In the “Natural Extremum” series of articles, I cited the “Equilibrium State Axiom” mentioned in Mathematical Methodology and Problem Solving Research (edited by Zhang Xiong and Li Dehu) and used it to solve several problems in mathematics and physics. The Equilibrium State Axiom states that the equilibrium state of a system is always achieved when the potential energy reaches an extremal (minimum) value. Simply put, nature always develops in the direction of lower potential energy, much like how “water flows to lower ground.” This is inherently correct in classical mechanics. However, at times, we may unconsciously imagine it as “the equilibrium state of a system is always achieved when the total energy reaches an extremal (minimum) value.” This, however, is incorrect. This article aims to explore this issue.
First, let us look at the origin of the Equilibrium State Axiom. Starting from the Principle of Least Action and considering a conservative system, every system should correspond to an action S that takes an extremal value: S=\int_{t_1}^{t_2} L(x,\dot{x})dt
where L(x,\dot{x}) is the Lagrangian function, which generally equals T-U, i.e., kinetic energy minus potential energy. From the Euler-Lagrange equations, the variation of S being zero yields: \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)=\frac{\partial L}{\partial x}
In a narrow sense, the equilibrium state refers to a “fixed point,” meaning that the kinetic energy is zero at this state. Since potential energy U is generally only a function of position and independent of velocity, the equilibrium condition yields: \frac{\partial U}{\partial x}=0
This indicates that the equilibrium state will be the case where the potential energy takes an extremal value.
However, a generalized equilibrium state is not necessarily a pure fixed point; it might simply mean that the velocity of a certain degree of freedom is zero. For example, on a disk rotating at a constant angular velocity around its center, there exist its own fixed points, meaning they do not move relative to the disk. But from our perspective, such a point possesses a certain kinetic energy \frac{1}{2} m\omega^2 r. If we were to use the extremum of total energy to handle this, an error would occur. The correct approach should be taking the extremum of “kinetic energy minus potential energy.” In mathematical terms, the action becomes: S=\int_{t_1}^{t_2} L(x,0)dt
Which gives: \frac{\partial L(x,0)}{\partial x}=0
Therefore, the equilibrium state is still the extremum of L(x,0), which is kinetic energy minus potential energy, rather than the total energy. In the previous article Variational Solution to the Rotating Spring Extension Problem, I accidentally made two mistakes: first, I added an extra sign to the potential energy, and then I used the wrong formula. Unexpectedly, “two negatives made a positive,” and I arrived at the correct answer. I have now corrected the original text and hope for the readers’ understanding.
Furthermore, I proceeded to reflect on a point in theoretical mechanics: in the Lagrangian function L=T-U, both T and U are energies, just in different forms. Why is it that their statuses are unequal? One must take its original value, while the other must take its negative. Are kinetic energy and potential energy truly so unequal? No; under a change of reference frame, they can be converted into each other. Consider the disk problem again. We say the equilibrium point still has kinetic energy \frac{1}{2} m\omega^2 r, but for a person measuring on the disk, they indeed cannot measure any kinetic energy. They can only measure a potential energy -\frac{1}{2} m\omega^2 r, which is what we call centrifugal potential energy. In this case, there is no need to introduce “kinetic energy minus potential energy”; everything can be viewed as a potential energy problem.
This further leads me to wonder: could we simply change L=T-U to L=-(-T+U)? Here, -T would be a special term of potential energy, reflecting the non-inertial motion of the system. What would such a physics look like?
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