# Implicit & explicit dependence derivative sum canonical ense

• binbagsss
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
binbagsss

## Homework Statement

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.

see above

## The Attempt at a Solution

see above [/B]

#### Attachments

• sm av energy.png
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binbagsss said:
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}##...

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

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

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

binbagsss said:

## Homework Statement

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.

see above

## The Attempt at a Solution

see above [/B]
Stephen Tashi said:
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.

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.

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 let's 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.

FactChecker
binbagsss said:
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.
Stephen Tashi said:
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 let's 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.

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} ##?

binbagsss said:
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} ##?

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/

## What is the difference between implicit and explicit dependence in derivatives?

Implicit dependence refers to a situation where a variable is not explicitly stated in the equation but is still affected by it. This means that the variable is dependent on the equation, even though it is not explicitly shown. On the other hand, explicit dependence refers to a variable that is explicitly stated in the equation and is directly affected by it.

## What is the importance of understanding implicit and explicit dependence in derivatives?

Understanding implicit and explicit dependence in derivatives is crucial for accurately modeling and predicting the behavior of complex systems. It allows scientists to identify and account for all factors that may influence a variable, leading to more accurate and reliable results.

## What is the canonical ensemble in statistical mechanics?

The canonical ensemble is a statistical mechanical model used to describe the physical properties of a system in thermal equilibrium with a heat bath. It assumes that the system is in contact with a heat reservoir at a constant temperature, and the total energy of the system is fixed.

## Why is the canonical ensemble important in statistical mechanics?

The canonical ensemble is important because it allows for the calculation of thermodynamic quantities, such as the average energy, entropy, and free energy of a system. It also provides a way to analyze the behavior of systems in thermal equilibrium, which is essential in understanding the macroscopic properties of matter.

## What is the sum of derivatives in the context of the canonical ensemble?

The sum of derivatives in the canonical ensemble refers to the calculation of the average values of thermodynamic quantities using the derivatives of the partition function. It is a crucial step in the mathematical derivation of thermodynamic properties of a system in thermal equilibrium and is used extensively in statistical mechanics.

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