Advanced Solid State Physics

For every system, on which we would like to perform experiments, using the extrinsic variables as the
independent ones isn’t a good idea. Usually in an experiment we hold the temperature under control.
To realize this, we can make a Legendre transformation to get the free energyF:

F = U − TS

But in an experiment we usually are also in control of the stress, the electric and the magnetic field we
apply to the material. So it is a good idea to perform a couple of additional Legendre transformations
to get the generalized Gibbs free energy:

G = U − TS − σijij − EkPk − HlMl

By taking the total derivative and using eqn. (61), we are left with:

dG = −SdT −ijdσij −PkdEk −MldHl

or, using the definition of a total derivative

dG =

(

∂G

∂T

)

dT +

(

∂G

∂σij

)

dσij +

(

∂G

∂Ek

)

dEk +

(

∂G

∂Hl

)

dHl

we can identify

−

(

∂G

∂T

)

= S −

(

∂G

∂σij

)

= ij

−

(

∂G

∂Ek

)

= Pk −

(

∂G

∂Hl

)

= Ml

So if we know the Gibbs free energy of a system in dependence of the intrinsic variables, we can deter-
mine the extrinsic ones with a simple derivation. But calculating the Gibbs free energy is the difficult
part. Remembering the course on statistical physics, we know how to calculate the free energy of a
canonical ensemble:

Z =

∑

j

exp

(−E

j

kBT

)

F = −kBTln(Z)

Here the sum over j sums over all microscopic states corresponding to a macroscopic state. But there
occurs the energy of the states in the sum. We can get the energy by solving the system quantum
mechanically. Calculating the Gibbs free energy is very similar, but the energy in the partition function
also depends on the magnetic and the electric fields. The calculations for the hamiltonians of electrons
in electromagnetic fields are part of chapter 4.
By taking the total derivative of the extrinsic variables we can get to materials constants or physical
properties. As an example we take the total derivative of the strain:

dij =

(

∂ij

∂σkl

)

1

dσkl +

(

∂ij

∂Ek

)

2

dEk +

(

∂ij

∂Hl

)

3

dHl +

(

∂ij

∂T

)

4

dT

Where the single derivations represent: