\frac{1}{\omega_{J}^{2}l}\partial_{x}^{2}\partial_{t}\phi.\tag{3} $$

I do not understand this definition of conjugate momenta. Can someone explain why is it defined like so?


2 Answers 2

  1. To properly define the momentum density $p(x,t)$ it is important to be able to distinguish between the dependence of the position field $q(x,t)$ and the velocity field $v(x,t)$. Therefore we cannot use the action $$S[q]~=~\left. \int\! dt~ L[q(\cdot,t),v(\cdot,t);t]\right|_{v=\dot{q}}, \tag{A}$$ which is only a functional of the position fields $q(x,t)$.

  2. Instead the fundamental object is the Lagrangian $$ L[q(\cdot,t),v(\cdot,t);t], \tag{B}$$ which in field theory is a functional.

  3. The momentum density is then defined as the functional/variational derivative $$ p(x,t)~:=~\frac{\delta L[q(\cdot,t),v(\cdot,t);t]}{\delta v(x,t)}\tag{C}$$ wrt. the velocity field $v(x,t)$.

  4. In OP's last eq. (3) there appears the notation of a 'same-spacetime' functional derivative $$\frac{\delta {\cal L}(x,t)}{\delta v(x,t)}\tag{D},$$ which is a somewhat-misleading-although-common notation for above definition (C). In particular (D) is not an actual variational derivative of the Lagrangian density ${\cal L}(x,t)$: If it were it would be infinite, since the 'numerator' and 'denominator' of eq. (D) are evaluated at the same spacetime point $(x,t)$.

  5. See also my related Phys.SE answers here, here, here & here.


It is the usual definition of conjugate momenta. The extra term comes from taking the $\textbf{variation}$ of the Lagrangian respect to $\partial_t\phi$ while allowing for mixed partial terms $\mathcal{L}[\partial_t\phi,\partial_x\partial_t\phi]$.

$$ \pi=\frac{\delta\mathcal{L}}{\delta[\partial_{t}\phi]}=\frac{\partial\mathcal{L}}{\partial[\partial_{t}\phi]}-\partial_{x}\frac{\partial\mathcal{L}}{\partial[\partial_{x}\partial_{t}\phi]} = c\partial_{t}\phi-\frac{1}{\omega_{J}^{2}l}\partial_{x}^{2}\partial_{t}\phi $$

The procedure to derive the conjugate momentum is analogous to that of deriving the Euler-Lagrange equations (Without asking for the variation to vanish), that's why the expressions are really similar.

  • $\begingroup$ Could you elaborate more on the steps for the second equality? In particular, I don't see why there should be a negative sign for the second term. Thanks $\endgroup$
    – kowalski
    Commented May 31, 2022 at 19:25

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