I am developing a mathematical model of a mechanical device consisting basically of coupled harmonic oscillators. It turns out that the system mass matrix is asymmetric. I seem to read somewhere that a mass matrix has to be symmetric, but I am not sure. So I would like to know whether it is possible for a mass matrix in this case to be asymmetric. If it can't, what are the physical implications of an asymmetric mass matrix in this case?

  • 3
    $\begingroup$ What do you mean by mass matrix? Tensor of effective mass? $\endgroup$ – Ruslan Jan 28 '14 at 11:58
  • 1
    $\begingroup$ Could you be more precise? What mass matrix are you talking about, an effective mass tensor as for a semiconductor? Or the mass matrix of analytical mechanics? $\endgroup$ – DaP Jan 28 '14 at 11:59
  • $\begingroup$ I mean a mass matrix in a simple dynamic system. $\endgroup$ – adipro Jan 28 '14 at 12:34
  • 2
    $\begingroup$ In the world of robotics and dynamical systems the mass matrix is always symmetric. It is also positive definite, a result of kinetic energy $$K=\frac{1}{2} \dot{q}^\top M \dot{q}$$ being always positive. $\endgroup$ – ja72 Jan 28 '14 at 17:48
  • $\begingroup$ @ja72, could you write your comment as an answer, please? $\endgroup$ – adipro Jan 28 '14 at 19:26

I agree with the answer of Lubos (and ja72 is right that there is also a positivity criterion). However, since you repeated the question again to ja72, let me explain the answer of Lubos more explicitly and with more details.

Suppose that in your system of harmonic oscillators the kinetic term is defined with a mass matrix $M$, i.e. (as in the answer of ja72):

$$K=\frac{1}{2} \dot{q}^T M \dot{q}$$

Now define the following to matrices $M_s=\frac{1}{2}(M+M^T)$ and $M_a=\frac{1}{2}(M-M^T)$. It is very easy to verify that $M=M_s+M_a$, $M_s^T=M_s$ and $M_a^T=-M_a$. Because of these properties, $M_s$ is called the symmetric part of $M$, while $M_a$ is called the antisymmetric part (this is also the terms Lubos used).

Now, let us notice that since $M_a$ is antisymmetric the identity $v^T M_a v=0$ holds for any vector $v$. Writing up the kinetic term again, we obtain

$$K=\frac{1}{2} \dot{q}^T M \dot{q}= \frac{1}{2} \dot{q}^T M_s \dot{q} + \frac{1}{2} \dot{q}^T M_a \dot{q}= \frac{1}{2} \dot{q}^T M_s \dot{q}$$

Hence $M_s$ defines exactly the same kinetic term as $M$! This means that without loss of generality we can assume that the mass matrix is symmetric. Or in other words: whenever you write down a kinetic energy with a a non-symmetric mass matrix $M$, you can just forget about the antisymmetric part, and write down the same kinetic energy with $M_s$ - it simply leads to the same system.

I hope that this more detailed explanation helps in understanding the answer (which was also previously given). Don't hesitate to ask, if something is still not clear.

Best, Zoltan

  • $\begingroup$ This is useful indeed, thank you. I see now that the antisymmetric part does not contribute to the kinetic energy. So I guess there is no way to know if my mass matrix is correct by looking at its symmetry? For example, let's say I arrive at a mass matrix $M = \begin{pmatrix} a & c & 0\\ c & b & 0\\ e & 0 & d \end{pmatrix}$. The kinetic energy for this matrix would be the same as that for matrix $M_s = \begin{pmatrix} a & c & e/2\\ c & b & 0\\ e/2 & 0 & d \end{pmatrix}$. However, can I tell whether $e$ should be there or not in $M$, in the first place? $\endgroup$ – adipro Jan 29 '14 at 11:03
  • $\begingroup$ Yes, both of your matrices give the same kinetic term! Exactly, as you said. (From this point of view, I don't think it makes sense to talk about where $e$ should be in the first place.) I'm happy that I could help by expanding the others' answers. Cheers, Z $\endgroup$ – Zoltan Zimboras Jan 29 '14 at 15:49

Whenever the mass is expressed as a real matrix $M$, the mass term or something else that matters is a bilinear expression. For example, the effective mass in condensed matter physics is the mass matrix $M$ so that the kinetic term of the Hamiltonian is written as $$ E_k = \frac{\hbar^2}{2m} \vec k \cdot M^{-1} \cdot \vec k $$ where $M^{-1}$ is the inverse matrix. One may always divide the latter matrix to the symmetric and antisymmetric part. The antisymmetric part doesn't affect the Hamiltonian (doesn't affect the physics) at all because it is being contracted with the symmetric tensor $k_i k_j$. So without a loss of generality, we may demand the matrix to be symmetric, and everyone does so.

In some contexts, like the mass terms for complex fields in quantum field theory, it is only the Hermitian part of the mass matrix that affects the Lagrangian or the Hamiltonian because the term is $\bar\psi \cdot M \cdot \psi$ with an extra bar (complex conjugation). In that case, we assume that the mass matrix is Hermitian for a reason analogous to the previous paragraph; the anti-Hermitian part doesn't contribute.

  • $\begingroup$ This does not answer the question, not one bit. The question uses the term "mass matrix" in the context of mechanical devices such as robotic arms. In particular, the question asks about physically coupled harmonic oscillators. $\endgroup$ – David Hammen Apr 23 '18 at 16:20
  • $\begingroup$ @DavidHammen, your complaint makes no sense at any level. It doesn't matter whether the mass matrix describes robotic arms or something else. The mathematics and the general physical reasons for symmetry, antisymmetry, or the role of the two parts are always the same. With a general enough understanding of "coupled linear oscillators", every example where we need mass matrices may be thought of as a system of coupled linear oscillators, and that's exactly what the question as well as my answer addressed. $\endgroup$ – Luboš Motl Apr 26 '18 at 15:41

In the world of robotics and dynamical systems the mass matrix is always symmetric. It is also positive definite, a result of kinetic energy

$$ K=\frac{1}{2} \dot{q}^\top M \dot{q} $$

being always positive.

  • $\begingroup$ So is it right to say that when $M \neq M^\top$, the kinetic energy of the system is incorrect? $\endgroup$ – adipro Jan 28 '14 at 23:21
  • $\begingroup$ Hi, since you asked this again, I added a more detailed answer. I hope that helps. Cheers! $\endgroup$ – Zoltan Zimboras Jan 29 '14 at 2:54
  • $\begingroup$ @adipro only the symmetric parts contribute to the KE. It is like this transformation happens internally $M \rightarrow \frac{1}{2} ( M + M^\top )$ $\endgroup$ – ja72 Jan 29 '14 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.