# Fundamental equation of thermodynamics for magnetocaloric materials

The fundamental equation of thermodynamics, as us chemists (and chemical engineers!) are used to seeing it, is

$$dG = - S~dT + V~dP + \sum_{i}\mu_i~dN_i$$

This gives the Gibbs free energy as a function of temperature, pressure, and composition, assuming there are no other relevant forces other than mechanical pressure.

The other day I watched a video on the magnetocaloric effect. Obviously, there are non-pressure magnetic forces acting in such systems. What's the proper form of the fundamental equation for magnetocaloric materials?

Suppose that the magnetocaloric material used is chemically pure and non-reactive during the magnetization process. Then we could get rid of the $\sum_{i}\mu_i~dN_i$ term. I suppose its also reasonable to assume that pressure is constant during magnetization / demagnetization process, and that the volume of the material is unchanged by magnetization so probably we could dispense with the $V~dP$ term as well (is that true?).

That leaves us with $dG = -S~dT + \rm{MAGNETIC~STUFF}$. The $\rm{MAGNETIC ~STUFF}$ term probably has a $B$ or $H$ or something like that in it to represent the imposed magnetic field, but what else goes in there?

• A quick google search led me to this paper. Maybe it has the answer you are looking for. I cannot access the paper from my home. – Papul Jun 25 '15 at 17:52
• Thanks for the link -- it is definitely relevant. Since I'm not a physicist, it's tough going for me, but if I am able to make enough sense of it over the next few days, I might submit an answer to my own question. It seems like they use $H$ in their treatment, so I guess the right term in the equation would either be $x dH$ or $H dx$, but I still don't know what $x$ is or should be. – Curt F. Jun 25 '15 at 18:22

$$dU=TdS+BdM+\mu dN$$
$$dU=TdS+BdM$$
$$y(0)=U(S,M)$$ $$dy(0)=dU=TdS+BdM$$ $$dy(1)=dA(T,M)=-SdT+BdM$$ $$dy(2)=dG(T,B)=-SdT-MdB$$ Now changing the order of the equation, we can find enthalpy, $$y(0)=U(M,S)$$ $$dy(0)=dU=BdM+TdS$$ $$dy(1)=dH(S,B)=-MdB+TdS$$
• Is M the magnetic moment or the magnetization? If the latter, it seems like in SI units it has units of Amperes per meter, or (dividing by Faraday's constant) moles per second per meter, and B has units of kg per second per Coulomb, or kg per second per mole. Then the magnetic force term has units of $kg/s^{2}/m$, which could also be written as Joules per cubic meter. But don't the other terms in the equation have units of Joules, or at least Joules per mol? – Curt F. Jun 26 '15 at 17:39
• M is magnetic moment while magnetization is density of magnetic moment. For dimensional consistency, B=intensity of magnetic field (Tesla or T), M=Magnetic moment (J/T). So, $B \times M = J ~or~ J/mol$ in terms of per mol basis – Osman Mamun Jun 26 '15 at 18:48