How is the zeroth term of the Minkowski force derived? I'm working through T.M. Helliwell's Special Relativity on my own, and one of the questions asks to show that the Minkowski force $K_\mu=dp_\mu/d\tau$ can be written $$K_\mu=\left(\gamma\frac{\mathbf{v\cdot F}}{c}, \gamma F_x, \gamma F_y, \gamma F_z\right).$$
The first second and third term seem trivial, and by inspection it seems that $dE/d\tau=\gamma \mathbf{v\cdot F}$. I've tried using the relations $E=\gamma mc^2$ and $E^2=(pc)^2 + (mc^2)^2$, but I can't get the appropriate form starting from either of those. How can I show that $dE/d\tau=\gamma\mathbf{v\cdot F}$?
 A: It's just the following (setting $c = 1$):
$$\frac{dE}{d\tau} = \frac{dE}{dt} \frac{dt}{d\tau} = (\mathbf{v} \cdot \mathbf{F})\, \gamma.$$

I guess you might also want to ask, why is $dE/dt = \mathbf{v} \cdot \mathbf{F}$ in special relativity? Well, if you start with
$$E^2 = p^2 + m^2$$
and take the time derivative of both sides, you have
$$E \, \frac{dE}{dt} = \mathbf{p} \cdot \frac{d\mathbf{p}}{dt}.$$
The second term on the right is $\mathbf{F}$, by definition. And 
$$\frac{\mathbf{p}}{E} = \frac{\gamma m \mathbf{v}}{\gamma m} = \mathbf{v}$$
by the definitions of relativistic momentum and energy. So solving for $dE/dt$ gives $\mathbf{v} \cdot \mathbf{F}$.
A: $K_{\mu}v^{\mu}$ is a Lorentz invariant. Evaluating it with a (+---) signature in the proper frame where $\textbf v = (c,0)$: 
$$K_{\mu}v^{\mu} = \gamma\frac{d}{dt}(E/c,\,\gamma\textbf{p})\cdot\gamma(c,\textbf{v})=\frac{d}{d\tau}(mc,\,\textbf{p})\cdot(c,0) = c^2\frac{dm}{d\tau}$$
If the force is (rest) mass preserving, $dm/d\tau$=0:
$$K_{\mu}v^{\mu} = \gamma^2\left( \frac{dE}{dt} - \textbf{v}\cdot\frac{d}{dt}\gamma\textbf{p}\right) = 0,\quad \Rightarrow \frac{dE}{dt} = \textbf{v}\cdot\textbf{F}$$
Therefore:
$$
K_\mu=\gamma\left(\frac{1}{c}\frac{dE}{dt}, F_x, F_y, F_z\right)=
\gamma\left(\frac{\mathbf{v\cdot F}}{c}, F_x, F_y, F_z\right)
$$
