How many optical and acoustical branches are in a primitive cell? I am reading Introduction to Solid-State Physics (by Kittel) and I don't understand how he counts the optical and acoustical branches in a primitive cell. 
It says that if there are $p$ atoms in a primitive cell then we have $3p$ branches, 3 acoustical branches and $3p-3$ optical branches. 
I understand the physical difference of an optical and acoustical branche. But I don't understand:


*

*How do you know there are $3p$ branches?

*How do you know only $3$ ($3p-3$) are acoustical (optical)?

 A: TL;DR: We have $3p$ branches in total, corresponding to $3p$ independent modes of vibration. We have $3$ acoustic branches, because we are in $3d$ space, AND, Because we are dealing with an Elastic Medium, and not a Fluid.
Long version: The following discussion is based on $\S69$, Statistical Physics ; Part 1 (3ed) , Landau. The discussion mirrors Kittel's treatment (but is more lucid, in my opinion) :
Let us say we have $p$  (assume for simplicity, identical) atoms per unit cell on a lattice. Each primitive cell is labelled by $n$ = $(n_1,n_2,n_3)$ (or, $r_n =n_i a_i$ , where $a_i$ are the lattice vectors) . We are looking for modes of Elastic vibration of this system. We will denote the displacement of the $i^{th}$ atom in the $n^{th}$ primitive cell as $u_{i}(n)$. 
The solid is elastic, so the equations of motions for the displacements of the atoms should look like a system of coupled oscillators :
$\ddot{u_{i}}(n)= -\sum_{j,m}\Lambda_{ij}(n,m)u_j(m)$, where the $\Lambda$'s are some positive definite "spring constants". It is easy to motivate from the (discrete) translation symmetry of the lattice that $\Lambda(n,m) = \Lambda(n-m)$.
We are looking for plane-wave solutions of the form $u_i(n) = e_i(k)e^{i(k.r_n - \omega t)}$ for some wave-vector $k$ , where $e_i(k)$ is the "Polarization vector" corresponding to the given mode of vibration. Note that the polarization has an index $i$ ; in general, we are allowed independent vibrations for the $p$ atoms in the primitive cell.
Plugging into the E.O.M. above, we obtain :
$-\omega^2e_i(k) = -\sum_{j,m}\Lambda_{ij}(n-m)e^{ik.(r_m-r_n)}e_j(k) = -\sum_j \Lambda_{ij}(k)e_j(k)$, 
where $\Lambda_{ij}(k) \equiv \sum_m \Lambda_{ij}(m) e^{-ir_m.k}$
The polarization vectors are themselves $3d$ vectors. We can make this explicit, by including another index $\alpha , \beta = 1,2,3$. Finally, we can write the above equation in the form of an eigenvalue equation: 
$\sum_{(j,\beta)}[\Lambda_{(i\alpha)(j\beta)} (k)-\omega^2 \delta_{ij} \delta_{\alpha\beta}]e_{j\beta}(k) = 0$
The eigenvalues are solutions of $det|\Lambda_{(i\alpha)(j\beta)} (k)-\omega^2 \delta_{ij} \delta_{\alpha\beta}| = 0$. We can see that for any $k$, the $\Lambda$'s are $3p \times 3p$ matrices. Hence, for any $k$, there are $3p$ distinct solutions for $\omega(k)$. These $3p$ distinct solutions correspond to the $3p$ distinct branches
Now, any $3d$ elastic material is known to have $3$ modes of vibration, which have the property that $\omega(k) \to 0$ as $k \to 0$. These are the so called acoustic modes. These represent the propagation of Sound in the elastic medium, which are mechanical waves observed macroscopically (ie, in the long wavelength, slow frequency limit). Naturally, these are the modes for which there is no relative motion between the p-atoms in a single unit cell, ie., the entire primitive cell undergoes center of mass displacement (that is, for these modes, $e_i$ is independent of $i$). 
This is why $\omega(k) \to 0$ as $k \to 0$ : For the acoustic modes, The $k \to 0$ limit corresponds to a parallel displacement of the entire lattice ; there is obviously no restoring forces in this case, and hence, $\omega(k) \to 0$. 
Why are there $3$ acoustic modes? It can be seen that this is because for a given $k$, there are $3$ distinct directions for $e$ (remembering that for acoustic modes, there is no $i$ dependence) : Two in the plane $\perp$ $k$  ie.  , $e.k = 0$ , and the remainder along $k$ ; these are what Kittel calls the TA and  LA branches respectively (pg $97$). 
$NOTE$ that apart from the $3-$ dimensionality of space, the elasticity of the medium is also a crucial ingredient in how many modes of sound we have. The two transverse modes of sound represent shear waves, whereas the remaining one is the familiar Longitudinal sound. In contrast, (normal) Fluids cannot support shear waves. That is why they have only one mode of sound per $k$. 
A: For a primitive cell with p atoms in 3d space, there are 3p degrees of freedom, say x, y, and z for each of the p atoms. This leads to 3p total harmonic modes and hence there are a total of 3p branches.
For an acoustic mode, the atoms within the primitive cell need to move exactly in phase, giving a dispersion relation where the frequency vanishes linearly with k in the long-wavelength limit. This can happen in 3 ways, where all the atoms in the lattice move in phase along x, y or z directions (or equivalently along any 3 orthogonal directions). Hence, there are 3 acoustic modes.
In all the remaining modes, all atoms within the primitive cell will not move in phase, giving 3p-3 optical modes.
Quoting Ashcroft-Mermin, "An acoustic mode is one in which all ions within a primitive cell move essentially in phase, as a unit, and the dynamics are dominated by the interaction between cells; an optical mode, on the other hand, is one in which the ions within each primitive cell are executing what is essentially a molecular vibratory mode, which is broadened put into a band of frequencies by virtue of the intercellular interactions".
