Stat mech derivation: Covarience of energy and pressure

[swmmr1928]

Instead it is $\frac {\partial {\sum\nolimits_{i=1}^n E_i p_i}} {\partial V}$
And the other $\sum\nolimits_{i=1}^n E_i p_i*(-\frac {\partial E_i} {\partial V})$
Can these be differentialted?
If both the probability and energy are functions of Volume, i can use product rule

 

[swmmr1928]

If I use the expression for U that expresses the probability in E_i and exp(-BE_i) and Q(N,V,T) then Product and quotient rule?

 

[BruceW]

Instead it is $\frac {\partial {\sum\nolimits_{i=1}^n E_i p_i}} {\partial V}$
And the other $\sum\nolimits_{i=1}^n E_i p_i*(-\frac {\partial E_i} {\partial V})$
Can these be differentialted?
If both the probability and energy are functions of Volume, i can use product rule

yep. you should differentiate the first one, using product rule. And the second one, you can just leave it as it is. It is fine like that.

 

[swmmr1928]

$p_i \frac {\partial E_i} {\partial V} + E_i \frac {\partial p_i} {\partial V}$
$p_i=exp(-B*E_i)/Q(N,T,V)=exp(-B*E_i)/exp(-B*E_i)$
$\frac {\partial p_i} {\partial V}=0$
$\frac {\partial {\sum\nolimits_{i=1}^n E_i p_i}} {\partial V}=\sum\nolimits_{i=1}^n p_i \frac {\partial E_i} {\partial V}$

 

[BruceW]

$p_i=exp(-B*E_i)/Q(N,T,V)=exp(-B*E_i)/exp(-B*E_i)$

The first equality is correct. But the second equality is definitely not correct. $Q$ is a completely separate sum, not the same sum. So you need to use another product rule here. You're getting there, though. So, try using product rule on the middle expression:
$exp(-B*E_i)/Q(N,T,V)$

 

[swmmr1928]

$\frac {\partial p_i} {\partial V}=exp(-B*E_i)*\frac {\partial} {\partial V} \frac{1} {Q} + 1/Q* \frac {\partial exp(-B*E_i)} {\partial V}$
$\frac {\partial exp(-B*E_i)} {\partial V}= exp(-B*E_i)*(-B)* \frac{\partial E_i}{ \partial V}$
This B will hopefully cancel the kT
Sum of $\frac {\partial} {\partial V} exp(B E_i)=exp(B E_i)*B*E_i$

 

[BruceW]

yep, that's it! keep going with it

 

[swmmr1928]

$\frac {\partial p_i} {\partial V}=B*E_i-(B/Q)*e^{-B*E_i}*\frac {\partial E_i} {\partial V}$

 

[BruceW]

$\frac {\partial p_i} {\partial V}=B*E_i-(B/Q)*e^{-B*E_i}*\frac {\partial E_i} {\partial V}$

that's not quite right. I agree with the second term on the right-hand side. But I don't agree with the first term on the right-hand side.
In your previous post, the first term on the right-hand side was this:
exp(-B*E_i)*\frac {\partial} {\partial V} \frac{1} {Q}
I'm pretty sure this does not become $BE_i$

 

[swmmr1928]

$e^{-B*E_i}*\frac {\partial} {\partial V} \frac {1} {\sum\nolimits_{j=1}^n e^{-B*E_j}}=e^{-B*E_i}*\frac {\partial} {\partial V} \sum\nolimits_{j=1}^n e^{B*E_j}= e^{-B*E_i}*\sum\nolimits_{j=1}^n B*e^{B*E_j}*\frac {\partial E_j} {\partial V}$
Edit: Maybe I should avoid expressing Q explicitly
$\frac {\partial} {\partial V}\frac {1} {Q(V)}= -\frac {\partial Q} {\partial V} * Q^{-2}$
$E_i*\frac {\partial p_i} {\partial V}=\frac {-E_i*e^{-B*E_i}} {Q}[{\frac {1}{Q} * \frac {\partial Q} {\partial V} + B*\frac {\partial E} {\partial V}}]$
$\frac {1} {Q} * \frac {\partial Q} {\partial V} = \frac {\partial ln(Q)} {\partial V}$
$E_i*\frac {\partial p_i} {\partial V}=\frac {-E_i*e^{-B*E_i}} {Q}[ \frac {\partial ln(Q)} {\partial V} + B*\frac {\partial E} {\partial V}]=-p_i E_i*[ \frac {\partial ln(Q)} {\partial V} + B*\frac {\partial E} {\partial V}]$

 

[BruceW]

$E_i*\frac {\partial p_i} {\partial V}=\frac {-E_i*e^{-B*E_i}} {Q}[ \frac {\partial ln(Q)} {\partial V} + B*\frac {\partial E} {\partial V}]=-p_i E_i*[ \frac {\partial ln(Q)} {\partial V} + B*\frac {\partial E} {\partial V}]$

yep. very nice. and you can use your equation for
$\frac {\partial E} {\partial V}$
(but remember, this should be $E_i$ not $E$). And also, in your relevant equations section on the first post, you have an equation for
$\frac {\partial ln(Q)} {\partial V}$
So then you should be able to write out the whole term as something 'nicer'. And then you can start to prove the equation that they asked for.

 

[swmmr1928]

$\frac {\partial ln(Q)} {\partial V}=\overline{P}*β$
$\frac {\partial E_i} {\partial V}=-\overline{P}$
$\sum\nolimits_{i=1}^n E_i p_i *(-\frac {\partial E_i} {\partial V}) - U*\overline{P}=kT*[{\overline{P}+\sum\nolimits_{i=1}^n p_i*\frac {\partial E_i} {\partial V}}-p_i*E_i*[\frac {\partial lnQ}{\partial V}+B*\frac {\partial E} {\partial V}]]$
$\sum\nolimits_{i=1}^n -E_i p_i \overline{P} - U*\overline{P}=1/B*[{\overline{P}-\sum\nolimits_{i=1}^n p_i*\overline{P}}-p_i*E_i*[\overline{P}*B-B\overline{P}]]$

 

[BruceW]

yep. Although, to keep notation consistent, It's probably best to write $\overline{p}$. Anyway, you can stat to calculate all the terms now, and see how they all go together to prove the statement that you were given to prove.

 

[swmmr1928]

$\sum\nolimits_{i=1}^n -E_i p_i \overline{P} - U*\overline{P}=1/B*[{\overline{P}-\sum\nolimits_{i=1}^n p_i*\overline{P}}]$Wrong:
$-\overline{P}[\sum\nolimits_{i=1}^n E_i p_i + U]=1/B*[{\overline{P}}-\sum\nolimits_{i=1}^n p_i*\overline{P}]$
$-\overline{P}[\sum\nolimits_{i=1}^n E_i p_i + U]=1/B*[{\overline{P}}-\overline{P}]$
$2*U*\overline{P} =0$
Edit: Do you see my error?

 

[swmmr1928]

I'll back up:
$E_i*\frac {\partial p_i} {\partial V}=-p_i E_i*[ \frac {\partial ln(Q)} {\partial V} + B*\frac {\partial E} {\partial V}]=-p_i E_i*[ B*\overline{P} - B*\overline{P}] =0$Right?

 

[BruceW]

$\frac {\partial E_i} {\partial V}=-\overline{P}$

This is not quite correct. The correct equation is in your post #22.

$\sum\nolimits_{i=1}^n E_i p_i *(-\frac {\partial E_i} {\partial V}) - U*\overline{P}=kT*[{\overline{P}+\sum\nolimits_{i=1}^n p_i*\frac {\partial E_i} {\partial V}}-p_i*E_i*[\frac {\partial lnQ}{\partial V}+B*\frac {\partial E} {\partial V}]]$
$\sum\nolimits_{i=1}^n -E_i p_i \overline{P} - U*\overline{P}=1/B*[{\overline{P}-\sum\nolimits_{i=1}^n p_i*\overline{P}}-p_i*E_i*[\overline{P}*B-B\overline{P}]]$

uggh.. It is kinda difficult to follow. In the previous posts, you were trying to find:
( \frac{\partial U}{\partial V} )_T
and you got to
( \frac{\partial U}{\partial V} )_T = \sum_i p_i ( \frac{\partial E_i}{\partial V} )_T + \sum_i ( \frac{\partial p_i}{\partial V} )_T E_i
And then you found that
E_i ( \frac{\partial p_i}{\partial V} )_T = -p_i E_i ( \overline{P} \beta + \beta ( \frac{\partial E_i}{\partial V} )_i )
Generally, it is best not to do too many steps at once, So maybe the next best step is to simply write out:
( \frac{\partial U}{\partial V} )_T
using these equations. (and don't be surprised if it contains any terms with $\overline{EP}$).

 

[swmmr1928]

$\frac {\partial U} {\partial V}_T=\sum\nolimits_{i=1}^n \frac {\partial E_i*p_i} {\partial V}\overset{Product rule}=\sum\nolimits_{i=1}^n E_i\frac {\partial p_i} {\partial V}+p_i\frac {\partial E_i} {\partial V}$

$E_i*\frac {\partial p_i} {\partial V}\overset{From post 40}=-p_i E_i*[ \frac {\partial ln(Q)} {\partial V} + B*\frac {\partial E_i} {\partial V}]\overset{using sub. 2 and 3}=$
$-p_i E_i B[ \overline{P} - P_i]=-p_i E_i B \overline{P} + P_i p_i E_i B\overset{sub. 1}=-p_i E_i B \overline{P} + B\overline{P E}$
$\frac {\partial U} {\partial V}_T=\sum\nolimits_{i=1}^n -p_i E_i β \overline{P} + β\overline{P E}+p_i\frac {\partial E_i} {\partial V}$

$\sum\nolimits_{i=1}^n -p_i E_i β \overline{P} + β\overline{P E}+p_i\frac {\partial E_i} {\partial V}\overset{sub. 3}=\sum\nolimits_{i=1}^n -p_i E_i β \overline{P} + β\overline{P E}-p_i P_i$

$\sum\nolimits_{i=1}^n -p_i E_i β \overline{P} + β\overline{P E}-p_i P_i \overset{sub. 4,5}= \sum\nolimits_{i=1}^n -β U \overline{P} + β\overline{P E}-\overline{P}$

Substitutions Used:
1) $E(E*P)=\sum\nolimits_{i=1}^n E_i P_i p_i$

2) $\overline{P}=kT(\frac{\partial{ln Q}}{\partial V})=(\frac{\partial{ln Q}}{\partial V})/B$

3) $P_i=-\frac {\partial E_i} {\partial V}$

4) $U=\sum\nolimits_{i=1}^n p_i E_i$

5) $\overline{P}=\sum\nolimits_{i=1}^n p_i P_i$

Back to the full equation:
$\overline{(E-U)(p-\overline{p})}=kT[(\frac{\partial U} {\partial V})_{N,T} + \overline{p}]$
$\overline{EP}-U*\overline{P}=kT[\sum\nolimits_{i=1}^n -β U \overline{P} + β\overline{P E}]$

 

[BruceW]

ah, very nice. nice and clear working, too. nice work. It's quite a lot of steps really. uh, in the last line you've left in the summation sign, but the summation has already happened. I'm guessing this is just because you copied and pasted the text and changed some of the text, but not all? (i.e. typing error). So anyway, looks good.