Relativistic Mechanics - Kinetic Energy

Kinetic Energy

The work-energy theorem says the change in kinetic energy is equal to the work done on the body. In special relativity:

\begin{align} \Delta K = W = (\gamma_1 - \gamma_0) m_0c^2.\end{align}
Derivation
\begin{align} \Delta K = W &= \int_{\mathbf{r}_0}^{\mathbf{r}_1} \mathbf{F} \cdot d\mathbf{r} \\ &= \int_{t_0}^{t_1} \frac{d}{dt}(\gamma m_0 \mathbf{v})\cdot\mathbf{v}dt \\ &= \left. \gamma m_0 \mathbf{v} \cdot \mathbf{v} \right|^{t_1}_{t_0} - \int_{t_0}^{t_1} \gamma m_0\mathbf{v} \cdot \frac{d\mathbf{v}}{dt} dt \\ &= \left. \gamma m_0 v^2 \right|^{t_1}_{t_0} - m_0\int_{v_0}^{v_1} \gamma v\,dv \\ &= m_0 \left( \left. \gamma v^2 \right|^{t_1}_{t_0} - c^2\int_{v_0}^{v_1} \frac{2v/c^2}{2\sqrt{1-v^2/c^2}}\,dv \right) \\ &= \left. m_0\left(\frac {v^2}{\sqrt{1-v^2/c^2}} + c^2 \sqrt{1-v^2/c^2} \right) \right|^{t_1}_{t_0} \\ &= \left. \frac {m_0c^2}{\sqrt{1-v^2/c^2}} \right|^{t_1}_{t_0} \\ &= \left. {\gamma m_0c^2}\right|^{t_1}_{t_0} \\ &= \gamma_1 m_0c^2 - \gamma_0 mc^2.\end{align}

If in the initial state the body was at rest (γ0 = 1) and in the final state it has speed v (γ1 = γ), the kinetic energy is

a result that can be directly obtained by subtracting the rest energy m0c2 from the total relativistic energy γm0c2.

Read more about this topic:  Relativistic Mechanics

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