# A question on the functional dependence of the Lagrangian density

I understand that in classical mechanics the state of a particle at a given instant in time is given by its position $q$ and its velocity at that point $\dot{q}$, and given that, for any given point $q$ we are free to choose any velocity $\dot{q}$. As such, in order for the Lagrangian to characterise the dynamics of such a (one-particle, one-dimensional) system, it must be a function of position and velocity, i.e. $\mathcal{L}=\mathcal{L}(q,\dot{q})$.

What I'm slightly unsure about, is the case when we generalise Lagrangians to fields. I get that such an object must be local (as the properties of a field are, in general, local) and hence we must define a Lagrangian density $\mathscr{L}$, from which we can calculate the "classical Lagrangian" as $$\mathcal{L}=\int d^{3}x\;\mathscr{L}$$ I'm slightly unsure, however, as to why the Lagrangian becomes a function of the spatial derivative $\nabla\phi$ also? Is it simply because in order to characterise the dynamics of a system at a given point $(t,\mathbf{x})$ we need to know not only the value of the field $\phi$ at that point, but also how it is changing (in analogue to when it was just changing in time, in the classical case, we now need to consider how it change as we move in space as well as time), at that point, hence the Lagrangian must be a function of $\phi$, $\partial_{t}\phi$ and $\partial_{i}\phi$, i.e. $\mathscr{L}=\mathscr{L}(\phi, \partial_{t}\phi, \partial_{i}\phi)$?

• This question (v1) is essentially a duplicate of physics.stackexchange.com/q/27019/2451 – Qmechanic Mar 27 '15 at 15:32
• I read the post and it doesn't seem to explicitly give any intuition as to why the Lagrangian density is dependent on spatial derivatives as well?! – Will Mar 27 '15 at 15:45

2. Another answer is that if the theory is supposed to be local, this puts strong conditions on the form of the action functional $S[\phi]$, cf. e.g. this Phys.SE post and links therein. A local $S[\phi]$ has to be a spacetime integral over a Lagrangian density ${\cal L}(x)$, which only depends on a single spacetime point $x^{\mu}$. However, intuitively speaking, ${\cal L}(x)$ is allowed to depend on infinitesimally close spacetime points via spacetime derivatives of the fields.