It is only if the free-body system is symmetrical that this is true. It does not follow that for a given free-body system in equilibrium it is always stationary. We know, for example, that at a given time after collision the velocity of each of the two particles is always zero, they are not at equilibrium. However, it can be shown that when both particles come together they can have different velocities and remain at equilibrium. The free energy is independent of both, and so the symmetry and velocity must remain unchanged by the time the collision occurs. So the above reasoning does not seem to rule out the free energy being zero at equilibrium.
So how can one explain the asymmetry of energy distribution in the free-body system at equilibrium? It is not hard to find a model that allows the free energy to be zero at equilibrium. This may be done with an equation analogous to Gibbs energy (I) and the free-body time variable (T), or it may be done by setting the free energy variable zero but leaving the time variable constant constant, and the free-body time constant constant unchanged. Gibbs energy can be estimated by taking the ratio of the distance from the center of force of the system to the center of mass of the particles. In the case of symmetric systems this can be estimated when the particles are very massive, so that the distance to the center of the gravitational force equals the distance to the center of mass of the particles (which is the same as the total mass of the system), and the distance to the center of mass is a constant. We can take the radius (r) of each particle and calculate the amount of energy contained in the system at rest. Because the particles move by conservation of energy and mass (see Gibbs energy above,) it can be assumed that the energy is concentrated at the mass center.
In asymmetric systems the time constant, T, can also be calculated using Gibbs. Suppose two different free-body systems (i.e. a one-dimensional (1D) system and a three-vector system). In the one-dimensional system the energy is distributed equally between all the potential energy centers; it is in equilibrium. In the three-vector system, the energy is spread uniformly between the mass centers, so the time constant is the same as in the one-dimensional system. For a given energy distribution with density, r, this will appear to be a one-dimensional system, as seen above. To estimate the energy distribution more precisely, each
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