What does 'massive' scalar field theory mean? I understand what scalar field theory is, it's just a theory that studies scalar fields.
But what does the 'massive' mean?, this might be a trivial question but I just wanted some clarification. Is it just a really big scalar field? If so why does the size really matter a field is presumed to cover all space anyways?
 A: The Lagrangian density $\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-V(\phi)$ for real $\phi $ has equation of notion $\partial_\mu\partial^\mu\phi=-V'(\phi)$. When the $\phi^2$ coefficient in $V$ is $\frac{m^2}{2}$ with $m\ge 0$, $\phi$ has mass $m$. For a complex field the Lagrangian $\partial_\mu^\ast\phi\partial^\mu\phi-V(\phi,\,\phi^\ast)$ obtains the same eom, but this time we look for $V$ to include $m^2\phi^\ast\phi$. A coupling to a vector field complicates matters, but more generally the squared mass of a real (complex) field is $\frac{\partial^2 V}{\partial^2\phi}$ ($\frac{\partial^2 V}{\partial \phi \partial \phi^\ast}$).
A: A couple of quintessential behaviors (as opposed to the definition given by JG) of a "massive" scalar field are:


*

*The dispersion relation is (for a free particle) $\omega = \pm c\,\sqrt{k^2 + \frac{m^2\,c^2}{\hbar^2}}$ rather than $\omega = k\, c$ and disturbances to the field propagate at less than the universal signalling speed limit $c$;

*The time invariant solutions for the field are solutions of the screened Poisson equation, so the field tends to be very confined. Time invariant solutions of scalar field equations with zero mass vary like $1/r$, with $r$ the distance from the "center" of the field. With a massive field, we get a variation of the form $\exp\left(-\frac{m\,c}{\hbar} r\right)/r$, which reverts to the $1/r$ variation as $m\to0$.


To understand the origin of the terminology better, witness that the Klein Gordon equation results from the Lagrangian density in J.G.'s answer when for a free particle. As in the linked Wiki article, the KG equation can be thought of as an expression of the classical Minkowski norm of the momentum 4-vector:
$$E^2 - p^2 = m_0^2$$
when we put $E=i\,\hbar\,\partial_t$, $p = -i\,\hbar\,\nabla$.
The dispersion relationship leads to another interesting insight into the mass notion for a lone, free particle. From classical mechanics and general relativity, the notions of inertial mass and gravitational mass are wonted to us, but a third interpretation is that rest mass measures what I call the particle's "stay-puttedness": the group velocity is:
$$v_g = \frac{\mathrm{d}\,\omega}{\mathrm{d}\,k} = \frac{c}{\sqrt{1+\frac{m^2\,c^2}{\hbar^2\,k^2}}}$$
Massless particles must always be observed to be travelling at speed $c$, as shown by the above (which becomes $v_g=c$ if $m=0$). They are always dispersionless. However, if $m$ is nonzero, you can slow a particle down, or "make it stay put" by making the momentum $\hbar\,k$ very small. You can see now what I mean by mass measures a particle's "stay puttedness". You can make a massive particle stay put, at least as far as the Heisenberg uncertainty principle lets you.
