When considering normal modes we often end up with a hamiltonian of the form

$$ H = \frac{1}{2} \dot{x}^T M \dot{x} + \frac{1}{2} x^T K x $$

where $x$ is a vector of degrees of freedom in the system, $K$ is a symmetric "potential" matrix (often positive definite), and $M$ is a positive definite (often diagonal) mass matrix. This Hamiltonian is analogous to $H= 1/2 m\dot{x}^2 + 1/2 k x^2$ for a one dimensional harmonic oscillator. The Hamiltonian gives rise to the equations of motion $$ M\ddot{x} = -Kx $$

Which is analogous to Newton's and Hooke's laws for a one dimensional harmonic oscillator: $m\ddot{x} = -kx$.

Where $x$ is a vector and $M$ is the "mass" matrix.

Normal modes can be found by examining the $M$ and $K$ matrices.

For networks of masses connected springs the kinetic energy in the Hamiltonian always looks something like:

$$ \frac{1}{2} \sum_i\sum_j \frac{1}{2} m_i \dot{x}_{i,j}^2 $$

Where $m_i$ is the mass of the $i^{th}$ mass and $\dot{x}_{i,j}$ is the velocity of the $i^{th}$ mass in the $j^{th}$ direction. When the kinetic part of the Hamiltonian looks like this it can always be written as above with $M$ diagonal.

My question: Is it ALWAYS the case, in all physical systems where we consider normal modes in this way, that the mass matrix (i.e. the matrix that gives energy due to the first derivative of the elements) is diagonal? Or are there physical examples where the mass matrix is merely positive definite?

  • $\begingroup$ A search for "mass matrix" "diagonal" "positive definite" in Google Books or Google Scholar indicates that the mass matrix is diagonal iff the masses are lumped. $\endgroup$ Jun 7, 2023 at 17:03
  • 1
    $\begingroup$ No. See this post and the associated comments. $\endgroup$ Jun 7, 2023 at 17:22

2 Answers 2


In general, you can always perform a change of basis and make $M$ non diagonal. It is more a question of whether the physically natural variables make $M$ diagonal.

A classic example where it is not the case is the linearization of the double pendulum about the lower equilibrium point. The exact Lagrangian is (choosing units of time s.t. $g=1$): $$ L=\frac{m_1l_1^2\dot \phi_1^2+m_2(l_1^2\dot\phi_1^2+2l_1l_2\cos(\phi_1-\phi_2)\dot\phi_1\dot\phi_2+l_2^2\dot\phi_2^2)}{2}+m_1l_1\cos\phi_1+m_2(l_1\cos\phi_1+l_2\cos\phi_2) $$ which after linearizing becomes: $$ L=\frac{(m_1l_1^2+m_2l_2^2)\dot\phi_1^2+2m_2l_1l_2\dot\phi_1\dot\phi_2+m_2l_2^2\dot\phi_2^2-(m_1+m_2)l_1\phi_1^2-m_2l_2\phi_2^2}{2} $$ giving a diagonal $K$ but non diagonal $M$.

In general, constrained systems are a good method for constructing systems with modes coupled by the kinetic energy.

Hope this helps.

  • $\begingroup$ Thanks, that makes sense. If the double-pendulum was considered in the $x$, $y$ coordinate system the kinetic term would be diagonal in the coordinate variables, but in polar coordinates, which are somehow more natural, there is coupling. But yes, also the point that there is always a transformation that will make the matrix diagonal. $\endgroup$
    – Jagerber48
    Jun 7, 2023 at 19:56

If the state vector $~\mathbf x~$ is constrained you obtain

$$\mathbf x=\mathbf f(\mathbf q)$$

where $~\mathbf q~$ is the neue state vector (generalized coordinates)

from here the Newton equation

$$ \mathbf M\,\ddot{\mathbf{x}}=-\mathbf K\,\mathbf x\quad\mapsto\\ \underbrace{\mathbf J^T\,\mathbf M\,\mathbf J}_{\mathbf M_J}\,\ddot{\mathbf{q}}=-\underbrace{\mathbf J^T\,\mathbf K\,\mathbf J}_{\mathbf K_J}\,\mathbf q\\ \mathbf J=\frac{\partial\mathbf x}{\partial \mathbf q}$$

in this case the mass matrix $~\mathbf M_J~$ is not diagonal


to obtain the normal modes you have to linearized the Jacobi matrix $~\mathbf J~$ at the equilibrium state $~\mathbf q_E~$ , thus the mass matrix $~\mathbf M_J~$ and the stiffness matrix $~\mathbf K_J~$ both are constant .


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