Question

Show that the relativistic expression for kinetic energy $\left(\gamma -1\right)\mathit{m}{\mathit{c}}^{\mathbf{2}}$is equivalent to the classical $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mu}}^{\mathbf{2}}$ when $\mathbf{u}\mathbf{}\mathbf{\ll }\mathbf{c}$

Short answer

The relativistic expression for kinetic energy become equivalent to classical expression at very low speed of particle.

Step-by-step solution

Relativistic Kinetic Energy

For a particle moving at relativistic speeds the rest mass of the particle gets converted into dynamic mass, which varies with the velocity of the particle. Thus, to calculate the kinetic energy of the particle, moving at relativistic speeds, relativistic correction need to be done.

Proof for relativistic kinetic energy to become classical kinetic energy at very low speed

The Lorentz factor for particle is given as:

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\left({\displaystyle \frac{u}{c}}\right)}^2}} \\ \gamma =1-{\left(\frac{u^2}{c^2}\right)}^{-1/2} \end{gathered}$$

The relativistic kinetic energy of a particle is given as:

$K=\left(\gamma -1\right)m{c}^2$

Here, c is the speed of light

Substituting the values in the above equation.

$K=\left({\left(1-\frac{u^2}{c^2}\right)}^{-1/2}-1\right)m{c}^2$

for $u\ll c$, expand the expression by using binomial theorem and neglect the higher terms.

$$\begin{gathered} K=\left(\left(1-\left(-\frac{1}{2}\right)\frac{u^2}{c^2}\right)-1\right)m{c}^2 \\ K=\left(1+\frac{u^2}{2c^2}-1\right)m{c}^2 \\ K=\left(\frac{u^2}{2c^2}\right)m{c}^2 \\ K=\frac{1}{2}m{u}^2 \end{gathered}$$

Therefore, the relativistic kinetic energy of particle at very low speed become classical kinetic energy of particle.