Chapter 41, Problem 31SP

To determine

To prove: The expression ofenergy as $E^2=p^2{c}^2+m^2{c}^4$ and the expression for kinetic energy as $\sqrt{m^2{c}^4+p^2{c}^2}-m{c}^2$ in relativistic mechanics.

Explanation of Solution

Given data:

The expression to prove that $E^2=p^2{c}^2+m^2{c}^4$.

The expression to show that $KE=\sqrt{m^2{c}^4+p^2{c}^2}-m{c}^2$.

Formula used:

Write the expression for relativistic total energy:

$E=\gamma m{c}^2$

Here, $m$ is the massand $c$ is the speed of light.

Write the expression for relativistic momentum $p$:

$p=\gamma mv$

Here, $m$ is the mass and $v$ is the speed.

Write the expression for $\gamma$:

$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $c$ is the speed of light and $v$ is the speed or velocity of the particle.

Write the expression for the total energy:

$E=KE+m{c}^2$

Here, $c$ is the speed of light, $KE$ is the kinetic energy, and $m$ is the mass of electron.

Explanation:

Recall the expression for $\gamma$:

$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Recall the expression for relativistic total energy:

$E=\gamma m{c}^2$

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$

$E=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}m{c}^2$

Square both sides

$E^2=\left(\frac{1}{1-\frac{v^2}{c^2}}\right)m^2{c}^4$…… (1)

Recall the expression for relativistic momentum p:

$p=\gamma mv$

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$

$p=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}mv$

Square both sides of the above equation

$p^2=\left(\frac{1}{1-\frac{v^2}{c^2}}\right)m^2{v}^2$

Multiply $c^2$ on both sides of the above equation

$p^2{c}^2=\left(\frac{1}{1-\frac{v^2}{c^2}}\right)m^2{v}^2{c}^2$…… (2)

Subtract equation (2) from equation (1)

$\begin{array}{c}{E}^2-p^2{c}^2=\left(\frac{1}{1-\frac{v^2}{c^2}}\right)m^2{c}^4-\left(\frac{1}{1-\frac{v^2}{c^2}}\right)m^2{v}^2{c}^2\\ =\left(\frac{1}{1-\frac{v^2}{c^2}}\right)m^2{c}^4\left(1-\frac{v^2}{c^2}\right)\\ =m^2{c}^4\end{array}$

Further solve

$E^2=m^2{c}^4+p^2{c}^2$

Hence, proved.

Now, rewrite the expression for the total energy in relativistic mechanics.

$E=\sqrt{m^2{c}^4+p^2{c}^2}$

Recall the expression for the total energy:

$E=KE+m{c}^2$

Equate both the above equations of $E$

$\begin{array}{c}KE+m{c}^2=\sqrt{m^2{c}^4+p^2{c}^2}\\ KE=\sqrt{m^2{c}^4+p^2{c}^2}-m{c}^2\end{array}$

Conclusion:

The required expressionsareobtained.