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While reading through the basic derivation of how kinetic energy is related to temperature, I stumbled upon equipartition theorem where $\frac{1}{2}mv^2 = \frac{1}{2}kT$ thus $\frac{3}{2}kT$ in 3-Dimensions for linear velocity components.

This was baffling since for linear velocities, axis can be chosen and despite however one chooses the axis, the total velocity is the same, while this is not true for the $\frac{1}{2}kT$ part. So, it seems for a single particle in 3D, the energy could be $\frac{1}{2}kT$ instead of $\frac{3}{2}kT$ depending on the choice of axis for linear velocity component.

Can anyone enlighten me on this confusion where the equipartition theorem does not seem to hold depending on choice of axis?

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2 Answers 2

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The usual way of deriving the equipartition theorem is through the manipulation of an ensemble average. Equilibrium ensemble probability densities in phase space depend on the canonical coordinates through the Hamiltonian. The kinetic part of the Hamiltonian of a system of $N$ particles in the 3D space has complete rotational symmetry, and therefore, at the level of probability density, it is impossible to have any unbalance between $x$, $y$ or $z$ directions of the velocity distribution.

Looking at the same thing in the language of time averages, the situation does not change. If time averages can be written as ensemble averages, the system must be ergodic, which means that no additional constant of motion exists, beyond energy. If the motion can be confined only along one coordinate direction, the system is not ergodic. That would be the case with the ideal gas. However, even if sometimes it is not explicitly stated, the application of statistical mechanics to the ideal gas implicitly assumes that some additional mechanism that ensures ergodicity is present, even though not appearing explicitly in the Hamiltonian (for example, one could imagine that a microscopic roughness of a thermalizing confining wall could be a sufficient source of chaotic motion to make the system ergodic).

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  • $\begingroup$ An ideal gas as described in textbooks (non-interacting point particles) never reaches thermodynamic equilibrium, cannot be associated to any temperature or pressure, or any macroscopic quantity, precisely because it is not strongly mixing. This is often neglected and this answer is valuable to remind people of that. However, ergodicity is not enough to reach equilibrium and weird confining walls don't help: The energy of each particle is still conserved, therefore there is no equipartition. To reach equilibrium one needs chaos, i.e. strong mixing, which can only arise if particles interact. $\endgroup$
    – QuantumBrick
    Commented Jul 7, 2023 at 7:22
  • $\begingroup$ @QuantumBrick Interaction with the walls is enough, provided it introduces mixing. I'll add an adjective to the confining wall to be more clear. Thanks for the comment. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jul 7, 2023 at 8:27
  • $\begingroup$ Interaction with the walls cannot introduce mixing, because the energy of each particle is conserved. This means that for e.g. a gas of $N$ particles in $d$ dimensions, there are at least $N$ conserved quantities regardless of wall geometry. The fact that the movement of each particle is chaotic is irrelevant, because phase space splits into a disjoint product of $N$ $d$-dimensional submanifolds. If $d \leq 2$ each submanifold is $d$-dimensional torus that separates chaotic and regular components, so mixing is impossible. If $d \geq 3$ there is Arnold diffusion within the submanifolds only. $\endgroup$
    – QuantumBrick
    Commented Jul 7, 2023 at 9:17
  • $\begingroup$ @QuantumBrick A thermalizing wall cannot be a rigid hard wall. It exchanges energy. Therefore there is no conservation of one-article energy. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jul 7, 2023 at 10:24
  • $\begingroup$ I didn't see you had added the word "thermalizing" in your answer before. It's worse now, because "thermalizing" needs a definition. Best case scenario the walls introduce random fluctuations by absorbing energy from certain particles and then then giving it back to other particles, and thermalization will indeed be achieved. It will be achieved because particles now interact, but through the wall. It's a contrived way to force interaction between the particles. There is no thermalization without interaction. $\endgroup$
    – QuantumBrick
    Commented Jul 7, 2023 at 11:14
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The $E=\frac{k_B T}{2}$ per degree of freedom is an average value. We do not consider the motion of a particular particle, but the ensemble average. Thus, if we have a single particle in free space, we can not associate a temperature with its motion. We need random motion to associate the kinetic energy per particle with a temperature.

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