Total Energy vs Average Energy (Thermo)

[laser1]

Homework Statement

Total Energy vs Average Energy (Thermo)

Relevant Equations

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When there is a probability involved with an energy state, e.g. with partition function, why is total energy the same as average energy (if it is). Just want to make sure - is this just a definition? Thanks

 

[Steve4Physics]

When there is a probability involved with an energy state, e.g. with partition function, why is total energy the same as average energy (if it is). Just want to make sure - is this just a definition?

IMO the question is unclear. Maybe you can give an example, making it clear to what the total and average energy values apply. And, of course, tell us what you think.

 

[laser1]

In part (b). In my notes there is a formula for average energy, by just using the definition of expected value, i.e. sum of all probabilities times values etc. etc.

Total energy IMO for this seems a strange term because it's based on probabilities. Honestly, my first thought was to simply put down the answer 6 epsilon (E1+E2+E3). Just wondering if "average energy" and "total energy" meant the same thing by "definition".

 

[TSny]

View attachment 351398
In part (b). In my notes there is a formula for average energy, by just using the definition of expected value, i.e. sum of all probabilities times values etc. etc.

Total energy IMO for this seems a strange term because it's based on probabilities. Honestly, my first thought was to simply put down the answer 6 epsilon (E1+E2+E3). Just wondering if "average energy" and "total energy" meant the same thing by "definition".

I agree that "total energy" is an odd phrase for this problem. At any instant, the energy of the system must be one of the values $E_1$, $E_2$, or $E_3$. (It will never be $E_1+E_2+E_3$.) I don't see any distinction between "energy of the system" and "total energy of the system".

For part (b), I think they are asking for the average energy of the system if it is in thermal equilibrium with a heat reservoir. The average energy is the same as the "expected energy".

 

[Steve4Physics]

@laser1, I will add this to what @TSny already said.

It looks like you considered equating the sum of the three energy level values with the total energy: $1\epsilon+2\epsilon+3\epsilon=6\epsilon$. This is wrong because:
- it would require the system to be simultaneously in all three levels;
- it ignores the degeneracy of level-2.
So you need to reflect on this!

My thermodynamics is somewhat shaky but if you have an expression for the partition function $Z$ (from part a) of the Post #3 problem) then the total energy (part b) of the problem) is given by $- \frac {\partial~lnZ} {\partial \beta}$ where $\beta = \frac 1{k_B T}$. This should be in your notes/text-book.

In the present context I think total/average energy are different names for the same thing because, for an isolated system, the time-averaged energy is constant and must always equal the total energy.

EDIT. I interpreted the question incorrectly. The system is not isolated, it is in thermal equilibrium with its surroundings (see Post #11). The system’s instantaneous total energy will randomly fluctuate between the three allowed values. The question in Post #3 asks for the ‘total energy’, but should (IMO) ask for the expected (average) value of energy, <E>.

 

[nasu]

@laser1 The enery of the system depends on the number of particles and not just the energy levels. The sum of the three energies has no physical meaning. What was the expression you obtained for the partition function in part (a)?

 

[Steve4Physics]

@laser1 The enery of the system depends on the number of particles and not just the energy levels

Yes. For an $N$-particle system, the average energy per particle, $\langle E \rangle = - \frac {\partial~lnZ} {\partial \beta}$, so the total system energy is $N \langle E \rangle$. Maybe that’s the distinction between ‘average’ and ‘total' queried in Post #1. @laser1, would that answer your question?

That's wrong so I've deleted it. (I need to avoid topics about which I'm ignorant!)

 

[laser1]

@laser1 The enery of the system depends on the number of particles and not just the energy levels. The sum of the three energies has no physical meaning. What was the expression you obtained for the partition function in part (a)?

$$Z=e^{-\beta\epsilon}+2e^{-2\beta\epsilon}+e^{-3\beta\epsilon}$$

 

[laser1]

Yes. For an $N$-particle system, the average energy per particle, $\langle E \rangle = - \frac {\partial~lnZ} {\partial \beta}$, so the total system energy is $N \langle E \rangle$. Maybe that’s the distinction between ‘average’ and ‘total' queried in Post #1. @laser1, would that answer your question?

yeah good point, N is not given though so I'm a bit hesitant. However, that would make sense!

 

[TSny]

I think the system is meant to be one object with three possible energies rather than N objects (particles) each with three possible energies. Of course, I could be wrong.

Here's a similar problem form Statistical Physics by F. Mandl:

In the back of the book, Mandl gives the following answer:

This answer corresponds to treating the system as a single object having three possible energies.

 

[hutchphd]

This answer corresponds to treating the system as a single object having three possible energies.

I believe you need to include that such a single system is in equilibrium (i.e. loosely coupled) to a thermal resevoir at temperature T (or β if you prefer). Otherwise the correspondence is confusing (to me at least).

 

[TSny]

I believe you need to include that such a single system is in equilibrium (i.e. loosely coupled) to a thermal resevoir at temperature T (or β if you prefer). Otherwise the correspondence is confusing (to me at least).

Yes. I took the equilibrium condition for granted. See post #4. But you are right to emphasize it.