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.

This entry was posted in Physics, Relativity, Special Relativity. Bookmark the permalink.

3 Responses to Relativistic acceleration due to a constant force

  1. Sion Draco says:

    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.

  2. Phil says:

    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.

  3. Eric Su says:

    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.

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