Chapter 26, Problem 60P

To determine

The derivation of energy momentum relation.

Answer to Problem 60P

The derivation is given below.

Explanation of Solution

Write the expression for total energy

  $E=K+E_0$                                                                                      (I)

Here, $E$ is the total energy, $K$ is the kinetic energy and $E_0$ is the rest energy.

Write the expression for relativistic kinetic energy

  $K=\left(\gamma -1\right)E_0$                                                                                       (II)

Here, $\gamma$ is the Lorentz factor.

Write the expression for Lorentz factor.

  $\gamma ={\left(1-v^2/c^2\right)}^{-1/2}$                                                                                      (III)

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

Write the expression for rest mass.

  $E_0=m{c}^2$                                                                                            (IV)

Here, $m$ is the rest mass.

Square (I)

  $E^2=K^2+E_0^2+2K{E}_0$                                                                             (V)

Square (II)

  $K^2={\gamma }^2{E}_0^2+E_0^2-2\gamma E_0$                                                                             (VI)

Multiply (II) by $2E_0$

  $2K{E}_0=2\left(\gamma -1\right)E_0^2$

Rearrange

  $2K{E}_0=2\gamma E_0^2-2E_0^2$                                                                                      (VII)

Add (VII) and (VI)

  $K^2+2K{E}_0={\gamma }^2{E}_0^2+E_0^2-2\gamma E_0+2\gamma E_0^2-2E_0^2$

Rearrange

  $K^2+2K{E}_0=\left({\gamma }^2-1\right)E_0^2$                                                                         (VIII)

Substitute for ${\gamma }^2$ and $E_0^2$

  $\begin{array}{c}{K}^2+2K{E}_0=\left[{\left(1-v^2/c^2\right)}^{-1}-1\right]{\left(m{c}^2\right)}^2\\ ={\left(1-v^2/c^2\right)}^{-1}\left[1-\left(1-v^2/c^2\right)\right]{\left(m{c}^2\right)}^2\end{array}$

Rearrange

  $K^2+2K{E}_0={\left(1-v^2/c^2\right)}^{-1}\left[v^2/c^2\right]{\left(m{c}^2\right)}^2$

Substitute ${\gamma }^2$ for ${\left(1-v^2/c^2\right)}^{-1}$

  $K^2+2K{E}_0={\gamma }^2\frac{v^2}{c^2}{\left(m{c}^2\right)}^2$

Rearrange

     $K^2+2K{E}_0={\gamma }^2{v}^2{m}^2{c}^2$                                                                          (IX)

Substitute (IX) in (V)

  $E^2=E_0^2+{\gamma }^2{v}^2{m}^2{c}^2$                                                                             (X)

Write the expression for relativistic momentum

  $p=\gamma mv$

Here, $p$ is the momentum.

Square the above equation

  $p^2={\gamma }^2{m}^2{v}^2$                                                                                         (XI)

Substitute (XI) in (X)

  $E^2=E_0^2+p^2{c}^2$

The above equation is the energy momentum relation.