Why is this equation true for the sound pressure a loudspeaker creates? In an answer, there is this equation:
$$p = \frac{\rho S_D}{2 \pi r} \, a$$
This describes that the sound pressure ($p$) is proportional to the acceleration ($a$) of the cone of a loudspeaker.
($\rho$ is the density of air, $S_D$ is the surface area of the cone, and $r$ is the distance from the cone)
I wonder, where does this equation come from? What's the theory behind it?
 A: To derive this result, there are various possible starting points.  Probably the most rigorous approach is to start from the pressure due to a baffled rigid piston.  In this case, at a distance $r$ along the axis of the piston we have
$$p(r,t) = \rho c \left( 1-e^{-ik\phi} \right) v(r,t) \; ,
$$ 
where $\rho$ is the density of air, $c$ is the speed of sound, 
$$
\phi \doteq \sqrt{r^2+r_0^2} - r \quad\text{and}\quad
v(r,t) = v_0 e^{i(\omega t-kr)} \; .
$$
Above, $v$ is the (oscillating) piston velocity and $r_0$ is the piston radius.  This textbook result can be found, for example, in Section 7.4 of Kinsler (4th Ed).  Note that the motion is assumed to be harmonic, such that $\omega$ is the oscillation frequency and $k$ is the wavenumber such that $\omega = ck$.  In the far-field, for which $\phi \ll 1$, this reduces to
$$
p(t) = i \rho c k \phi \, v(t) \; .
$$
The presence of the $i$ shows that the velocity is out of phase with the pressure.  Since the motion is harmonic, we can rewrite this in terms of the acceleration using $a = \partial_t v = i \omega v$.  Then, the complex pressure becomes
$$
p(t) = \rho \phi \, a(t) \; .
$$
Taking $r \gg r_0$ gives $\phi \sim r_0^2/(2r)$, or
$$p(t) = \frac{\rho r_0^2}{2r} \, a(t) \; .$$
The area of the cone is $S_D=\pi r_0^2$, so we are left with
$$p(t) = \frac{\rho S_D}{2\pi r} \, a(t) \; .$$
