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

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?

• The $\gamma$ factor should be there but its effect can often be neglected. Commented Dec 6, 2019 at 9:41
• @Frobenius Did you compare Griffiths eqn. with Tong's? Where is the factor $\gamma$ in Griffiths? Commented Dec 6, 2019 at 17:51

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 $$$$\mathbf{B}' \boldsymbol{=}\boldsymbol{-}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right) \tag{06}\label{06}$$$$ Equation \eqref{06} corresponds to Griffiths' equation.

Based on equations \eqref{04a},\eqref{04b} we have proved that $$$$\mathbf{B}\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{=\!=\!=\!\Longrightarrow}}\quad \mathbf{B}' \boldsymbol{+}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right)\boldsymbol{=0} \tag{06.1}\label{06.1}$$$$ But we can prove the validity of its inverse $$$$\mathbf{B}' \boldsymbol{+}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right)\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{=\!=\!=\!\Longrightarrow}}\quad \mathbf{B}\boldsymbol{=0} \tag{06.2}\label{06.2}$$$$ So these conditions are equivalent $$$$\boxed{\:\:\mathbf{B}\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{\Longleftarrow\!=\!=\!\Longrightarrow}}\quad \mathbf{B}' \boldsymbol{+}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right)\boldsymbol{=0}\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tag{06.3}\label{06.3}$$$$ Equation \eqref{06.2} is valid because $$$$\mathbf{B}' \boldsymbol{+}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}'\right)\boldsymbol{=}\gamma^{\boldsymbol{-}1}\mathbf{B}_{\boldsymbol{\perp}} \boldsymbol{+}\mathbf{B}_{\boldsymbol{\parallel}} \tag{06.4}\label{06.4}$$$$ where $$\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 $$$$\mathbf{E}' \boldsymbol{=}\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}' \tag{08}\label{08}$$$$

Based on equations \eqref{04a},\eqref{04b} we have proved that $$$$\mathbf{E}\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{=\!=\!=\!\Longrightarrow}}\quad \mathbf{E}' \boldsymbol{-}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}'\right)\boldsymbol{=0} \tag{08.1}\label{08.1}$$$$ But we can prove the validity of its inverse $$$$\mathbf{E}' \boldsymbol{-}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}'\right)\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{=\!=\!=\!\Longrightarrow}}\quad \mathbf{E}\boldsymbol{=0} \tag{08.2}\label{08.2}$$$$ So these conditions are equivalent $$$$\boxed{\:\:\mathbf{E}\boldsymbol{=0}\quad\stackrel{\eqref{04a},\eqref{04b}}{\boldsymbol{\Longleftarrow\!=\!=\!\Longrightarrow}}\quad \mathbf{E}' \boldsymbol{-}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}'\right)\boldsymbol{=0}\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tag{08.3}\label{08.3}$$$$ Equation \eqref{08.2} is valid because $$$$\mathbf{E}' \boldsymbol{-}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}'\right)\boldsymbol{=}\gamma^{\boldsymbol{-}1}\mathbf{E}_{\boldsymbol{\perp}} \boldsymbol{+}\mathbf{E}_{\boldsymbol{\parallel}} \tag{08.4}\label{08.4}$$$$ where $$\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 $$$$\begin{matrix} \hphantom{c}\mathbf{E}&\boldsymbol{-\!-\!\!\!\longrightarrow}&\boldsymbol{-}c\mathbf{B}\\ c\mathbf{B}&\boldsymbol{-\!-\!\!\!\longrightarrow}&\hphantom{\boldsymbol{-}c}\mathbf{E} \end{matrix} \tag{09}\label{09}$$$$ These replacements must be done in the primed system also $$$$\begin{matrix} \hphantom{c}\mathbf{E}'&\boldsymbol{-\!-\!\!\!\longrightarrow}&\boldsymbol{-}c\mathbf{B}'\\ c\mathbf{B}'&\boldsymbol{-\!-\!\!\!\longrightarrow}&\hphantom{\boldsymbol{-}c}\mathbf{E}' \end{matrix} \tag{09'}\label{09'}$$$$ In the aforementioned we met pairs of dual equations or expressions, that is under a duality transformation they are transformed one to the other : $$$$\begin{matrix} \eqref{04a}&\stackrel{\mathtt{duality}}{\boldsymbol{\longleftarrow\!\!\!-\!\!\!\longrightarrow}}&\eqref{04b}\\ \eqref{06}&\stackrel{\mathtt{duality}}{\boldsymbol{\longleftarrow\!\!\!-\!\!\!\longrightarrow}}&\eqref{08}\\ \eqref{06.3}&\stackrel{\mathtt{duality}}{\boldsymbol{\longleftarrow\!\!\!-\!\!\!\longrightarrow}}&\eqref{08.3}\\ \eqref{06.4}&\stackrel{\mathtt{duality}}{\boldsymbol{\longleftarrow\!\!\!-\!\!\!\longrightarrow}}&\eqref{08.4} \end{matrix} \tag{10}\label{10}$$$$

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

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

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

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

• @Dhruv Maroo : Many thanks for your attention. I apologize since I rejected your edit. I like the double line as I have it in my answer. Commented Dec 8, 2019 at 10:04
• Can you possibly give me a hint where I find the proof of the two vector equations you gave to describe the generalized Lorentz boost transformation given in the figure? I think the equations should correspond to the Lorentz matrix given in this question link, but I cannot verify that. Can I exclude a typo in your 2 equations? Thanks a lot. Commented Jul 21, 2022 at 15:33
• @Roland Salz : I joined Physics SE as diracpaul in June'15 and I quit the site in Sep'15 for personal reasons. I came back as Frobenius in Mar'16. Under my answers as former diracpaul now you could see the name user82794. My 2015 answer here Two sets of coordinates each in frames O and O′ - Lorentz transformation gives the details for your question about the Lorentz transformation along an arbitrary direction. Commented Jul 21, 2022 at 17:47
• Thank you very much for your many hints. I'm familiar with the boost along the x-axis. I transformed your equations into a Lorentz matrix, with CAS, but something is not correct with it. It is not quite identical with the one (which seems to be correct) I mentioned in the link above. So actually, for a day or two I have been trying to find where the error is (of course it can be in my own calculations). Tomorrow morning I will look at your links and continue searching. Thanks a lot for your kind help. Commented Jul 21, 2022 at 18:53
• Frobenius: Thanks to your detailed derivations I have resolved all my issues. One last question: what graphics software do you use for your figures? They look perfect. Commented Jul 24, 2022 at 18:14

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