Thermodynamics (statistical): Boltzmann distribution derivation

[@lseinjr1]

(for *mobile* users) 0:14 Introduction 0:30 Canonical ensemble 3:44 Two constraints 5:21 Number of ways to achieve distribution (Ω) 8:03 Stirling's approximation 9:36 ln Ω 14:34 Lagrange's method of undetermined multipliers 16:15 Derivative of ln Ω 18:02 Apply Stirling's approximation twice 20:53 Derivative of N ln N 27:14 Full derivative equations 30:36 Use of definition of N 33:20 Canonical ensemble partition function Q 35:08 Boltzmann distribution

[@nvrhanna]

Thanks!

[@veronicanoordzee6440]

After 11:08: it should be the summation of ln(n,!), not just of n,! It's not so easy to copy something you've already written down ;-)

[@aww2historian]

Yes please fix! Got super confused after that

[@Asphalt9_talks]

I had fun watching this. Really helped me understand entropy from this video. Helped me get over a lot of frustration in understanding where did entropy come from. Thanks a lot.

[@lseinjr1]

Glad it helped!

[@matteopiccioni196]

very good derivation, i followed it up the end, and i got it a lot. Yes at 11:30 there is a mistake, i would say only an oversight, because then proceeding the derivation you correct yourself.

[@guest_of_randomness]

it's always the random video from youtube that save my day! thank you so much, you are a great explainer !

[@londonmackerel7462]

I have doubts about minutes 21 to 22 of this video. Here N is differentiated with respect to nk. However, N is a constant (isn't it?) so its derivative should be zero. That is, even if an nk might vary, N will not as the other n terms vary to keep N constant. It seems to me that If you treat N as a constant and so with a derivative of zero, you derive the Boltzman distribution more easily.

[@GoogleAccount-mz6vn]

I think we didn't consider the constraint because at that point we have had already imposed the restriction by deciding to use the extra terms (in the final derivative) presented by Lagrange's Method of Undetermined Multipliers. We shouldn't have imposed it twice since the substractions of the second and third terms of the final derivative already was there to account for it. chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/21%3A_The_Boltzmann_Distribution_Function/21.02%3A_Lagrange's_Method_of_Undetermined_Multipliers

[@tedsheridan8725]

With Lagrange optimization, the constraints are considered elsewhere. If F = x+y+z, dF/dx = 1 regardless of the constraint that x+y+z = 7

[@tanhevg]

Thanks for making this video. This is really helpful.

[@lseinjr1]

Glad it was helpful!

[@LibertyAzad]

Took me a while to find such a clear explanation. But I still wonder about the thermodynamic vs. empirical temperature. I guess at the end you were referring to thermodynamic temperature. Thank you for your videos!!!

[@pavelmorales5491]

Thank you, a good detailed explanation! PS: In 25:09 you missed a one but did not affected further calculation.

[@purielmafiacity6193]

Essa aula é muito boa, agradeço por sua explicação. 🎉

[@SlowRiderDucati]

Hey Dude, Thanks it finally makes sense!

[@tedsheridan8725]

I've been watching several related videos, this is the first one that explained it clearly. Thanks! One thing I'm still confused on - the discrete nature of energy levels of the various systems. Is energy discrete here because of quantized energy levels in the QM sense? Or are we taking continuous energy and putting them into discrete bins of arbitrary width?

[@lseinjr1]

Great question! No assumption is made that the energy levels of individual systems are discrete. Even in QM, energy is not necessarily quantized - an electron has quantized energy levels when it is "bound" by the nucleus, but not when it is unbound. It is true that there are a limited number of energy levels available for the N systems in the Boltzmann distribution, but that is because there are only a finite number (N) systems involved. The total energy of the system is the sum of the energies of the N individual systems. Does that answer your question?

[@paulg444]

Great video , would be nice to do a follow on where momentum and position are accounted for and the sum of momenta of each microstate is conserved.

[@jianqiuzhang4607]

Really appreciate you. This helps me a lot!

[@AlphaNumeric123]

Excellent content presentation. The best resource I've found on this topic. Thank you so much.

[@lorianamesia331]

Nice picture. I love the painter who did it

[@100rubmighty]

Your Lagrange multiplier method is incorrect: (1) The constraint is \sum n_j = N, the Lagrange multiplier method requires \alpha*(\sum n_j - N). The multiplier is applied to the constraint. Similarly for E_tot (2) You would have realized this when you supplement the equations with partial derivatives w.r.t. \alpha and \beta. These are supposed to enforce the constraints. (3) Fortunately for you, when you take the derivative w.r.t. n, the N and E_tot terms drop out. But the mistake checks itself out only for the purpose of the video. Any other conclusions will be wrong.

[@lseinjr1]

No. 1. Since Σ n_j = N, dN = 0 = Σ n_j dn_j, so α Σ n_j dn_j = 0 as well. Similarly, Σ n_j E_j = E_total, so dE = Σ n_j E_j dn_j = 0 as well, so β Σ n_j E_j dn_j = 0. Do you have a citation for your method?

[@Sky-pg6xy]

Where did the natural log go when he wrote the n’s as a summation?

[@lseinjr1]

See the *NOTE* in the description box.

[@williamwolfe8708]

Thanks for the content. Was interesting to see all the energy spent in changing colors. At this level of abstraction, and expected education level of the viewer, the probability the colors were helpful was approximately 0. Just kidding around -- enjoyed the nice video, thanks!

[@NicolasSchmidMusic]

noooo colors always help :)

[@tedsheridan8725]

Actually the colors help a great deal.

[@tunneltrance]

thank you for the knowledge

[@leonardofidelcordovacastil6717]

what happened with the sumation of all ln(n_{j}) in 11:16 instant?

[@-theveniesa9508]

can you please say weather, the individual systems {n(i)} can be taken as electronic energy levels in an atom

[@lseinjr1]

No, they can not. There we have to use "Fermi-Dirac" statistics instead of Boltzmann statistics.

[@-theveniesa9508]

@@lseinjr1 thankyou

[@leonardofidelcordovacastil6717]

many many thanks for this content

[@josephmoore4764]

32:17 Shouldn't the sum over e^a = j*e^a or however many energy states we have?

[@lseinjr1]

In the Boltzmann distribution, all the states are non-degenerate (i.e., j=1).

[@alis5893]

Fantastic. Thank you

[@lseinjr1]

Glad you liked it!

[@siarez]

From what I know about Lagrange multipliers, you can solve for the exact value of the multiplier. How come here it ends up being a variable?

[@lseinjr1]

It is because the derivation is completely general, for *any* temperature. Once you fix the temperature (T), we can solve for β= 1/kBT, so long as T not equal to 0 (which is guaranteed by the Third Law).kB is the Boltzmann constant. Does that make sense?

[@siarez]

@@lseinjr1 My difficulty is that there should be no degree of freedom (aka undetermined variable) left after we find the optima. But here β seems to be left undetermined. In the problem setup at the beginning of the video, there is no notion of temperature in the picture. It feels like it is pulled out of the hat at the last minute.

[@lseinjr1]

In a sense, you are correct that temperature was "pulled out of the hat at the last minute." I wanted to keep the video to a manageable length, and restrict my attention to the one derivation. It would take another video of twenty (20) minutes to show exactly how we know that beta is equal to the value we said. I have such a video in progress. As a prerequisite, we need the notion of the canonical ensemble partition function for translation, which you can see here: kzbin.info/www/bejne/r17Fk3yGZtWUp6s

[@siarez]

First, many thanks for the video. At @32:00, when we remove the sum term behind e^-α, shouldn't we multiply e^-α by the number of energy levels?

[@lseinjr1]

Great question! I should not have written the summation sign in front of the e^-a when I rewrote the exponential. Since exp (-α) is a constant, we can pull it "through" the summation sign, just as we do for integrals. Good catch!

[@alexz26389]

Excellent video. At the end you rewrite \beta in terms of T. Is this a definition of temperature? If so is it possible to show it's equivalent to other definitions?

[@lseinjr1]

Yes. Perhaps the most useful definition of temperature is that, at the same temperature, two (different) gases will have the same average kinetic energy.

[@asifesedov9381]

Why you ignored sun " ln" nj in 14:00 cuz, i 5hink you missed ln there

[@lseinjr1]

This was already noted in the Comments.

[@apoorvmishra6992]

Sir I have found your video very helpful but just in the last part could you please provide some insight on using what results beta was found out to be 1/kBT?

[@lseinjr1]

That is a great question! I am working on a video to show this result theoretically, but the reasoning in it is very involved, and it will take almost as long as the entire first Boltzmann video. We can also find the result through experimentation. Suppose we are performing the NMR experiment, with just two (2) energy levels (call them E1 and E2)- with the field or against the field. If we know the field strength, we know the separation between E1 and E2. The signal strength will be proportional to the energy difference E2 - E1. Then we change the field strength, to get new E2 - E1. The signal strength will decrease as the energy difference E2 - E1 increases. If we plot several of these experiments, assuming that the state populations are Boltzmann distributed and knowing the value of the Boltzmann constant, we can then calculate beta. We could also use electronic or vibrational spectroscopy to do the same sorts of things. Does that make sense?

[@apoorvmishra6992]

Thank you very much for taking time to reply. Yes it does make sense to verify the result for beta experimentally via above methods.

[@paulg444]

@ 13:00 what happened to the log operator on the njs ?

[@lseinjr1]

That was a typo on my part. The log operator "reappears" at 17:16. My apologies!

[@chanwoo1143]

I must ask. How did you get the top most/first equation at 14:54? Did you use Lagrange's method of undetermined multipliers to get to this equation?? If so, then how?? This baffles me ~

[@lseinjr1]

Yes.

[@chanwoo1143]

@@lseinjr1 How can I get the logic behind this?.. I don't get it..

[@lseinjr1]

Langrange's method is a way of solving a system of equations with additional constraints (requirements). Here, there are two (2) constraints: the first is that if we add up all the nj's, their total is N - the total number of systems in the ensemble is a constant. That is the reason for α∑nj. α is the first Lagrange multiplier The second constraint is that the total energy must be a constant. The energy of *each* system is Ejnj, and therefore, the total energy of the ensemble is a constant, ∑Ejnj. β is the second Lagrange multiplier. Does that help?

[@chanwoo1143]

@@lseinjr1 Yes thank you very much for the explanation and for the video. That helps a lot What you have written is pretty much what is already given in the video as far as I remember. I understand that differentiating against something which is a constant will yield 0. The equation that was written at 14:54 makes sense in the mathematical sense(equation-wise?). I understand that you used this to derive boltzman distribution in the end. hmm.. I am more concerned with how you came up with this equation from the proceeding explanation given in the video. I am sorry for making my question possibly confusing. If you feel so, you don't need to answer. Anyways, again, thank you ! Best of luck!

[@lseinjr1]

The two (2) constraints were mentioned earlier in the video, when I described the characteristics of the *canonical ensemble*. (There are several different types of ensembles that are used in thermodynamics, the "microcanonical" and "grand canonical" ensembles being two (2) examples. The first constraint is mentioned at 3:41, and the second one is mentioned at 3:49. The equation at 14:39 attempts to find a relative maximum )the most probable distribution) by setting *all* first derivatives equal to 0. Since there are many nk's, each of the equations on the board i(at 14:39) is actually a *system* of equations, each relative to different variable nk. Keeping track of many variables with different indices (j and k) is certainly confusing. It can be helpful to imagine that there are just three (30 values of j and k, and to work out the derivatives using numbers rather than variables.

[@mohamedessa1840]

l want this prove but for degenert energe

[@lseinjr1]

The Boltzmann distribution is non-degenerate.

[@arunnavinjoseph9262]

You are changing pens too frequently.Please try to avoid that.