Why the Lagrangian must involve derivative of field?

[ndung200790]

Please teach me this:
Why the Lagrangian in QFT must involve derivative of field? Is it correct that because fermions and bosons(meaning all things) obey Dirac and Klein-Gordon equations,then the corresponding Lagrangians include the derivative of field?
(I know that the derivative has a important role in Gauge Local Symmetries(leading to comparator that reserve the Gauge Symmetry of the derivative))
Thank you very much for your kind helping.

 

[tom.stoer]

w/o spatial derivatives the field values at neighboring points would be uncoupled, there would be no waves, no propagation in space.

 

[dextercioby]

The Lagrangian contains derivatives, because the Hamiltonian contains momenta.

 

[haushofer]

Well, try to derive the propagator of the field if there are no derivatives in the Lagrangian, as tom.stoer notes, and you'll see the necessity :)

It's a same kind of question as "why are there time derivatives in the Lagrangian of Newtonian systems"? What's wrong with an action like, say

[tex] L[x] = A x_i x^i + B (x_i x^i)^2 + \ldots [/tex]

for coefficients A,B,...

 

[jrlaguna]

In physical terms: derivatives of the field give the notion of "neighborhood" and, therefore, "smoothness".

Imagine that you have a field in the line, f(x), characterized by a certain energy function. A very simple energy function would be ∫|df/dx|²dx, which is zero for flat fields, and small for "smooth" fields. It's the same for Lagrangians.

 

[ndung200790]

So,''physical'' means ''smoothness''?

 

[jrlaguna]

No, physical do not mean smooth. But if field values at nearby points did not have a tendency to "stay together", we would have no notion of space at all. The reason for which we have created the notion of "neighborhood" is because nearby points tend to be more correlated than far-away points. And this is achieved with the derivative terms in the lagrangian.