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Writting total energy from temperature

  1. Oct 20, 2008 #1
    Hi all,

    I have to compute the entropy, temperature and show that the total energy can be written as

    E(T,N) = 3/2*N*hbar*omega*coth(hbar*omega/(2*kT))

    I have found that the temperature can be written as

    1/T = k/(hbar*omega)*ln([E/N*hbar*omega+3/2]/[E/N*hbar*omega-3/2])

    by differentiating the entropy with respect to E. Next, I would like to show that the energy can be written as shown above:

    1. (hbar*omega)/(k*T) = ln([E/N*hbar*omega+3/2]/[E/N*hbar*omega-3/2])

    2. exp((hbar*omega)/(k*T)) = ([E/N*hbar*omega+3/2]/[E/N*hbar*omega-3/2])

    but this last term doesnt give me any hyperbolic cotangens???

    Any help or suggestions appreciated. Thanks in advance

    Best
    M
     
  2. jcsd
  3. Oct 20, 2008 #2

    tiny-tim

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    Hi mhellstrom! :smile:

    Hint: cothx = (e2x + 1)/(e2x - 1) :wink:
     
  4. Oct 20, 2008 #3
    Re: writing total energy from temperature

    thanks - I have tried to isolate E but cannot. I can see how my expression to the right looks like coth... Could u guide me one more step I hope that is all I need :smile:
     
  5. Oct 20, 2008 #4

    tiny-tim

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    Hi mhellstrom! :smile:

    Hint: if e2x = A/B,

    then (e2x + 1)/(e2x - 1) = (A + B)(A - B) :wink:
     
  6. Oct 21, 2008 #5
    did you mean
    (A+B)/(A-B) ?

    A = E and B = 3/(2N)*hbar*omega
    So insert this into the equation

    exp (hbar*omega/(kT))= (A/B +1) / (A/B-1)
    exp (hbar*omega/(kT))= (A+B)/(A-B)

    I am still confused how to get A isolated and how to transform the right hand side into coth?

    How to proceed

    Thanks very much

    Best
    M
     
  7. Oct 21, 2008 #6

    tiny-tim

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    I'm not following you. :confused:

    Apply (A+B)/(A-B) to:
     
  8. Oct 21, 2008 #7
    Hi

    What I mean is that I have rewritten the expression as

    exp((hbar*omega)/(k*T)) = (E+3/2*N/(hbar*omega)/(E-3/2*N/(hbar*omega)

    which is (A+B)/(A-B) I presume...than I would like to isolate E... but here I am still lost

    Thanks for helping me...

    Best regards
     
  9. Oct 21, 2008 #8

    tiny-tim

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    erm … if that's (A+B)/(A-B), then what is A/B? :redface:
     
  10. Oct 21, 2008 #9
    u are right - I have not understood that hint;

    A/B = exp(hbar*omega/(4*k*T)) / exp(-hbar*omega/(4*k*T))

    is A/B? but how one can rearrange it to coth(X) I still don't understand....
     
  11. Oct 21, 2008 #10

    tiny-tim

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    cothx = (exp2x + 1)/(exp2x - 1)

    Put A = E, B = 3/2*N/(hbar*omega) :smile:
     
  12. Oct 21, 2008 #11
    ahhh - I did that later: A = E, B = 3/2*N/(hbar*omega). Ok I see how to solve it now Thanks u very much couldnt see the forest for trees ;-)

    Thanks for the hint and yr help all the best

    M
     
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