# Find the probability that a system has energy $\epsilon$ at $T = \infty$

A system of $8$ identical distinguishable particles is in equilibrium in a heat bath at temperature $T$ Each particle has $5$ states with equally spaced energy levels ${\epsilon}_{j}=j{\epsilon}$ for $j=0,1,2,3,4$.

What is the entropy of a single particle at $T=\infty$?

$(\mathrm{a}) \, 3.69k_B$

$(\mathrm{b}) \, 1.61k_B$

$(\mathrm{c}) \, 1.30k_B$

$(\mathrm{d}) \, k_B$

$(\mathrm{e}) \, 0$

I know that the entropy can be calculated via the Gibbs' entropy: $$S_G=-k_B\sum_{j}p_j\ln(p_j)\tag{1}$$ which is a generalization of the Boltzmann entropy: $$S_B=-N\,k_B\sum_{j}p_j\ln(p_j)=k_B\ln\Omega\tag{2}$$

In the answer it states that

$\color{red}{\fbox{$\text{At}\,T=\infty \,\text{all probabilities are equal}$}}$.

which I don't understand.

Using $(2)$ with $N=5$ and $p_j=\left(\dfrac{1}{5}\right) \, \forall\, j$
$$S_B=-N\,k_B\sum_{j}p_j\ln(p_j)=-5k_B\frac{1}{5}\ln\left(\frac{1}{5}\right)\approx 1.609437912k_B\implies (\mathrm{b})$$

Now here is the real question that I have, there is a reason why I put the former question & answer first which will soon become clear:

A system of $8$ identical distinguishable particles is in equilibrium in a heat bath at temperature $T$ Each particle has $5$ states with equally spaced energy levels ${\epsilon}_{j}=j{\epsilon}$ for $j=0,1,2,3,4$ $\color{blue}{\text{(everything is the same as the previous question)}}$.

What is the probability that the system has energy $\epsilon$ at $T = \infty$?

$(\mathrm{a}) \, 0.2$

$(\mathrm{b}) \, 3.03\times 10^{-2}$

$(\mathrm{c}) \, 2.05\times 10^{-5}$

$(\mathrm{d}) \, 2.56\times 10^{-6}$

$(\mathrm{e}) \, 0$

My Attempt:

Given that

$$p_j=\frac{\exp{\left(\frac{-{\Large\epsilon}_j}{k_B\, T}\right)}}{Z}\tag{3}$$

where the partition function $Z$ is given by $$Z=\sum\limits_j\exp{\left(\frac{-{\epsilon}_j}{k_B\, T}\right)}\tag{4}$$

But since $T=\infty$ I won't be able to make any use of $(3)$ or $(4)$.

So I would say that the answer is $(\mathrm{a})$ as I made use of quote from the previous question marked in $\color{red}{\mathrm{red}}$. If all probabilities are equally likely then $\dfrac15$ seems correct to me.

End of attempt.

Now the answer to that question states that

The probability is $p_0\cdot p_1^{N-1}$ times the number of permutations of one particle in the $j=0$ and $(N-1)$ in the $j=1$ states, which is $N$.

I have no idea what the quote is talking about but the correct numerical answer is $2.0480000000000007\times 10^{-5}$, which is $(\mathrm{c})$.

Could anyone please provide me with some hints or show how the author was able to reach the answer of $(\mathrm{c})$?

Thank you.

## EDIT:

I've just realized the correct answer is $$\frac{8}{5^8}=\frac{8}{390625}\approx 2.0480000000000007\times 10^{-5}$$ which I just found by using $$\frac{\text{Number of microstates of the system that are assigned energy } \Large\epsilon}{\text{Number of all possible microstates}}$$ thanks to the comment below by @JánLalinský.

Now I would just like to understand the other way of calculating it which was outlined in the final quotation.

Regards.

• The condition $T=\infty$ here means all microstates are equally probable. Probability of specified energy $\epsilon$ can be calculated as (number of microstates of the system that are assigned energy $\epsilon$ ) / (number of all possible microstates). – Ján Lalinský Jan 29 '17 at 12:58
• @JánLalinský Thanks for your reply. But why does $T=\infty$ mean all microstates are equally probable? – user138066 Jan 29 '17 at 13:06
• The condition $T=\infty$ is unphysical. Its purpose is to consider limit of probabilities as function of $T$, when $T\rightarrow \infty$. When you take Boltzmann probabilities and take the limit of probabilities of all microstates, you'll find out the limit is the same for all of them. What this means is the higher the temperature, the closer the probabilities will be. This is true for any system with finite number of states. – Ján Lalinský Jan 29 '17 at 13:20
• @JánLalinský So as $T\propto S$ and the largest entropy is when all microstates are equally likely. Is that correct? – user138066 Jan 29 '17 at 13:24

Tip: Aren't you confusing the probabilty of a particle in a $\varepsilon$ state with the probability of the system ($8$ particles) to be in a $\varepsilon$ state? I mean, you CAN use your (3) and (4) expressions in order to find the probability of a particle of being in an $\varepsilon$ state (as $T=\infty$ stands for the limit $T\rightarrow \infty$): $$p_j=\frac{e^{\frac{-\varepsilon }{k_BT}}}{1+e^{\frac{-\varepsilon}{k_B T}}+e^{\frac{-2\varepsilon}{k_B T}}+e^{\frac{-3\varepsilon}{k_B T}}+e^{\frac{-4\varepsilon}{k_B T}}},$$ which for $T\rightarrow\infty$ goes to $1/5$, which is not what they are asking for but a correct result. (In fact it can be intuitive to think that if $T\rightarrow \infty$ there is so much thermal energy so the particles have no preference of being in any energy state, so they are equiprobable, imagine a giant trying to be in every step of a human stairs)
- So now in order to find the correct answer you should consider how many configurations (microstates) are possible that give a sum of $\varepsilon$ for the system. (i.e one particle in the $\varepsilon _1$ and all the other on the $\varepsilon _0.)$