# Implicit & explicit dependence derivative sum canonical ense

1. Jan 28, 2017

### binbagsss

1. The problem statement, all variables and given/known data

Hi,

I am trying to follow the working attached which is showing that the average energy is equal to the most probable energy, denoted by $E*$,

where $E*$ is given by the $E=E*$ such that:

$\frac{\partial}{\partial E} (\Omega (E) e^{-\beta E}) = 0$

MY QUESTION: the third equality, i.e. the second line

I have it explained the first term is taking care of the explicit dependence and the second term is taking care of the implicit dependence.

I'm pretty confused, I have never seen an example like this before. The only thing I can see is that if there is implicted and explicit dependene you do the chain rule, getting a product of terms, not a sum. I.e. letting $f(E(\beta))$ denote the function we are taking the deriviate of, I would conclude :

$\frac{\partial}{\partial \beta} f(E(\beta)) = \frac{\partial E}{\partial \beta}\frac{\partial}{\partial E}$...

I have never seen a sum of terms obtained from differentiation of explicit and implicit dependence of some variable.

Can some please expalin and tell me why the chain rule is not correct here? or (Links to any material on this also appreciated, thanks )

context is canonical ensemble, statistical mechanics.

2. Relevant equations
see above

3. The attempt at a solution
see above

#### Attached Files:

• ###### sm av energy.png
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2. Jan 29, 2017

### Stephen Tashi

Do you mean $\frac{\partial}{\partial \beta} f(E(\beta)) = \frac{\partial f}{\partial \beta}\frac{\partial E}{\partial \beta}$ ?

Perhaps the idea is to apply $\log(pq) = \log(p) + \log(q)$ before differentiating.

3. Jan 29, 2017

### binbagsss

Ive tried the log property.
I get
$\frac{\partial}{\partial\beta}(log(\Omega(E*(\beta))) + \frac{\partial}{\partial\beta}(-\beta E*)$
$=\frac{\partial}{\partial E}(log (\Omega)\frac{\partial E*(\beta)}{\partial\beta} - E*-\beta\frac{\partial E*}{\partial \beta}$, So I have the second term, can't see how I am going to get the first term.

4. Jan 30, 2017

### Stephen Tashi

I think the pattern involves a function of two variables $w(\beta,E^*)$ with the "side condition" that $E*$ is a function of $\beta$. The chain rule has the form:
$\frac{dw}{d\beta} = \frac{\partial w}{\partial \beta} + \frac{\partial w}{\partial E^*} \frac{\partial E^*}{\partial \beta}$.

It would have been clearer to use the notation $\frac{d}{d\beta}Z$ instead of $\frac{\partial}{\partial\beta} Z$.

To illustrate what I have in mind with a simple example:
Let $f(g) = g^2$ and let $g(x) = 2x + 1\$. Define $g^*$ as the solution to $\frac{df}{dg} = 0\$ Then $g^* = 0$. To get $g^* = 0$, the only choice for $x$ is $x = -1/2\$. So, $g^*$ is not a function of $x$ (or at least it isn't the same function of $x$ that $g$ is).

Now lets change the definition of $f$ to a function of two variables $f(g,x) = g^2 - gx^2$ and keep the definition that $g(x) = 2x + 1$. Let $g^*$ be the solution to $\frac{\partial f}{\partial g} = 0$.
$\frac{\partial f}{\partial g} = 2g - x^2$. So $g^* = x^2/2$, making $g*$ a function of $x$ and not the same function of $x$ as $g(x) = 2x -1$.

Now, as an example, define $w(x) = f(x, g*) = (g^*)^2 + g^*x$

One interpretation of $\frac{\partial w}{\partial x}$ is "Take the partial derivative of $W$ with respect to the first argument". By that definition $\frac{\partial w}{\partial x} = g^*$

Another interpretation of $\frac{\partial w}{\partial x}$ is "Write $w$ explicitly as function of $x$ alone (i.e. $w(x,g*) = (g^*)^2 + g^*x = (x^2/2)^2 + (x^2/2)x$ and then take the derivative with respect to $x$". A better notation for this concept would be $\frac {dw}{dx}$, using "$d$" instead of "$\partial$". This could also be done by the chain rule:
$\frac{dw}{dx} = \frac{\partial w}{\partial x} + \frac{\partial w}{\partial g*}\frac{\partial g*}{\partial x}$ where the last term has the meaning given in the earlier paragraph.

5. Jan 31, 2017

### binbagsss

okay thank you, that has cleared up a few things.
In particular the point that $g*$ is a different function of $x$ then $g$ is.
However I am a little confused with $f(g,x)$ and $f(g*,x)$, $f$ is the same function of $g$ and $g*$ right? you just replace $g$ with $g*$? so can you write :

$\frac{dw}{dx} = \frac{\partial w}{\partial x} + \frac{\partial w}{\partial g*}\frac{\partial g*}{\partial x}$ as $\frac{dw}{dx} = \frac{\partial w}{\partial x} + \frac{\partial w}{\partial g}\frac{\partial g*}{\partial x}$?

6. Jan 31, 2017

### Stephen Tashi

Yes, as a function of its second argument, $f(x,g^*)$ , symbolically defined by an expression using the symbol "$g^*$" as $f(x,g^*) = (g^*)^2 - g^*x^2$ is the same function as $f$ defined using a different symbol like "$y$" and writing $f(x,y) = y^2 - yx^2$.

Physics uses an ambiguous notation for functions. Notation like $f(x,g^*) = (g^*)^2 - g^*x^2$ can be interpreted two different ways. On the one hand, $f(x,g^*) =( g^*)^2 - g^* x^2$ might denote a function whose domain is pairs of real numbers. However, if we have in mind the "side condition" that $g^* = x^2/2$ then the same notation stands for a different function who domain is the set of real numbers (as opposed to pairs of real numbers).

A forum article on such ambiguity is: https://www.physicsforums.com/insights/partial-differentiation-without-tears/