The Equilibrium Macrostate: Reif's Statistical Physics

[Kashmir]

Reif, statistical physics

"The equilibrium macrostate of a system can be completely specified by very few macroscopic parameters. For example, consider again the isolated gas of $N$ identical molecules in a box. Suppose that the volume of the box is $V$, while the constant total energy of all the molecules is $E$. If the gas is in equilibrium and thus known to be in its most random situation, then the molecules must be uniformly distributed throughout the volume $V$ and must, on the average, share equally the total energy $E$ available to them. A knowledge of the macroscopic parameters $V$ and $E$ is, therefore, sufficient to conclude that the average number $\bar{n}_{s}$ of molecules in any subvolume $V_{s}$ of the box is $\bar{n}_{s}=N\left(V_{s} / V\right)$, and that the average energy $\bar{\epsilon}$ per molecule is $\bar{\epsilon}=E / N$. If the gas were not in equilibrium, the situation would of course be much more complicated. The distribution of molecules would ordinarily be highly nonuniform and a mere knowledge of the total number $N$ of molecules in the box would thus be completely insufficient for determining the average number $\bar{n}_{s}$ of molecules in any given subvolume $V_{s}$ of the box."

The author defines equilibrium state corresponding to the "**most random situation**" .

I understand why on the average the particles are uniformly distributed throughout the volume, because the number of combinations for a uniform distribution is maximum.

But I'm having trouble understanding why using this definition of equilibrium as the most random situation how can I deduce that the average energy of a gas particle in equilibrium is $\bar{\epsilon}=E / N$.
Where does the idea of "most random situation" enter the argument?

I'm sorry if this question is elementary. Please tell me where to go to learn if this question is too basic.

Thank you.

 

[PeroK]

But I'm having trouble understanding why using this definition of equilibrium as the most random situation how can I deduce that the average energy of a gas particle in equilibrium is $$\bar{\epsilon}=E / N$$.
Where does the idea of "most random situation" enter the argument?

I'm sorry if this question is elementary. Please tell me where to go to learn if this question is too basic.

Thank you.

The average energy per particle is always $\bar{\epsilon}=E / N$, by the definition of average!

 

[Kashmir]

The average energy per particle is always $\bar{\epsilon}=E / N$, by the definition of average!

Hello sir,
The author says just before this paragraph:

"The time independent equilibrium situation which is ultimately attained corresponds therefore to the most random distribution of the total energy of the gas over all the molecules. Each molecule has then, on the average, the same energy and thus also the same speed"

The author suggests that the most random situation implies that the average energy is E/N.
I'm not able to see the link between the most random distribution and average energy.

 

[PeroK]

The author suggests that the most random situation implies that the average energy is E/N.
I'm not able to see the link between the most random distribution and average energy.

What definition of "average" are you using?

 

[Kashmir]

What definition of "average" are you using?

I'm following the authors definition, given before this paragraph, that is $\frac{1}{T} \int E \cdot d t$.

Also the way the author gives his argument on another page to find the average energy per particle is:
"Hence the basic question becomes:
How is the fixed total energy of the gas distributed over the individual
molecules? It is possible that one group of molecules might have very
high energies while another group might have very low energies. But
this kind of situation is quite special and would not persist as the molecules collide with each other and thus exchange energy
The time independent equilibrium situation which is ultimately attained corresponds therefore to the most random distribution of the total energy of the gas over all the molecules.
Each molecule has then, on the
average, the same energy"

Please note how the author is stressing the 'most random distribution ' in his proof.

I don't understand how each particle having average energy as E/N corresponds to the most random distribution?

(By random the author means "A situation, which can be obtained in many different ways, is said to be random or disordered". The most random therefore means the situation that can be obtained in most different ways. )

 

[PeroK]

I'm following the authors definition, given before this paragraph, that is $\frac{1}{T} \int E \cdot d t$.

That's the average over time for the system. That's not the average per particle.

Are you sure you are not way out of your depth with this textbook?

 

[Kashmir]

That's the average over time for the system. That's not the average per particle.

Are you sure you are not way out of your depth with this textbook?

$\frac{1}{T} \int E \cdot d t$ the E is energy of one particle ,i should have written it as $\frac{1}{T} \int e \cdot d t$

 

[PeroK]

$\frac{1}{T} \int E \cdot d t$ the E is energy of one particle ,i should have written it as $\frac{1}{T} \int e \cdot d t$

It's still a time average and completely independent of how $e(t)$ is calculated.

 

[vanhees71]

Well, assuming ergodicity (i.e., that time-averaging corresponds to ensemble-averaging) you can define the thermodynamical energy of a particle via the time average along the trajectory of the particle. You have to integrate over a time interval large compared to the "typical microscopical time scales" but small compared to the "typical macroscopical time scales". This coarse grains over the random thermal fluctuations. Of course this assumes that you have these separations of time scales. For thermal equilibrium this time average becomes constant of course.

 

[Kashmir]

Well, assuming ergodicity (i.e., that time-averaging corresponds to ensemble-averaging) you can define the thermodynamical energy of a particle via the time average along the trajectory of the particle. You have to integrate over a time interval large compared to the "typical microscopical time scales" but small compared to the "typical macroscopical time scales". This coarse grains over the random thermal fluctuations. Of course this assumes that you have these separations of time scales. For thermal equilibrium this time average becomes constant of course.

Hello sir,
You said "For thermal equilibrium this time average becomes constant of course" and I think you meant the average then equals $E/N$.

I am not sure how the average equals $E/N$.

The author first defined the equilibrium as the most random situation, and then using this definition he tries to prove that the average energy of each particle is $E/N$.This is what the author writes about a gas in isolated box having energy $E$ :
"Hence the basic question becomes:
How is the fixed total energy of the gas distributed over the individual
molecules? It is possible that one group of molecules might have very
high energies while another group might have very low energies. But
this kind of situation is quite special and would not persist as the molecules collide with each other and thus exchange energy.
The time independent equilibrium situation which is ultimately attained corresponds therefore to the most random distribution of the total energy of the gas over all the molecules.
Each molecule has then, on the
average, the same energy"

The author clearly is suggesting that because the equilibrium is the most random situation therefore the average energy per particle is $E/N$.

Can you please help me understand that having an average energy of each particle as $E/N$ is the most random distribution.

Please help me.

 

[PeroK]

Well, assuming ergodicity (i.e., that time-averaging corresponds to ensemble-averaging) you can define the thermodynamical energy of a particle via the time average along the trajectory of the particle. You have to integrate over a time interval large compared to the "typical microscopical time scales" but small compared to the "typical macroscopical time scales". This coarse grains over the random thermal fluctuations. Of course this assumes that you have these separations of time scales. For thermal equilibrium this time average becomes constant of course.

Nevertheless, if a system of $N$ particles has total energy $E$, then the average energy of a particle is $\dfrac E N$, by definition. I don't understand how that could be a source of confusion.

Invoking the concept of a time-based average seems to miss the point entirely.

 

[Kashmir]

Nevertheless, if a system of $N$ particles has total energy $E$, then the average energy of a particle is $\dfrac E N$, by definition. I don't understand how that could be a source of confusion.

Invoking the concept of a time-based average seems to miss the point entirely.

Author says about the average:

"This does not mean that each molecule has the same energy at anyone time. The energy of anyone molecule fluctuates quite appreciably in the course of time as a result of its collisions with other molecules. But when each molecule is observed over a sufficiently long time interval $\tau$, its average energy over that time interval is the same as that of any other molecule"

And also author gives the definition of average as( here average of number density)
" $[\bar{n}(t)]_{r} \equiv \frac{1}{\tau} \int_{t}^{t+\tau} n\left(t^{\prime}\right) d t^{\prime} .$"

 

[vanhees71]

It depends on which average you want to calculate. If you just want the average energy per particle, you are right. If you want an ensemble average in the sense of statistical mechanics it may be different. In thermal equilibrium, however it's clear that each particle has the same average energy, which is $U/N$ ($U$ total internal energy).

 

[Kashmir]

It depends on which average you want to calculate. If you just want the average energy per particle, you are right. If you want an ensemble average in the sense of statistical mechanics it may be different. In thermal equilibrium, however it's clear that each particle has the same average energy, which is $U/N$ ($U$ total internal energy).

I don't know what ensemble average means. I'm just on the first chapter of this undergraduate book, and I'm following his definition of average i.e time-average.

How is it clear "that each particle has the same average energy, which is U/N (U total internal energy)" in equilibrium?

 

[PeroK]

I don't know what ensemble average means. I'm just on the first chapter of this undergraduate book, and I'm following his definition of average i.e time-average.

How is it clear "that each particle has the same average energy, which is U/N (U total internal energy)" in equilibrium?

Perhaps you should start by analysing a system of two particles. Try to do it yourself. Not rely on someone else to do it and explain it to you.

 

[Philip Koeck]

This is what the author writes about a gas in isolated box having energy $E$ :
"Hence the basic question becomes:
How is the fixed total energy of the gas distributed over the individual
molecules? It is possible that one group of molecules might have very
high energies while another group might have very low energies. But
this kind of situation is quite special and would not persist as the molecules collide with each other and thus exchange energy.
The time independent equilibrium situation which is ultimately attained corresponds therefore to the most random distribution of the total energy of the gas over all the molecules.
Each molecule has then, on the
average, the same energy"

The author clearly is suggesting that because the equilibrium is the most random situation therefore the average energy per particle is $E/N$.

Can you please help me understand that having an average energy of each particle as $E/N$ is the most random distribution.

Please help me.

I think what Reif is explaining is actually very simple, maybe simpler than it sounds.
You could imagine that half the molecules, maybe the half that is in the left half of the volume, together have 3/4 of the total kinetic energy E, whereas the other half, in the right half of the volume, have a total of E/4 together. This would also mean that the average energy per molecule is different in the two halves (by a factor 3).
However, this is not equilibrium and the only (simple) way you can keep the gas like that is by inserting an insulating wall between the two halves.
Without the wall the molecules will collide with each other and if you wait long enough each half will have a total energy of E/2 with relatively small fluctuations around that value.
Then the gas is in equilibrium and on average each molecule will have the energy E/N with relatively large fluctuations around that value.