What is the physical meaning of the zeroth component of the Minkowski force 4-vector? In David Griffiths' Intro to Electrodynamics, section on ED and relativity, he poses a simple problem regarding the zeroth component of the Minkowski force 4-vector for a charge $q$ moving with velocity $v$. I think this is
$$\frac{q \mathbf{E} \cdot \mathbf{v}}{\sqrt{c^2-v^2}}.$$
What I cannot figure out is the physical meaning of this component, assuming I got it right.
 A: The Minkowski force is simply $dp^\mu/d\tau$. In the nonrelativistic limit, $p^0$ is energy and $\tau$ is time, so
$$F^0 \approx \frac{dE}{dt} = \mathbf{F} \cdot \mathbf{v}.$$
That is, it is the rate of work being done on the particle by the force. (I've dropped a factor of $c$ here for simplicity.)
A: If  $\:\mathbf{f}\:$ is the force 3-vector, like the Lorentz force, applied on a particle of rest mass  $\:m_{o}\:$ moving with velocity 3-vector $\:\mathbf{w}\:$, then the  force 4-vector  is
\begin{equation}
\mathbf{F}=\left(\gamma_{w} \mathbf{f} , \gamma_{w}\dfrac{ \mathbf{f}\boldsymbol{\cdot}\mathbf{w} }{c}\right), \quad \gamma_{w}=\dfrac{1}{\sqrt{1-\dfrac{w^2}{c^2}}}
\tag{001}
\end{equation}
Now $\: \mathbf{f}\boldsymbol{\cdot}\mathbf{w} \:$ is the rate of work $\:W\:$done on the particle by the force, rate with respect to time $\:t\:$ :
\begin{equation}
\mathbf{f}\boldsymbol{\cdot}\mathbf{w}=\dfrac{\mathrm{d}W}{\mathrm{d}t}
\tag{002}
\end{equation} 
If $\:\tau\:$ is the proper time, that is the time in the rest frame of the particle, then
\begin{equation}
\gamma_{w}=\dfrac{\mathrm{d}t}{\mathrm{d}\tau}
\tag{003}
\end{equation} 
and
\begin{equation}
\gamma_{w}\left(\mathbf{f}\boldsymbol{\cdot}\mathbf{w}\right)=\dfrac{\mathrm{d}W}{\mathrm{d}\tau}
\tag{004}
\end{equation} 
So the zeroth ("time") component of the force 4-vector times $\:c\:$  is the rate of work $\:W\:$ done on the particle by the force, rate with respect to proper time $\:\tau\:$. 
In the case of  Lorentz force and particle with charge $\:q\:$
\begin{equation}
\mathbf{f} =q\left(\mathbf{E}+\mathbf{w}\boldsymbol{\times}\mathbf{B}\right)
\tag{005}
\end{equation}
and
\begin{equation}
\gamma_{w}\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{w}}{c}=\gamma_{w}\dfrac{q\left(\mathbf{E}+\mathbf{w}\boldsymbol{\times}\mathbf{B}\right)\boldsymbol{\cdot}\mathbf{w}}{c}=\dfrac{q\mathbf{E}\boldsymbol{\cdot}\mathbf{w}}{\sqrt{c^2-w^2}}
\tag{006}
\end{equation}
identical to the expression in question. 
