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...