I've seen textbooks saying that for a given equation of state, for example

$$ M = \frac{C_cH}{T}. $$

Where $M$ is the magnetic moment of a paramagnetic material and $H$ is the magnetic intensity. In this case the work is

$$ dW = -HdM.$$

But how do I derive this expression? Or the expression of work for any equation of state?

  • $\begingroup$ Define the terms $\endgroup$
    – Bob D
    Commented Apr 25, 2023 at 15:03

1 Answer 1


Starting with the Maxwell's equation you multiply the 1st with E the second with H add them and you get Poynting's theorem according to which

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \mathbf J \cdot \mathbf E + \mathbf E \cdot \frac{\partial \mathbf D}{\partial t} + \mathbf H \cdot \frac{\partial \mathbf B}{\partial t}$$ Integrate this over all space and assuming that the fields go to zero at least as fast as $1/r$ the integral over the div is zero then and you have

$$0 = \int_{\infty}dV \mathbf J \cdot \mathbf E + \int_{\infty}dV \mathbf E \cdot \frac{\partial \mathbf D}{\partial t} + \int_{\infty}dV \mathbf H \cdot \frac{\partial \mathbf B}{\partial t} \tag{1}\label{1}$$

In the first term the current $\mathbf J $ can be split in to two terms $\mathbf J = \mathbf J_f+\mathbf J_e$ in which $\mathbf J_f$ represents the current induced by the fields and $\mathbf J_e$ is the externally imposed current by mechanical or chemical means, it corresponds to the EM description of a "generator: that does the work. In other words $\int_{\infty}dV \mathbf J_f\cdot \mathbf E$ is the dissipation and $\int_{\infty}dV \mathbf J_e \cdot \mathbf E $ is the rate of working of the sources.

Now assume that the medium is lossless, $\mathbf J_f =0$ and you get $$- \int_{\infty}dV \mathbf J_e \cdot \mathbf E= \int_{\infty}dV \mathbf E \cdot \frac{\partial \mathbf D}{\partial t} + \int_{\infty}dV \mathbf H \cdot \frac{\partial \mathbf B}{\partial t} \tag{2}\label{2}$$ or in a short $\delta t$ time interval

$$\delta w = - \int_{\infty}dV \mathbf J_e \cdot \mathbf E \delta t= \int_{\infty}dV \mathbf E \cdot \delta \mathbf D + \int_{\infty}dV \mathbf H \cdot \delta \mathbf B \tag{3}\label{3}$$

In general, in the presence of externally induced polarization we can write that $$\mathbf D = \epsilon_0 \mathbf E +\mathbf P\\ \mathbf B = \mu_0 (\mathbf H +\mathbf M)$$ so the $\eqref 3$ can be written as

$$\delta w = \delta w_e +\delta_m$$ where

$$ \delta w_e= \int_{\infty}dV \epsilon_0 \mathbf E \cdot \delta \mathbf E + \int_{\mathcal V}dV \epsilon_0 \mathbf E \cdot \delta \mathbf P \tag{4}\label{4}$$ $$\delta w_m = \int_{\infty}dV \mu_0\mathbf H \cdot \delta\mathbf H + \int_{\mathcal V}dV \mu_0 \mathbf H \cdot \delta \mathbf M \tag{5}\label{5}$$ where $\mathcal V$ is the volume containing the polarizable matter.

Sometimes it is claimed, erroneously, that in $\eqref{4}$ and $\eqref{5}$ the fields $\mathbf E$ and $\mathbf H$, resp. represent the vacuum fields but that claim is wrong. Instead we have to include the bias fields $\mathbf E_0$ and $\mathbf H_0$ created by the (charge and current) sources that induce the polarization. Then the terms $\delta w_e$ and $\delta w_m$ represent the amount of additional work expended by the sources to maintain the charges and currents that induced the polarization before the material was placed in the field. Without getting into vector analytical details, see Stratton one can show that integrals can be transformed to contain the bias fields themselves as follows: $$ \delta w_e= \int_{\infty}dV \epsilon_0 \mathbf E_0 \cdot \delta \mathbf E_0 + \int_{\mathcal V}dV \epsilon_0 \mathbf E_0 \cdot \delta \mathbf P \tag{6}\label{6}$$ $$\delta w_m = \int_{\infty}dV \mu_0\mathbf H_0 \cdot \delta\mathbf H_0 + \int_{\mathcal V}dV \mu_0 \mathbf H_0 \cdot \delta \mathbf M \tag{7}\label{7}$$

In this formula there are two terms, $\delta w_m =\delta w'_m+\delta w''_m$, the first being $\delta w'_m = \int_{\mathcal V}dV \mu_0 \mathbf H_0 \cdot \delta \mathbf H_0 $ is the amount of work it takes to modify the bias field itself and has no direct thermodynamics consequence, the second term is $\delta w''_m = \int_{\mathcal V}dV \mu_0 \mathbf H_0 \cdot \delta \mathbf M $ and is the isothermal work it takes to polarize the material. Notice that this term is not a spatial density, instead it is a spatial integral of a density $\mu_0 \mathbf H_0 \cdot \delta \mathbf M$. When measuring the work at the current source one measures $\delta w_m$ or $\delta w''_m$ and that is represented by the spatial integral, the energy density is not measured and it is just as ambiguous, for lacking a better term, as is a scalar or vector potential. Only the spatial integral of the energy density is meaningful. Nevertheless one can employ in calculations the energy density directly cautioning that some other split leading to the same spatial integral is just as good as any.

This is all true irrespective of the constitutive equation between polarization and field. In your question you asked about $\mathbf M = \chi_m(T) \mathbf H$ where $\chi_m (T) = \frac{C}{T}$. For fixed $T$ everything above is relevant as long as $C$ is indeed a material constant and does not depend on the field.

And because the calculations assumed an isothermal system that is in thermal equilibrium with its environment the reversible work on the system is equal the change in free energy of the system.

  • $\begingroup$ Amazing answer! $\endgroup$
    – Alex
    Commented Nov 5, 2023 at 19:55

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