Why is it that in presence of a long-range force Goldstone excitations are absent? Page 15 of this note states, 

If a continuous symmetry of the Lagrangian is spontaneously
  broken, and if there are no long-range forces, then exists a
  zero-frequency excitation at zero momentum.

i.e., in presence of long-range interactions symmetry breaking is not accompanied by the emergence of Goldstone bosons. It also says that

The absence of long-range forces, which may tend to couple
  spins at large distances, is necessary for the existence of a
  mode with $\omega\to 0$ as $k\to 0$.

It's worth pointing out that similar stuff is also mentioned on page 432 of Chaikin and Lubensky's Principles of Condensed Matter Physics:

In this case (i.e., in the absence of long-range forces), the Goldstone theorem implies that there is a mode whose frequency goes continuously to zero as the wavenumber goes to zero.


What does the author mean by the phrase "if there are no long-range forces"? Does it refer to the presence of a gauge field as in case of superconductivity? 
Is this phrase trying to point out the familiar fact that in presence of a gauge field the Goldstone bosons are "eaten up" in the unitary gauge? But I doubt this because once the symmetry is broken, the gauge field acquires a mass and it's no longer a long-range force.
 A: In a theory - 


*

*One can have continuous global symmetries (such as flavor symmetry). These do not have any massless gauge bosonic fields associated with them. 

*One can have continuous gauge symmetries (such as the U(1) electromagnetic symmetry). These have massless gauge bosons associated to them. These mediate long range forces.
Goldstone's theorem then states - 


*

*When a continuous global symmetry is spontaneously broken, there exists a zero-frequency excitation at zero momentum. This is a massless d.o.f. which is called the Goldstone boson.

*When a continuous gauge symmetry is spontaneously broken, the massless goldstone boson associated to the gauge symmetry becomes massive. 
In essence what happens is that the Goldstone mode of the spontaneous symmetry breaking is "eaten up" by the gauge boson and it appears as the longitudinal mode which makes it massive. In terms of representation theory, 
$$2 (\text{massless spin-1 d.o.f}) + 1 (\text{Goldstone/longitudinal mode}) = 3 (\text{massive spin-1 d.o.f})$$
