[haz658]

       I should first apologize for abbreviating "regarding" in the subject title, however, I ran out of letters to use and that word was marginalized by the importance of the others. Though it is likely evident from the information provided on this website, this is my first real post. I apologize for any improprieties with respect to my post and request that you please inform me of them so that I can alter my approach in the future.
         Both of my questions stem from Chapter 13 (or 31, if you possess the one-volume text), "The Boltzmann Distribution", in Engel and Reid's Thermodynamics, Statistical Thermodynamics, and Kinetics (or, Physical Chemistry, in the one-volume edition). My more perplexing question arises from page 312 in the two-volume edition, equation 13.34. The authors are demonstrating the dominance of the Boltzmann distribution by taking the natural log of the ratio of the weight of the Boltzmann distribution (dominant configuration) to that of another configuration, which has been modified by a change in the system. This expression is eventually determined to equal sums featuring occupation numbers and fractional changes in occupations numbers.

$$\begin {eqnarray} ln \left (\frac{W_{max}} {W} \right) & = & \sum_{n} a_n \alpha_n \mathbf {ln a_n} + \sum_{n} \frac {a_n} {2} (\alpha^{2}_{n}) \\ & = & \sum_{n} a_n \alpha_n \mathbf {(ln a_0 - \beta \varepsilon_n)} + \sum_{n} \frac {a_n} {2} (\alpha^{2}_{n}) \end{eqnarray} /$$

My first question is how the authors were able to convert ln(a_n) to (ln(a₀) - beta * epsilon_n), where a_n is the occupation number of the nth energy level, beta is a Lagrange multiplier used earlier in the analysis, and which remains in the expression for the Boltzmann distribution, and epsilon_n is the energy of the nth energy level. What earlier equation or piece of information was necessary to make this change?

     My second question seems, to me, anyway, to possess a simpler answer. On pages 305-306 of the same chapter, the authors are in the midst of deriving the Boltzmann distribution by showing that the Boltzmann distribution can be found as the maximum of the curve relating the natural log of the weight of an energy configuration to the occupation number. When the derivative of the natural log of weight with respect to occupation number equals zero, the original curve is at a maximum. The authors then use the derivative of their previously obtained expression for weight, with respect to occupation number, to derive the Boltzmann distribution. They soon make the claim that the derivative of the total number of particles over which the energy is distributed, with respect to the occupation number of the nth energy level, is equal to 1, since, when you sum the occupation numbers for all n, you obtain the total number of particles in the system.

$$ \frac {dN} {da_{n}} = 1 /$$

It seems to me, though, that this derivative is actually determining the change in the total particle population with respect to the change in occupation number. Since there is no change to the total population of particles, I would reason that the derivative of this term should equal zero. I am quite certain that I am wrong, since this change would seemingly result in a different expression of the Boltzmann distribution.
              I would like to thank, in advance, anyone who is willing to offer their help in these matters. It is quite possible that the answers are obvious, but I have been struggling with these problems for several days, and I do not believe that I will find the answers on my own. Also, I apologize for any issues regarding my post. I tried to limit my focus to the information that I believe is necessary to solve the problem. If you need me to alter the content of this post in any way, I will do so. This includes summarizing, expounding, reformatting, clarifying, etc. Thanks again.
                                                                                                          -Matt

[haz658]

I'm not sure if anyone is interested, since, from what little I have read of the contents of this forum, this does not seem to be one of the recommended books for physical chemistry; but I found the answer to my first question. The authors derived the following equality on page 314 (in the two, separate-volume edition), two sections after they inserted it into the equation I listed previously (as the subject of my first question), without any explanation of its source.

$$ ln a_n = ln a_0 - \beta \varepsilon_n /$$

They obtained this equality while demonstrating the relationship between the weight (probabilistic) of a configuration of energy with respect to a system of particles and the energy available to that system. They also used an equation they derived earlier while showing that the Boltzmann distribution can be found when the derivative of the natural log of the weight of an energy configuration, with respect to the occupation number of energy level n, equals zero. Here is the mathematical expression of the derivative I am describing, followed by the useful result they obtained after inserting Lagrange multipliers and terms to consider the lack of change in the number of particles in the system and energy available to the system.

$$ \frac {dlnW} {da_n} = 0 /$$ for the Boltzmann distribution

$$ a_n = Ne^{\alpha} e^{- \beta \varepsilon_n} /$$

Using this last equation, they obtained the first equation listed in this post by taking the ratio of a_n to a₀, where the energy of energy level 0 equals 0.

I still have not found an answer to my second question from the original post. If anyone has a thought regarding that question or, better yet, an answer, I would be extremely interested to read it, as well as grateful for the help.

[Juan R.]

    My second question seems, to me, anyway, to possess a simpler answer. On pages 305-306 of the same chapter, the authors are in the midst of deriving the Boltzmann distribution by showing that the Boltzmann distribution can be found as the maximum of the curve relating the natural log of the weight of an energy configuration to the occupation number. When the derivative of the natural log of weight with respect to occupation number equals zero, the original curve is at a maximum. The authors then use the derivative of their previously obtained expression for weight, with respect to occupation number, to derive the Boltzmann distribution. They soon make the claim that the derivative of the total number of particles over which the energy is distributed, with respect to the occupation number of the nth energy level, is equal to 1, since, when you sum the occupation numbers for all n, you obtain the total number of particles in the system.



It seems to me, though, that this derivative is actually determining the change in the total particle population with respect to the change in occupation number. Since there is no change to the total population of particles, I would reason that the derivative of this term should equal zero. I am quite certain that I am wrong, since this change would seemingly result in a different expression of the Boltzmann distribution.

Good that you solved by yourself the first question. About the second. I have not that book to look the context of the equation that you give, but notice that

dN / d (Sum_n a_n) = 1

or better {*}

dN = d (Sum_n a_n)

At equilibrium in a closed system the occupation numbers are constant

dN = Sum_n d(a_n) = d(a_n)

{*} because dN = 0, the fraction looks so odd as (0/0)=1

[haz658]

Thank you very much for your help. This answer didn't occur to me because the authors' derivation of the Boltzmann distribution, of which the troublesome derivative was a part, commenced with taking the derivative of the natural log of the weight with respect to the occupation number of a particular energy level, with the weight noticeably changing with a change in occupation number. But it certainly fits and explains their result, whether or not this is the method they used. Thanks again for taking the time to help me and provide the answer to my question. I am extremely grateful. Though I doubt you would ever require my assistance in this forum, or that the help you would potentially require would fall in an area in which I am knowledgeable, if such circumstances do align, I would be happy to oblige.
                                                                                     -Matt

[McCoy]

[/quote]

{*} because dN = 0, the fraction looks so odd as (0/0)=1
[/quote]

that one is undefined!

[McCoy]

  \sum_{n} \frac {a_n} {2} (\alpha^{2}_{n}) \\ & = & \sum_{n} a_n \alpha_n \mathbf {(ln a_0 - \beta \varepsilon_n)} + \sum_{n} \frac {a_n} {2} (\alpha^{2}_{n}) \end{eqnarray} }[/img]
  


It seems to me, though, that this derivative is actually determining the change in the total particle population with respect to the change in occupation number. Since there is no change to the total population of particles, I would reason that the derivative of this term should equal zero.

No, dN/dan is not equal to zero...it's equal to one. you are correct in saying that both dN and dan = 0, but this doesn't mean that dN/dan = 0. Think about it...for example differentiate 2x^2 + 4x and assume that it's at max at first derivative. you got 4x + 4 =0, right? okay , now divide 4x + 4 by itself....you get 1. That's the authors' argument , I think.

[haz658]

 Thank you very much for your reply. I believe that my confusion stemmed from not recognizing that dN = da_n. The authors established that
                                      
but I did not see how that implied that dN = da_n. Juan R. helped relieve some of that confusion, but only with the constraint that the occupation numbers (a_n's) are constant in a closed system. Constant occupation numbers seems to be the only situation for which
                                                                  
must be true. Otherwise, it would seem that N must equal a_n, which is certainly not true.
       Thanks again.      

[McCoy]

Thank you very much for your reply. I believe that my confusion stemmed from not recognizing that dN = da_n. The authors established that
                                      
but I did not see how that implied that dN = da_n. Juan R. helped relieve some of that confusion, but only with the constraint that the occupation numbers (a_n's) are constant in a closed system. Constant occupation numbers seems to be the only situation for which
                                                                  
must be true. Otherwise, it would seem that N must equal a_n, which is certainly not true.
       Thanks again.      

I think no one has suggested that N=sum(a_n) has something to do with deltaN being equal to delta a_n. you stated above in your first post that you don't understand why deltaN/delta an = 1, which in my opinion is correct. It's correct because N= an1 + an2 +an3.., so the derivative of N with respect to any of the an is 1....meaning delta N = delta an (and that's when an have their most probable values).
As far as  occupation numbers are concerned, yes they are conserved but sum(dan)= dan is not correct, sum (dan) = 0 is the correct one .
Okay, now I've the book in front of me here and the only thing I didn't like about their derivation is the derivative sign...I prefer the curly d since they are taking partial derivatives. Another thing I'd have been happy with is label the an they considered to be configurational index differently to the other occupational numbers.Because it's important later in the derivation steps.
 So here is what they have so far:
dlnW/dan = d(NlnN)/dan - sum (d(anln(an)/dan)). So I’m going to label occupation numbers to rid confusion later.

   dlnW/dan = d(NlnN)/dan - sum (d(ailn(ai)/dan)). Remember ai is still occupation number, the I label just means that we are summing up occupation numbers over i energy level.
So dlnW/dan = lnN + 1 - sum (d(ailn(ai)/dan)).  For the part under summation, i  may or may not be independent of n. If they are independent,  i.e i not equal n and delta ai/delta an = 0. However if I = n, then delta ai/delta an = 1. The last part (i.e din =1, the kronecker delta) is the only condition that makes - sum (d(ailn(ai)/dan)) = ln (ai) + 1 = ln(an) + 1.  It has nothing to do with sum (dan) being equal to dan.

Anyway bye.

[haz658]

    You have greatly improved my understanding of this derivation, for which I am very grateful. One question continues to trouble me, though if you choose not to explain further, I will continue to appreciate the help you have already provided.
             You removed any confusion regarding the mathematical progression to demonstrate that dN/da_n = 1 (using partial derivatives for n equals some number i energy level). Conceptually, though, why would there be any change in N with respect to the change in the number of particles in the ith energy level, if none of the populations for the other energy levels are changing? The only means by which there could be a change in population for a_i would be if there is an increase or decrease in the total population (N). This is why I originally believed dN/da_n should equal zero, because it seems that there shouldn't be any change in N unless more particles or molecules are added to the population.
               I realized after finishing the above portion of the message that I have one additional question. Can you explain how you determined that a_n have their most probable values when dN = da_n?
          Thanks again for all of your help.

[McCoy]

2. for the most probable values stuff, it's the other way around...not how you stated it in your last post....to have the dominant configuration, an must have their most probable values.
1. N is constant, one varies ai and see how dominant configuration will behave/change with respect to an

By the way, I realized that I'm wrong and is Juan.R is right. sorry for wasting your time.