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For end masses $m$, central mass $M$ and two equal springs of stiffness $k$, use the normal modes $v_0=(1,1,1{)}^T$, $v_s=(1,0,-1{)}^T$ and $v_a=(1,-2m/M,1{)}^T$. Their squared frequencies are $0$, ${\omega }_s^2=k/m$ and ${\omega }_a^2=k/m+2k/M$. The prescribed velocity decomposes asThusThe zero mode is a uniform center-of-mass motion, with conserved total momentum $mu$ and velocity $mu/(M+2m)$. The other two modes have fixed centre of mass. Dividing their velocity coefficients by their nonzero frequencies gives the sinusoidal displacement amplitudes; for the zero mode the correct time dependence is linear rather than a bounded sine. This distinguishes translation of the complete system from its internal vibration.

Put $M=m_1+m_2$, define the center of mass $\mathbf{R}=(m_1{\mathbf{x}}_1+m_2{\mathbf{x}}_2)/M$, and let $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$ be relative position. Adding the two Newton's second law equations givesThe center of mass has constant velocity, by conservation of total momentum.

Subtracting the particle accelerations instead givesThus relative motion is a one-particle problem with the reduced mass $\mu$, while center-of-mass motion is uniform. The same reduced mass appears in the kinetic energy decomposition $T=\frac{1}{2}M|\dot{\mathbf{R}}{|}^2+\frac{1}{2}\mu |\dot{\mathbf{r}}{|}^2$.

Use displacements $x=(x_1,x_2,x_3{)}^T$ of the left oxygen, carbon and right oxygen along a common positive direction. In this harmonic model the kinetic energy and elastic potential energy areThus the generalized eigenvalue problem for small oscillations is $Kv={\omega }^2\mathsf{M}v$, whereThe normal modes of three masses connected by two equal springs give the three independent solutionsThe second row follows by holding the centre fixed and moving the end masses oppositely. In the remaining internal mode the end displacements are equal, and zero displacement of the centre of mass gives $2m{v}_1+M{v}_2=0$, yielding the third row; substitution verifies its frequency. Both nonzero modes have fixed centre of mass. In the symmetric stretch the two bond-length changes are equal; in the antisymmetric stretch they are opposite.

The zero mode occurs because a uniform translation changes neither spring length, so $K{v}_0=0$. It represents free center-of-mass motion, not an additional restoring vibration. Its general time amplitude is $A+Bt$. Only one translational zero mode exists in the restricted one-dimensional model; transverse and rotational motion are outside its specified coordinates.

For the initial impulse, all displacement amplitudes at zero time vanish. Resolve the initial velocity into these three mode shapes:Writing ${\omega }_s=\sqrt{k/m}$ and ${\omega }_a=\sqrt{k/m+2k/M}$ gives the impulse on an end mass of a free symmetric three-mass chain:In particularwhere $V_{\mathrm{cm}}=mu/(M+2m)$. The initial derivatives are $(u,0,0)$, as required. The carbon dioxide model therefore drifts uniformly while undergoing both internal stretches. Its total momentum is the initial $mu$; omitting the zero mode would incorrectly hold the centre of mass fixed.

Separate the center-of-mass motion from the relative motion. Two equal quark masses give reduced mass $\mu =m_q/2$. Replacing the hydrogen Coulomb strength by $a>0$ gives the Coulomb binding energies and length scaleThe factor of two in the reduced mass matters: using $m_q$ directly would double the binding energy. These are relative binding energies measured from the free quark–antiquark threshold; the rest energy $2m_q{c}^2$ is excluded.

Separate center-of-mass motion and replace $m$ by the reduced mass $\mu =mM/(m+M)$ in the relative-coordinate formulas. Then $r_n$ is the proton-electron separation, energy levels and the Rydberg constant scale by $\mu /m$, and transition wavelengths increase by $m/\mu$. The electron's own orbital radius about the center of mass is $M{r}_n/(M+m)$; the proton radius is $m{r}_n/(M+m)$. The relative speed retains $\alpha c/n$, while the electron's individual speed acquires $M/(M+m)$. Replacing the mass in the radius alone but leaving energies unchanged would be inconsistent.