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 \[\varepsilon(t)=\sqrt{p^{2}c^{2}+m^{2}c^{4}}= \sqrt{\mu^{2}+\tilde{t}^{2}},\quad\text{where}\; \mu=mc^2,\;\tilde{t}=eEct.\] 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 $\dot{p}=\gamma^{3}m\dot{v}$: \[a=\left.\frac{dv'}{d\tau}\right|_{v'=0}= \gamma\left.\frac{d}{dt}\frac{u+v}{1+uv} \right|_{u+v=0}=\gamma \frac{dv/dt}{1-\beta^{2}}= \gamma^{3}\dot{v}=\frac{\dot{p}}{m}= \frac{eE}{m}=\frac{c^{2}}{\rho}.\]

The uniformly accelerated observer will never observe the region \[D_{+}=\big\{(x,t)\;|\;c(t-t_{0})>(x-x_{0}+\rho)\big\},\] as light from there will never reach him. That means the considered observer will never see the part of evolution of any fixed or uniformly moving particle, starting from some moment of time \[t_{h}(x)=t_{0}+\tfrac{1}{c}(x-x_{0}+\rho),\] in which it enters $D_+$ by crossing its boundary, the future event horizon. As the observer is constantly accelerating, its proper time slows down, so from his point of view the particle will asymptotically approach the $x$ coordinate, never reaching it. He will only see the rest of the particle's worldline if he stops accelerating (this always happens, though, in a realistic situation). Likewise, an observer in \[D_{-}=\big\{(x,t)\;|\;c(t-t_{0})<-(x-x_{0}+\rho)\big\}\] will never receive light emitted by the uniformly accelerating particle. That is, any uniformly moving observer only sees the accelerating particle starting from some finite time. Before that the particle lies beyond the past event horizon.