Relativistic acceleration due to a constant force

Consider a body with mass $m$ that is accelerated using a constant force $\vec{F}$ acting on the body. Let the body start at rest, such that the momentum $\vec{p} = 0$ for $t = 0$ . The force is given by $\displaystyle \vec{F} = \frac{d\vec{p}}{dt}$ . So if the force is constant, then the momentum is simply linear in time, thus $\vec{p} = \vec{F} t$ . As the momentum is given by $\displaystyle \vec{p} = \frac{m_0 \vec{v}}{\sqrt{1 - v^2/c^2}}$ , we obtain the equation

(1) $\displaystyle F t \hat{e}_F = \frac{m_0 v \hat{e}_F}{\sqrt{1 - v^2/c^2}}$ ,

where $\hat{e}_F$ is the unit vector in the direction of the force. Then

(2) $\displaystyle \left( \frac{F}{m_0 c} \right) t = \frac{v/c}{\sqrt{1 - v^2/c^2}}$

(3) $\displaystyle \frac{v}{c} = \frac{ \left( \displaystyle \frac{F}{m_0 c} \right) t}{\sqrt{1 + \left( \displaystyle \frac{F}{m_0 c} \right)^2 t^2}}$

Note that

(4) $\displaystyle \frac{v}{c} = \frac{m_0 c}{F} \frac{d}{dt} \left( \sqrt{1 + \left( \displaystyle \frac{F}{m_0 c} \right)^2 t^2} - 1 \right)$

The travelled path of the body is then given by

(5) $s(t) = \displaystyle c \left( \frac{m_0 c}{F} \right) \left( \sqrt{1 + \left( \displaystyle \frac{F}{m_0 c} \right)^2 t^2} - 1 \right)$

Thus $s(t) \simeq \displaystyle \frac{1}{2} \left( \frac{F}{m_0} \right) t^2$ – the Classical result. The motion of an accelerating body, due to a constant force; the result:

(6) $\left[ \begin{array}{rclrcl} s(t) &=& \displaystyle c \left( \frac{m_0 c}{F} \right) \left( \sqrt{1 + \left( \displaystyle \frac{F}{m_0 c} \right)^2 t^2} - 1 \right) && \simeq & \displaystyle \frac{1}{2} \left( \frac{F}{m_0} \right) t^2 \\\\ v(t) &=& \displaystyle \frac{ \displaystyle c \left( \frac{F}{m_0 c} \right) t}{\displaystyle \sqrt{1 + \left( \displaystyle \frac{F}{m_0 c} \right)^2 t^2}} && \simeq & \displaystyle \left( \frac{F}{m_0} \right) t \\\\ a(t) &=& \displaystyle \frac{\displaystyle c \left( \frac{F}{m_0 c} \right) }{\displaystyle \sqrt{1 + \left( \displaystyle \frac{F}{m_0 c} \right)^2 t^2}^3} & & \simeq & \displaystyle \frac{F}{m_0} \end{array} \right.$

3 Responses to Relativistic acceleration due to a constant force

Sion Draco says:

November 28, 2010 at 3:35 am

These are the classical equations of hyperbolic motion, it is clear that the Rybzyck idiot doesn’t know how to integrate an ordinary differential equation.

Phil says:

April 9, 2015 at 1:31 pm

Looks complicated! How would the aforementioned be used to work out the time of flight of a relativistic electron? The simple answer is: distance x momentum/energy for ALL speeds.

Eric Su says:

March 14, 2018 at 1:34 am

The approximation works only if v is small. Relativistic result is different from Newtonian result. In fact, relativistic result can not be correct because of conservation of momentum.
http://vixra.org/abs/1803.0005