3rd order derivatives in the lagrangian

RedX

I heard that in classical field theory, terms in the Lagrangian cannot have more than two derivatives acting on them. Why is this?

In quantum field theory, I read somewhere that having more than two derivatives on a term in the Lagrangian leads to a violation of Poincare invariance. Is this true?

One thing I derived is that, for a scalar field, if you accept the canonical commutation relations as true:

[tex]
[\phi(x,t),\Pi(y,t)]=i\delta^3(x-y)
[/tex]

then unless your canonical momentum [tex]\Pi(x,t) [/tex] is equal to [tex]\dot{\phi}(x,t) [/tex], then the commutation relations of the Fourier components of [tex]\phi(x,t) [/tex] no longer obey equations like:

[tex]
[a(k,t),a^\dagger(q,t)]=\delta^3(k-q)
[/tex]

or using a different normalization scheme:

[tex]
[a(k,t),a^\dagger(q,t)]=\delta^3(k-q)(2\pi)^32E_k
[/tex]

GreyBadger

P.17 of Zee states this is because 'we don't know how to quantize actions with more than two time derivatives'. Why this is mathematically I do not know though (and was wondering the same thing myself...).

RedX

I found this in a book by some guy named Pierre Ramond, "Field Theory a Modern Primer".

"Third we demand that S [the action] leads to classical equations of motions that involve no higher than 2nd-order derivatives. Classical systems described by higher order differential equations will typically develop non-casual solutions. A well-known example is the Lorentz-Dirac equation of electrodynamics. It is a 3rd-order differential equation that incorporates the effects of radiation reaction and shows non-casual effects such as preacceleration of particles yet to be hit by radiation."

But this bugs me. I thought as long as your Lagrangian density is Lorentz-invariant, then the equations of motion will be Lorentz-invariant. So how can an equation that is Lorentz-invariant be non-causal?

genneth

The 2nd derivatives are actually first derivatives --- just integrated by parts. It is easier to consider non-field theory, but just a single particle. The Lagrangian is a function of position and velocity, and a "third derivative" would actually be a dependence on the 2nd derivative. Then see: http://www.tcm.phy.cam.ac.uk/~gz218/...-theories.html

In general, higher derivative theories require some exceptional fine-tuning to make sense.