 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2018/1/b/solution

For a uniform constant electric field and zero magnetic field, the mechanical momentum changes linearly even though the velocity does not:

p (t) = p_{0} + e E t, p_{0} = m γ_{0} v_{0}, γ_{0} = (1 - v_{0}^{2} / c^{2})^{- 1/2} .

Invert

p = mγ v

using

γ = 1 + p^{2} / (m^{2} c^{2})

. The result for arbitrary initial direction is

v (t) = \frac{γ _{0} v _{0} + ( e E / m ) t}{1 + ∣ γ _{0} v _{0} + ( e E / m ) t ∣ ^{2} / c ^{2}} .

Equivalently,

v = c p / m^{2} c^{2} + p^{2}

. This gives

v (0) = v_{0}

and guarantees

∣ v ∣ < c

at every finite momentum. The transverse momentum stays constant, but its velocity component generally decreases because the Lorentz factor increases. Thus an initial transverse velocity must not be treated as constant during the electric acceleration; this is the relativistic motion in a constant electric field with transverse momentum effect.

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