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According to the second law of thermodynamics the entropy of a closed system always increases or stays the same. In case of the presence of potentials the system will evolve to a state with less Gibbs free energy.

Now imagine a huge collection of randomly positioned small massive marbles in a volume of free space, making the entropy of the collection a maximum.

We then arrange the marbles in a pattern. Which means the marbles show a coherent, non-random whole, with less entropy than the random case.

Now at first sight, no potentials seem involved. So it seems the energy of both arrangements is the same. The rest energies of the particles is the same in both cases.

On closer inspection, the marbles are gravitationally interacting, no matter how weak. If in outer space, waiting long enough, they will move towards their center of mass.

My question is, will a random distribution have a lower potential energy than an ordered state?

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Now imagine a huge collection of randomly positioned small massive marbles in a volume of free space, making the entropy of the collection a maximum.

If the configuration is precisely defined, we are dealing with one specific state, having probability $1$, which means that the entropy is zero. In this sense there is no difference between a random distribution of marbles and their ordered state.

The entropy becomes a meaningful concept, if we do not know in which exact state the system is - i.e., if many random states are possible. In thermodynamic equilibrium we assume that all the states with the same energy are equally probable.

On closer inspection, the marbles are gravitationally interacting, no matter how weak. If in outer space, waiting long enough, they will move towards their center of mass.

If the marbles are not able to give away energy, but only exchange energy among themselves, they will not assemble closer to their center-of-mass, but rather continue flying around, colliding. Essentially, we have here a non-ideal gas with hard core repulsive interaction at distances less or equal the radius of a marble, and weak attractive interaction at larges distances.

My question is, will a random distribution have a lower potential energy than an ordered state?

Assuming that marbles can lose kinetic energy, e.g., it is transferred into their internal energy in collisions or emitted as electromagnetic waves (via some mechanism), then the system will relax to a lower energy state. If this energy state is unique, then we will arrive to it regardless of the initial configuration - provided that we wait long enough. If there are more than one ground state, we will eventually have equal probability of finding a system in any of them. However, the time for equalizing these probabilities may be prohibitively long, in which case one may get an impression that the system is stuck in one or another ground state - we then talk about phase transitions (see Anderson's More is different. for an excellent discussion).

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  • $\begingroup$ If the marbles are stationary in space initially, won't they start moving towards their CoM because of gravity? Isn't there a gravitational potential energy, no matter how small? $\endgroup$
    – MatterGauge
    Commented Jun 16, 2022 at 10:11
  • $\begingroup$ If all marbles on the edge of the volume, then the potential energy is higher then when all near the center. $\endgroup$
    – MatterGauge
    Commented Jun 16, 2022 at 10:14
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    $\begingroup$ @Felicia The marbles will start moving, converting their potential energy into kinetic energy. But they will not stop in the configuration with the lowest potential energy but simply continue moving - like a pendulum that does not just stop in its lowest position, but continue swinging... unless it loses energy, e.g., due to friction. $\endgroup$
    – Roger V.
    Commented Jun 16, 2022 at 10:21
  • $\begingroup$ But suppose they have a fixed position in a volume. How do you know that an ordered configuration has the same gravitational potential energy as a random one. $\endgroup$
    – MatterGauge
    Commented Jun 16, 2022 at 14:04
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    $\begingroup$ @Felicia If the marbles are distinguishable (classical) then there are multiple endstates - at least as many as there are permutations of the marbles. Then there are also states that differ by rotation, etc. But I don't think that any of the final states are separated by high energy barriers, i.e., I doubt that there are distinct phases (like polarizations of ferromagnet). $\endgroup$
    – Roger V.
    Commented Jun 16, 2022 at 14:43

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