It is a well-known thing that there is an analogy between mechanical vibrating systems and electrical RLC circuits (see this link). My question is whether there is circuit analogue of a vibrating molecule. That is, is there an $RLC$ circuit whose eigenfrequencies coincide with those of the mechanical system below?

In the above figure we can see a vibrating system. Each black circle represents a body of mass $m$ moving in a 2D plane. The green lines represent springs with spring constants $k$, which connect the bodies. The overall shape of the molecule is such that the system has three-fold rotational symmetry and mirror symmetries.

The above system has $3 \times 2 = 6$ degrees of freedom. By solving the equations of motion of the system we find that 3 out of the 6 eigenfrequencies are zero. These frequencies are related to the $(i)$ rotation of the whole molecule around its center of mass and the translation of the whole system along the $(ii)$ $x$ and $(iii)$ $y$ directions. The other three eigenfrequencies are non-zero, two of them are degenerate with $\omega_{2,3}^2 = \frac{3 k}{2m}$ and the last one is $\omega_1^2 = \frac{3 k}{m}$.

Correspondingly the analogue $RLC$ circuit should have $6$ degrees of freedom with the same structure of eigenvalues.


1 Answer 1


There is no precise electrical analogue using only RLC components, because these components are linear. Mechanical systems are almost never linear. Assuming ideal springs, one has a Lagrangian

$$ L_M = \sum_{i = 1,2,3} \frac{1}{2}m\dot{x}_i^2 + \frac{1}{2}m\dot{y}_i^2 - \frac{k}{2}\left( \sqrt{(x_i - x_{i+1})^2 +(y_i -y_{i+1})^2} - R_0\right)^2 $$

The idea is to map positions to voltages and currents to momenta, but this is not possible using only linear components since we have a nonlinearity in the potential. This will not really simplify under a nonlinear change of variables (e.g. in terms of center of mass, three separations and an angle) - this will simply move the nonlinearity to the kinetic term.

The best we can hope to do with only inductors and capacitors (resistance is not helpful here, as it violates energy conservation) is emulate a quadratic approximation of this action about its ground state. This is not a particularly glaring approximation, considering that the 'intermolecular force is quadratic' approximation is only valid for very small displacements from equilibrium.

Let's change variables to $$ \vec{x}_n = \frac{R_0}{\sqrt{3}} \begin{pmatrix} \cos\left( \frac{2\pi}{3} n\right) \\ \sin \left(\frac{2\pi}{3}n \right) \end{pmatrix} + \vec{\xi}_n$$

Taking $\xi_n$ small, some work will hopefully get you

$$L_M \approx \sum_n \frac{1}{2}m \left(\vec{\dot{\xi}} \right)^2 - \frac{k}{2} \xi_i^2 + \text{constant} $$

which maps onto the electrical circuit

$$ L_E = \sum_{i = 1,2,3} \frac{1}{2}m\dot{q_x}_i^2 + \frac{1}{2}m\dot{q_y}_i^2 - \frac{k}{2}({q_x}_i - {q_x}_{i+1})^2 - \frac{k}{2}({q_y}_i - {q_y}_{i+1})^2 $$


Where $\phi_1, \phi_2, \phi_3$ are understood as the voltages at the nodes, the inductors have inductance $m$ and the capacitors have capacitance $1/k$.

Now suppose we did want to simulate the full nonlinearity in an electrical system. It may be possible to recreate the dynamics using op-amps for subtraction and diodes for squaring, but it would be very complicated.

  • 1
    $\begingroup$ I have experimented with a Lagrangian like this and I think that it is incorrect. For example, it cannot describe the coupling between x_1 and y_2, which is there in reality (the springs have a nonzero rest length). $\endgroup$
    – zltn.guba
    Nov 13, 2022 at 19:21
  • 2
    $\begingroup$ See edit - I don't think it's possible to do what you have in mind in a straightforward way. In particular, RLC circuits are linear ODEs with constant coefficients; a cursory look at the double pendulum will show that mechanical systems are decidedly not that. $\endgroup$ Nov 13, 2022 at 22:12
  • $\begingroup$ Thank you for the clarification. I will set this as the accepted answer for now. $\endgroup$
    – zltn.guba
    Nov 13, 2022 at 23:11
  • $\begingroup$ just throwing in some thoughts, can you perform the subtraction with destructive interference? $\endgroup$
    – antimony
    Nov 14, 2022 at 0:16
  • $\begingroup$ What do you mean? $\endgroup$ Nov 14, 2022 at 0:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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