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Printing error or?

  1. Aug 19, 2009 #1
    Hi guys, I have a question which equation i need to prove:

    <N> = 1 / exp (E-U)/KT +1

    from Zg = 1 + Exp (-E-U)/KT

    where <N> = kt 9 d (ln Zg) / dU

    I have <N> = 1 / exp (U-E)/KT +1

    is the Zg they have given me incorrect? I find my answer very simliar, is it a printing error?

    My working just consists of putting in the numbers.

    <N> = 1/KT. KT . 1 / Zg

    Thanks.
     
  2. jcsd
  3. Aug 19, 2009 #2

    Cyosis

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    How did you obtain this expression from the general expression [itex]Z_g=\sum_N \sum_{s(N)} \exp(\frac{N \mu-\epsilon_s}{k_B T})[/itex]? The U in your expression represents the chemical potential [itex]\mu[/itex]?

    This is wrong the 9 shouldn't be in there and the t should be a capital T.

    Differentiating isn't just putting numbers in an expression so your wording here is confusing me. Secondly from the looks of it you forgot to use the chain rule.
     
  4. Aug 19, 2009 #3
    Hi thanks for you reply.
    Zg is given in the question.
    The 9 was a type. I will amend my original post.

    I originally used the chain rule but I got something silly like (1+ exp(thinge))/exp (thinge).

    I meant putting in the numbers as, it should be straight forward.

    for the chain rule i used d ln Zg/ U times DZg /U

    would that be incorrect?
     
  5. Aug 19, 2009 #4
    ps I#m posting on behalf of Zophixan as I discovered I know him! haha
     
  6. Aug 19, 2009 #5

    Cyosis

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    Okay. Using the chain rule yields:

    [tex]
    \frac{d \ln Z_g}{dU}=\frac{d \ln Z_g}{dZ_g}\frac{dZ_g}{dU}
    [/tex]

    What you are doing is:

    [tex]
    \frac{d \ln Z_g}{dU}=\frac{d \ln Z_g}{U}\frac{dZ_g}{dU}
    [/tex]

    Which is incorrect.
    Try to work it out and show me the calculations. With the given Zg both you and the book are wrong, are you sure the minus in front of the U is there in the book?
     
  7. Aug 19, 2009 #6
    Hi, it is an exam past paper. I assume it is correct, the minus is definately there.

    dlnZg/Zg should be 1/exp(thinge)

    d Zg/du = 1/KT

    wich when combined with the original formual, just gets me back to what I got originally. Should I assume that it's a misprint in the paper?

    Thanks.
     
  8. Aug 19, 2009 #7

    Cyosis

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    While there may be a misprint in the given Zg, what you're doing is wrong regardless of said misprint. Both your derivatives are incorrect. Please write out your steps in full without using things like 'thingie' and incomplete steps.

    Hint:
    [tex]\frac{d \ln Z_g}{dU}=\frac{1}{Z_g}\frac{d Z_g}{dU}
    [/tex]

    and use the chain rule to evaluate the derivative on the right hand side.
     
    Last edited: Aug 19, 2009
  9. Aug 19, 2009 #8
    Sorry about that:

    dlnZg/Zg

    first Zg = 1 + Exp (-E-U)/KT, rewrite: Zg = 1 + Exp (-E+U)/KT

    then Ln Zg = 1 + Exp (-E+U)/KT

    to differentiate ln x, it becomes 1/x

    therefore: 1 / 1 + Exp (-E+U)/KT

    d Zg/du = 1/KT

    Zg = 1 + Exp (-E+U)/KT

    differentiate 1 = 0

    differentiate exp (-E+U)/KT split into -E/KT (constant so disappears) +U/KT

    =(exp (-E+U)/KT) . 1/KT

    KT. 1/KT (exp (-E+U)/KT) times (1 / 1 + Exp (-E+U)/KT)

    = (exp (-E+U)/KT)
    1 + Exp (-E+U)/KT

    Which is gibberish to me.

    next post your method for clarity sake.
     
  10. Aug 19, 2009 #9
    d ln Zg/dU times d Zg / d ln Zg

    = d Zg / dU

    d ln Zg/Du = (1 / 1 + Exp (-E+U)/KT) times KT

    d Zg/ ln Zg = errr?
     
  11. Aug 19, 2009 #10

    Cyosis

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    This is a mess. First, use the correct notation. You are not interested in [tex]\frac{d \ln Z_g}{Z_g}[/tex], but in [tex]\frac{d \ln Z_g}{dU}[/tex].

    Here you just change the - in front of the U into a plus? How can these be the same if you randomly change a sign? This is wrong. The first equation is the one you listed in your first post, the second equation is the one, I suspect, on the exam?

    Here you claim [itex]\ln Z_g=Z_g[/itex], this is simply wrong, [tex]\ln Z_g=\ln (1+\exp(\frac{-E-U}{kT}))[/tex], which is something totally different.

    This is correct.

    This is equal to what exactly? Also the way you wrote it down means that the exponent is not in the denominator although it surely should so don't forget brackets. This is what you really meant, 1 / (1 + Exp ((-E+U)/KT)). But again, this is equal to what?

    Wrong, the exponent does not disappear while differentiating. [tex] \frac{d e^{f(x)}}{dx}=e^{f(x) }\frac{df(x)}{dx} \neq \frac{df(x)}{dx}[/tex]

    Correct, but again the definition of Zg you use here is not the same as the one in your original post. So which one is correct?

    Correct, again if and only if the Zg you listed in post 1 is incorrect, which I think it is.

    Ignoring the fact that you have placed your brackets incorrectly again, or omitted them altogether, this is correct. This expression will also lead to the correct answer if manipulated properly, however the way you got there is more or less gibberish.

    I have no idea what you're doing here.

    Try to answer the questions in this post and correct the mistakes I have pointed out.
     
  12. Aug 19, 2009 #11
    Sigh, a mod has deleted my mates account, which was a valid account made from my computer hence the IP address. Regardless, I'll post on his behalf now anyway.

    He says:

    Sorry, I'm really tired today , I will check my working, I really appreciate your help.

    I say:
    I'm going to have a crack at this problem too :-p but my maths (as evident from all other threads which I have started) is not top notch.

    Ta.
     
  13. Aug 20, 2009 #12
    Posting on behalf of Zophixan:



    <N> = 1 / exp (E-U)/KT +1

    from Zg = 1 + Exp (-(E-U))/KT

    where <N> = kt d (ln Zg) / dU

    He left out some brackets, hence why it makes no sense. Unfortunately i can't do the question either.

    however here is my working for you to ponder. Bear with me, this is my first latex:

    [tex]

    \frac{1}{1 + Exp (-(E-U))/KT}

    [/tex]

    =

    [tex]

    \frac{d Ln Zg}{dZg}

    [/tex]

    and
    [tex]

    \frac{d Zg}{dU}

    [/tex]

    =

    [tex]

    \frac{1}{KT} . exp (-(e-u)/KT )

    [/tex]
     
    Last edited: Aug 21, 2009
  14. Aug 21, 2009 #13

    Cyosis

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    This is looking a lot better.

    While you guys said there was definitely a minus in front of the U it seems I was right after all. This means there is no misprint in the the book.

    Note that [itex]\exp(-(E-U))=\exp(-E+U)[/itex], which is exactly the plus in front of the U I was talking about. Using brackets and notations properly is extremely important especially when you're asking someone to help you. We can't ready our mind after all! Anyhow enough preaching.

    The first derivative is correct, the second is not. Remember that differentiating a constant makes the constant disappear. Do you see how to fix your answer (it is almost correct)?

    After you have differentiated the constant correctly you get an expression you may think looks nothing like the answer, but is in fact equal to the answer. The trick to put it in the correct form is to multiply by '1'. In other words find a factor that when multiplied with the numerator yields 1. To make sure the equation holds you will then have to multiply the denominator with the same factor.
     
    Last edited: Aug 21, 2009
  15. Aug 21, 2009 #14
    [tex]


    \frac{1}{1 + Exp (-(E-U))/KT}


    \frac{1}{KT} . exp (-(e-u)/KT )


    [/tex]
    I have edited the previous post to reflect what I think is the correct derivation. The factor which I tried is [tex]\1 + Exp (-(E-U))/KT [/tex]however what I don't understand is, surely multiplying by 1, will just result in the same thing? I can cancel out the top with what \1 + Exp (-(E-U))/KT but since i'm multiplying it by the top, its the same!
     
  16. Aug 21, 2009 #15

    Cyosis

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    That's the entire idea behind multiplying by 1, it has to be the same! You cannot just alter an expression. Multiply top and bottom by [itex]\exp(+(E-U)/(kT))[/itex] and see what happens.
     
    Last edited: Aug 21, 2009
  17. Aug 24, 2009 #16
    Yay, thank you, I got it, me and my friend are very grateful!
     
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