[QUE/ME-14002] ME-PROBLEM

Show that the principal moments of inertia of a linear chain consisting of two kinds of atoms \(a,b\), with origin chosen to coincide with the centre of mass, is given by \[I_1=I_2=\frac{1}{M}\sum_{i\ne j}m_im_j \,d_{ij}^2, \qquad I_3=0.\] where the summation includes each pair of \(i,j\) atoms once and \(d_{ij}\)is the distance between atoms in the pair and \(M\) is the total mass. Verify that this gives correct answer for a triatomic molecule.

Let the positions of the masses be \(x_i\) measured from some origin on the line joining the atoms. The moment of inertia about the origin, \(I_0\) is given by \[ I_0 = \sum_i m_i x_i^2.\] Letting \(I_\text{cm}\) denote the moment of inertia about the centre of mass. Using the parallel axes theorem we have \[I_0= I_\text{cm} + M X^2 , \] where \[ X = \frac{1}{M} \sum_i m_i x_i\] is the position of the centre of mass and \(M\) is the total mass. Therefore we get \begin{eqnarray} I_\text{cm} &=& I_0 - M X^2 = \frac{1}{M}\Big\{M \sum_i x_i^2 - \Big(\sum_im_ix_i\Big)^2\Big\} \\ &=& \frac{1}{M}\Big\{\Big(\sum_jm_j\Big) \sum_i (m_i x_i^2) - \Big(\sum_im_ix_i\Big)\Big(\sum_j m_j x_j\Big) \Big\}\\ &=& \frac{1}{M}\sum_{ij}\Big\{ m_jm_i (x_i^2 -x_ix_j) \Big\}\\ &=& \frac{1}{M}\sum_{i\ne j}\Big\{ m_jm_i (x_i^2 -x_ix_j) \Big\}\\ &=& \frac{1}{2M}\sum_{i\ne j}\Big\{ m_jm_i (x_i^2 -2x_ix_j + x_j)^2\Big\}\label{EQ01} \end{eqnarray} In the last step, for \(S_{ij}\) symmetric under exchange \(i\leftrightarrow j\), we have used \[\sum_{ij} S_{ij} T_{ij} =\frac{1}{2}\sum_{ij} S_{ij} \big(T_{ij} + T_{ji}\big)\] In the sum in \eqRef{EQ01}, each pair {ij} is counted twice, hence we get \begin{eqnarray} I_\text{cm} &=& \frac{1}{M}\sum_\text{pairs}\Big\{ m_jm_i (x_i^2 -2x_ix_j + x_j)^2\Big\}\\ &=& \frac{1}{M} \sum_\text{pairs} (x_i-x_j)^2. \end{eqnarray} where now each pair is counted once.