# Why can we always define the Lagrangian Density this way?

Learning some field theory, and many authors just claim "it is nice to express the Lagrangian as an integral $L = \int\mathcal{L}\,\mathrm{d}^3x$." Now I understand when dealing with fields, the sum's over all the particles turn into integrals and hence $L$ is an integral over space, but is that ALWAYS the case? Could I possibly have a Lagrangian who wasn't the integral of anything?

• A single-particle Lagrangian is not typically the integral of anything. I'm not sure I understand the question. – ACuriousMind Nov 3 '16 at 22:38
• does the OP maybe mean that you always define the action $S$ as an integral over time / integral over time and space of a lagrangian density? – Quantumwhisp Nov 3 '16 at 23:22

However, there exist ideas that a local Lagrangian formulation is not always the most efficient description, cf. e.g. the amplituhedron formulation of planar pure ${\cal N}=4$ SYM by Arkani-Hamed et al.
Ultimately, we want to work with relativistic field theories. To achieve that, the usual Lagrangian $L=L(\mathbf{q}(t), \dot{\mathbf{q}} (t), t)$ is not good enough. The Euler-Lagrange equations are: $$\frac{\partial L}{\partial q_j} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_j} \right) = 0$$ You could also write it as $$\frac{\partial L}{\partial \mathbf{q}} - \frac{d}{dt} \left( \frac{\partial L}{\partial \left( \frac{d\mathbf{q}}{dt} \right) } \right) = 0$$ This is obviously not Lorentz invariant. We need some Lorentz-scalar $\mathcal{L} = \mathcal{L} (\phi (x), \partial_\mu \phi (x))$ which satisfies: $$\frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \right) = 0$$
This $\mathcal{L}$ is usually called the Lagrangian density and it satisfies: $$L = \int d^3 x \mathcal{L}$$