Lets begin with the absorption.
The probability of ANY absorption taking place is given by the the Einstein's B coefficient, which is \begin{equation}
B_{n^{\prime}n^{\prime\prime}}=\frac{8 \pi^3}{3h^2}[|\int \psi_{n^\prime}\mu_{X} \psi_{n^{\prime\prime}}d \tau|^2 + |\int \psi_{n^\prime}\mu_{Y} \psi_{n^{\prime\prime}}d \tau|^2 + |\int \psi_{n^\prime}\mu_{Z} \psi_{n^{\prime\prime}}d \tau|^2]
\end{equation}
Above equation is in complete generality, and the wave function $\psi$ is the complete wave function. $\mu_{X}$ is the dipole moment component (in X, using Cartesian coordinates). Integration is over all space.
RULE: if the dipole moment components for a given transition are zero, the absorption will not take place. (By components I mean the integral. If the integral goes to zero then no absorption for that component, if all three integrals are zero then no absorption at all) This determines the selection rules for all absorption processes (and also induced emission, but not discussing here !).
Back to absorption, the wave-function for the molecule can be expressed as
\begin{equation}
\psi= \psi_{E} \psi_{V} \psi_{R} \psi_{T}
\end{equation}
where $\psi_{E}, \psi_{V}, \psi_{R}$ and $\psi_{T}$ are the electronic, vibrational, rotational and translational wave-functions. For vibrational analysis the electronic wave function does not matter and we can safely exclude it. (This is the wave function needed for the calculation of the integral. For general understanding consider the wave function as a mathematical function whose integral gives you information about the molecule.)
The translational selection rule is that the change in the quantum number should be $\Delta n =0$ that is the quantum number does not change (while translation in space).
[At the stage we have separated the electronic and translational wave functions, and we deal with only vibrations, hence $\psi_{v}$.]
Coming to rotational and vibrational part,
If $X_{\alpha}$, $Y_{\alpha}$ and $Z_{\alpha}$ is the coordinate which moves with the molecule, then a new coordinate can be introduced which moves and also rotates with the molecule. This is say $x_{\alpha}$, $y_{\alpha}$ and $z_{\alpha}$.
The relation between $X_{\alpha}$ and $x_{\alpha}$ are governed by angles expressed as trigonometric relations, also known as direction cosines.
Advantage of this transformation is that we can separate vibration from rotation and discuss vibration.
The electric moment of the full system of molecule has three components in Cartesian coordinates, \begin{equation}
\mu_{x}=\sum_{\alpha}e_{\alpha}x_{\alpha}
\end{equation}
\begin{equation}
\mu_{y}=\sum_{\alpha}e_{\alpha}y_{\alpha}
\end{equation}
\begin{equation}
\mu_{z}=\sum_{\alpha}e_{\alpha}z_{\alpha}
\end{equation}
Now, during vibration the charged entities are moving. To be more clear I would say the nuclei are moving. The dipole moment is dynamic and changing with the vibration. The changing dipole moment can be expressed a Taylor expansion so as to include the change from equilibrium position. This expansion is carried out using the normal coordinates( normal coordinates are coordinates of the atoms which change with the vibration, when all atoms move with same frequency but different amplitudes).\begin{equation}
\mu_{x}=\mu_{x}^{0}+\mu_{x}^{1}Q_{1}+\mu_{x}^{2}Q_{2}+...
\end{equation}
where $\mu_{x}^{n}$ is the nth derivative. Generally, only the first derivative is taken and hence, \begin{equation}
\mu_{x}=\mu_{x}^{0}+ \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1}
\end{equation}
Above result is now used with the vibrational wave function.
\begin{equation}
\int \psi_{n^\prime}\mu_{x} \psi_{n^{\prime\prime}}d\tau= \int \psi_{n^\prime} [ \mu_{x}^{0}+ \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1} ]\psi_{n^{\prime\prime}}d\tau
\end{equation}
\begin{equation}
\int \psi_{n^\prime}\mu_{x} \psi_{n^{\prime\prime}}d\tau=\mu_{x}^{0} \int \psi_{n^\prime} \psi_{n^{\prime\prime}} d \tau + \int \psi_{n^\prime} [ \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1} ]\psi_{n^{\prime\prime}}d \tau
\end{equation}
The first term would go to zero since the vibrational wave functions are orthogonal. In the second term, the condition of non-zero derivative of dipole moment is required to have this term being non-zero. Hence the change if dipole moment with vibration is utmost necessary.
Other comments: With similar approach one can derive, for UV and visible absorption that the dipole is required; and for microwave absorption we need a permanent dipole moment.
References:
Physical Chemistry by Atkins, Paula.;
Modern Spectroscopy by Hollas ;
(More advanced reference. Molecular vibrations by Wilson, Decius, Cross.)
