Hey there! Sign in to join this conversationNew here? Join for free

implicit explicit dependence derivaitve, context statistical mechanics canonical ense Watch

    • Thread Starter
    Offline

    3
    ReputationRep:
    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.

    Many thanks in advance.
    Attached Images
     
    Offline

    17
    ReputationRep:
    (Original post by xfootiecrazeesarax)
    ..
    Consider differentitating \beta f(E(\beta)), You get f(E(\beta)) + \beta \dfrac{\partial E}{\partial \beta} \dfrac{d}{dE} f(E(\beta)) (product rule).

    What you've got is similar but a little more compicated. But the extra term is coming because in the process of diffing the whole thing you need to differentiate the product \beta E_\star(\beta))
    in the e^{-\beta E_\star(\beta))} term.

    Edit: and explictly,the chain rule doesn't work in the way you seem to expect here, because we can't write the whole expression as a function of E(\beta) because of that single \beta in the product that doesn't depend on E.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by DFranklin)
    Consider differentitating \beta f(E(\beta)), You get f(E(\beta)) + \beta \dfrac{\partial E}{\partial \beta} \dfrac{d}{dE} f(E(\beta)) (product rule).

    What you've got is similar but a little more compicated. But the extra term is coming because in the process of diffing the whole thing you need to differentiate the product \beta E_\star(\beta))
    in the e^{-\beta E_\star(\beta))} term.

    Edit: and explictly,the chain rule doesn't work in the way you seem to expect here, because we can't write the whole expression as a function of E(\beta) because of that single \beta in the product that doesn't depend on E.

    Thank you for your reply
    I get (using log additive properties) \frac{\partial}{\partial\beta}(l  og(\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.
    Offline

    17
    ReputationRep:
    (Original post by xfootiecrazeesarax)
    Thank you for your reply
    I get (using log additive properties) \frac{\partial}{\partial\beta}(l  og(\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 can't read your latex, but I explained how you get the first term in the previous post.

    From what I can make out from your LaTeX, your response doesn't seem to have any attempt to use what I said in the previous post. Until you do so, I don't think I can help.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by DFranklin)
    I can't read your latex, but I explained how you get the first term in the previous post.

    From what I can make out from your LaTeX, your response doesn't seem to have any attempt to use what I said in the previous post. Until you do so, I don't think I can help.
    Okay, well from the form of the chain rule you provided, it is pretty clear that it gives the correct answer as the solution does not compute anything more explicitly.

    I have a couple of questions concerning notation however:
    - so the chain rule is  \frac{\partial}{\partial \beta} f (E*(\beta)) + \frac{\partial (E*(\beta))}{\partial \beta} \frac{\partial}{\partial E*}f(E*(\beta))

    Q1) Shouldn't one write  f(E*(\beta)),\beta) rather than  f(E*(\beta)) to distinguish that there is dependence on  \beta alone without  E* ? It is clearer than that f takes the form of the chain rule you provided rather than the 'usual' one I initially went for?
    Q2) concerning partial and total derivatives, I see you changed the  \frac{\partial}{\partial E} f to  \frac{d}{d E} f , to me f has both dependence on  E and  \beta , so I would have thought these need to be partials, however it would have made sense to me, since  E* only depends on  \beta to change \frac{\partial}{\partial \beta} E to  \frac{d}{d \beta} f ?
    Offline

    17
    ReputationRep:
    (Original post by xfootiecrazeesarax)
    Okay, well from the form of the chain rule you provided,
    It appears you're confused. I didn't give you a form of the chain rule, I showed you how to differentiate a particular expression involving a product between \beta and a function f of E_\star(\beta).

    Q1) Shouldn't one write  f(E*(\beta)),\beta) rather than  f(E*(\beta)) to distinguish that there is dependence on  \beta alone without  E* ?
    Again, f is just a function of E_*(\beta). But the product \beta f(E_*(\beta)) obviously also has an explicit \beta-dependence due to the first term (\beta) in the product.

    Q2) concerning partial and total derivatives, I see you changed the  \frac{\partial}{\partial E} f to  \frac{d}{d E} f , to me f has both dependence on  E and  \beta
    No, it doesn't. See above.

    Edit: since you're obviously confused about all of this, I should perhaps add: I don't believe the first line I wrote has any ambiguities about what depends on what etc. However, that was in reference to a simple product, and as I said, what you have is a lot more complicated. The reason for the extra term is because you end up differentiating a product term involving \beta, but because it's more complicated you'll need to be careful about what depends on what.

    At postgraduate level I just expect to be giving hints and nudges, not needing to provide a step-by-step solution (which would be a lot of work).
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Will you be richer or poorer than your parents?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.