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The scalar field lagrangian 
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#1
Apr1405, 02:26 AM

P: 1

Hi,
I have a question about a statement I've seen in many a Quantum Field Theory book (e.g. Zee). They say that the general form of the Lagrangian density for a scalar field, once two conditions are imposed: (1) Lorentz invariance, and (2) At most two time derivatives, is: L = 1/2(d\phi)^2  V(\phi) where V(\phi) is a polynomial in \phi. Why is this? I can understand how the conditions restrict the kinetic energy term to being what it is, but I don't understand why V has to be _polynomial_ in \phi. 


#2
Apr1405, 03:11 AM

Sci Advisor
P: 1,677

Good question.
The reason is dimensional analysis. The action must be dimensionless to be a lorentz invariant, so the lagrangian has to have dimension mass^4 . So you can simply power count your fields to tabulate all renormalizable interactions. You've probably seen this before.. You know the spinor field has dimension 3/2, scalar fields dimension 1 etc So for scalar fields you can only have a (phi)^3, or b phi^4 where a is dimension 1 and b is dimensionless.. Anything higher than that would lead to negative mass dimension coupling constants and a badly nonrenormalizable theory. 


#3
Apr1405, 10:53 AM

Sci Advisor
HW Helper
P: 11,896

It's not a requirement for max.2 timederivatives.Think about Weyl gravity.The Hamiltonian formalism can be externded to an arbitrary # of time derivatives in the "kinetic" term...
Daniel. 


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