In Ashcroft and Mermin's first chapter, equation (1.13), they write the Lorentz force law as (due only to a magnetic field) $$\textbf{F} = \frac{-e}{c} \textbf{v} \times \textbf{H}$$ rather than as $$\textbf{F} = \frac{-e}{c} \textbf{v} \times \textbf{B}$$ as I would have expected in Gaussian units. Why is this? I don't think that $\textbf{B} = \textbf{H}$ in Gaussian?
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3$\begingroup$ If you look directly on the page for gaussian units they will only be the same if there is no magnetization which is likely the assumption here. $\endgroup$– TriatticusCommented Apr 9, 2023 at 17:09
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$\begingroup$ @Triatticus I suppose that's an assumption which is a bit odd in an arbitrary material though no? Even in a metal? $\endgroup$– EE18Commented Apr 9, 2023 at 17:38
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$\begingroup$ It will certainly depend on the magnetic suceptibility of the material, and I don't personally own that book so I don't have the actual context. $\endgroup$– TriatticusCommented Apr 9, 2023 at 18:57
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$\begingroup$ @Triatticus Oy, I forgot most materials were nonmagnetic and was confusing with the dielectric case. Thanks for clarifying. If you supply an official answer I'd be happy to give it a green check. Thanks again! $\endgroup$– EE18Commented Apr 10, 2023 at 1:43
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$\begingroup$ @Mauricio Can you please provide a reference for that? I checked my textbooks and can't find that statement anywhere. $\endgroup$– EE18Commented Apr 11, 2023 at 14:50
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2 Answers
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$H$ does equal $B$ in Gaussian units. Their units are given different names to honor two physicists. In any event, any force law is better written in terms of $B$, even when $H$ is equal to $B$ in magnitude. If $\mu$ does not equal 1, then the force should not be given in terms of $H$.
answered Apr 11, 2023 at 11:55
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$\begingroup$ Wikipedia (en.wikipedia.org/wiki/…) does not agree with you (see the extra term including $\textbf{M}$ in the table). $\endgroup$– EE18Commented Apr 11, 2023 at 14:52
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$\begingroup$ I interpreted the question to be in free space where M=0. In my last sentence, when $\mu$ does not equal one, M will not equal zero, H will not equal B, and should not be used in the force equation. $\endgroup$ Commented Apr 18, 2023 at 1:15
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The relation between the $\mathbf B$ and the $\mathbf H$ field is given in gaussian-cgs units as $$\mathbf B=\mathbf H+4\pi \mathbf M$$ where $\mathbf M$ is the magnetization. In the case where $\mathbf M=0$, we have $\mathbf B =\mathbf H$, so we write the Lorentz force (in the absence of electric field $\mathbf E$ and polarization $\mathbf P$) on a point charge $q$ as $$\mathbf F=\frac{q}{c}\;\mathbf v \times\mathbf B=\frac{q}{c}\;\mathbf v \times\mathbf H$$ it does not matter if we use $\mathbf B$ or $\mathbf H$.
For more details on what the equation looks like in matter when all fields ($\mathbf M,\mathbf E,\mathbf H,\mathbf P$) are present, see Eq. 14a of https://arxiv.org/abs/1312.3383 (M. Mansuripur, A.R. Zakharian 2013).
Note that in gaussian-cgs units, $\mathbf B$, $\mathbf H$ and $\mathbf M$ could be measured in the same units, sadly this is not often the case. Due to historical reasons, $\mathbf B$ is measured in gauss, $\mathbf H$ in oersted and $\mathbf M$ in emu/cm$^3$, with weird conversion units in between.