Relativistic kinetic energy and force.

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granpa
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classically, an electron accelerating from rest in a uniform electric field will have a kinetic energy proportional to the distance 'd' from its point of origin.

will this continue to hold even when the electron is moving at relativistic velocity?

I understand that the formula for relativistic kinetic energy is

Equation for relativistic kinetic energy: E_k = mγc² − mc² = mc²(1/√(1 − (v/c)²) − 1)


so basically kinetic energy is proportional to gamma - 1

so gamma(d) ≡ d + 1?
 
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granpa, There are several ways of writing the force and acceleration, but one thing you can rely on, even in relativity, is energy conservation. The total energy of a charge particle in an E field is E = γmc2 + eΦ where Φ is the electrostatic potential, and the value of this will be the same both when the particle is at rest and when it is moving relativistically. (This ignores only the radiation it will emit.)
 
Thank you. Thats was very helpful.

I've also been told that the rate of change of rapidity with respect to proper time will be constant.
I guess that would be the proper acceleration.

Equation alpha = Δw/Δt = c Δη/Δτ = c² Δγ/Δx, relating rapidity, proper acceleration and gamma.


gamma(x) = x + 1

a = dg(x)/dx = 1
 
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if instead of a uniform electric field we use an inverse square law

F = 1/r^2

KE = integral F dr = 1/r

KE = gamma(r) - 1 = 1/r

gamma(r) = 1 + 1/r