# About electromagnetic work in first principle of thermodynamics

Reading L. E. Reichl, A Modern Course in Statistical Physics, first principle of thermodynamics is stated as

$$dU = \delta Q − \delta W + \sum_{j=1}^{v} μ_j dN_j$$

with

$$\delta W = P dV − J dL − σ dA − \mathbf{E} ⋅ d\mathbf{P} − \mathbf{H} ⋅ d\mathbf{M} − \phi de$$

Where $$P$$ is the pressure, $$V$$ volume, $$J$$ tension, $$L$$ length, $$\sigma$$ surface tension, $$A$$ area, $$\mathbf{E}$$ electric field, $$\mathbf{P}$$ electric polarization, $$\mathbf{H}$$ magnetizing field, $$\mathbf{M}$$ magnetization, $$\phi$$ electric potential and $$e$$ is the charge.

• As spotted out in the answer by @hyportnex, there's a substantial incoherence regarding global/local terms in this formula, to this end please consider what is meant here for $$\mathbf{P}$$ as what in electromagnetics would be referred to as $$\mathbf{P}\,dV$$ where (this latter) $$\mathbf{P}$$ is the electric dipole moment per unit volume. Same reasoning applies to the use of $$\mathbf{M}$$ in place of $$\mathbf{M}\,dV$$, where (this latter) $$\mathbf{M}$$ is the magnetic dipole moment per unit volume.

But from Poynting's theorem (see D.J. Griffiths, Introduction to electrodynamics, p357-358 for a proof) we know that

$$\frac{dW_{el}}{dt}=-\int_{V}\frac{1}{2}\frac{\partial}{\partial t}\bigg{(}\epsilon_0\,E^2+\frac{1}{\mu_0}B^2\bigg{)}dV-\int_{\partial V}(\mathbf{E}\times\mathbf{B})\cdot d\mathbf{A}$$

Where $$W_{el}$$ is the work done on charges done by electromagnetic force, $$\mathbf{B}$$ is the magnetic field.

Is it possible to derive, from this last espression, the electromagnetic terms in $$\delta W$$?

• I'm aware that not all Poynting's vector terms may fell into $$\delta W$$, since a part of them may be enclosed in $$dU$$ as potential, if this is the case, may this division be done explicitely in the answer, thanks

The Poynting formula you have is for vacuum. For ponderable matter you can only derive the result for $$\delta W$$ if in the volume integral you start with $$\mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}+\mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t}$$.
Assume now that you integrate for all space and then you can ignore the surface integral over $$\partial V$$ containing the radiation term. Next multiply both sides with $$\delta t$$, so per unit volume the EM energy content is $$\delta \tilde w_{em} = \mathbf{E} \cdot \delta \mathbf{D}+\mathbf{H} \cdot \delta \mathbf{B}$$. This is energy density. Introduce the constitutive relationships: $$\mathbf {D} = \epsilon_0 \mathbf {E} + \mathbf {P}$$ and $$\mathbf {B} = \mu_0 (\mathbf {H} + \mathbf {M})$$, an dyou can write for the energy density $$\delta \tilde w_{em} = \epsilon_0 \mathbf{E} \cdot \delta \mathbf{E}+\mathbf{E} \cdot \delta \mathbf{P}+\mu_0 \mathbf{H} \cdot \delta \mathbf{H}+\mu_0\mathbf{H} \cdot \delta \mathbf{M}\\ =\frac{1}{2}\epsilon_0 \delta |\mathbf{E}|^2+\mathbf{E} \cdot \delta \mathbf{P}+\frac{1}{2}\mu_0 \delta |\mathbf{H}|^2+\mu_0\mathbf{H} \cdot \delta \mathbf{M}$$ This can be rewritten as follows: $$\delta \tilde w_{mat}=\mathbf{E} \cdot \delta \mathbf{P}+\mu_0\mathbf{H} \cdot \delta \mathbf{M}$$ where $$\delta \tilde w_{mat} = \delta \big(\tilde w_{em} - \frac{1}{2}\epsilon_0 \delta |\mathbf{E}|^2-\frac{1}{2}\mu_0 \delta |\mathbf{H}|^2\big)$$ represents the energy density without the, for lack of better term, "vacuum fields" $$\mathbf{H}$$ and $$\mathbf{E}$$ that create, so to speak, the material polarization fields $$\mathbf{M}$$ and $$\mathbf{P}$$.
Now to get back to your original equation, that one cannot be proven for it mixes energy densities with total energies, for example the terms $$pdV$$ or $$\phi de$$ represent total work (or energy) for the whole volume while a term like $$EdP$$ is density per unit volume. Anyhow, you may say that to create the polarizations $$\mathbf{M}$$ and $$\mathbf{P}$$ in the fields $$\mathbf{H}$$ and $$\mathbf{E}$$ you must expend $$\delta \tilde w_{mat}$$ amount of work.
• Very useful answer, although three things are still not clear to me: 1. Maxwell's equations in matter do not add any information with respect to the vacuum ones, since they simply reflect a convenient division of charge and current into free and non-free parts, accordingly all electromagnetic energy we could talk about should be already present (for an all-space-volume) in $\epsilon_0 \mathbf{E}\cdot\frac{\partial \mathbf{E}}{\partial t}+\frac{1}{\mu_0} \mathbf{B}\cdot\frac{\partial \mathbf{B}}{\partial t}$ - so, to my present understanding, the starting point of... – Giorgio Pastasciutta Jan 28 '20 at 15:07
• $\delta \tilde w_{em} = \mathbf{E} \cdot \delta \mathbf{D}+\mathbf{H} \cdot \delta \mathbf{B}$ is something i cannot motivate 2. Identification of $\mathbf{E}\cdot\delta\mathbf{P}+\mu_0\mathbf{H}\cdot\delta\mathbf{M}$ terms as work done on the material seems just something axiomatic 3. Is really the case to consider integration for all space in Poynting's formula when the volume of integration related to first principle is actually well defined and equal to the material or the part of it considered? – Giorgio Pastasciutta Jan 28 '20 at 15:07
• The incoherence you spotted out about global/local energies in first law is correct but i think it can be circumvented by considering (as i think author meant to) $\mathbf{P}$ as $\mathbf{P}\,dV$ and $\mathbf{M}$ as $\mathbf{M}\,dV$ - i'll make soon an edit on this – Giorgio Pastasciutta Jan 28 '20 at 15:08
• Regarding Poynting's theorem the point of integrating over all space is that this way we do not need to handle the $E \times H$ energy passing through the surface, but move that to infinity where it can be ignored, after that the volume integral, by definition, of only ponderable matter having nonzero polarization gets automatically restricted over it. In other words, this is just a proof that the integral of $E \times H$ is zero if outside the enclosing surface has no polarizable matter. After all, both $E$ and $H$ are static so there should not be any radiative field propagation. – hyportnex Jan 28 '20 at 15:21
• to motivate the term $E\delta D$ recall that the "source" of $E$ is $\phi$ and the "source" of $D$ is $\rho$ (charge density), so the term is completely analogous, and derivable from the energy $\phi\delta\rho$, and vice versa. If you also consider magnetic dipole energy, with a bit more work, you get the analogous term for $H\delta M$ . – hyportnex Jan 28 '20 at 15:52