References: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.17;
F. Mandl, Statistical Physics, Second Edition, (John Wiley & Sons, 1988) – Section 1.4.
Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 5.3.
We can derive thermodynamic identities for a magnetic system in which pressure and volume are constant. In such a system, work is done by an external electrical power source such as a battery, with the work resulting in changes to the magnetization and magnetic field present in the system, rather than changes in pressure or volume.
As an example, consider a solenoid with a (very long, so we can approximate the field produced as equivalent to that produced by an infinite solenoid) length and turns per unit length. Suppose that the interior of the solenoid is filled with a cylinder of magnetic material with a magnetization (magnetic dipole moment per unit volume) of . By applying Ampère’s law to a loop with one edge inside and the opposite edge outside the solenoid, and using the fact that the magnetic field outside a solenoid is zero (see Griffiths referenced above, example 5.9), we have
where is the free current enclosed by the loop. By choosing the loop edge to have a unit length we have , where is the current in the wire. The integrand is defined to be the auxiliary field :
Since both and are parallel to the axis of the solenoid, the integrand is non-zero only on the edge of the loop inside the solenoid and we have
I’m using to represent magnetic field to distinguish it from , which represents enthalpy.
This is true for a steady current, but now suppose we vary the current to change the magnetic field strength . According to Faraday’s law, a change in the magnetic flux through a loop causes a back-emf to be produced which opposes the change in flux, according to
The work done by the battery is therefore the work required to overcome this back-emf and thus keep the current flowing at its new value. The total magnetic flux through the solenoid is the flux through a single turn (which is , where is the cross-sectional area of the solenoid) multiplied by the total number of turns, which is . Therefore
The power (work per unit time) generated by the battery is the voltage times the current, which is , (negative because the battery’s voltage is opposite to the back-emf) so the work done in time is
where is the volume of the solenoid.
From 2, we have
The first term can be written as
In a vacuum, the energy density in a magnetic field is
Thus, 12 is the change in the magnetic field energy in the entire solenoid, assuming there is a vacuum inside the solenoid. [Actually, in a linear magnetic material, , where is the permeability of the material, and it is only in a vacuum that . Thus, 14 is the magnetic energy density in a vacuum.]
The second term in 11 is the work required to change the magnetization of the sample inside the solenoid:
where is the total magnetization of the sample.
Assuming that only the energy and magnetization change, we can work out a thermodynamic identity for magnetic matter. By analogy with the earlier derivation, we can imagine a system in which the entropy changes by either the energy changing or the magnetization changing. The entropy change due to an increase in energy at temperature is
How does the entropy change when the magnetization increases? A larger magnetization usually means that the dipoles in the sample are more ordered, so we’d expect an increase in to cause a decrease in . Thus we’d expect
From 16, we’d like to be something times . To find the ‘something’, we can look at the units. has units of . From 2, has units of a which works out to . The derivative therefore has units of . Multiplying this by gives a quantity with units of which are the units of , so we propose
The total entropy change is therefore
This gives a thermodynamic identity of
As we might expect, the last term is just the work done on the system as given by 16.
By analogy with the definition of enthalpy for a pressure-volume system, the enthalpy here is
This is the energy required to create a magnetic system from scratch, with an internal energy and magnetization in a field . Presumably the second term is negative, since aligning the dipoles with the external field reduces the energy of the system.
The corresponding thermodynamic identity is
The Helmholtz free energy is still defined as
since there is no reference to pressure or volume. The thermodynamic potential is
The Gibbs free energy is thus defined as
Its thermodynamic identity is