Difference between revisions of "Technical warm-up"

The particle's equation of motion is $\dot{p}=eE=const$, so on choosing conveniently the zero point of time, we have $p(t)=eEt$. Then energy is \[\varepsilon(t)=\sqrt{p^{2}c^{2}+m^{2}c^{4}}= \sqrt{\mu^{2}+\tilde{t}^{2}},\quad\mbox{где}\; \mu=mc^2,\;\tilde{t}=eEct.\] As $p=\varepsilon v/c^{2}$, for velocity we get \[\beta(t)\equiv \frac{v(t)}{c}=\frac{cp}{\varepsilon}= \frac{\tilde{t}}{\sqrt{\mu^2+\tilde{t}^2}}.\] Integrating and restoring initial conditions $(t_{0},x_{0})$, we obtain \[(x+\rho-x_{0})^{2}-c^{2}(t-t_{0})^{2}=\rho^{2}, \qquad\mbox{где}\quad \rho=\frac{mc^{2}}{eE}.\] This is a hyperbola branch $x>0$, with asymptotes on the light cone, symmetry axis $OX$ and center $t=t_{0}$, $x=x_{0}-\rho$. Now calculate the $4$-acceleration of the particle: from $\dot{p}^{1}=\dot{p}=eE$, and $\dot{p}^{0}=\dot{\varepsilon}/c=eEv/c$ we obtain \[a_{\mu}a^{\mu}= \left(\frac{du^0}{ds}\right)^2- \left(\frac{du^1}{ds}\right)^2= \gamma^{2} \left[(\dot{u}^{0})^{2}-(\dot{u}^{1})^{2}\right]= \frac{\gamma^{2}}{(mc)^{2}} \left[\frac{\dot{\varepsilon}^{2}}{c^{2}} -\dot{p}^{2}\right]= -\frac{a^{2}}{c^{2}}.\] It is instructive to show that this acceleration is also the acceleration in the momentarily comoving frame. The latter is derived using the velocity composition law $v'=\frac{v+u}{1+vu}$, the expression for the proper time interval $d\tau=\gamma^{-1}dt$, and relation $\dot{p}=\gamma^{3}m\dot{v}$: \[a=\left.\frac{dv'}{d\tau}\right|_{v'=0}= \gamma\left.\frac{d}{dt}\frac{u+v}{1+uv} \right|_{u+v=0}=\gamma \frac{dv/dt}{1-\beta^{2}}= \gamma^{3}\dot{v}=\frac{\dot{p}}{m}= \frac{eE}{m}=\frac{c^{2}}{\rho}.\]