1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Tip: Put commas insider double-dollar-signs for better formatting. $\endgroup$ – Qmechanic Jul 9 at 8:06
1
$\begingroup$
  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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – GRrocks Jul 9 at 10:32
  • $\begingroup$ 1. No, a linear term is known as a source term. 2. No, mass is the same. 3. Anybodies guess. $\endgroup$ – Qmechanic Jul 9 at 10:46
  • $\begingroup$ 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 $(\partial_\mu\phi)^2-m^2\phi^2/2$'. $\endgroup$ – GRrocks Jul 9 at 13:12
  • $\begingroup$ Do you have a link? $\endgroup$ – Qmechanic Jul 9 at 13:16
  • $\begingroup$ I couldn't find a proper link, but it's available on Libgen-libgen.io/book/index.php?md5=CF453EB042CFD110297E562EB618CF5D $\endgroup$ – GRrocks Jul 9 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.