I was wondering why physical systems "like" to go to the minimum of potential energy and I found this question, that tries to justify the minumum total potential energy principle. I was also reading some notes on classical mechanics that treat oscillations in the following way:

Suppose we have a system with $n$ generalised coordinates $q^i$. The Lagrangian of the system is

$$ L=~~\frac12 g_{ij}(q)\dot q^i\dot q^j-V(q).\tag{1}$$

I know that the Euler-Lagrange equation yields a system of second order differential equations in which it can be seen that there is a conservative field that can be expressed in terms of its potential. This field would "point" towards the minumum of potential energy.

In the notes they say that $g_{ij}(q)$ is a metric, and a symmetric array which may depend upon the configuration coordinates. I know that this $g_{ij}(q)$ should be related to the masses of the particles in the system, but I sincerely don't know why the word "metric" is used there. Is there any other name for that term $g_{ij}(q)$?

Also, if we take the Taylor series expansion about $x^i:=q^i-q^i_0$, where

$$\dfrac{\partial V}{\partial q^i}(q_0)=0,\tag{2}$$

we get

$$L\approx \frac12 g_{ij}(q_0)\dot q^i\dot q^j - \frac12 \dfrac{\partial^2 V}{\partial q^i\partial q^j}(q_0)x^ix^j.\tag{3}$$

From here it is clear what I said about the Euler Lagrange equations showing the thing about the field, so this would be a justification for the minimum total potential energy principle, wouldn't it?

  • 2
    $\begingroup$ The $g_{ij}$ is also known as the "mass-mixing matrix" or similar things in quantum contexts. I'm not sure what else you're asking. $\endgroup$
    – ACuriousMind
    Mar 29, 2015 at 19:05
  • $\begingroup$ @ACuriousMind thanks, yes, the question is not very clear but that helps. I will take a look at that mass-mixing matrix. Thanks a lot! $\endgroup$ Mar 29, 2015 at 19:15
  • $\begingroup$ @ACuriousMind also, do you know why it is called a metric as well? $\endgroup$ Mar 29, 2015 at 19:17
  • $\begingroup$ $\uparrow$ Which notes? Which page? $\endgroup$
    – Qmechanic
    Mar 29, 2015 at 19:48
  • 1
    $\begingroup$ The Lagrangian (1) is a special case of a NLSM. $\endgroup$
    – Qmechanic
    Mar 29, 2015 at 19:52

1 Answer 1


The kinetic term of the Lagrangian is proportional to $$g_{ij}v^iv^j$$ where the $v$s are the generalised velocities. Writing them as the time derivative of the generalised coordinates, i.e. $v^i\dot q^i$, taking the square root, and multiplying by a small time lapse $\epsilon$ you get $$\sqrt{g_{ij}\dot q^i\dot q^j}\epsilon,$$ which is a first order approximation of the distance travelled by the point in configuration space $Q$ described by the coordinates $q$ during the time lapse $\epsilon$, provided that the chosen metric on $Q$ is precisely the one given by $g_{ij}$. This is easily seen to be a genuine metric, since the kinetic energy is a positive definite quadratic form.

This association works the other way around, namely given a metric on a manifold, this can be used to define a kinetic term for a Lagrangian (cf. Jacobi-Maupertuis principle).

As for why physical systems tend to the minimum of the energy, this is a fact that gets deduced from the experience and therefore assumed as a postulate for Nature. Granted this, the impossibility of having perpetual motion, free energy and other unobserved phenomena, leads to the existence of a lowest energy state (both on a classical and quantum level).

  • $\begingroup$ Could you help me in other issue I have? [Here](physics.usu.edu/torre/6010_Fall_2010/Lectures.html) in the first page, when the Lagrangian is expanded, they only expand the potential energy, right? $\endgroup$ Mar 30, 2015 at 16:42

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