chemical potential vs pressure and temperature; difficulty with Fermi gasesby jfizzix Tags: chemical, difficulty, fermi, gases, potential, pressure, temperature 

#1
Apr2612, 04:51 PM

P: 219

For any system where the thermodynamic limit exists, we know that the internal energy U, the entropy σ, the total particle number N and the total volume V are all extensive. Because of this, we know that the Euler relation holds true
U = PV + [itex]\tau[/itex]σ + [itex]\mu[/itex]N and that the chemical potential is just the Gibbs free energy per particle. G = [itex]\mu[/itex] N We can use the differential relation of the Gibbs free energy dG = V dP σ d[itex]\tau[/itex] + [itex]\mu[/itex] dN to find expressions for the derivatives of the chemical potential. d[itex]\mu[/itex]/d[itex]\tau[/itex] = 1/N * dG/d[itex]\tau[/itex] =σ/N d[itex]\mu[/itex]/dP = 1/N * dG/dP = V/N Since N, [itex]\sigma[/itex], and V are all positive, it appears that the chemical potential must increase with pressure at constant temperature, and decrease with temperature at constant pressure. My question is this: the 1D fermi gas is a case where the chemical potential actually increases with temperature for small temperatures, though we would think in all cases the chemical potential should be decreasing with temperature. How do we reconcile this result with the previous assumptions about the chemical potential? James 



#2
Apr2712, 02:01 AM

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P: 3,375

I suspect you are considering a Fermi gas with fixed particle number instead of fixed pressure or equivalently U. 



#3
Apr2712, 02:58 AM

P: 1

Excuse me,but I don't very understand how it is done with this:
dμ/dτ = 1/N * dG/dτ =σ/N dμ/dP = 1/N * dG/dP = V/N I am a little confused. 



#4
Apr2712, 04:14 AM

Sci Advisor
P: 3,375

chemical potential vs pressure and temperature; difficulty with Fermi gases
He is using e.g.
[itex](\partial \mu/\partial T)_{P,N}=(\partial S/\partial N)_{T,P}=S/N[/itex] The first equality is a Maxwell relation, the second one follows from S(N,p,T) being an extensive function whence it must be proportional to N. 



#5
Apr2712, 08:53 PM

P: 219

For the 1D (or 2D or 3D) Fermi gas we can calculate the chemical potential in the following way. In a nutshell, we say that the number of particles is constant whether the temperature is zero or not. First, we can find the number of particles at zero temperature in terms of the Fermi energy. Starting at the ground state, we put in fermions at each successive orbital until all N fermions have been placed. The highest energy (that of the final fermion) is the Fermi energy. Next we can calculate the number of particles at nonzero temperature by integrating over the density of states. For a nonzero temperature, the integral is harder to evaluate, but can be carried out for small temperatures with what is known as the Sommerfeld expansion/approximation. Setting these two expressions for the number of particles equal to each other, we can calculate the chemical potential in terms of the Fermi energy. for small temperatures the chemical potential of the 1D Fermi gas increases in the following way with temperature: [itex]\frac{\mu}{E_{F}}[/itex] ≈ 1 + [itex]\frac{\pi^{2}}{12}[/itex]*([itex]\frac{k_{b}T}{E_{F}})^{2}[/itex] In the 2D case, the chemical potential is just the Fermi energy at low temperatures, and in the 3D case, the chemical decreases with temperature in the same way that it increases for the 1D gas. Kittel and Kroemer has a nice derivation, as well as Schroeder's book. 



#6
Apr2712, 08:59 PM

P: 219

U=U(S,V,N) H=H(S,P,N) (for the enthalpy) F=F(T,V,N) (for the Helmholtz free energy) Since P and T are intensive, and G is extensive, we can write G as N times some other function of P and T. Taking the derivative of G with respect to N, we get that [itex]\mu[/itex] is that same function of P and T. 



#7
May1012, 02:02 PM

P: 219

in 1D, the density of states decreases with increasing energy as [itex]1/\sqrt{\epsilon}[/itex]. This means the number of accessible states decreases with increasing particle number (and hence increasing Fermi energy). As such, the entropy must decrease with the addition of more fermions, which means that the chemical potential increases with increasing temperature. I'm satisfied as to why the chemical potential decreases using the maxwell relations, but why then can we not say G = [itex]\mu[/itex]*N. Thanks to all who helped, James 



#8
May1012, 02:27 PM

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#9
May1012, 02:43 PM

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#10
May1112, 03:21 AM

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P: 3,375

Ok,
I think I understand it now: To lowest order in the Sommerfeld expansion independent of dimension [itex]\mu(T,P)=E_F(T,P)[/itex] so [itex] \partial \mu /\partial T_P=\partial E_F /\partial T_P= (\partial P/\partial T)_{E_F}(\partial E_F/\partial P)_T[/itex] As the Fermi energy is a monotonously increasing function of the density, only, in the first derivative on the right we can alternatively keep the density constant. But P always increases with increasing temperature at constant density, so this term is positive. Considering the second derivative, the density (and E_F) increases with p at constant T, so this term is also positive. So to lowest order, the change of mu with T is always negative as it should be. The small higher order corrections will not change this. Or stated differently, when T is increasing, density or E_F has to disminish for P to remain constant. 



#11
May1212, 12:26 AM

P: 219

The Fermi energy [itex]E_{f}[/itex] does not depend explicitly on temperature; only on particle number [itex]N[/itex] and the space in which the particles are contained (i.e. in 1D, the length; in 2D, the area; in 3D, the volume). One can express the Fermi energy in terms of N and length and get the chemical potential in terms of N, length, and temperature. Integrating this with respect to N gives you the Helmholtz free energy [itex]F(T,V,N)[/itex] since [itex]μ(T,V,N)=(∂F/∂N)_{T,V}[/itex]. we can then take the derivative of F with respect to volume to get the pressure [itex]P(T,V,N)=(∂F/∂V)_{T,N}[/itex]. We can also take the derivative of F with respect to temperature to get the entropy [itex]S(T,V,N)=(∂F/∂T)_{V,N}[/itex]. With the state equation of the pressure, we can substitute this into our expression for the entropy to get the entropy in terms of N,P, and T. If you take the derivative of this entropy with respect to N, you find it is negative.. which means that where [itex](∂μ/∂T)_{P,N}=−(∂S/∂N)_{T,P}[/itex] the chemical potential must increase with temperature for the 1D degenerate Fermi gas. Now, in 2D and 3D (taking higher order terms) the chemical potential decreases with temperature, but the question persistently remains... How do we say that G is not μ*N? Is G no longer extensive for a 1D Fermi gas? Thanks again, James 



#12
May1212, 06:08 AM

Sci Advisor
P: 3,375

Think of the ideal gas equation which is also the limit of the Fermi equation of state at low densities or high temperature: N/V=P/RT. 



#13
May1512, 10:58 PM

P: 219

We start with the Fermi energy of a 1D Fermi gas. [itex]E_{f} = (\frac{\hbar^2 \pi^2}{8m})(\frac{N}{L})^2 = \alpha_{1} (\frac{N}{L})^2[/itex] and the formula for the chemical potential at low temperature where the Fermi energy has already been substituted [itex]\mu(\tau , L, N) ≈ \alpha_{1} (\frac{N}{L})^2 + \frac{\pi^2 \tau^2}{12 \alpha_{1}}(\frac{L}{N})^2[/itex] We integrate the chemical potential with respect to N from 0 to N to obtain the Helmholtz free energy F. Recall that in 1D... [itex]dF =  P dL  \sigma d\tau + \mu dN[/itex] so that [itex]F( \tau, L, N) = \int_{0}^{N}{\mu (\tau, L, N) dN} = \frac{\alpha_{1}}{3}\frac{N^3}{L^2}  \frac{\pi^2 \tau^2}{12 \alpha_{1}} \frac{L^2}{N}[/itex] Now we take the derivative of the Helmholtz free energy with respect to length to find the pressure in 1D [itex]P = (\frac{\partial F}{\partial L})_{\tau,N} = \frac{2 \alpha_{1}}{3}\frac{N^3}{L^3} + \frac{\pi^2 \tau^2}{6 \alpha_{1}} \frac{L}{N} ≈ \frac{2 \alpha_{1}}{3}\frac{N^3}{L^3} [/itex] Substituting this expression for the pressure into our original expression for the chemical potential, we find [itex] \mu(\tau , P, N) ≈ \alpha_{1} (\frac{3 P}{2 \alpha_{1}})^{2/3}+ \frac{\pi^2 \tau^2}{12 \alpha_{1}}(\frac{2 \alpha_{1}}{3 P})^{2/3}[/itex] In short, for small temperatures, and in one dimension, the chemical potential does increase with temperature at constant pressure and particle number. I could have made a mistake with the math, but I think it's all there. James 



#14
May1612, 02:18 AM

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P: 3,375

Obviously what is happening here is that at small N and fixed L and T the Sommerfeld expansion breaks down. 



#15
May1612, 07:33 AM

P: 219

I actually forgot that infinity there.. I'll have to get back to you on how to do it right. James 



#16
May1612, 08:01 AM

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P: 3,375

You really should calculate the Sommerfeld expansion of p in terms of E_F and T.




#17
May1612, 10:13 PM

P: 219

I just substitute the Fermi pressure along with factors of N and [itex]\tau[/itex] into the expression, and that should do the trick.. more to follow... James 



#18
May1712, 08:29 PM

P: 219

So.. in 1D,
The fermi energy [itex]E_{f}[/itex] can be expressed in terms of the Fermi pressure [itex]P_{f}[/itex] using the expression for the internal energy [itex]U[/itex] at absolute zero. If we differentiate the internal energy [itex]U[/itex] with respect to the length of the 1D box [itex]L[/itex] and add a negative sign, we have an expression for the pressure. [itex]P = (\frac{\partial U}{\partial L})_{\sigma,N}[/itex] In 1D at absolute zero... [itex]U = \frac{1}{3} N E_{f}[/itex] where [itex]E_{f} = \frac{\hbar^{2}}{8 m}(\frac{N \pi}{L})^{2}[/itex] From these expressions, we can express the Fermi energy in terms of the pressure at absolute zero. [itex]E_{f} = (\frac{3 \pi}{2})^{\frac{2}{3}}(\frac{\hbar^{2}}{8 m})^{\frac{1}{3}} P^{\frac{2}{3}} = C_{1} P^{\frac{2}{3}}[/itex] Then the expression for the chemical potential becomes [itex]\mu ≈ C_{1} P^{\frac{2}{3}} + \frac{\pi^{2}}{12} \frac{\tau^{2}}{C_{1} P^{\frac{2}{3}}}[/itex] With this as our form for the chemical potential, we see that it does decrease with temperature at constant pressure and particle number. I'm pretty sure I've got it right this time, but feel free to correct me. Does this make sense? James 


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