The reduced mass of a phonon mode is implicitly contained in the normalization and metric of the phonon displacements.

The scalar product and with that the norm in the displacents' space is defined as:

$$(U,V)\equiv \sum _k{m}_k{\mathbf{U}}_k\cdot {\mathbf{V}}_k$$

where

$k$ is the atomic index and

$m_k$ are the mass and the displacement of the atom

$k$.

Each normal mode

$\nu$ is thus normalized according to the condition

$${(u^{\nu },u^{\nu })}^{\frac{1}{2}}=\sqrt{\sum _k{m}_k({\mathbf{u}}_k^{\nu }\cdot {\mathbf{u}}_k^{\nu })}=1$$

With this metric one can then define the coordinate

$Q_{\nu }$ of for a generic displacement pattern

$\left\{U_k\right\}$ along normal node

$\{u_k^{\nu }\}$ as:

$$Q_{\nu }=\frac{\sum _k{m}_k({\mathbf{U}}_k\cdot {\mathbf{u}}_k^{\nu })}{\sqrt{\sum _k{m}_k({\mathbf{u}}_k^{\nu }\cdot {\mathbf{u}}_k^{\nu })}}$$

This has units

$[m{]}^{\frac{1}{2}}[l]$ but

$Q_{\nu }$ is also the value of the adimensional number representing the component of the

$U_k$ pattern along the

$u_k^{\nu }$ mode. Indicating the dimensional value as

${\overline{Q}}_{\nu }$ and the numerical value as

${\tilde{Q}}_{\nu }$ and considering that the modes are orthonormal we have

$${\overline{Q}}_{\nu }={\tilde{Q}}_{\nu }\times \sqrt{\sum _k{m}_k({\mathbf{u}}_k^{\nu }\cdot {\mathbf{u}}_k^{\nu })}$$

A way to define the reduced mass

${\mu }_{\nu }$ could then be :

$${\mu }_{\nu }^{\frac{1}{2}}=\frac{\sqrt{\sum _k{m}_k({\mathbf{u}}_k^{\nu }\cdot {\mathbf{u}}_k^{\nu })}}{\sqrt{\sum _k{\mathbf{u}}_k^{\nu }\cdot {\mathbf{u}}_k^{\nu }}}$$

or

$${\mu }_{\nu }=\frac{\sum _k{m}_k({\mathbf{u}}_k^{\nu }\cdot {\mathbf{u}}_k^{\nu })}{\sum _k{\mathbf{u}}_k^{\nu }\cdot {\mathbf{u}}_k^{\nu }}$$

and if we redefine the normal mode coordinate as

$${\overline{Q}}_{\nu }^{\prime }=L_{\nu }{\tilde{Q}}_{\nu }$$.

$L_{\nu }=\sqrt{\sum _k{\mathbf{u}}_k^{\nu }\cdot {\mathbf{u}}_k^{\nu }}$ is a length and

${\tilde{Q}}_{\nu }$ the adimensional numerical value representing the projection of our pattern on mode

$\nu$; it is easy to see that

$${\overline{Q}}_{\nu }=\sqrt{{\mu }_{\nu }}{\overline{Q}}_{\nu }^{\prime }$$
In this way for example the potential energy term of the normal mode oscillator would become

$$\frac{1}{2}{\omega }_{\nu }^2{\overline{Q}}_{\nu }^2\to \frac{1}{2}{\mu }_{\nu }{\omega }_{\nu }^2{{\overline{Q}}_{\nu }^{\prime }}^2$$