What is the significance of having higher order derivatives in an action describing some system? For example, suppose I have the following two actions \begin{align} S_1&\propto \int \text{d}^4x \psi \partial \psi\\ S_2&\propto \int \text{d}^4x \psi \square \psi. \end{align} The field $\psi$ may or may not have multiple components (and describe a non-zero spin particle), I'm not sure if that is important to answer the question. What is the significance of $S_1$ being of first order in derivatives and $S_2$ being second order? To my knowledge the higher the order of derivatives contained in the Lagrangian, the more restrictions we place on the field when varying the action and ignoring whatever bounder terms may arise. I don't know why this is so, and this probably not the only significance of the order.
Asked
Active
Viewed 74 times
1
-
Related: https://physics.stackexchange.com/q/4102/ , https://physics.stackexchange.com/q/18588/2451 and links therein. – Qmechanic Oct 17 '17 at 03:53
1 Answers
0
Note that you can rewrite $S_2$ in terms of first order derivatives in the following way:
$$\psi\square\psi=\psi\partial_\mu\partial^\mu\psi=\partial_\mu(\psi\partial^\mu\psi)-\partial_\mu\psi\partial^\mu\psi.$$
Throwing away the first term (total derivative implies boundary term), you are left with:
$$S_2\propto\int\mathrm{d}^4x\partial\psi\partial\psi.$$

Demosthene
- 1,369
-
Yes, my question concerns those boundary conditions and the restrictions they may place on the field. – SigmaAlpha Oct 17 '17 at 09:09
-
@SigmaAlpha Well, my understanding is that those surface terms won't impose dynamical restrictions on your field (precisely because they don't enter the pertubative calculations), but can still contribute non-perturbative effects (e.g. instantons). – Demosthene Oct 17 '17 at 12:54