Generalizing the Lorentz force law in 4-vector notation

Question

In Goldstein's Classical Mechanics chapter on special relativity page 298 the Lorentz force law: $$\textbf{f}=q\left\{-\nabla\phi + \frac{1}{c}\frac{\partial \textbf{A}}{\partial t}+\textbf{v}\times(\nabla\times\textbf{A})\right\} \tag{1}$$ Is generalised in 4 vector notation as $$\frac{dP_{\mu}}{d\tau}=q\left[\frac{\partial(u^{\nu}A_{\nu})}{\partial x^{\mu}}-\frac{dA_{\mu}}{d\tau}\right] \tag{2}$$

Im trying to to replicate this by going from $(2)$ to $(1)$ but im having trouble managing to do it here is what I've done so far.

Firstly to define things $$A_{\mu}=(\frac{\phi}{c},-\textbf{A})$$ $d\tau$ is the proper time given as $$dt=\frac{d\tau}{\sqrt{a-\frac{v^2}{c^2}}}=\gamma d\tau$$

hence the 4-velocity $u^{\nu}$ is given as:

$$u^0=\frac{dct}{d\tau}=\gamma c, \ \ u^i = \frac{dx^i}{d\tau}=\gamma v^i$$

and finally

$$\frac{\partial}{\partial x^{\mu}}=\frac{\partial}{\partial ct}+\nabla$$

Where, $\nabla$, is the 3 dimensional del operator. So starting from the first term in $(2)$ and expanding $u^{\nu}A_{\nu}$ I get

$$u^{\nu}A_{\nu}=u^0\frac{\phi}{c} - u^iA_{i}=\gamma \phi- \gamma v^iA_{i}$$

Applying $\frac{\partial}{\partial x^{\mu}}$, I get, $$ \begin{align} \frac{\partial(u^{\nu}A_{\nu})}{\partial x^{\mu}} &= \frac{\partial }{\partial ct}(\gamma \phi- \gamma v^iA_{i})+\nabla(\gamma \phi- \gamma v^iA_{i}) \\ &=\frac{\partial }{\partial ct}(\gamma \phi) - \frac{\partial }{\partial ct} (\gamma v^iA_{i}) + \gamma\nabla \phi-\gamma\nabla(v^iA_{i}) \end{align} \tag{3}$$

and for $\frac{dA_{\mu}}{d\tau}$ I get

$$\frac{dA_{\mu}}{d\tau}=\gamma (\mathbf{v} \cdot \nabla)A_{\mu} +\gamma \frac{\partial A_{\mu}}{\partial t} \tag{4}$$

with $(3)$ I know we can write $\nabla(v^iA_{i}) = \nabla(\textbf{v}\cdot\textbf{A}) = \textbf{v}\times (\nabla \times \textbf{A}) +(\textbf{v}\cdot \nabla)\textbf{A}$ since $\textbf{v}$ is only time dependent.

putting $(3)$ and $(4)$ into $(2)$ and using the vector identity, I got

$$\frac{\partial(u^{\nu}A_{\nu})}{\partial x^{\mu}} - \frac{dA_{\mu}}{d\tau} = \frac{\partial }{\partial ct}(\gamma \phi) - \frac{\partial }{\partial ct} (\gamma v^iA_{i}) + \gamma\nabla \phi-\gamma(\textbf{v} \times(\nabla \times \textbf{A})) -\gamma (\textbf{v} \cdot \nabla)\frac{\phi}{c} -\gamma \frac{1}{c}\frac{\partial \phi}{\partial t} +\frac{\partial \textbf{A}}{\partial t}$$

Where I've broken up $A_{\mu}$ in the last two terms. However, I get a bit stuck from this point on, in the first two terms does the partial time derivative skip the $\gamma's$ and the velocity terms? I'm also unsure what I do with the $\gamma (\textbf{v} \cdot \nabla)\frac{\phi}{c}$ term?

Answer 1

You've got several issues here, the way I see it.

1. Goldstein's derivation is not very standard today; it's somewhat outdated. The expression you will find in most books on electrodynamics and relativity is: $$ \frac{dP^{\mu}}{d\tau}=qu^{\nu}\left.F_{\nu}\right.^{\mu} $$ where $ F_{\nu\mu}:=\frac{\partial A_{\mu}}{\partial x^{\nu}}-\frac{\partial A_{\nu}}{\partial x^{\mu}} $. So you will find it difficult to compare with other bibliographic sources.
2. Your Eq.(3) is inconsistent. You've got a 4-vector on the LHS and a scalar on the RHS. You would have to distinguish the 0-term, and then the 3-vector term, which is the one that gives you the Lorentz-force law: $$ \frac{dP_{k}}{d\tau}=q\left(\frac{\partial(u^{\nu}A_{\nu})}{\partial x^{k}}-\frac{dA_{k}}{d\tau}\right) $$ The expression I get from your prescription Eq.(2) is, $$ \frac{d}{d\tau}P^{k}=-q\gamma\frac{\partial\phi}{\partial x^{k}}+q\gamma\left(\boldsymbol{v}\times\boldsymbol{\nabla}\times\boldsymbol{A}\right)^{k}+q\gamma\left(\boldsymbol{v}\cdot\boldsymbol{\nabla}\right)A^{k}-q\gamma\left(\boldsymbol{v}\cdot\boldsymbol{\nabla}+\frac{\partial}{\partial t}\right)A^{k} $$ Where I have substituted what you seem to have (correctly) implied. Namely: that the total derivative and the partial derivative are related by, $$ \frac{d}{dt}=\boldsymbol{v}\cdot\boldsymbol{\nabla}+\frac{\partial}{\partial t} $$ although you seem to be in doubt about whether this is correct: It is. This is sometimes called a "material derivative", and it's a consequence of the chain rule in calculus. When you have, $$ f\left(t,x\left(t\right)\right) $$ the correct way to obtain how much $f$ changes with $t$ requires you to add a term depending on the drift: $$ \frac{d}{dt}f=\frac{\partial}{\partial t}f+\dot{x}\cdot\nabla f $$ Your vector identity is also correct. As it's the fact that the $\gamma$'s are unaffected by the spatial differential operators. I think you won't have problems filling in the details. You drop the $\gamma$'s on both sides and cancel the $\boldsymbol{v}\cdot\boldsymbol{\nabla}$'s, and you're there.

I hope that helped. If something wasn't clear, feel free to tell me.

Answer 2

$\boldsymbol{\S}\:$A. Preliminary

Consider $\,\mathbf f\,$ to be a 3-vector representing a force applied on a particle of rest mass $\,m_{0}\,$ moving with velocity 3-vector $\,\mathbf u$. If this force is preserving the rest mass $\,m_{0}\,$ of the particle then we can $''$build$''$ a Lorentz 4-vector $\,\mathbf F\,$ as follows \begin{equation} \mathbf F \boldsymbol{=}\left(\gamma_{\mathrm u}\mathbf f, \gamma_{\mathrm u}\dfrac{\mathbf f\boldsymbol{\cdot}\mathbf u}{c}\right) \tag{A-01}\label{A-01} \end{equation} where $\,\gamma_{\mathrm u} \boldsymbol{=}1\bigg{/}\sqrt{1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}}\,$. The 4-vector $\,\mathbf F\,$ of above equation is built to be a Lorentz 4-vector from its very definition as the differential of the 4-momentum Lorentz 4-vector $\,\mathbf P\,$ with respect to the proper time $\,\tau$, a Lorentz invariant scalar \begin{equation} \mathbf F \stackrel{\rm def}{\boldsymbol{\equiv\!\!\!\equiv}}\dfrac{\mathrm d\mathbf P}{\mathrm d\tau} \boldsymbol{=}\dfrac{\mathrm d\left(m_{0} \mathbf U\right)}{\mathrm d\tau}\boldsymbol{=}m_{0} \dfrac{\mathrm d\mathbf U}{\mathrm d\tau} \tag{A-02}\label{A-02} \end{equation} We see that for the 4-momentum vector $\,\mathbf P\boldsymbol{=}m_{0} \mathbf U\,$ to be a Lorentz 4-vector we need the rest mass $\,m_{0}\,$ to be invariant because in turn the velocity 4-vector $\,\mathbf U\boldsymbol{=}\mathrm d\mathbf X/\mathrm d\tau\,$ is a Lorentz one. To see a method to build Lorentz 4-vectors from other ones take a look in my answer here "Is mass still a scalar in special relativity?".

As concerns to the electromagnetic Lorentz 3-force $\,\mathbf f\boldsymbol{=}q\left(\mathbf E\boldsymbol{+}\mathbf u \boldsymbol{\times}\mathbf B\right)\,$ the 4-vector $\,\mathbf F\,$ was built in the early years of Special Relativity as in equation \eqref{A-01} but it's property as a Lorentz 4-vector was proved based on the Lorentz transformation of the 3-vectors $\,\left(\mathbf E,\mathbf B,\mathbf u\right)\,$ and the hypothesis of the electric charge $\,q\,$ invariance and NOT by the definition \eqref{A-02} and the hypothesis of the rest mass $\,m_{0}\,$ invariance. May be this fact is an indirect verification that the electromagnetic Lorentz 3-force preserves the rest mass $\,m_{0}\,$ of the particles on which is applied. For details see my answer here "Are magnetic fields just modified relativistic electric fields?".

$\boldsymbol{-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!}$

$\boldsymbol{\S}\:$B.The Proof

For convenience we'll start with the electromagnetic Lorentz 3-force per unit electric charge \begin{equation} \mathbf f \boldsymbol{=} \mathbf E\boldsymbol{+}\mathbf u \boldsymbol{\times}\mathbf B \boldsymbol{=}\boldsymbol{-}\boldsymbol{\nabla}\phi \boldsymbol{+}\dfrac{\partial \mathbf A}{\partial t}\boldsymbol{+}\mathbf u \boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right) \tag{B-01}\label{B-01} \end{equation} in order not to transfer to all subsequent equations the charge $\,q\,$ and the necessary braces or parentheses or brackets etc.

By this 3-vector $\,\mathbf f\,$ we'll build the 4-vector $\,\mathbf F\,$ of equation \eqref{A-01} and we'll prove that the latter is identical to the expression in the brackets of the rhs of equation (02) of the question.

Now, in order not to have confusion about signs $\,\boldsymbol{\pm}\,$, covariant and contravariant vectors etc we clarify the following : if we have the coordinate frame $\,\left(\rm x,y,z\right)$, the potentials, the velocity and the force \begin{equation} \mathbf A \boldsymbol{=}\left(\rm A_x,A_y,A_z\right)\,, \quad \mathrm A_t\boldsymbol{=}\phi/c\,,\quad \mathbf u \boldsymbol{=}\left(\rm u_x,u_y,u_z\right)\,,\quad \mathbf f \boldsymbol{=}\left(\rm f_x,f_y,f_z\right) \tag{B-02}\label{B-02} \end{equation} then using the $\boldsymbol{+---}$ sign convention we have for tensors \begin{align} &\begin{bmatrix} \:\:x^1 \:\:\vphantom{\dfrac{a}{b}}\\ x^2 \vphantom{\dfrac{a}{b}}\\ x^3 \vphantom{\dfrac{a}{b}}\\ x^0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \:\:\rm x\:\: \vphantom{\dfrac{a}{b}}\\ \rm y \vphantom{\dfrac{a}{b}}\\ \rm z \vphantom{\dfrac{a}{b}}\\ ct \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \boldsymbol{-}x_1 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}x_2 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}x_3 \vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}}x_0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{B-03a}\label{B-03a}\\ &\begin{bmatrix} \:\:A^1\:\: \vphantom{\dfrac{a}{b}}\\ A^2 \vphantom{\dfrac{a}{b}}\\ A^3 \vphantom{\dfrac{a}{b}}\\ A^0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \:\:\rm A_x \:\:\vphantom{\dfrac{a}{b}}\\ \rm A_y \vphantom{\dfrac{a}{b}}\\ \rm A_z \vphantom{\dfrac{a}{b}}\\ \mathrm A_t \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \boldsymbol{-}A_1 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}A_2 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}A_3 \vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}}A_0 \vphantom{\dfrac{a}{b}} \end{bmatrix}\,,\qquad \begin{bmatrix} \:\:U^1 \:\:\vphantom{\dfrac{a}{b}}\\ U^2 \vphantom{\dfrac{a}{b}}\\ U^3 \vphantom{\dfrac{a}{b}}\\ U^0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \rm \gamma_u u_x \vphantom{\dfrac{a}{b}}\\ \rm \gamma_u u_y \vphantom{\dfrac{a}{b}}\\ \rm \gamma_u u_z \vphantom{\dfrac{a}{b}}\\ \rm \gamma_u c\hphantom{_z}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \boldsymbol{-}U_1 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}U_2 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}U_3 \vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}}U_0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{B-03b}\label{B-03b}\\ &\begin{bmatrix} \:\:\dfrac{\partial \hphantom{\boldsymbol{x_1}}}{\partial x_1}\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \hphantom{\boldsymbol{x_2}}}{\partial x_2} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \hphantom{\boldsymbol{x_3}}}{\partial x_3} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \hphantom{\boldsymbol{x_0}}}{\partial x_0} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \:\:\dfrac{\partial \hphantom{\boldsymbol{\rm x}}}{\partial \rm x}\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \hphantom{\boldsymbol{\rm y}}}{\partial \rm y} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \hphantom{\boldsymbol{\rm z}}}{\partial \rm z} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hphantom{\boldsymbol{\rm x}}\dfrac{\partial \hphantom{\boldsymbol{ct}}}{\partial \rm ct} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \boldsymbol{-}\dfrac{\partial \hphantom{\boldsymbol{x^1}}}{\partial x^1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \boldsymbol{-}\dfrac{\partial \hphantom{\boldsymbol{x^2}}}{\partial x^2} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \boldsymbol{-}\dfrac{\partial \hphantom{\boldsymbol{x^3}}}{\partial x^3} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hphantom{\boldsymbol{-}}\dfrac{\partial \hphantom{\boldsymbol{x^0}}}{\partial x^0} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{B-03c}\label{B-03c} \end{align} From \eqref{A-01} and \eqref{B-01} we have \begin{equation} \mathbf F \boldsymbol{=} \begin{bmatrix} \:\:F^1\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ F^2\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ F^3 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ F^0 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \rm \gamma_u f_x \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm \gamma_u f_y \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm \gamma_u f_z \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm \gamma_u\dfrac{f_x u_x\boldsymbol{+}f_y u_y\boldsymbol{+}f_z u_z}{c} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \boldsymbol{-} F_1\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \boldsymbol{-} F_2\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \boldsymbol{-} F_3\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hphantom{\boldsymbol{-}} F_0\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{B-04}\label{B-04} \end{equation} The $''$space$''$ component $\,F^1\,$ is expressed as \begin{equation} F^1 \boldsymbol{=}\gamma_\mathrm u \mathrm{f_x} \boldsymbol{=}\underbrace{ \boldsymbol{-}\gamma_\mathrm u\dfrac{\partial \phi}{\partial \rm x}}_{\boxed{\,1\,}}\boldsymbol{+}\underbrace{\gamma_\mathrm u\dfrac{\partial \rm A_x}{\partial t}}_{\boxed{\,2\,}}\boldsymbol{+}\underbrace{\gamma_\mathrm u\left[\mathbf u \boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]_{\rm x}}_{\boxed{\,3\,}}\boldsymbol{=}\boxed{\,1\,}\boldsymbol{+}\boxed{\,2\,}\boldsymbol{+} \boxed{\,3\,} \tag{B-05}\label{B-05} \end{equation} We transform gradually using \eqref{B-03a},\eqref{B-03b} and \eqref{B-03c} \begin{align} \boxed{\,1\,} & \boldsymbol{=}\boldsymbol{-}\gamma_\mathrm u\dfrac{\partial \phi}{\partial \rm x}\boldsymbol{=}\boldsymbol{-}\gamma_\mathrm u c\dfrac{\partial \mathrm A_t}{\partial \rm x}\boldsymbol{=}U^0\dfrac{\partial A_0}{\partial x^1}\boldsymbol{=}\dfrac{\partial \left(U^0 A_0\right)}{\partial x^1} \tag{B-06a}\label{B-06a}\\ \boxed{\,2\,} & \boldsymbol{=}\boldsymbol{+}\gamma_\mathrm u\dfrac{\partial \rm A_x}{\partial t}\boldsymbol{=}\boldsymbol{-}\gamma_\mathrm u c\dfrac{\partial A_1}{\partial (ct)}\boldsymbol{=}\boldsymbol{-}U^0\dfrac{\partial A_1}{\partial x^0} \tag{B-06b}\label{B-06b} \end{align} Preparing $\boxed{\,3\,}$ note that \begin{equation} \boldsymbol{\nabla}\boldsymbol{\times}\mathbf A \boldsymbol{=} \begin{bmatrix} \mathbf e_{\rm x} & \mathbf e_{\rm y} & \mathbf e_{\rm z} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \hphantom{\boldsymbol{\rm x}}}{\partial \rm x} & \dfrac{\partial \hphantom{\boldsymbol{\rm x}}}{\partial \rm y} & \dfrac{\partial \hphantom{\boldsymbol{\rm x}}}{\partial \rm z}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm A_x & \rm A_y & \rm A_z\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \dfrac{\partial \rm A_z}{\partial \rm y}\boldsymbol{-}\dfrac{\partial \rm A_y}{\partial \rm z} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \rm A_x}{\partial \rm z}\boldsymbol{-}\dfrac{\partial \rm A_z}{\partial \rm x} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \rm A_y}{\partial \rm x}\boldsymbol{-}\dfrac{\partial \rm A_x}{\partial \rm y} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm x} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm y} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm z} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{B-07}\label{B-07} \end{equation} so ^{\begin{align} & \mathbf u \boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right) \boldsymbol{=} \nonumber\\ &\begin{bmatrix} \mathbf e_{\rm x} & \mathbf e_{\rm y} & \mathbf e_{\rm z} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm u_x & \rm u_y & \rm u_z\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm x} & \left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm y} & \left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm z}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \rm u_y\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm z} \boldsymbol{-}\rm u_z\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm y}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm u_z\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm x} \boldsymbol{-}\rm u_x\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm z} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm u_x\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm y} \boldsymbol{-}\rm u_y\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)_{\rm x} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}\boldsymbol{=} \nonumber\\ & \begin{bmatrix} \rm u_y\left(\dfrac{\partial \rm A_y}{\partial \rm x}\boldsymbol{-}\dfrac{\partial \rm A_x}{\partial \rm y}\right) \boldsymbol{-}\rm u_z\left(\dfrac{\partial \rm A_x}{\partial \rm z}\boldsymbol{-}\dfrac{\partial \rm A_z}{\partial \rm x} \right)\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm u_z\left(\dfrac{\partial \rm A_z}{\partial \rm y}\boldsymbol{-}\dfrac{\partial \rm A_y}{\partial \rm z}\right) \boldsymbol{-}\rm u_x\left(\dfrac{\partial \rm A_y}{\partial \rm x}\boldsymbol{-}\dfrac{\partial \rm A_x}{\partial \rm y} \right) \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \rm u_x\left(\dfrac{\partial \rm A_x}{\partial \rm z}\boldsymbol{-}\dfrac{\partial \rm A_z}{\partial \rm x}\right) \boldsymbol{-}\rm u_y\left(\dfrac{\partial \rm A_z}{\partial \rm y}\boldsymbol{-}\dfrac{\partial \rm A_y}{\partial \rm z} \right) \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}\boldsymbol{=} \nonumber\\ & \begin{bmatrix} \dfrac{\partial \left(\rm u_x\rm A_x\boldsymbol{+}\rm u_y\rm A_y\boldsymbol{+}\rm u_z\rm A_z\right)}{\partial \rm x}\boldsymbol{-}\left(\rm u_x\dfrac{\partial \rm A_x}{\partial \rm x}\boldsymbol{+}\rm u_y\dfrac{\partial \rm A_x}{\partial \rm y} \boldsymbol{+}\rm u_z\dfrac{\partial \rm A_x}{\partial \rm z} \right)\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \left(\rm u_x\rm A_x\boldsymbol{+}\rm u_y\rm A_y\boldsymbol{+}\rm u_z\rm A_z\right)}{\partial \rm y}\boldsymbol{-}\left(\rm u_x\dfrac{\partial \rm A_y}{\partial \rm x}\boldsymbol{+}\rm u_y\dfrac{\partial \rm A_y}{\partial \rm y} \boldsymbol{+}\rm u_z\dfrac{\partial \rm A_y}{\partial \rm z} \right)\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \left(\rm u_x\rm A_x\boldsymbol{+}\rm u_y\rm A_y\boldsymbol{+}\rm u_z\rm A_z\right)}{\partial \rm z}\boldsymbol{-}\left(\rm u_x\dfrac{\partial \rm A_z}{\partial \rm x}\boldsymbol{+}\rm u_y\dfrac{\partial \rm A_z}{\partial \rm y} \boldsymbol{+}\rm u_z\dfrac{\partial \rm A_z}{\partial \rm z} \right)\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}\boldsymbol{=} \nonumber\\ & \begin{bmatrix} \dfrac{\partial \left(\mathbf u\boldsymbol{\cdot}\mathbf A\right)}{\partial \rm x}\boldsymbol{-}\mathbf u\boldsymbol{\cdot}\boldsymbol{\nabla}\rm A_x\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \left(\mathbf u\boldsymbol{\cdot}\mathbf A\right)}{\partial \rm y}\boldsymbol{-}\mathbf u\boldsymbol{\cdot}\boldsymbol{\nabla}\rm A_y\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\partial \left(\mathbf u\boldsymbol{\cdot}\mathbf A\right)}{\partial \rm x}\boldsymbol{-}\mathbf u\boldsymbol{\cdot}\boldsymbol{\nabla}\rm A_x\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \left[\mathbf u \boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]_{\rm x} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \left[\mathbf u \boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]_{\rm y} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \left[\mathbf u \boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]_{\rm z} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}\boldsymbol{=}\boldsymbol{\nabla}\left(\mathbf u\boldsymbol{\cdot}\mathbf A\right)\boldsymbol{-}\left(\mathbf u\boldsymbol{\cdot}\boldsymbol{\nabla}\right)\mathbf A \tag{B-08}\label{B-08} \end{align}} Consequently \begin{align} &\boxed{\,3\,} \boldsymbol{=}\gamma_\mathrm u\left[\mathbf u \boldsymbol{\times}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]_{\rm x} \boldsymbol{=} \nonumber\\ & \dfrac{\partial \left(\gamma_\mathrm u\rm u_x\rm A_x\boldsymbol{+}\gamma_\mathrm u\rm u_y\rm A_y\boldsymbol{+}\gamma_\mathrm u\rm u_z\rm A_z\right)}{\partial \rm x}\boldsymbol{-}\left(\gamma_\mathrm u\rm u_x\dfrac{\partial \rm A_x}{\partial \rm x}\boldsymbol{+}\gamma_\mathrm u\rm u_y\dfrac{\partial \rm A_x}{\partial \rm y} \boldsymbol{+}\gamma_\mathrm u\rm u_z\dfrac{\partial \rm A_x}{\partial \rm z} \right) \nonumber\\ & \dfrac{\partial \left(U^1A_1\boldsymbol{+}U^2A_2\boldsymbol{+}U^3A_3\right)}{\partial x^1}\boldsymbol{-}\left(U^1\dfrac{\partial A_1}{\partial x^1}\boldsymbol{+}U^2\dfrac{\partial A_1}{\partial x^2}\boldsymbol{+}U^3\dfrac{\partial A_1}{\partial x^3} \right) \tag{B-09}\label{B-09} \end{align} Inserting in equation \eqref{B-05} the expressions of $\,\boxed{1},\boxed{2},\boxed{3}\,$ from equations \eqref{B-06a},\eqref{B-06b},\eqref{B-09} respectively we have ^{\begin{equation} F^1 \boldsymbol{=}\dfrac{\partial \left(U^0A_0\boldsymbol{+}U^1A_1\boldsymbol{+}U^2A_2\boldsymbol{+}U^3A_3\right)}{\partial x^1}\boldsymbol{-}\left(U^0\dfrac{\partial A_0}{\partial x^1}\boldsymbol{+}U^1\dfrac{\partial A_1}{\partial x^1}\boldsymbol{+}U^2\dfrac{\partial A_1}{\partial x^2}\boldsymbol{+}U^3\dfrac{\partial A_1}{\partial x^3} \right) \tag{B-10}\label{B-10} \end{equation}} Since \begin{equation} U^\rho \boldsymbol{=}\dfrac{\mathrm d x^\rho}{\mathrm d \tau}\qquad \left(\rho\boldsymbol{=}0,1,2,3\right) \tag{B-11}\label{B-11} \end{equation} we have for the expression in parentheses of equation \eqref{B-10} ^{\begin{equation} \!\!\!\!\!\!\!\!\!\!\!\! U^0\dfrac{\partial A_1}{\partial x^0}\boldsymbol{+}U^1\dfrac{\partial A_1}{\partial x^1}\boldsymbol{+}U^2\dfrac{\partial A_1}{\partial x^2}\boldsymbol{+}U^3\dfrac{\partial A_1}{\partial x^3} \boldsymbol{=} \dfrac{\partial A_1}{\partial x^0}\dfrac{\mathrm d x^0}{\mathrm d \tau}\boldsymbol{+}\dfrac{\partial A_1}{\partial x^1}\dfrac{\mathrm d x^1}{\mathrm d \tau}\boldsymbol{+}\dfrac{\partial A_1}{\partial x^2}\dfrac{\mathrm d x^2}{\mathrm d \tau}\boldsymbol{+}\dfrac{\partial A_1}{\partial x^3}\dfrac{\mathrm d x^3}{\mathrm d \tau}\boldsymbol{=}\dfrac{\mathrm d A_1}{\mathrm d \tau} \tag{B-12}\label{B-12} \end{equation}} and equation \eqref{B-10} yields \begin{equation} F^1 \boldsymbol{=}\dfrac{\partial \left(U^\nu A_\nu\right)}{\partial x^1}\boldsymbol{-}\dfrac{\mathrm d A_1}{\mathrm d \tau} \tag{B-13}\label{B-13} \end{equation} Similarly for $\,F^2,F^3\,$ so \begin{equation} F^\mu \boldsymbol{=}\dfrac{\partial \left(U^\nu A_\nu\right)}{\partial x^\mu}\boldsymbol{-}\dfrac{\mathrm d A_\mu}{\mathrm d \tau} \qquad \left(\mu\boldsymbol{=}1,2,3\right) \tag{B-14}\label{B-14} \end{equation} For the covariant components $\,F_\mu \boldsymbol{=}\boldsymbol{-}F^\mu\:\left(\mu\boldsymbol{=}1,2,3\right)$, see equation \eqref{B-04} \begin{equation} F_\mu \boldsymbol{=}\dfrac{\mathrm d A_\mu}{\mathrm d \tau}\boldsymbol{-}\dfrac{\partial \left(U^\nu A_\nu\right)}{\partial x^\mu} \qquad \left(\mu\boldsymbol{=}1,2,3\right) \tag{B-15}\label{B-15} \end{equation} For the $''$time$''$ component $\,F_0 \boldsymbol{=}F^0\,$ we have ^{\begin{align} F_0 & \boldsymbol{=}\gamma_{\mathrm u}\dfrac{\mathbf f\boldsymbol{\cdot}\mathbf u}{c}\boldsymbol{=}\boldsymbol{-}\gamma_{\mathrm u}\mathbf u\boldsymbol{\cdot}\boldsymbol{\nabla}\left(\dfrac{\phi}{c}\right)\boldsymbol{+}\gamma_{\mathrm u}\mathbf u\boldsymbol{\cdot}\dfrac{\partial \mathbf A}{\partial (ct)} \boldsymbol{=}\boldsymbol{-}\gamma_{\mathrm u}\mathbf u\boldsymbol{\cdot}\boldsymbol{\nabla}A_0\boldsymbol{+}\gamma_{\mathrm u}\mathbf u\boldsymbol{\cdot}\dfrac{\partial \mathbf A}{\partial x^0} \nonumber\\ & \boldsymbol{=}\left(U^1\dfrac{\partial A_0}{\partial x^1}\boldsymbol{+}U^2\dfrac{\partial A_0}{\partial x^2}\boldsymbol{+}U^3\dfrac{\partial A_0}{\partial x^3} \right)\boldsymbol{-}\dfrac{\partial \left(U^1A_1\boldsymbol{+}U^2A_2\boldsymbol{+}U^3A_3\right)}{\partial x^0} \nonumber\\ & \boldsymbol{=}\left(U^0\dfrac{\partial A_0}{\partial x^0}\boldsymbol{+}U^1\dfrac{\partial A_0}{\partial x^1}\boldsymbol{+}U^2\dfrac{\partial A_0}{\partial x^2}\boldsymbol{+}U^3\dfrac{\partial A_0}{\partial x^3} \right)\boldsymbol{-}\dfrac{\partial \left(U^0A_0\boldsymbol{+}U^1A_1\boldsymbol{+}U^2A_2\boldsymbol{+}U^3A_3\right)}{\partial x^0} \nonumber\\ & \boldsymbol{\implies}\quad F_0\boldsymbol{=}\dfrac{\mathrm d A_0}{\mathrm d \tau}\boldsymbol{-}\dfrac{\partial \left(U^\nu A_\nu\right)}{\partial x^0} \tag{B-16}\label{B-16} \end{align}} that is equation \eqref{B-15} is valid for $\,\mu\boldsymbol{=}0\,$ also, so finally \begin{equation} \boxed{\:\:F_\mu \boldsymbol{=}\dfrac{\mathrm d A_\mu}{\mathrm d \tau}\boldsymbol{-}\dfrac{\partial \left(U^\nu A_\nu\right)}{\partial x^\mu} \vphantom{\tfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \:\:}\qquad \left(\mu\boldsymbol{=}0,1,2,3\right) \tag{B-17}\label{B-17} \end{equation}