# 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.

• You don't seem very sure. Doesn't the textbook have anything to say about this? Can you post the problem as written in the textbook? – sammy gerbil Aug 14 '16 at 18:51

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.)
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 $$\mathbf{F}=\left(\gamma_{w} \mathbf{f} , \gamma_{w}\dfrac{ \mathbf{f}\boldsymbol{\cdot}\mathbf{w} }{c}\right), \quad \gamma_{w}=\dfrac{1}{\sqrt{1-\dfrac{w^2}{c^2}}} \tag{001}$$ 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 $\:t\:$ :
$$\mathbf{f}\boldsymbol{\cdot}\mathbf{w}=\dfrac{\mathrm{d}W}{\mathrm{d}t} \tag{002}$$ If $\:\tau\:$ is the proper time, that is the time in the rest frame of the particle, then $$\gamma_{w}=\dfrac{\mathrm{d}t}{\mathrm{d}\tau} \tag{003}$$ and $$\gamma_{w}\left(\mathbf{f}\boldsymbol{\cdot}\mathbf{w}\right)=\dfrac{\mathrm{d}W}{\mathrm{d}\tau} \tag{004}$$ 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 $\:q\:$
$$\mathbf{f} =q\left(\mathbf{E}+\mathbf{w}\boldsymbol{\times}\mathbf{B}\right) \tag{005}$$ and $$\gamma_{w}\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{w}}{c}=\gamma_{w}\dfrac{q\left(\mathbf{E}+\mathbf{w}\boldsymbol{\times}\mathbf{B}\right)\boldsymbol{\cdot}\mathbf{w}}{c}=\dfrac{q\mathbf{E}\boldsymbol{\cdot}\mathbf{w}}{\sqrt{c^2-w^2}} \tag{006}$$