Is it a typo in David Tong's derivation of spin-orbit interaction?

Question

A few lines below equation 7.8 D. Tong writes

The final fact is the Lorentz transformation of the electric field: as electron moving with velocity $\vec{v}$ in an electric field E will experience a magnetic field $\vec{B}=\frac{\gamma}{c^2}(\vec{v}\times\vec{E})$.

The note says that it was derived in another note but I couldn't find this expression.

Is this coefficient $\gamma/c^{2}$ correct? Griffiths derives this to be $-1/c^2$ and I did not find anything wrong there. See Griffiths electrodynamics, third edition, equation 12.109.

Then I looked at this book which uses Griffiths' expression in Sec. 20.5, but uses $\vec{p}=m\vec{v}$ instead to $\vec{p}=\gamma m \vec{v}$ to derive the same result. Which one is correct and why?

Answer 1

In above Figure-01 an inertial system $\:\mathrm S'\:$ is translated with respect to the inertial system $\:\mathrm S\:$ with constant velocity
\begin{align} \boldsymbol{\upsilon} & \boldsymbol{=}\left(\upsilon_{1},\upsilon_{2},\upsilon_{3}\right) \tag{02a}\label{02a}\\ \upsilon & \boldsymbol{=}\Vert \boldsymbol{\upsilon} \Vert \boldsymbol{=} \sqrt{ \upsilon^2_{1}\boldsymbol{+}\upsilon^2_{2}\boldsymbol{+}\upsilon^2_{3}}\:\in \left(0,c\right) \tag{02b}\label{02b} \end{align}

The Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}\right)\boldsymbol{\upsilon}\boldsymbol{-}\dfrac{\gamma\boldsymbol{\upsilon}}{c}c\,t \tag{03a}\label{03a}\\ c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma\left(c\,t\boldsymbol{-} \dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{03b}\label{03b}\\ \gamma & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{03c}\label{03c} \end{align}

For the Lorentz transformation \eqref{03a}-\eqref{03b}, the vectors $\:\mathbf{E}\:$ and $\:\mathbf{B}\:$ of the electromagnetic field are transformed as follows \begin{align} \mathbf{E}' & \boldsymbol{=}\gamma \mathbf{E}\boldsymbol{-}\dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\mathbf{E}\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}\,\boldsymbol{+}\,\gamma\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}\right) \tag{04a}\label{04a}\\ \mathbf{B}' & \boldsymbol{=} \gamma \mathbf{B}\boldsymbol{-}\dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\mathbf{B}\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}\boldsymbol{-}\!\dfrac{\gamma}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right) \tag{04b}\label{04b} \end{align} Now, if in system $\:\mathrm S\:$ we have $\:\mathbf{B}\boldsymbol{=0}$, then from \eqref{04a}-\eqref{04b} \begin{align} \mathbf{E}' & \boldsymbol{=}\gamma \mathbf{E}\boldsymbol{-}\dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\mathbf{E}\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon} \tag{05a}\label{05a}\\ \mathbf{B}' & \boldsymbol{=} \boldsymbol{-}\dfrac{\gamma}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right) \tag{05b}\label{05b} \end{align} Equation \eqref{05b} corresponds to Tong's equation (it remains to explain the minus sign).

From equations \eqref{05a}-\eqref{05b} we have \begin{align} \mathbf{B}' & \boldsymbol{=} \boldsymbol{-}\dfrac{\gamma}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right) \boldsymbol{=}\boldsymbol{-}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\gamma\mathbf{E}\right) \nonumber\\ & \boldsymbol{=} \boldsymbol{-}\dfrac{1}{c^2}\Biggl(\boldsymbol{\upsilon}\boldsymbol{\times}\left[\gamma \mathbf{E}\boldsymbol{-}\dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\mathbf{E}\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}\right]\Biggr) \boldsymbol{=}\boldsymbol{-}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right) \nonumber \end{align} that is \begin{equation} \mathbf{B}' \boldsymbol{=}\boldsymbol{-}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right) \tag{06}\label{06} \end{equation} Equation \eqref{06} corresponds to Griffiths' equation.

Based on equations \eqref{04a},\eqref{04b} we have proved that \begin{equation} \mathbf{B}\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{=\!=\!=\!\Longrightarrow}}\quad \mathbf{B}' \boldsymbol{+}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right)\boldsymbol{=0} \tag{06.1}\label{06.1} \end{equation} But we can prove the validity of its inverse \begin{equation} \mathbf{B}' \boldsymbol{+}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right)\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{=\!=\!=\!\Longrightarrow}}\quad \mathbf{B}\boldsymbol{=0} \tag{06.2}\label{06.2} \end{equation} So these conditions are equivalent \begin{equation} \boxed{\:\:\mathbf{B}\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{\Longleftarrow\!=\!=\!\Longrightarrow}}\quad \mathbf{B}' \boldsymbol{+}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right)\boldsymbol{=0}\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tag{06.3}\label{06.3} \end{equation} Equation \eqref{06.2} is valid because \begin{equation} \mathbf{B}' \boldsymbol{+}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right)\boldsymbol{=}\gamma^{\boldsymbol{-}1}\mathbf{B}_{\boldsymbol{\perp}} \boldsymbol{+}\mathbf{B}_{\boldsymbol{\parallel}} \tag{06.4}\label{06.4} \end{equation} where $\mathbf{B}_{\boldsymbol{\parallel}},\mathbf{B}_{\boldsymbol{\perp}}$ the components of $\mathbf{B}$ parallel and normal to the velocity vector $\boldsymbol{\upsilon}$ respectively.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

$\textbf{ADDENDUM}$

If in system $\:\mathrm S\:$ we have $\:\mathbf{E}\boldsymbol{=0}$, then from \eqref{04a}-\eqref{04b} \begin{align} \mathbf{E}' & \boldsymbol{=}\gamma\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}\right) \tag{07a}\label{07a}\\ \mathbf{B}' & \boldsymbol{=} \gamma \mathbf{B}\boldsymbol{-}\dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\mathbf{B}\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon} \tag{07b}\label{07b} \end{align} so that \begin{align} \mathbf{E}' & \boldsymbol{=} \gamma\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}\right)\boldsymbol{=} \left(\boldsymbol{\upsilon}\boldsymbol{\times}\gamma\mathbf{B}\right) \nonumber\\ & \boldsymbol{=} \boldsymbol{\upsilon}\boldsymbol{\times}\left[\gamma \mathbf{B}\boldsymbol{-}\dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\mathbf{B}\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}\right] \boldsymbol{=}\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}' \nonumber \end{align} that is \begin{equation} \mathbf{E}' \boldsymbol{=}\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}' \tag{08}\label{08} \end{equation}

Based on equations \eqref{04a},\eqref{04b} we have proved that \begin{equation} \mathbf{E}\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{=\!=\!=\!\Longrightarrow}}\quad \mathbf{E}' \boldsymbol{-}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}'\right)\boldsymbol{=0} \tag{08.1}\label{08.1} \end{equation} But we can prove the validity of its inverse \begin{equation} \mathbf{E}' \boldsymbol{-}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}'\right)\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{=\!=\!=\!\Longrightarrow}}\quad \mathbf{E}\boldsymbol{=0} \tag{08.2}\label{08.2} \end{equation} So these conditions are equivalent \begin{equation} \boxed{\:\:\mathbf{E}\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{\Longleftarrow\!=\!=\!\Longrightarrow}}\quad \mathbf{E}' \boldsymbol{-}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}'\right)\boldsymbol{=0}\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tag{08.3}\label{08.3} \end{equation} Equation \eqref{08.2} is valid because \begin{equation} \mathbf{E}' \boldsymbol{-}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}'\right)\boldsymbol{=}\gamma^{\boldsymbol{-}1}\mathbf{E}_{\boldsymbol{\perp}} \boldsymbol{+}\mathbf{E}_{\boldsymbol{\parallel}} \tag{08.4}\label{08.4} \end{equation} where $\mathbf{E}_{\boldsymbol{\parallel}},\mathbf{E}_{\boldsymbol{\perp}}$ the components of $\mathbf{E}$ parallel and normal to the velocity vector $\boldsymbol{\upsilon}$ respectively.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

The duality transformation of the electromagnetic field is produced by the replacements \begin{equation} \begin{matrix} \hphantom{c}\mathbf{E}&\boldsymbol{-\!-\!\!\!\longrightarrow}&\boldsymbol{-}c\mathbf{B}\\ c\mathbf{B}&\boldsymbol{-\!-\!\!\!\longrightarrow}&\hphantom{\boldsymbol{-}c}\mathbf{E} \end{matrix} \tag{09}\label{09} \end{equation} These replacements must be done in the primed system also \begin{equation} \begin{matrix} \hphantom{c}\mathbf{E}'&\boldsymbol{-\!-\!\!\!\longrightarrow}&\boldsymbol{-}c\mathbf{B}'\\ c\mathbf{B}'&\boldsymbol{-\!-\!\!\!\longrightarrow}&\hphantom{\boldsymbol{-}c}\mathbf{E}' \end{matrix} \tag{09'}\label{09'} \end{equation} In the aforementioned we met pairs of dual equations or expressions, that is under a duality transformation they are transformed one to the other : \begin{equation} \begin{matrix} \eqref{04a}&\stackrel{\mathtt{duality}}{\boldsymbol{\longleftarrow\!\!\!-\!\!\!\longrightarrow}}&\eqref{04b}\\ \eqref{06}&\stackrel{\mathtt{duality}}{\boldsymbol{\longleftarrow\!\!\!-\!\!\!\longrightarrow}}&\eqref{08}\\ \eqref{06.3}&\stackrel{\mathtt{duality}}{\boldsymbol{\longleftarrow\!\!\!-\!\!\!\longrightarrow}}&\eqref{08.3}\\ \eqref{06.4}&\stackrel{\mathtt{duality}}{\boldsymbol{\longleftarrow\!\!\!-\!\!\!\longrightarrow}}&\eqref{08.4} \end{matrix} \tag{10}\label{10} \end{equation}

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

Equations \eqref{06} and \eqref{08} are the following equations \eqref{12.109} and \eqref{12.110} respectively \begin{equation} \boxed{\:\:\overset{\boldsymbol{-\!\!\!\!\!-}}{\mathbf{B}} \boldsymbol{=}\boldsymbol{-}\dfrac{1}{c^2}\left(\mathbf{v}\boldsymbol{\times}\overset{\boldsymbol{-\!\!\!\!\!-}}{\mathbf{E}}\right)\boldsymbol{.}\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tag{12.109}\label{12.109} \end{equation}

\begin{equation} \boxed{\:\:\overset{\boldsymbol{-\!\!\!\!\!-}}{\mathbf{E}} \boldsymbol{=}\mathbf{v}\boldsymbol{\times}\overset{\boldsymbol{-\!\!\!\!\!-}}{\mathbf{B}}\,\boldsymbol{.}\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tag{12.110}\label{12.110} \end{equation} as shown in ''Introduction to Electrodynamics'' by David J.Griffiths, 3rd Edition 1999.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

Answer 2

$\vec{p}=\gamma m\vec{v}$ is the technically correct equation, but for non-relativistic particles where $|\vec{v}|\ll c$, the Lorentz factor becomes \begin{equation} \gamma=\frac{1}{\sqrt{1-v^2/c^2}}\approx 1, \end{equation} and so can be neglected.

For your reference, I had a quick look and I believe Eq. (6.45) of his EM notes is where this is derived.

Not sure about the negative sign in Griffiths though.