We commonly encounter the first law of thermodynamics in the form $\mathrm{d}E = T\mathrm{d}S - p\mathrm{d}V + \mu\mathrm{d}N$, which notes the contribution of thermal, pressure-volume, and chemical work to the total energy respectively. More generally, we can write $$\mathrm{d}E = \sum_i f_i\mathrm{d}X_i,$$ where $f_i$ denotes a generalized force and $X_i$ a generalized displacement.

The pattern for all these types of work seems to be rather obvious---the generalized force is intensive, whereas the generalized displacement is extensive---but this fails when we try to take into account magnetic work, which contributes the term $-\mathbf{m}\cdot\mathrm{d}\mathbf{B}$, with $\mathbf{m}$ the magnetic moment and $\mathbf{B}$ the magnetic field (see Hill, Chandler, Blundell, etc.). Here the pattern is reversed: the magnetic moment $\mathbf{m}$ is extensive, whereas the magnetic field $\mathbf{B}$ is intensive!

  • How can we rationalize the unusual form of the magnetic work term?

Blundell makes an argument for this for the case of electric dipoles; I summarize it here. The relevant work term here is the electrical work $-\mathbf{p}_\text{E}\cdot\mathrm{d}\mathbf{E}$, with $\mathbf{p}_\text{E}$ the electric dipole moment and $\mathbf{E}$ the electric field.

The potential energy of the dipole in the electric field is $-\mathbf{p}_\text{E}\cdot\mathbf{E}$. If the electric field changes, the potential energy changes by $\mathrm{d}(-\mathbf{p}_\text{E}\cdot\mathbf{E}) = -\mathbf{p}_\text{E}\cdot\mathrm{d}\mathbf{E}-\mathbf{E}\cdot\mathrm{d}\mathbf{p}_\text{E}$. However, the change in the electric field will also stretch the dipole, increasing its dipole moment and doing work $q\mathbf{E}\cdot\mathrm{d}\mathbf{l} = \mathbf{E}\cdot\mathrm{d}\mathbf{p}_\text{E}$, so the net work supplied to the system is $-\mathbf{p}_\text{E}\cdot\mathrm{d}\mathbf{E}$.

I am somewhat skeptical of this argument. It seems that I could repeat the argument for other types of work---say, $PV$ work---and conclude that I should also have a $V\mathrm{d}P$ term in my energy expression.

  • Is this argument correct? If it is, why isn't there a $V\mathrm{d}P$ work term in the first law?

  • What are the conditions under which this argument holds? Do the generalized force and generalized displacement have to interact with each other in a particular manner?

It also seems to me to be of limited applicability, because the convenient cancellation of magnetic-dipole and electric-dipole terms are not replicable for other, more complicated interactions.

  • Is there a general, methodical approach for identifying the form of the terms in the first law for arbitrary types of work?
  • 2
    $\begingroup$ In a way, magnetic work is an oxymoron. Magnetic force is perpendicular to velocity, hence it doesn't do work. $\endgroup$ Nov 16, 2017 at 8:12
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    $\begingroup$ @Ivan_Neretin Of course, magnetic fields do work on spins, they just do not work on moving charges. $\endgroup$
    – hyportnex
    Nov 16, 2017 at 20:00
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    $\begingroup$ Both are meaningful: $mdB$ is the work done on a dipole by changing the field, $Bdm$ is the work done on the field by changing the magnetization of the dipole. $B$ is closer in spirit to a potential, hence an intensive quantity, while piling more and more dipoles on top of each other is like having more matter, hence extensive. $\endgroup$
    – hyportnex
    Nov 16, 2017 at 20:14

1 Answer 1


"Magnetic work" generally refers to work done by a constant-current supply connected to a solenoid. Whether this work is done upon the thermodynamic system of interest depends on whether the system is chosen to include, in addition to the dipoles, the self-field energy of the solenoid and/or the mutual-field energy of solenoid and dipoles, defined below. For details, consider that a stationary microscopic dipole $\boldsymbol\mu$ sits in the magnetic field $\bf {B}_{sol}$ inside a solenoid driven by a constant-current supply. Integrals herein are over all space. If the field changes by ${\bf {dB}}_{sol}$ while $\boldsymbol\mu$ remains fixed, then the current supply does work only on the solenoid’s self-field energy, $\int{dV}{B_{sol}^2({\bf r})/{2\mu_0}}$. Also, the mutual-field energy of solenoid and dipole $\int{dV}{{\bf{B}}_{sol}({\bf r}) \cdot {\bf{B}}_{dip}({\bf r}) / { \mu_0}}$, and the dipole energy change by equal and opposite amounts, +$\boldsymbol\mu \cdot {\bf{dB}}_{sol} $ and -$\boldsymbol\mu \cdot {\bf{dB}}_{sol} $, respectively. In other words, the mutual-field energy does work -$\boldsymbol\mu \cdot {\bf{dB}}_{sol} $ on the dipole. If the mutual-field energy is not part of the system, then work $\boldsymbol\mu \cdot {\bf{dB}}_{sol} $ was done on the system. Otherwise, no.

In the converse case wherein the dipole changes (by $\Delta \boldsymbol{\mu}$) while the field remains fixed, the current supply does work +${\bf{B}}_{sol} \cdot \Delta \boldsymbol{\mu}$ on the mutual-field energy, and the same amount of energy leaves the dipole and flows ultimately as heat into a thermal reservoir. Namely, the dipole energy changes by -${\bf{B}}_{sol} \cdot \Delta \boldsymbol{\mu}$. If the mutual-field energy is part of the thermodynamic system, then work was done on the system. Otherwise, no. n.b. The change in dipole orientation necessarily comes about via scattering with some quantum excitation that can take energy and angular momentum from the dipole. The change in dipole orientation may be as large as the dipole moment itself, not a differential quantity.


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