Electromagnetism in Gaussian Units

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\begin{document} \title{Electromagnetism (EM) in Gaussian Units} The ubiquity of SI units in undergraduate and graduate EM texts requires the astronomy graduate student to understand the conversion of classical EM results from SI to cgs (Gaussian) units. These results were hard-earned in our undergraduate courses, so the present page is to help you avoid having to re-derive these results from the Gaussian version of Maxwell's Equations (because who honestly wants to mess around with vector potentials again?). This page is not meant to be comprehensive. Instead, we show by example how to use the (many) conversion factors required to re-express SI formulas in Gaussian units. The primary reference for this page is the excellent Wikipedia article {\bf Gaussian Units}, linked above. Throughout this page, we will use the subscripts/superscripts SI and G to differentiate between SI or Gaussian quantities, respectively. These subscripts/superscripts will be omitted when it is clear from context.

\section*{Units} The most important thing to note is the units of electric charge. The cgs unit of electric charge is the ``electrostatic unit, or esu, where 1 esu = 1 statC = 1 g$^{1/2}$ cm$^{3/2}$ s$^{-1}$. The most important value to memorize in cgs units is the electron charge: $$\boxed{e = 4.8 \times 10^{-10} \, {\rm g}^{1/2} {\rm cm}^{3/2} {\rm s}^{-1}.}$$ In Gaussian units, the electric and magnetic fields have the same base units. The Lorentz force law provides a mnemonic for recalling the base units of the two fields. The {\bf Lorentz Force Law in Gaussian units} is $$\vec{F} = Q\bigg[\vec{E} + \frac{\vec{v}}{c}\times \vec{B}\bigg] .$$ Notice that the electric field term is the same as in SI units, $\vec{F} = Q\vec{E}$. That means, $$[F] = [Q][E]$$ so $$[B] = [E] = \frac{[F]}{[Q]} = \frac{{\rm g} \, {\rm cm}\, {\rm s}^{-2}}{{\rm g}^{1/2} \, {\rm cm}^{3/2} \,{\rm s}^{-1}} = {\rm g}^{1/2} {\rm cm}^{-1/2} {\rm s}^{-1}.$$ When discussing magnetic fields, this derived unit is called a Gauss (G), so that $$1 {\rm G} = 1 {\rm g}^{1/2} {\rm cm}^{-1/2} {\rm s}^{-1}.$$ In the context of electric fields, the derived unit is called a statV/cm, but the underlying cgs units are the same. If you have the units of charge squared away, it is easy to verify that Coulomb's law no longer requires coefficients to give units of force. {\bf Coulomb's Law in Gaussian units} is $$\vec{F} = \frac{Q_1 Q_2}{r^2}\hat{r}.$$ Dimensionally, we can confirm $$\bigg[\frac{Q_1 Q_2}{r^2}\bigg] = \frac{{\rm g} \, {\rm cm}^{3} {\rm s}^{-2}}{{\rm cm}^2} = {\rm g} \, {\rm cm}\, {\rm s}^{-2} = [F].$$

\section*{Conversion Factors} An extensive list of conversion factors is given in table 1 of \href{https://en.wikipedia.org/wiki/Gaussian_units}{Gaussian units (current version)}. Conversion factors we will use in the follow examples are: \begin{table}[h] \begin{center} \begin{tabular}{lll} \toprule Quantity & Symbol & Conversion Factor\\ \midrule Electric charge & $Q$ & \Large{ $\frac{Q_\g}{Q_\si} = \frac{1}{\sqrt{4\pi\epsilon_0^\si}}$ }\\ \midrule Electric field & $E$& \Large{$\frac{E_\g}{E_\si}$} \normalsize{$ = \sqrt{4\pi\epsilon_0^\si}$} \\ \midrule Magnetic field & $B$& \Large{$\frac{B_\g}{B_\si} = \sqrt{\frac{4\pi}{\mu_0^\si}}$ } \\ \midrule Magnetic dipole moment & $m$& \Large{$\frac{m_\g}{m_\si} = \sqrt{\frac{\mu_0^\si}{4\pi}}$ } \\

\bottomrule \end{tabular} \caption{Recall that the speed of light in SI units is $c = 1 / \sqrt{\epsilon_0^\si \mu_0^\si}$.} \end{center} \end{table}

\section*{Larmor Formula} The most important equation in any Radiative Processes course is the Larmor Formula and it is also one of the most simple formulas to convert. In mks units, this is given by $$P_\si = \frac{\mu_0^\si Q^2_\si a^2}{6\pi c}.$$ From the table, $Q_\si = \sqrt{4\pi\epsilon_0^\si} Q_\g$, which gives $$P_\g = \frac{\mu_0^\si \big(\sqrt{4\pi\epsilon_0^\si} Q_\g\big)^2 a^2}{6\pi c} = \mu_0^\si \epsilon_0^\si\frac{4\pi Q_\g^2 a^2}{6\pi c} = \frac{1}{c^2}\frac{2}{3}\frac{Q_\g^2 a^2}{c}.$$ Dropping subscripts for clarify, then the {\bf Larmor formula in Gaussian units} is $$\boxed{P = \frac{2}{3} \frac{Q^2 a^2}{c^3}}$$

\section*{Energy Density in the Fields} Recall that in mks units, the energy density stored in the electric and magnetic fields is $$u_\si = \frac{\epsilon_0^\si}{2}E^2_\si + \frac{1}{2\mu_0^\si}B^2_\si.$$ From the table, $$E_\si^2 = \frac{E^2_\g}{4\pi\epsilon_0^\si}$$ and $$B_\si^2 = B^2_\g \frac{\mu_0^\si}{4\pi}$$ so $$u_\g = \frac{\epsilon_0^\si}{2} \bigg(\frac{E^2_\g}{4\pi\epsilon_0^\si}\bigg) + \frac{1}{2\mu_0^\si}\bigg(B^2_\g \frac{\mu_0^\si}{4\pi}\bigg) = \frac{E_\g^2}{8\pi} + \frac{B_\g^2}{8\pi}.$$ Therefore, {\bf the energy density stored in the electric field in Gaussian units} is $$\boxed{u_E = \frac{E^2}{8\pi}}$$ and the {\bf energy density stored in the magnetic field in Gaussian units} is $$\boxed{u_B = \frac{B^2}{8\pi}}.$$

\section*{Magnetic Field of a Spinning Sphere of Charge} In the above examples, only one substitution was necessary because the quantity on the left-hand-side of the equal sign was not an electric or magnetic quantity. The magnetic field of a dipole is more complicated. Recall that in SI units (and in spherical coordinates) the vector potential of the perfect dipole $$\vec{m}_\si = m_\si \hat{z},$$ located at the origin, is \beq \vec{A}_\si = \frac{\mu_0^\si}{4\pi}\frac{\vec{m}_\si\times\hat{r}}{r^2} = \frac{\mu_0^\si}{4\pi}\frac{m_\si \sin\theta}{r^2}\hat{\phi} \eeq (see equation 5.87 of Griffiths). The magnetic field is \beq \vec{B}_\si = \nabla \times \vec{A}_\si = \frac{\mu_0^\si}{4\pi r^3}m_\si\bigg[2\cos(\theta)\hat{r} + \sin(\theta)\hat{\theta}\bigg]. \label{eq:bdip} \eeq Conversion of this formula requires two substitutions: $$B_\si = B_\g \sqrt{\frac{\mu_0^\si}{4\pi}}$$ and $$m_\si = m_\g\sqrt{\frac{4\pi}{\mu_0^\si}}.$$ Then equation \ref{eq:bdip} becomes \beq \vec{B}_\g \sqrt{\frac{\mu_0^\si}{4\pi}} = \frac{\mu_0^\si}{4\pi r^3}\bigg(m_\g\sqrt{\frac{4\pi}{\mu_0^\si}}\bigg) \bigg[2\cos(\theta)\hat{r} + \sin(\theta)\hat{\theta}\bigg]. \eeq Multiplying both side by $\sqrt{\mu_0^\si/4\pi}$ gives \beq \vec{B}_\g \frac{\mu_0^\si}{4\pi} = \frac{\mu_0^\si}{4\pi} \frac{m_\g}{r^3} \bigg[2\cos(\theta)\hat{r} + \sin(\theta)\hat{\theta}\bigg]. \eeq So the {\bf magnetic dipole moment in Gaussian units} for a perfect dipole located at the origin and oriented along the z-axis is \beq \boxed{\vec{B} = \frac{m_\g}{r^3} \bigg[2\cos(\theta)\hat{r} + \sin(\theta)\hat{\theta}\bigg]} \label{eq:bg} \eeq

A classic example with applications to astrophysics is a uniform ball of total charge Q, radius R, and constant angular velocity $\vec{\omega} = \omega \hat{z}$. The charge density of the sphere is $$\rho = \frac{3 Q}{4 \pi R^3}.$$ The sphere can be broken up into rings of volume $$dV = 2\pi r^2 \sin\theta dr d\theta$$ and charge $$dq = \rho dV$$. The current due to each ring is $$dI = \frac{\omega}{2\pi}dq = \frac{\omega}{2\pi}\rho 2\pi r^2 \sin\theta dr d\theta = \omega\rho r^2 \sin\theta dr d\theta.$$ The area enclosed by the ring is $$\pi r^2 \sin^2\theta$$ so the magnetic dipole moment due to each ring is $$dm = dI \pi r^2 \sin^2\theta = \pi r^2 \sin^2\theta \omega\rho r^2 \sin\theta dr d\theta = \pi \omega\rho r^4 \sin^3\theta dr d\theta.$$ Then the total magnetic moment is \begin{align} m &= \int_0^\pi \int_0^R \pi \omega\rho r^4 \sin^3\theta dr d\theta \\ &= \pi\omega\rho\bigg[\int_0^\pi\sin^3\theta d\theta\bigg]\bigg[\int_0^R r^4 dr\bigg] \\ &= \frac{4\pi}{3} \omega \rho \frac{R^5}{5}. \end{align} Replacing $$\rho = 3 Q / 4\pi R^3$$ then gives $$m_\si = \frac{1}{5}\omega Q_\si R^2.$$ Replacing SI units with the appropriate conversion factors, $$m_\g\sqrt{\frac{4\pi}{\mu_0^\si}} = \frac{1}{5}\omega \sqrt{4\pi\epsilon_0^\si} Q_\g R^2$$ or $$m_\g = \frac{1}{5}\omega \sqrt{\mu_0^\si\epsilon_0^\si} Q_\g R^2 = \frac{1}{5}\frac{\omega}{c} Q_\g R^2$$ Plugging this into equation \ref{eq:bg} then gives the {\bf magnetic field outside a spinning ball of charge in Gaussian units}: \beq \boxed{\vec{B} = \frac{\omega Q R^2}{5cr^3} \bigg[2\cos(\theta)\hat{r} + \sin(\theta)\hat{\theta}\bigg]} \eeq which has a strength at the pole ($\theta = 0, r = R$) of \beq B_0 = \frac{2}{5}\frac{\omega Q}{cR}. \eeq <\latex>