The equations of motion in relativistic notation
It doesn’t do much good to find electric and magnetic fields from Maxwell’s equations unless we know what the fields do when we have them. You may remember that the fields are required to find the forces on charges, and that those forces determine the motion of the charge. So, of course, part of the theory of electrodynamics is the relation between the motion of charges and the forces.  For a single charge in the fields $\FLPE$ and $\FLPB$, the force is \begin{equation} \label{Eq:II:26:23} \FLPF=q(\FLPE+\FLPv\times\FLPB). \end{equation} This force is equal to the mass times the acceleration for low velocities, but the correct law for any velocity is that the force is equal to $d\FLPp/dt$. Writing $\FLPp=m_0\FLPv/\sqrt{1-v^2/c^2}$, we find that the relativistically correct equation of motion is \begin{equation} \label{Eq:II:26:24} \ddt{}{t}\biggl(\frac{m_0\FLPv}{\sqrt{1-v^2/c^2}}\biggr)= \FLPF=q(\FLPE+\FLPv\times\FLPB). \end{equation}  We would like now to discuss this equation from the point of view of relativity. Since we have put our Maxwell equations in relativistic form, it would be interesting to see what the equations of motion would look like in relativistic form. Let’s see whether we can rewrite the equation in a four-vector notation.  We know that the momentum is part of a four-vector $p_\mu$ whose time component is the energy $m_0c^2/\sqrt{1-v^2/c^2}$ divided by $c$. So we might think to replace the left-hand side of Eq. (26.24) by $dp_\mu/dt$. Then we need only find a fourth component to go with $\FLPF$. This fourth component must be related to the rate-of-change of the energy, or the rate of doing work, which is $\FLPF\cdot\FLPv$. We would then like to write the right-hand side of Eq. (26.24) as a four-vector like $(\FLPF\cdot\FLPv/c,F_x,F_y,F_z)$. But this does not make a four-vector.  The time derivative of a four-vector is no longer a four-vector, because the $d/dt$ requires the choice of some special frame for measuring $t$. We got into that trouble before when we tried to make $\FLPv$ into a four-vector. Our first guess was that the time component would be $cdt/dt=c$. But the quantities \begin{equation} \label{Eq:II:26:25} \biggl(c,\ddt{x}{t},\ddt{y}{t},\ddt{z}{t}\biggl)=(c,\FLPv) \end{equation} are not the components of a four-vector. We found that they could be made into one by multiplying each component by $1/\sqrt{1-v^2/c^2}$. The “four-velocity” $u_\mu$ is the four-vector \begin{equation} \label{Eq:II:26:26} u_\mu=\biggl(\frac{c}{\sqrt{1-v^2/c^2}}, \frac{\FLPv}{\sqrt{1-v^2/c^2}}\biggr). \end{equation} So it appears that the trick is to multiply $d/dt$ by $1/\sqrt{1-v^2/c^2}$, if we want the derivatives to make a four-vector.  Our second guess then is that \begin{equation} \label{Eq:II:26:27} \frac{1}{\sqrt{1-v^2/c^2}}\,\ddt{}{t}(p_\mu) \end{equation} should be a four-vector. But what is $\FLPv$? It is the velocity of the particle—not of a coordinate frame! Then the quantity $f_\mu$ defined by \begin{equation} \label{Eq:II:26:28} f_\mu=\biggl(\frac{\FLPF\cdot\FLPv/c}{\sqrt{1-v^2/c^2}}, \frac{\FLPF}{\sqrt{1-v^2/c^2}}\biggr) \end{equation} is the extension into four dimensions of a force—we can call it the “four-force.” It is indeed a four-vector, and its space components are not the components of $\FLPF$ but of $\FLPF/\sqrt{1-v^2/c^2}$.  The question is—why is $f_\mu$ a four-vector? It would be nice to get a little understanding of that $1/\sqrt{1-v^2/c^2}$ factor. Since it has come up twice now, it is time to see why the $d/dt$ can always be fixed by the same factor. The answer is in the following: When we take the time derivative of some function $x$, we compute the increment $\Delta x$ in a small interval $\Delta t$ in the variable $t$. But in another frame, the interval $\Delta t$ might correspond to a change in both $t'$ and $x'$, so if we vary only $t'$, the change in $x$ will be different. We have to find a variable for our differentiation that is a measure of an “interval” in space-time, which will then be the same in all coordinate systems. When we take $\Delta x$ for that interval, it will be the same for all coordinate frames. When a particle “moves” in four-space, there are the changes $\Delta t$, $\Delta x$, $\Delta y$, $\Delta z$. Can we make an invariant interval out of them? Well, they are the components of the four-vector $x_\mu=(ct,x,y,z)$ so if we define a quantity $\Delta s$ by  \begin{equation} \label{Eq:II:26:29} (\Delta s)^2=\frac{1}{c^2}\,\Delta x_\mu\Delta x_\mu=\frac{1}{c^2} (c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2) \end{equation} 
 \begin{align} \label{Eq:II:26:29} (\Delta s)^2&=\frac{1}{c^2}\,\Delta x_\mu\Delta x_\mu\\[1ex] &=\frac{1}{c^2}(c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2)\notag \end{align}  —which is a four-dimensional dot product—we then have a good four-scalar to use as a measure of a four-dimensional interval. From $\Delta s$—or its limit $ds$—we can define a parameter $s=\int ds$. And a derivative with respect to $s$, $d/ds$, is a nice four-dimensional operation, because it is invariant with respect to a Lorentz transformation.  It is easy to relate $ds$ to $dt$ for a moving particle. For a moving point particle, \begin{equation} \label{Eq:II:26:30} dx=v_x\,dt,\quad dy=v_y\,dt,\quad dz=v_z\,dt, \end{equation} and  \begin{equation} \label{Eq:II:26:31} ds=\sqrt{(dt^2/c^2)(c^2-v_x^2-v_y^2-v_z^2)} =dt\sqrt{1-v^2/c^2}. \end{equation} 
 \begin{align} ds&=\sqrt{(dt^2/c^2)(c^2-v_x^2-v_y^2-v_z^2)}\notag\\[1ex] \label{Eq:II:26:31} &=dt\sqrt{1-v^2/c^2}. \end{align} 
 So the operator  \begin{equation*} \frac{1}{\sqrt{1-v^2/c^2}}\,\ddt{}{t} \end{equation*} is an invariant operator. If we operate on any four-vector with it, we get another four-vector. For instance, if we operate on $(ct,x,y,z)$, we get the four-velocity $u_\mu$: \begin{equation*} \ddt{x_\mu}{s}=u_\mu. \end{equation*} We see now why the factor $\sqrt{1-v^2/c^2}$ fixes things up.  The invariant variable $s$ is a useful physical quantity. It is called the “proper time” along the path of a particle, because $ds$ is always an interval of time in a frame that is moving with the particle at any particular instant. (Then, $\Delta x=$ $\Delta y=$ $\Delta z=$ $0$, and $\Delta s=\Delta t$.) If you can imagine some “clock” whose rate doesn’t depend on the acceleration, such a clock carried along with the particle would show the time $s$.  We can now go back and write Newton’s law (as corrected by Einstein) in the neat form \begin{equation} \label{Eq:II:26:32} \ddt{p_\mu}{s}=f_\mu, \end{equation} where $f_\mu$ is given in Eq. (26.28). Also, the momentum $p_\mu$ can be written as \begin{equation} \label{Eq:II:26:33} p_\mu=m_0u_\mu=m_0\,\ddt{x_\mu}{s}, \end{equation} where the coordinates $x_\mu=(ct,x,y,z)$ now describe the trajectory of the particle. Finally, the four-dimensional notation gives us this very simple form of the equations of motion: \begin{equation} \label{Eq:II:26:34} f_\mu=m_0\,\frac{d^2x_\mu}{ds^2}, \end{equation} which is reminiscent of $F=ma$. It is important to notice that Eq. (26.34) is not the same as $F=ma$, because the four-vector formula Eq. (26.34) has in it the relativistic mechanics which are different from Newton’s law for high velocities. It is unlike the case of Maxwell’s equations, where we were able to rewrite the equations in the relativistic form without any change in the meaning at all—but with just a change of notation.  Now let’s return to Eq. (26.24) and see how we can write the right-hand side in four-vector notation. The three components—when divided by $\sqrt{1-v^2/c^2}$—are the components of $f_\mu$, so  \begin{equation} \label{Eq:II:26:35} f_x=\frac{q(\FLPE+\FLPv\times\FLPB)_x}{\sqrt{1-v^2/c^2}}= q\biggl[ \frac{E_x}{\sqrt{1-v^2/c^2}}+ \frac{v_yB_z}{\sqrt{1-v^2/c^2}}- \frac{v_zB_y}{\sqrt{1-v^2/c^2}} \biggr]. \end{equation} 
 \begin{align} \label{Eq:II:26:35} f_x&=\frac{q(\FLPE+\FLPv\times\FLPB)_x}{\sqrt{1-v^2/c^2}}\\[1.1ex] &=q\biggl[ \frac{E_x}{\sqrt{1-v^2/c^2}}\,+\notag\\[.5ex] &\phantom{=q\biggl[\;}\frac{v_yB_z}{\sqrt{1-v^2/c^2}}- \frac{v_zB_y}{\sqrt{1-v^2/c^2}} \biggr].\notag \end{align}  Now we must put all quantities in their relativistic notation. First, $c/\sqrt{1-v^2/c^2}$ and $v_y/\sqrt{1-v^2/c^2}$ and $v_z/\sqrt{1-v^2/c^2}$ are the $t$-, $y$-, and $z$-components of the four-velocity $u_\mu$. And the components of $\FLPE$ and $\FLPB$ are components of the second-rank tensor of the fields $F_{\mu\nu}$. Looking back in Table 26–1 for the components of $F_{\mu\nu}$ that correspond to $E_x$, $B_z$, and $B_y$, we get3 \begin{equation*} f_x=q(u_tF_{xt}-u_yF_{xy}-u_zF_{xz}), \end{equation*} which begins to look interesting. Every term has the subscript $x$, which is reasonable, since we’re finding an $x$-component. Then all the others appear in pairs: $tt$, $yy$, $zz$—except that the $xx$-term is missing. So we just stick it in, and write \begin{equation} \label{Eq:II:26:36} f_x=q(u_tF_{xt}-u_xF_{xx}-u_yF_{xy}-u_zF_{xz}). \end{equation} We haven’t changed anything because $F_{\mu\nu}$ is antisymmetric, and $F_{xx}$ is zero. The reason for wanting to put in the $xx$-term is so that we can write Eq. (26.36) in the short-hand form \begin{equation} \label{Eq:II:26:37} f_\mu=qu_\nu F_{\mu\nu}. \end{equation} This equation is the same as Eq. (26.36) if we make the rule that whenever any subscript occurs twice (as $\nu$ does here), you automatically sum over terms in the same way as for the scalar product, using the same convention for the signs.  You can easily believe that (26.37) works equally well for $\mu=y$ or $\mu=z$, but what about $\mu=t$? Let’s see, for fun, what it says: \begin{equation*} f_t=q(u_tF_{tt}-u_xF_{tx}-u_yF_{ty}-u_zF_{tz}). \end{equation*} Now we have to translate back to $E$’s and $B$’s. We get  \begin{equation} \label{Eq:II:26:38} f_t=q\biggl(0+ \frac{v_x}{\sqrt{1-v^2/c^2}}\,E_x/c+ \frac{v_y}{\sqrt{1-v^2/c^2}}\,E_y/c+ \frac{v_z}{\sqrt{1-v^2/c^2}}\,E_z/c \biggr), \end{equation} 
 \begin{equation} \begin{aligned} f_t=q\biggl(0&+\frac{v_x}{\sqrt{1-v^2/c^2}}\,E_x/c\\[.5ex] &+\frac{v_y}{\sqrt{1-v^2/c^2}}\,E_y/c\\[.25ex] &+\frac{v_z}{\sqrt{1-v^2/c^2}}\,E_z/c\biggr), \end{aligned} \label{Eq:II:26:38} \end{equation}  or \begin{equation*} f_t=\frac{q\FLPv\cdot\FLPE/c}{\sqrt{1-v^2/c^2}}. \end{equation*} But from Eq. (26.28), $f_t$ is supposed to be \begin{equation*} \frac{\FLPF\cdot\FLPv/c}{\sqrt{1-v^2/c^2}}= \frac{q(\FLPE+\FLPv\times\FLPB)\cdot\FLPv/c}{\sqrt{1-v^2/c^2}}. \end{equation*} This is the same thing as Eq. (26.38), since $(\FLPv\times\FLPB)\cdot\FLPv$ is zero. So everything comes out all right.  Summarizing, our equation of motion can be written in the elegant form \begin{equation} \label{Eq:II:26:39} m_0\,\frac{d^2x_\mu}{ds^2}=f_\mu=qu_\nu F_{\mu\nu}. \end{equation} Although it is nice to see that the equations can be written that way, this form is not particularly useful. It’s usually more convenient to solve for particle motions by using the original equations (26.24), and that’s what we will usually do. 