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This isn't a duplicate of already-answered questions, but rather a follow-up of this answer. The author presents a field-theoretical argument whereby a problematic run-away particle creation is triggered whenever third-order terms are included.

However, isn't it a much simpler explanation to attribute the lack of higher-order derivatives to the fact that the potential only needs the second derivative to be nullified to exhibit stability (i.e, at extrema and saddle points)?

PS: I'm only trying to get an intuition for the problem, and would rather avoid elaborate field-theoretic derivations.

Qmechanic
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Tfovid
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  • Related: https://physics.stackexchange.com/q/18588/2451 , https://physics.stackexchange.com/q/4102/2451 and links therein. – Qmechanic Jul 09 '19 at 08:01
  • I already mentioned the first link in the OP, as for the second, I'm afraid it's field-theoretic and doesn't help in developing an intuition. – Tfovid Jul 09 '19 at 08:07
  • ... my focus is really whether the equilibrium points of the potential are all we need since they only require second derivatives. – Tfovid Jul 09 '19 at 09:06

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Short answer-it can. You will have to modify your E-L equations accordingly(because now the variation of action will contain terms containing higher order derivatives).

While the Lagrangian CAN have higher order derivatives, it can only have a FINITE number of them. If it had an infinite number of derivatives as arguments, then you essentially have the Taylor series for that function, and your Lagrangian is now NOT LOCAL(because you can use that Taylor series to write the Lagrangian as a function of both the field value at that point, and the field value at any other point). A standard example would be $$\phi(x)\phi(x+a)$$-the second term can be known only if you know $\phi$ and all it's derivatives. So, to prevent action at a distance, it must have a finite number of higher order derivatives.

A simple example would be including terms like $\eta^{\mu\nu}\partial_\mu\partial_\nu \phi$, which is essentially a divergence and wouldn't change the physics. But now your Lagrangian IS dependent on second order derivatives, so you'll have to use the appropriate EL equations.

The question now is, why does it USUALLY depend only on first order derivatives? Well, one can think of it this way-you want your theory to reduce, in some limit, to the usual Newtonian case. And that is a 2nd order differential equation. So it would make sense if your theory's differential equation(the E-L equations) were second order too, and that can happen only if you restrict $L$ to 1st order derivatives(modulo divergence terms). Anything else and you equation has a higher order.

GRrocks
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