# I Boltzmann Distribution Derivation Question

1. Oct 17, 2016

### kidsasd987

Hello, I have a question about Boltzmann Distribution.

I wonder why partial N of Nj is 1 and partial U of Nj=Ej. because N is constant, partial N of Nj has to be 0 and Partial Nj of U has to be 0 as well.

They are constants so, to make sense of the equation, alpha and beta have to be 0 but this is still absurd.

Can anyone tell me why this is true?

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2. Oct 17, 2016

### BvU

With partial N of Nj you seem to mean $\partial N\over \partial N_j$, right ?

Well, if $\ \ A = x+y+z\ \$ then what is $\ \ {\partial A\over \partial x}\ \$ according to you ?

3. Oct 17, 2016

### kidsasd987

if A is const, isnt it 0 where x is 0? or y,z is depedent on x where partial_x (y+z)=-1.
N is not a function. it is a constant, according to the second and third screen shot. on the second screen shot, it says we are multiplying constants by zero.

Well I thought, it would make sense if N is a function because we are dealing with some unknown N number of particles. then, partial_Nj (N)=1. but since the slide says Partial_Nj(N) is 0, so.. I got lost.

Last edited: Oct 17, 2016
4. Oct 18, 2016

### BvU

N is $\sum N_j$ and as such has a partial derivative with respect to $N_j$. Look up the definition of partial derivative. That doesn't change because a constraint ($N$ is constant) is imposed.
That would be the sheet with "13/28" and it's nonsense. (I understand your question a lot better now )

If you look up the method of Lagrange multipliers perhaps it'll become a bit clearer, in the sense that at an extremum a necessary condition is that the gradients are linearly dependent. Hence these Lagrange multipliers $\alpha$ and $\beta$.
It certainly isn't true that all $\ \ {\partial \phi\over \partial N_j} \ \$ and $\ \ {\partial \psi\over \partial N_j} \ \$ are zero.

5. Oct 18, 2016

### kidsasd987

Thanks a lot. I spent quite a lot of time on figuring out what went wrong in the note and what I did wrong.
Thank you!