D4b - motion in uniform electric field

\vec{E} = (\begin{matrix} E_{x} & 0 & 0 \end{matrix}) \vec{B} = (\begin{matrix} 0 & 0 & 0 \end{matrix}) F^{α}_{β} = (\begin{matrix} 0 & E_{x} / c & 0 & 0 \\ E_{x} / c & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})

$μ = 2, 3 :$

\frac{d u^{μ}}{d τ} = \frac{q}{m} \sum_{ν} F^{μ}_{ν} u^{ν} = 0

hence, $u^{2}, u^{3}$ are constants
for $μ = 0, 1 :$

\frac{d}{d τ} (\begin{matrix} u^{0} \\ u^{1} \end{matrix}) = \frac{q}{m} \frac{E_{x}}{c} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} u^{0} \\ u^{1} \end{matrix})

to find the eigenvalues:

| \begin{matrix} - λ & 1 \\ 1 & - λ \end{matrix} | = 0 ⟹ λ = \pm 1 λ_{1} = 1 ⟹ {\vec{e}}_{1} = (\begin{matrix} 1 \\ 1 \end{matrix}) λ_{2} = - 1 ⟹ {\vec{e}}_{2} = (\begin{matrix} 1 \\ - 1 \end{matrix}) \vec{u} (τ) = a e^{λ_{1}} {\vec{e}}_{1} + b e^{λ_{2}} {\vec{e}}_{2} (\begin{matrix} u^{0} \\ u^{1} \end{matrix}) = a (\begin{matrix} 1 \\ 1 \end{matrix}) \exp (\frac{q E_{x}}{m c} τ) + b (\begin{matrix} 1 \\ - 1 \end{matrix}) \exp (- \frac{q E_{x}}{m c} τ)

initial conditions: $τ = 0$ , $u^{1} = 0, u^{0} = c$
this gives $c = a + b$ and $0 = a - b ⟹ a = b = c / 2$

u^{0} = γ c = \frac{c}{2} [\exp (- \frac{q E_{x}}{m c} τ) + \exp (- \frac{q E_{x}}{m c} τ)] = c \cosh [\frac{q E_{x} τ}{m c}] u^{1} = γ v_{x} = \frac{c}{2} [\exp (- \frac{q E_{x}}{m c} τ) - \exp (- \frac{q E_{x}}{m c} τ)] = c \sinh [\frac{q E_{x} τ}{m c}]

to find $v_{x} :$

v_{x} = \frac{u_{1}}{u_{0}} c = c \tanh [\frac{q E_{x} τ}{m c}]

image: ChatGPT

in coordinate time:$$t = \int_{0}^{\tau}\gamma(\tau) d\tau = \frac{mc}{qE_{x}} \sinh \left(\frac{qE_{x}\tau}{mc}\right)$$
using the identity:

\tanh^{2} x = 1 - {sech}^{2} x = \frac{\cosh^{2} x - 1}{\cosh^{2} x} = \frac{\sinh^{2} x}{1 + \sinh^{2} x}

v_{x} = c \sqrt{\frac{{(\frac{q E_{x} t}{m c})}^{2}}{1 + {(\frac{q E_{x} t}{m c})}^{2}}} = \frac{1}{γ (t)} \frac{q E_{x} t}{m}

considering the transverse motion:

\frac{d u^{2}}{d τ} = 0 ⟹ u^{2} = const = γ v_{y}

since $γ (τ)$ changes, $v_{y}$ changes:

v_{y} (τ) = \frac{1}{γ (τ)} v_{y} (τ = 0) = \frac{v_{y} (0)}{\cosh (\frac{q E_{x} τ}{m c})}

therefore, the transverse component of the 3-vector decreases as the particle accelerates
considering the 4-momentum:

p^{2} = γ m v_{y} = m v_{y} (0) = const

therefore, momentum is conserved as the relative mass increases so the velocity decreases