Validity of the Boltzmann Distribution

In summary: If you set the masses equal, say 15 units. Set the velocities equal but at different directions by choosing ##v_x## and ##v_y## components so the sums of the squares are equal. I used ##v_x=15,v_y=-35## for one and ##v_x=-35,v_y=15## for the other.After every collision the velocities will cancel out and the particles will be at rest.
  • #1
SteveMaryland
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I (mechanical engineer) have researched this question but can't get to an answer. My question concerns the validity of the Boltzmann distribution.

We start with "particles in a box". These particles (at t-zero) may exhibit a range of energies. We place this box of particles in a heat bath for a "long time" until we have thermal equilibrium - no net heat or mass transfer through the box.

Boltzmann says that in this equilibrium condition, a specific energy gradient (distribution) among the particles in the box will occur, and that this specific energy distribution will result in max system entropy per Lagrange.

Doesn't this violate the concept of spontaneous process? If hot and cold particles are placed in thermal contact, thermal energy will, spontaneously, diffuse until all energy gradient goes to zero. "Nature abhors a gradient".

But now Boltzmann says that an energy gradient will persist, even among a system of particles held under equilibrium conditions.

I understand that this distribution satisfies the Lagrange criteria for max entropy but it still does not make any sense. By extension, Boltzmann is saying that energy gradients will always persist, even in a universe that has reached max entropy.

It seems to me that the only way a system of particles can reach max entropy is when each of the particles has the SAME energy - all system energy becomes uniformly diffused among the particles, energy gradient decays to zero, and energy distribution curve decays to a singularity. How could it be otherwise?
 
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  • #2
Hello @SteveMaryland ,
:welcome: ##\qquad ## !
It isn't clear to me what you mean with 'energy gradient'. The customary gradient of a scalar function is a derivative, so e.g. w.r.t. position. The energies you are referring to are not varying with position.

##\ ##
 
  • #3
SteveMaryland said:
It seems to me that the only way a system of particles can reach max entropy is when each of the particles has the SAME energy - all system energy becomes uniformly diffused among the particles, energy gradient decays to zero, and energy distribution curve decays to a singularity. How could it be otherwise?
I would think all particles having the same energy would be the lowest possible entropy i.e. the most order, not the most disorder.
 
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  • #4
BvU said:
Hello @SteveMaryland ,
:welcome: ##\qquad ## !
It isn't clear to me what you mean with 'energy gradient'. The customary gradient of a scalar function is a derivative, so e.g. w.r.t. position. The energies you are referring to are not varying with position.

##\ ##
energy gradient = the range of energies among particles in an isolated population. If held to adiabatic conditions, why do such particles not equilibrate to a common energy?
 
  • #5
bob012345 said:
I would think all particles having the same energy would be the lowest possible entropy i.e. the most order, not the most disorder.
If I put a hot thing into thermal contact with a cold thing, and this system of 2 things then spontaneously equilibrates to a common temperature, then the entropy of the system increases, not decreases.

Why don't Boltzmann's particles in a box behave the same way?
 
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  • #6
SteveMaryland said:
If I put a hot thing into thermal contact with a cold thing, and this system of 2 things then spontaneously equilibrates to a common temperature, then the entropy of the system increases, not decreases.

Why don't Boltzmann's particles in a box behave the same way?
Because if you look at the microscopic detail of the two things, it is the average temperature of both that reaches an equilibrium, not the entire distributions of each.
 
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  • #7
Microscopic vs macroscopic? The "laws of physics" vary with scale??

Yes, I have heard that argument - which says that the "macroscopic" world is irreversible, but the "microscopic" world is reversible. Well, then where is the demarcation between micro- and macroscopic? What defines micro and macro in the universe?

I can't buy the idea that the laws of physics vary with scale, any more than with reference frame.
 
  • #8
SteveMaryland said:
Microscopic vs macroscopic? The "laws of physics" vary with scale??

Yes, I have heard that argument - which says that the "macroscopic" world is irreversible, but the "microscopic" world is reversible. Well, then where is the demarcation between micro- and macroscopic? What defines micro and macro in the universe?

I can't buy the idea that the laws of physics vary with scale, any more than with reference frame.
Then let's consider a small toy universe of perfectly elastic collisions of particles in a box. Start with two particles moving at random directions but with the same speed. If I understand, you are asserting they will always maintain the same speeds after every collision or should equal out the speeds.

Here is a simple 2D visual collision calculator.

http://www.sciencecalculators.org/mechanics/collisions/

If you set the masses equal, say 15 units. Set the velocities equal but at different directions by choosing ##v_x## and ##v_y## components so the sums of the squares are equal. I used ##v_x=15,v_y=-35## for one and ##v_x=-35,v_y=15## for the other to start. I used random starting positions ##x_1=0,y_1=230## and ##x_2=300,y_2=230## to start with. Then turn on the velocity vectors. It does not give numbers but shows the magnitude. Do they stay equal? Imagine you had three particles, then more...Note, it is fun to set the speeds very high and see what happens. Have fun...
 
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  • #9
SteveMaryland said:
energy gradient = the range of energies among particles in an isolated population
I do not understand. And I think no one does. If you want to communicate, a common language is rather essential.

SteveMaryland said:
bob012345 said:
I would think all particles having the same energy would be the lowest possible entropy i.e. the most order, not the most disorder.
If I put a hot thing into thermal contact with a cold thing, and this system of 2 things then spontaneously equilibrates to a common temperature, then the entropy of the system increases, not decreases.
So that is extremely unlikely (extremely being the understatement of the century). So the situation you propose (all particles the same kinetic energy) is extremely unlikely -- it has near zero entropy.

Why don't Boltzmann's particles in a box behave the same way?
They do, as any textbook tries to make clear.

##\ ##
 
  • #10
SteveMaryland said:
It seems to me that the only way a system of particles can reach max entropy is when each of the particles has the SAME energy - all system energy becomes uniformly diffused among the particles, energy gradient decays to zero, and energy distribution curve decays to a singularity. How could it be otherwise?
The case that all particles have the same kinetic energy is so unlikely that it lasts only until the very first collision. And there are a lot of collisions happening :smile:

(and "so unlikely" ##\equiv## has such a low entropy)

##\ ##
 
  • #11
SteveMaryland said:
How could it be otherwise?

Well, reality proves that it is otherwise. Kinetic energy distributions can be measured and rather nicely (*) obey the MB formula.

(* another understatement !)

##\ ##
 
  • #12
BvU said:
Well, reality proves that it is otherwise. Kinetic energy distributions can be measured and rather nicely (*) obey the MB formula.

(* another understatement !)

##\ ##
So what Boltzmann is saying is that the total energy E of a set of confined particles under adiabatic conditions will continuously share the same E among themselves, and the "most likely" distribution of E is the Boltzmann Distribution? And never a uniform distribution?

And this energy transfer process (among the particles) never stops? Until it reaches the Boltzmann Distribution?

I'm trying to reconcile that with the fact that thermal energy spontaneously (=always) transfers among particles until the energy is uniformly distributed among them - and then the transfer stops.
 
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  • #13
SteveMaryland said:
So what Boltzmann is saying is that the total energy E of a set of confined particles under adiabatic conditions will continuously share the same E among themselves, and the "most likely" distribution of E is the Boltzmann Distribution? And never a uniform distribution?
Yes

And this energy transfer process (among the particles) never stops?
yes

Until it reaches the Boltzmann Distribution?
Even then it just goes on and on.

I'm trying to reconcile that with the fact that thermal energy spontaneously (=always) transfers among particles until the energy is uniformly distributed among them - and then the transfer stops.
I am not aware of such a 'fact'. Collisions keep happening and a particle has a different kinetic energy after each collision. But on a macroscopic scale the distribution of energies is constant.

https://en.wikipedia.org/wiki/Kinetic_theory_of_gases

##\ ##
 
  • #14
Well that is amazing. Thank you all for assisting me. I will continue to work on my understanding of this.

But this leaves another problem - where does "micro" end and "macro" begin? Surely the laws of physics cannot be a function of scale.
 
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  • #15
I also like the picture/movie here (lemma https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution )
The distribution rapidly tends towards the MB distribution.
As you can see, there are fluctuations wrt the theoretical prediction.

But this is for a few hundred particles. For a few quadrillion the fluctuations are as good as zilch. And that's a very small volume of gas under standard conditions.

The laws of physics do not depend on scale.

##\ ##
 
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  • #16
BvU said:
I also like the picture/movie here (lemma https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution )
The distribution rapidly tends towards the MB distribution.
As you can see, there are fluctuations wrt the theoretical prediction.

But this is for a few hundred particles. For a few quadrillion the fluctuations are as good as zilch. And that's a very small volume of gas under standard conditions.

The laws of physics do not depend on scale.

##\ ##
Correct me if I'm wrong but even a two particle model would eventually populate the MB distribution over time.
 
  • #17
BvU said:
I also like the picture/movie here (lemma https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution )
The distribution rapidly tends towards the MB distribution.
As you can see, there are fluctuations wrt the theoretical prediction.

But this is for a few hundred particles. For a few quadrillion the fluctuations are as good as zilch. And that's a very small volume of gas under standard conditions.

The laws of physics do not depend on scale.

##\ ##
I think MB applies only to constrained particles which are at a density high enough to exchange kinetic energy. Because if they were not close enough, they could not exchange energy by collision and their initial energy distribution would be maintained and could not devolve to an MB distribution.

Is the set of all particles in the material universe a constrained or an unconstrained set?

For "unconstrained" particles such as those far from boxes and g-fields, I do not think MB energy distribution applies. MB is a rarified condition of matter.

Radiation, not collision, is the dominant mode of energy distribution in the universe.

So in an expanding universe, energy diffusion (by radiation) would dominate such that the terminal state ("heat death") would be characterized by every particle in the universe having acquired (by radiation) the same KE. Or approximately the same KE, and a very low KE.

And if "same KE" is a low-entropy condition, then maybe "heat death" resets entropy to zero and thus marks the beginning of a new universe...
 
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  • #18
The equilibrium state of a system is always that of maximum entropy. It follows from kinetic theory through the detailed-balance property of transition probabilities and only requires the unitarity of quantum-mechanical time evolution: The total entropy of a closed system is never decreasing, i.e., in the stationary limit of any process it must take the maximum value.

To get the equilibrium distribution fortunately you just need this maximum-entropy principle and not the entire kinetic theory. For the distribution to be stationary it must be a function of the conserved (additive) quantities of the system, i.e., energy, momentum, angular momentum, and conserved charges (or particle number in non-relativistic contexts).
 
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Related to Validity of the Boltzmann Distribution

What is the Boltzmann Distribution?

The Boltzmann Distribution is a mathematical model that describes the distribution of particles in a system at a given temperature. It is based on the principles of statistical mechanics and is used to calculate the probability of a particle occupying a certain energy state in a system.

How is the Boltzmann Distribution derived?

The Boltzmann Distribution is derived from the laws of thermodynamics and statistical mechanics. It is based on the assumption that particles in a system will distribute themselves among different energy states in a way that maximizes the system's entropy, or disorder.

What is the significance of the Boltzmann Distribution?

The Boltzmann Distribution is significant because it allows scientists to make predictions about the behavior of particles in a system at a given temperature. It is used in a wide range of fields, including physics, chemistry, and engineering, to understand and analyze the properties of materials and systems.

What are the limitations of the Boltzmann Distribution?

The Boltzmann Distribution assumes that particles in a system are in thermal equilibrium, meaning they are constantly exchanging energy with each other. It also assumes that the particles are non-interacting. These assumptions may not hold true in all systems, leading to limitations in the accuracy of the model.

How is the validity of the Boltzmann Distribution tested?

The validity of the Boltzmann Distribution can be tested by comparing its predictions to experimental data. If the model accurately predicts the behavior of particles in a system, then it is considered valid. Additionally, the assumptions of the model can be tested and validated through various experiments and simulations.

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