# Can you have a purely position(field) dependent Lagrangian(density)?

Let us start with the usual Euler Lagrange equations, and impose $$L=L(q)$$ only-$$\frac{\partial L}{\partial q}=\frac{d}{dt}(\partial L/\partial \dot{q})$$, and the RHS becomes zero, implying $$\partial L/\partial q =0 \implies L(q)=const$$, so in fact the Lagrangian is something trivial. In the field theory case, a similar thing happens obviously $$\frac{\partial L}{\partial \phi}=\partial_\mu(\frac{\partial}{\partial(\partial_\mu\phi)}L)$$, and imposing $$L=L(\phi)$$, one gets $$\partial L/\partial \phi=0$$, a similar result.

Now things get worse. Take, say, $$L=\phi(x^\mu)$$(imagine the appropriate dimensional constant is unity). Clearly, $$\partial L/\partial \phi\neq0$$ here!Then, on switching the partial derivatives on the RHS of the E-L equations(with $$L=\phi$$), we get $$\frac{\partial \phi}{\partial \phi}=\partial_\mu(\frac{\partial}{\partial(\partial_\mu\phi)}\phi)=(\frac{\partial}{\partial(\partial_\mu\phi)}\partial_\mu \phi)$$, which is correct, it simply says $$1=1$$. But had we NOT switched the partial derivatives, we would have gotten $$\frac{\partial \phi}{\partial \phi}=\partial_\mu(\frac{\partial}{\partial(\partial_\mu\phi)}\phi)\implies 1=0$$

A similar problem can be seen in the point particle case too.

My question is, why does the order of partial derivatives screw everything up here? On the one hand, the E-L equations imply $$L$$ will be a constant if we choose it to depend only on $$q$$ or $$\phi$$. But simply switching the order of derivatives now allows us to consider a non-constant $$L$$, as the example shows.

What am I missing here?

• Tip: Put commas insider double-dollar-signs for better formatting. – Qmechanic Jul 9 at 8:06

1. The Lagrangian density $${\cal L}=\phi$$ has no stationary solution, i.e. it is inconsistent as a theory.
2. The spacetime derivative in EL equation is a total spacetime derivative $$\frac{d}{dx^{\mu}}$$ rather than an explicit spacetime derivative $$\frac{\partial}{\partial x^{\mu}}$$, see e.g. my Phys.SE answer here.
3. The derivatives $$\frac{d}{dx^{\mu}}$$ and $$\frac{\partial}{\partial \partial_{\nu}\phi}$$ do not commute. This explains the discrepancy in OP's 2 ways of calculating.
• Great, thanks. A few clarifications. 1. This doesn't prevent me from adding linear terms in $\phi$ to my $L$, right? Such terms will not change the actual behavior of the system (a point particle analogue would be that a term linear in $q$ along with the harmonic potential provides a constant global force which merely shifts the equilibrium position) 2. Say I add such a linear term to the usual Klein Gordon Lagrangian, and then complete the square. Do I have to redefine my mass now? (as it is supposed to be the coefficient of the quadratic term) 3. Why is this misleading notation so prevalent? – GRrocks Jul 9 at 10:32
• Okay, this is slightly confusing. In Classical theory of Gauge fields by Valery Rubakov, page 13, the author says that the 'theory with the general Lagrangian $(\partial_\mu\phi)^2-m^2\phi^2/2+b\partial^\mu\partial_\mu\phi+c\phi\partial^\mu\partial_\mu\phi+k\phi$ is equivalent to the theory with the theory with the Lagrangian $(\partial_\mu\phi)^2-m^2\phi^2/2$'. – GRrocks Jul 9 at 13:12