Is the Boltzmann energy distribution an instance of energy diffusion?

In summary, the equilibrium condition for confined particle diffusion of a solute in a solvent is reached when the solute spatial density is uniform (= zero density gradient), and entropy is max. However, per Boltzmann, when confined particles reach equilibrium, a non-zero energy gradient) persists indefinitely, even as max entropy is reached.
  • #1
SteveMaryland
16
2
I (mechanical engineer) have researched this question but can't get to an answer.

The equilibrium condition for confined particle diffusion of a solute in a solvent is reached when the solute spatial density is uniform (= zero density gradient), and entropy is max.

But per Boltzmann, when confined particles reach equilibrium, a non-zero energy gradient) persists indefinitely, even as max entropy is reached.

My assumption here is that both are cases of diffusion - one of species, one of thermal energy.

If both processes are cases of diffusion, why, at max entropy, does solute diffusion reach a zero density gradient, but energy diffusion reaches a non-zero energy gradient?
 
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  • #2
Hi,

SteveMaryland said:
But per Boltzmann, when confined particles reach equilibrium, a non-zero energy gradient) persists indefinitely, even as max entropy is reached.
Could you clarify this ? What gradient are you referring to ?
Any links or references ?

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  • #3
Capture.JPG
Boltzmann predicts that a non-zero range (= distribution, = gradient) of particle energies will persist "forever" in the steady-state. Since this is an energy-diffusion process, why do the particle energies not "diffuse" to a single uniform value for all particles the same way a solute/solvent particle density gradient diffuses to zero in the steady-state?
 
  • #4
SteveMaryland said:
non-zero range (= distribution, = gradient)
That is not correct. A non-zero gradient means that there is a change when you look at another position.

The distribution as found comes out as a result of averaging and is not position dependent: the outcome is the same in all places. Ergo a zero gradient.

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  • #5
But isn't this a pretty simple outcome of the diffusion equation,
$$\partial_t \rho(t,\vec{x})=D \Delta \rho(t,\vec{x}).$$
Solving this for a finite volume with the appropriate boundary conditions, this gives ##\rho=\text{const}## for ##\partial_t \rho=0##.
 
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  • #6
BvU said:
That is not correct. A non-zero gradient means that there is a change when you look at another position.

The distribution as found comes out as a result of averaging and is not position dependent: the outcome is the same in all places. Ergo a zero gradient.

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What then is the "heat death" of the universe? My understanding of the "heat death" is that condition reached where all of the thermal energy in the universe has (spontaneously) diffused to the point where there are no more (= zero) thermal gradients anywhere - all thermal energy has diffused to the limit of uniform distribution, just like ink will (spontaneously) diffuse in a solution to the limit of completely uniform distribution.

To what extent does the universe resemble the Maxwell-Boltzmann box of particles?

If there IS a resemblance, then that means that thermal energy will never "completely" diffuse (like ink in water) but rather will "stop" at the M-B distribution - which in turn means there will always be vestigial thermal gradients in the universe and "heat death" will never be reached.
 
  • #7
You have lost me completely. And it seems to me you still think that there is a gradient where there is none.

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  • #8
Alright, I will restate the question without using the word gradient - it's the same question:

What then is the "heat death" of the universe? My understanding of the "heat death" is that condition reached where all of the thermal energy in the universe has (spontaneously) diffused to the limit of uniform distribution, just like ink will (spontaneously) diffuse in a solution to the limit of completely uniform distribution.

To what extent does the universe resemble the Maxwell-Boltzmann box of particles? If there IS a resemblance, then that means that thermal energy will never "completely" diffuse (like ink in water) but rather will "stop" at the M-B distribution - which in turn means there will always be vestigial thermal gradients differences in the universe and "heat death" will never be reached.
 
  • #9
@SteveMaryland: forget about the heat death of the universe for the moment. Are you familiar with energy conservation? (That orbits of closed classical systems are confined to a constant energy "level set"?) From a purely classical perspective, the Boltzmann distribution emerges from a sort of "diffusion" over a space that very closely resembles the constant energy surface (modulo ##O(N^{\alpha})## additional conserved quantities, where ##\alpha < 1##), after projecting the resulting uniform distribution over that surface to a (much) lower-dimensional subspace. Consider the uniform distribution on a sphere: after projecting it onto a 2D cross section, is it uniform? No: instead, it looks something like ##\rho(r) \propto \frac{1}{\sqrt{1-r^2/R^2}}##. Something similar happens in higher (i.e. ##N\sim 10^{23}##) dimensions, modulo subtleties pertaining to how "well separated" the dimensions are from each other in the Hamiltonian (i.e. you can anticipate the Maxwell Boltzmann distribution for a dilute gas coupled to a thermal solid along a smooth boundary, but it's slightly more difficult to guess an a priori accurate, effective, statistical mechanical Hamiltonian for, say, the Na ions in salt, ignoring the Cl ions and valence electrons.)

There are many systems in nature that satisfy the conditions for thermodynamic equilibrium, at least to reasonable precision, and many systems that do not. Usually (but not always), a macroscopic system that isn't in equilibrium in its entirety can be thought of as consisting of small unit cells that are each in local equilibrium, and to some extent vice versa (you might find it interesting to come up with a few criteria that might determine 'local thermodynamic equilibrium'.) For example, in fluid mechanics and dynamical models of phase-transitions, it is often assumed that a local temperature is well defined for each infinitesimal control volume even though there might be significant heat gradients over large scales; because heat spreads diffusively over short distances, the equilibration time scale tends to decrease rapidly with length scale.
 

Related to Is the Boltzmann energy distribution an instance of energy diffusion?

1. What is the Boltzmann energy distribution?

The Boltzmann energy distribution, also known as the Maxwell-Boltzmann distribution, is a probability distribution that describes the distribution of kinetic energies of particles in a gas at a given temperature. It is a fundamental concept in statistical mechanics and is used to explain the behavior of gases.

2. How is the Boltzmann energy distribution related to energy diffusion?

The Boltzmann energy distribution is an instance of energy diffusion, meaning that it describes the random movement of energy from one particle to another in a gas. This diffusion of energy is what leads to the distribution of kinetic energies described by the Boltzmann distribution.

3. Is the Boltzmann energy distribution applicable to all types of particles?

The Boltzmann energy distribution is applicable to all types of particles, as long as they are in thermal equilibrium. This means that the particles are in constant interaction and have reached a state where their average kinetic energy is proportional to the temperature of the system.

4. Can the Boltzmann energy distribution be used to predict the behavior of individual particles?

No, the Boltzmann energy distribution cannot be used to predict the behavior of individual particles. It only describes the average distribution of kinetic energies in a gas and does not take into account the specific movements of individual particles.

5. How is the Boltzmann energy distribution derived?

The Boltzmann energy distribution is derived from the principles of statistical mechanics, specifically the Boltzmann factor. This factor takes into account the number of microstates that a system can have at a given energy level, and when applied to a gas, leads to the Boltzmann distribution.

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