# Canonical ensemble

1. Sep 12, 2017

### Pushoam

1. The problem statement, all variables and given/known data

The probability that the system has the energy ε i.e.P(ε).
The system could have any energy between 0 and E.
So, P(ε) = 1/(no. of possible systems with different energies)
I cannot understand how P (ε) is related to the no. of possible microstates the reservior could have with energy ( E - ε).

In eq.4.10, it is differentiating wrt. E. But, E is constant. So, dE is 0. What about this?
I have taken the image from Blundell's book.
2. Relevant equations

3. The attempt at a solution

2. Sep 12, 2017

### Staff: Mentor

This is not correct. The probability is P(ε) = (# of possible arrangements where the energy of the system is ε) / (total # of arrangements)
The numerator is thus $\propto \Omega (E - \epsilon)$.

E is fixed for a particular situation, but multiplicity is still a function of E, i.e., it's value depends on E.

3. Sep 12, 2017

### Pushoam

The canonical ensemble applies to a system coupled to a large environment with which it can exchange energy (i.e., an energy “reservoir”). Suppose that we can describe the whole system + reservoir by a microcanonical ensemble. Then, the probability of the system being in a particular configuration of energy E is simply proportional to the number of ways of depositing the remaining energy U-E in the reservoir. How?
The no. of possible arrangements of the smaller system with energy $\epsilon$ is 1, while the no. of possible arrangements of the smaller system and the bigger system is $\Omega (E - \epsilon)$.
I think you mean by "# of possible arrangements where the energy of the system is ε)" the no. of possible arrangements of the smaller system and the bigger system is $\Omega (E - \epsilon)$ instead of no. of possible arrangements of the smaller system with energy $\epsilon$ is 1.Why so?

4. Sep 12, 2017

### Staff: Mentor

I don't understand your question. The total energy is $E$, which is distributed among the systems ($E_s = \epsilon$) and the reservoir ($E_r = E - \epsilon$). The total number of microstates for each value of $\epsilon$ will thus depend $\Omega_s(E_s)$ and $\Omega_r(E_r)$.

$\Omega (E - \epsilon)$ is the multiplicity of the reservoir, not system+reservoir (using the notation above, $\Omega_r (E - \epsilon)$).

See above.

5. Sep 12, 2017

### Pushoam

The total no. of microstates of the microcanical system (when the energy of the smaller system is $\epsilon$) is $\Omega_s(E_s)\Omega_r(E_r)$.
The smaller system could be allowed to have a no. of different energies corresponding to different microstates.
Let's say that the smaller system is allowed to have n different energies. Then, there are n microstates.
Now, the probabililty that the smaller system has energy $\epsilon$ is P(ε) = 1/n. Is this correct?

6. Sep 12, 2017

### Staff: Mentor

The number of possible energy states for the small system is irrelevant (and often taken to be infinity). What is important is the number of microstates for a given value of the energy ε. As the text states, it is assumed (without loss of generality) that the energy states of the small system are not degenerate, and therefore $\Omega_s(\epsilon) = 1 \forall \epsilon$.

That would be correct if the reservoir was not there. What is important here is that the reservoir has different probabilities of being in states of different energies, and some energy states of the reservoir are much more probable than others. In other words, you have to calculate the conditional probability $P(\epsilon | E_r = E - \epsilon)$.

7. Sep 12, 2017

### Pushoam

This is because the reservoir can have a no. of microstates corresponding to a given energy.Right?
This I am not getting.
I want to calculate the probability that the smaller system will have energy $\epsilon$.
Whenever the small system has energy $\epsilon$, the reservoir has to have the energy $E - \epsilon$.
I think $P(\epsilon | E_r = E - \epsilon)$ means : the probability that the smaller system has energy $\epsilon$ , when it is given that the reservoir has the energy $E - \epsilon$. But, in this case $P(\epsilon | E_r = E - \epsilon)= 1$.

Is this correct?

8. Sep 13, 2017

### Pushoam

I think I have got it now.
The smaller system and the reservoir together can be described by a Microcanonical ensemble as its energy E is constant.
Acc. to the fundamental assumption of stat-mech , each microstate corresponding to Microcanonical ensemble is equally probable.
Let's say that the smaller system is allowed to have n no. of different energies and so n no. of microstates.
Then the probabililty that the smaller system is with energy $\epsilon$ and the reservoir has energy E- $\epsilon$ is
denoted by P($\epsilon$).
Now, $P(\epsilon) = \frac {(\text{ no. of microstates of the smaller system with energy} ~ \epsilon )* (\text{no. of microstates of reservoir with energy }(E- \epsilon))} {\text{total no. of microstates of the microcanonical system}}$

$\\ ~~~~~~~~~~~~~= \frac {\Omega_s (\epsilon) \Omega_r (E-\epsilon)}{\Sigma_i \Omega_s(\epsilon _i) \Omega_r(E-\epsilon _i)}$
$~Since, \Omega_s (\epsilon)= 1 \text{given in the problem }~, \\P(\epsilon) ∝ \Omega_r (E-\epsilon)$

Is this correct?

9. Sep 13, 2017

### Pushoam

Let's take $P(\epsilon) = M \Omega_r(E- \epsilon)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(0)$ , where M is an appropriate constant.
So, we have to calculate, $\Omega_r(E-\epsilon)$.
Temperature of the microcanonical system is defined as :
$\frac1 {k_bT} = \frac{d \ln \Omega(E)} {dE}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1) \\~here, \Omega (E) = \Sigma_i\Omega_s(\epsilon_i)\Omega_r(E-\epsilon_i)$
Temperature of the reservoir (which is equal to the temperature of the microcanonical system) when its energy is (E - \epsilon) is given by
$\frac1 {k_bT} = \frac{d \ln \Omega_r(E -\epsilon)} {d(E- \epsilon)}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$
Is this correct or is the definition of tem. by eq.(1) valid only for a microcanonical system?

Assuming that (2) is correct.
$\frac {d(E- \epsilon)}{k_bT}= \frac {d \Omega_r(E -\epsilon)} { \Omega_r(E -\epsilon)}$
Integrating both sides,
$\frac {(E- \epsilon)}{k_bT} = \ln \Omega_r(E -\epsilon) +C_1$
$\Omega_r(E -\epsilon) = N e^{\frac{- \epsilon}{k_bT}} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$
Where, C_1 and N are the appropriate constants.

Using Eq(0) and (3) ,
$P(\epsilon) = Ae^{\frac{- \epsilon}{k_bT}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)$
Where A is an appropriate constant.

Is this correct?

10. Sep 13, 2017

### Staff: Mentor

As you figured out afterwards, this is not correct. $P(\epsilon | E_r = E - \epsilon)$ is a function of $\epsilon$, because the multiplicity of the reservoir is a function of $E_r$ (and hence of $\epsilon$).

Correct.

In the micro canonical ensemble, temperature is not defined, since $E$ is fixed.

I'm not used to seeing it derived this way, but I don't see anything wrong.

11. Sep 13, 2017

### Pushoam

Thank you, thanks a lot.