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I'm learning about diffusion speeds of particles in aqueous solution and the fundamental concept is thermal energy.

The notes I'm working from say that "every degree of freedom comes to thermal equilibrium with an energy proportional to temperature"

What does it mean for a degree of freedom to come to thermal equilibrium?

My best guess: Thermal equilibrium definition from wikipedia: two physical systems in thermal equilibrium when there's no net flow of thermal energy between them when connected by a path permeable to heat."

So I'd say that it means that at thermal equilibrium, every particle, regardless of which direction (degree of freedom) it's moving in, is moving with an energy proportional to temperature.

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It's a matter of probability distribution

It's not that every particle has energy proportional to the temperature, but statistically, particles are more likely to have speeds around the average temperature. Indeed, velocities are distributed according to Boltzmann distribution.

A liquid at Thermal Equilibrium

By studying the speed of thermally interacting molecules in a liquid, and by using the Boltzmann distribution, someone who has done a bit of Statistical Physics can derive another distribution called Maxwell-Boltzmann distribution (see image below from Wikipedia) which holds for thermally interacting particles which can exchange energy. Consider a pool of water at thermal equilibrium. Maybe, you expect to find few particles with kinetic energies $T<<K_bT$, few particles with energies $T>>K_b T$ and many with kinetic energy around the value $K_b T$. And this is indeed the case (image below)

Image from Wikipedia

Image taken from Wikipedia

A system not at Thermal Equilibrium

A situation of systems not at equilibrium for instance is a very weird pool where half the particles have zero speed and half the particles have a lot of kinetic energy. The particles are not interacting and hence their speeds don't change as a result of collisions with each other. Then the probability distribution does not follow the "special one" (Maxwell-Boltzmann distribution) and the system cannot be said to be at equilibrium.

To Recap:

In the case you described of the two systems at thermal equilibrium it means that both physical systems have molecules with speeds distributed according to the same probability distribution. This probability distribution is parametrised by the temperature. Hence if the systems have the same temperature, they have the same probability distribution and are at thermal equilibrium.

It is not entirely true that there is not heat exchange if the two systems are put in contact: indeed there might be small fluctuations, however on average the net heat exchange will be:

$$\langle Q\rangle_{time} = 0$$

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  • $\begingroup$ +1 Good answer, but fantastic visualisation. One suggestion I would have would be to start off with a distribution sufficiently different from the Maxwell-Boltzmann one, so that the "relaxation" to this distribution is more apparent. I wonder if that's easy to do. $\endgroup$
    – Philip
    Jul 19, 2021 at 16:09
  • $\begingroup$ Thank you very much for your comment. The image was from Wikipedia, I thought I referenced it, but I must have removed the reference while editing. It's fixed now. I agree with you that the Maxwel-Boltzmann is just a special case. Not knowing the knowledge of the OP, I decided to keep the explanation easy and I though the distribution of kinetic energies in a system to be a more intuitive and familiar concept than the distribution of a single degree of freedom. I will try and add a section later to outline the more general case! $\endgroup$
    – Lorenzo
    Jul 19, 2021 at 16:21
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First of all, in general, a degree of freedom is not the same as a direction in which a particle can move. All those directions are of course degrees of freedom, but for example if a gas molecule consists of two atoms, there are also rotational and vibrational degrees of freedom. As a general rule, the number of degrees of freedom is equal to the number of coordinates in phase space needed to fully determine the state of the system.

Having said that, every degree of freedom has some kind of energy associated with it. If the degree of freedom is a momentum, this is kinetic energy, for a rotation this would be rotation energy and for a vibration the energy of the oszillator used to describe the vibration. If the degree of freedom is in a thermal equilibrium with some heat reservoir of temperature $T$, it's associated energy will indeed in most cases be proportional to $T$.

By what I found in my notes from lectures I attended, it is not trivial to determine what "most cases" means exactly. If I am mistaken here and anyone notices, please leave a comment saying so and I will fix the answer. In a classical (as opposed to quantum-mechanical) system, "most cases" definitely include every degree of freedom the phase space coordinate of which appears quadratically in the energy of that degree of freedom. A counter-example would be a quantum-mechanical harmonic oscillator at low temperatures, where the energy is proportional to the angular frequency.

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  • $\begingroup$ Lorenzo's answer describes the idea of an equilibrium in great detail, but does not mention degrees of freedom which is why I tried to cover that aspect of the question here. $\endgroup$
    – nu.
    Jul 19, 2021 at 16:13

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