I got stuck in deriving the velocity of the particle from the acceleration equation. Here are the details of the problem.(adsbygoogle = window.adsbygoogle || []).push({});

The acceleration of a particle with a relativistic momentum is

[itex]\vec{a} = \frac{\vec{F}}{\gamma m} - \frac{\vec{v}}{\gamma m c^2}\left(\vec{F} \cdot \vec{v}\right)[/itex].

By integrating the above equation, the velocity of the particle can be calculated if the force applied and the mass of the particle are given. The greek letter [itex]\gamma[/itex] in this case is [itex]\left(1 - \frac{v^2}{c^2}\right)^{-1/2}[/itex]

Considering that a constant force [itex]\it{F}[/itex] is applied to a mass [itex]\it{m}[/itex] initially at rest, the velocity of the particle is

[itex]\int \vec{a} dt = \int \left[\frac{\vec{F}}{\gamma m} - \frac{\vec{v}}{\gamma m c^2}\left(\vec{F} \cdot \vec{v}\right)\right] dt \\

\vec{v} = \int \left[\frac{\vec{F}}{\gamma m} - \frac{\vec{v^2}}{\gamma m c^2}\left(\vec{F} \right)\right] dt[/itex].

Since [itex]F[/itex] and [itex]m[/itex] are constants in time, they are removed from the integral. Gamma is taken out from the two terms. Then,

[itex]\vec{v} = \frac{\vec{F}}{m} \int \frac{1}{\gamma} \left(1 - \frac{v^2}{c^2}\right) dt[/itex].

Using the definition of gamma, the integrand is

[itex]\vec{v} = \frac{\vec{F}}{m} \int \left(1 - \frac{v^2}{c^2}\right)^{3/2} dt[/itex].

This is where I get stuck. I don't know how to integrate it since the integrand is not a direct function of [itex]t[/itex] but of [itex]v[/itex] which is a function of [itex]t[/itex]. I tried having [itex]v = at[/itex], but I don't know if this is correct.

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# Derivation of Velocity from Acceleration in Special Relativity

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