Recent content by mhellstrom

yes of course... Thanks for the help. The partition function is also given in the exercise q_{rot} = \frac{T}{\omega}*(1+\frac{1}{3}(\frac{\omega}{T}+\frac{1}{15}(\frac{\omega}{T})^2+...) I presume the internal energy is given as E = -N(dLn Zrot / d beta) I would really...

Hi all, I have to show that the heat capacity can be expressed as Cv = Nk(1+1/45(Om/T)^2 + ...) where the internal energy is given as E = NkT*(1-(Om/(3T)-1/45(Om/T)^2) Normally I would just differentiate but if I do this I get something completely different - how to proceed any...

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 couldn't see the forest for trees ;-) Thanks for the hint and yr help all the best M

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...

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

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 - 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:

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...

thanks, if I want to estimate the variance of the elongation var = 1/N sum (xi-x_mean)2 I know the mean is x_mean = m*g/k which I insert into the expression and integrate from minus to plus infinity var = \int(m*g-k*x-m*g/k)^2 dx Could anyone give a hint if this is on the right...

hi, I am a little bit puzzled where my mistake is... I differentiate my expression for the potential energy in order to find a stationary point d(E_p) = m*g - k*x setting this equal to zero and solving for x x = m*g/k than I set this into the equation for the potential energy as I presume...

Hi, thanks for the answer. So the mean is when m*g = k*x solving for x x = m*g/k which results in the mean elongation of the spring is <dis> = 0.5*m*g/k Is this correct? Thanks in advance all the best

Hi all, I have to determine the potential energy of a hanging spring with a mass m in the end and spring constant k. I try to write down the force in the system F = m*g + k*x and integrate the force in order to get the potential energy E_p = m*g*x+0.5*k*x*x Does this look correct...