In John Jackson’s authoritative textbook
Classical Electrodynamics you have to wait until section 11.10 before he says
"one should properly speak of the electromagnetic field Fμv rather than E or B separately".
Simply explained because chapter 11 is the chapter that introduces the unification of space and time into space-time and the unification of electric phenomena and magnetic phenomena into electromagnetic phenomena. In a physics textbook, an author makes sure the readers have the background to work usefully and precisely with the material as it is given. This frequently takes the form of a history-of-physics approach of sneaking up on modern physics by discussing pre-modern physics.
The Lorentz force law is just that. It's not the Lorentz field law. Look at the opening sentence on Wikipedia and permit me to change point charge to charged particle: the Lorentz force is the combination of electric and magnetic force on a charged particle due to electromagnetic fields. Yes it goes on to say electric field E and a magnetic field B, but you know from Jackson that one should talk of the electromagnetic field Fuv.
But it's not
wrong to speak of the E and B fields just as long as you know that their particular values are dependent on the choice of coordinates.
Well, first of all, Jackson's textbook is a bit dated in concepts and notation. Therefore it won't trivially square with certain modern sources until you take the differences of notation involved. Likewise it won't square with Maxwell at all until you apply many more differences of notation. I'll use Cartesian coordinates throughout and (-+++) sign conventions for $$\eta_{\mu\nu}$$.
Jackson deals with
Classical Electrodynamics and in this context, this ignores quantum effects. So we are limited to Maxwell + Einstein.
Relativistic dynamics is about the replacement of $$d\vec{p}/dt \; = \; \vec{f} \; = \; m \, \vec{a} \; = \; m \, d^2 \vec{r} / dt^2$$ which comprises Newtonian dynamics for a classical particle. Relativity requires that we think about time as well as space, so we replace relative location ($$\vec{r}$$) with a 4-vector:
$$R^{\mu} = \begin{pmatrix} c \Delta t \\ \vec{r} \end{pmatrix} = \begin{pmatrix} c \Delta t \\ \Delta r_x\\ \Delta r_y\\ \Delta r_z \end{pmatrix}$$
For a particle moving slower than the speed of light, the inner product of the 4-position of the particle (relative to some point on the particle's world line) is negative.
$$R^2 = R_{\mu} R^{\mu} = \eta_{\mu\nu} R^{\mu} R^{\nu} = -c^2 (\Delta t)^2 + (\Delta r_x)^2 + (\Delta r_y)^2 + (\Delta r_z)^2 = \vec{r}^2 -c^2 (\Delta t)^2 < 0$$
This quantity $$R^2$$ analogous to a (squared) Euclidean length in many ways, specifically it does not change value under a change of inertial coordinates (Rotations or Lorentz transforms or translations of space or time). So for a general slower-than light path we model the time experienced by the particle as:
$$\Delta \tau = \int_{t_0}^{t_1} \sqrt{ - \frac{(d R)^2}{c^2 (dt)^2} } \, dt \; = \; \int_{t_0}^{t_1} \sqrt{ 1 - \left( \frac{ d\vec{r} }{c dt} \right)^2 } \, dt \; = \; \int_{t_0}^{t_1} \sqrt{ 1 - \left( \frac{\vec{v} }{ c} \right)^2 } \, dt \; = \; \int_{t_0}^{t_1} \frac{1}{\gamma(\vec{v})} \, dt $$
So for unaccelerated motion we have the relations: $$ \Delta t = \gamma(\vec{v}) \, \Delta \tau, \quad \left. R^2 \right|_{t_0}^{t_1} = (\Delta \vec{r})^2 - c^2 ( \Delta t)^2 = ( \vec{v}^2 - c^2 ) ( \Delta t)^2 = -c^2 ( \Delta \tau )^2$$
The relativistic momentum for a massive particle is:
$$P^{\mu} = \begin{pmatrix} E/c \\ \vec{p} \end{pmatrix} = \begin{pmatrix} \gamma(\vec{v}) m c \\ \gamma(\vec{v}) m \vec{v} \end{pmatrix} = m \begin{pmatrix} \gamma(\vec{v}) c \\ \gamma(\vec{v}) \vec{v} \end{pmatrix} = m \begin{pmatrix} \frac{c}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \\ \frac{v_x}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \\ \frac{v_y}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \\ \frac{v_z}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \end{pmatrix} = m V^{\mu}$$
Because it is unchanged by the Lorentz transform, the inner product of 4-velocity (or 4-momentum) with itself is a scalar invariant of special relativity:
$$V^2 = V_{\mu} V^{\mu} = \eta_{\mu\nu} V^{\mu} V^{\nu} = - \left( \frac{c}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \right)^2 + \left( \frac{v_x}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \right)^2 + \left( \frac{v_y}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \right)^2 + \left( \frac{v_z}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \right)^2 = \frac{- c^2 + v_x^2 + v_y^2 + v_z^2}{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}} = -c^2
P^2 = - m^2c^2$$
Thus the two important equations of relativistic kinematics of a free particle are: $$E^2 = (mc^2)^2 + (\vec{p} c)^2 \\ E \vec{v} = c^2 \vec{p}$$ or with 4-vectors: $$0 = c^2 + V^2 \\ m V^{\mu} = P^{\mu}$$
From comparison of $$V^{\mu}$$ and $$R^{\mu}$$ we see that $$V^{\mu} = \gamma(\vec{v}) \frac{d R^{\mu}}{d t} = \frac{d R^{\mu}}{d \tau}$$ which suggests why it transforms under the Lorentz transform like $$R^{\mu}$$. It also suggests that the proper analogue of an acceleration vector is:
$$\alpha^{\mu} = \frac{d V^{\mu}}{d \tau} = \gamma(\vec{v}) \frac{d V^{\mu}}{d t} = \gamma(\vec{v}) \begin{pmatrix} \frac{ d \gamma(\vec{v}) c}{dt} \\ \frac{ d \gamma(\vec{v}) \vec{v} }{dt} \end{pmatrix} = \gamma(\vec{v}) \begin{pmatrix} \frac{ d \gamma(\vec{v})}{dt} c \\ \frac{ d \gamma(\vec{v}) }{dt} \vec{v} + \gamma(\vec{v}) \frac{ d \vec{v} }{dt} \end{pmatrix} = \gamma(\vec{v}) \begin{pmatrix} \gamma^3(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c^2} c \\ \gamma^3(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c^2} \vec{v} + \gamma(\vec{v}) \vec{a} \end{pmatrix} = \begin{pmatrix} \gamma^4(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c} \\ \gamma^4(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c^2} \vec{v} + \gamma^2(\vec{v}) \vec{a} \end{pmatrix}$$
Then
$$\alpha^2 = - \gamma^8(\vec{v}) \frac{ \left( \vec{a} \cdot \vec{v} \right)^2 }{c^2} + \gamma^8 (\vec{v}) \frac{ \left( \vec{a} \cdot \vec{v}\right)^2 }{c^4} \vec{v}^2 + 2 \gamma^6(\vec{v}) \frac{ \left( \vec{a} \cdot \vec{v} \right)^2 }{c^2} + \gamma^4(\vec{v}) \vec{a}^2 = \gamma^6(\vec{v}) \frac{ \left( \vec{a} \cdot \vec{v} \right)^2 }{c^2} + \gamma^4(\vec{v}) \vec{a}^2 =\gamma^4(\vec{v}) \left( \vec{a}^2 + \frac{ \left( \vec{a} \cdot \vec{v} \right)^2 }{c^2 - \vec{v}^2} \right) $$
$$ \alpha^{\mu} V_{\mu} = -\gamma^5(\vec{v}) \left( \vec{a} \cdot \vec{v} \right) + \gamma^5(\vec{v}) \left( \vec{a} \cdot \vec{v} \right) \frac{ \vec{v}^2}{c^2} + \gamma^3(\vec{v}) \left( \vec{a} \cdot \vec{v} \right) = \gamma^3(\vec{v}) \left( \vec{a} \cdot \vec{v} \right) \left( -\gamma^2(\vec{v}) + \gamma^2(\vec{v}) \frac{ \vec{v}^2}{c^2} + 1 \right) = 0$$ (for $$\alpha^{\mu}$$ and $$V^{\mu}$$ corresponding to the same event on the same particle's path.
Geometrically, with the convention of the inner product we are using, $$\alpha^{\mu}$$ and $$V^{\mu}$$ are orthogonal and in many ways the acceleration 4-vector is analogous to the curvature vector of a path through Euclidean space while the velocity 4-vector is analogous to the tangent vector.
So we have our analog with Newton's second law:
$$\frac{d P^{\mu}}{d\tau} \; = \; f^{\mu} \; = \; m \, \alpha^{\mu} \; = \; m \, \frac{d^2 R^{\mu}}{ d\tau^2}$$
So if we apply the relativistic force law for a massive particle of charge q, $$ f^{\mu} = q F_{\nu}^{\mu} V^{\nu}$$ we get:
$$\gamma(\vec{v}) \begin{pmatrix} \frac{1}{c} \frac{d E}{dt} \\ \frac{d \vec{p}}{dt} \end{\pmatrix} = m \begin{pmatrix} \gamma^4(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c} \\ \gamma^4(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c^2} \vec{v} + \gamma^2(\vec{v}) \vec{a} \end{pmatrix} = f^{\mu} = q F_{\nu}^{\mu} V^{\nu} = q \begin{pmatrix} 0 & \frac{1}{c} E_x & \frac{1}{c} E_y & \frac{1}{c} E_z \\ \frac{1}{c} E_x & 0 & B_z & -B_y \\ \frac{1}{c} E_y & -B_z & 0 & B_x \\ \frac{1}{c} E_z & B_y & -B_x & 0 \end{pmatrix} \begin{pmatrix} \gamma(\vec{v}) c \\ \gamma(\vec{v}) v_x \\ \gamma(\vec{v}) v_y \\ \gamma(\vec{v}) v_z \end{pmatrix} = q \gamma(\vec{v}) \begin{pmatrix} \frac{1}{c} \vec{E}\cdot \vec{v} \\ \vec{E} + \vec{v} \times \vec{B} \end{pmatrix}$$
Or $$\frac{dE}{dt} = q \vec{E}\cdot \vec{v} \\ \vec{f} = \frac{d \vec{p}}{dt} = q \left( \vec{E} + \vec{v} \times \vec{B} \right)$$
So the dynamics of energy and momentum exchange in terms of the electromagnetic tensor $$F_{\mu\nu} = \eta_{\mu \xi} F_{\nu}^{\xi}$$ has exactly the same number of components (6) as $$\vec{E} \oplus \vec{B}$$ (6), in detail they are the
same components, and they lead to the same physical prediction of changes of momentum and changes of energy. Thus Farsight's discovery of the Minkowski quote was not a revelation about electromagnetic phenomena, but a unification of $$\vec{E} \oplus \vec{B}$$ into a mathematical object that transforms under Lorentz boosts of the coordinate frame. This was analogous the "force-screw of mechanics" that Minkowski was talking about. Screw theory is a mathematical topic about 6-dimensional objects and one of these objects is called the "force-screw" for it's two 3-dimensional components. Similarly, the electromagnetic tensor permits dissection into two 3-dimensional components, the $$\vec{E}$$ and $$\vec{B}$$ vectors of Maxwell's theory.
