# What is meant by a local Lagrangian density?

1. What is meant by a local Lagrangian density?

2. How will a non-local Lagrangian look like?

3. What is the problem that we do not consider such Lagrangian densities?

• – joshphysics Jan 24 '14 at 4:00

## 1 Answer

What is a local Lagrangian density?

A classical field theory on Minkowski space $\mathbb R^{d,1}$ is specified by a space $\mathcal C$ of field configurations $\phi:\mathbb R^{d,1}\to T$, and an action functional $S:\mathcal C\to\mathbb R$. The set $T$ is called the target space of the theory, and is often a vector space. If there exists a function $L:\mathcal C\times\mathbb R\to \mathbb R$ for which \begin{align} S[\phi] = \int_{\mathbb R} dt\, L[\phi](t), \end{align} then we call $L$ a lagrangian for the theory. If, further, there exists a function $\tilde L$ such that \begin{align} L[\phi](t) = \int_{\mathbb R^d} d^d\mathbf x \,\tilde L[\phi](t, \mathbf x) \end{align} then we call $\tilde L$ a langrangian density for the theory. Finally, if there exists a positive integer $n$ and a function $\mathscr L$ such that \begin{align} \tilde L[\phi](t, \mathbf x) = \mathscr L(t,\mathbf x,\phi(t, \mathbf x), \partial\phi(t,\mathbf x), \dots, \partial^n\phi(t,\mathbf x)) \end{align} then we say that the lagrangian density is local. In other words, the lagrangian density is local provided its value at a given spacetime point depends only on that point, the value of the field at that point, and a finite number of its derivatives at that same point.

An example of a non-local Lagrangian density.

Consider $T = \mathbb R$, namely a theory of a single real scalar field. Let $\mathbf a\in\mathbb R^d$ be given, and define a Lagrangian density by \begin{align} \tilde L[\phi](t,\mathbf x) = \phi(t,\mathbf x) + \phi(t, \mathbf x+\mathbf a). \end{align} This Lagrangian density is not local because the value of the Lagrangian at a given point $(t,\mathbf x)$ depends on the value of the field at that point and on the value of the field at the point $(t,\mathbf x+\mathbf a)$. If we were to Taylor expand the second term $\phi(t,\mathbf a)$ about $\mathbf x$, then we would see that the Lagrangian density depends on an infinite number of derivatives of the field, thus violating the definition of a local Lagrangian density.

What's the issue with theories with non-local Lagrangian densities?

I'm no expert on this, so I'll divert to another user. I will say, however, that people do study theories with non-local Lagrangian densities in practice, so there's nothing a priori "wrong" with them, but they might generically exhibit some pathology that you might prefer not to have.

Perhaps most relevant, though, if you're taking QFT from a high energy theorist, for example, is that the Lagrangian density of the Standard Model is local, so there's no need to consider non-local beasts if one is studying the Standard Model.

• Nice example! I think that the problem with non-local Lagrangian densities is mostly due to the fact that they need some further structure in addition to the natural local geometry and coupling constants (e.g., you introduced a vector ${\bf a}$ in your example) and these structures have never been experimentally observed. Another way to construct non-local Lagrangian densities is employing some smearing function as in the Bopp's model of (non-quantum) electron (discussed in the Feynman Lectures on Physics). – Valter Moretti Jan 24 '14 at 7:58
• @V.Moretti Thanks! I see; that's interesting. I'll take a look at Bopp's model. – joshphysics Jan 24 '14 at 8:02
• Here is: feynmanlectures.info/docroot/II_28.html#Ch28-S4 Unfortunately there is not an exhaustive discussion, nor a Lagrangian treatment just the basic non-local idea. – Valter Moretti Jan 24 '14 at 8:09