Homework Help: Writting total energy from temperature

1. Oct 20, 2008

mhellstrom

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. Oct 20, 2008

tiny-tim

Hi mhellstrom!

Hint: cothx = (e2x + 1)/(e2x - 1)

3. Oct 20, 2008

mhellstrom

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

4. Oct 20, 2008

tiny-tim

Hi mhellstrom!

Hint: if e2x = A/B,

then (e2x + 1)/(e2x - 1) = (A + B)(A - B)

5. Oct 21, 2008

mhellstrom

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

6. Oct 21, 2008

tiny-tim

I'm not following you.

Apply (A+B)/(A-B) to:

7. Oct 21, 2008

mhellstrom

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

8. Oct 21, 2008

tiny-tim

erm … if that's (A+B)/(A-B), then what is A/B?

9. Oct 21, 2008

mhellstrom

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....

10. Oct 21, 2008

tiny-tim

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

Put A = E, B = 3/2*N/(hbar*omega)

11. Oct 21, 2008

mhellstrom

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