Chapter 28, Problem 6TP

To determine

The expression for the Relativistic energy of a particle in terms of momentum.

Answer to Problem 6TP

Option $\left(\text{a}\right)$

  $E=\sqrt{\left(p^2{c}^2+m_0^2{c}^4\right)}$.

Explanation of Solution

Given:

The Relativistic energy is $E$,

Themomentum is $p$,

The rest mass is $m_0$

The speed of light is $c$.

Formula used:

The Relativistic energy is given by

  $E=m{c}^2$   ....... (I)

The Relativistic mass is given by

  $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$   ....... (II)

The Momentum is given by

  $p=mv$   ....... (III)

Substituting the value of m in equation (III)

  $p=\frac{m_{0}v}{\sqrt{1-\frac{v^2}{c^2}}}$   ........ (IV)

Calculation:

Substitute the value of $m$ in equation (I)

  $E=\frac{m_0{c}^2}{\sqrt{1-\frac{v^2}{c^2}}}$   ........ (V)

Square both sides of the equation (V)

  $\begin{array}{c}{E}^2=\frac{m_0^2{c}^4}{\left(1-\frac{v^2}{c^2}\right)}\\ =m_0^2{c}^2\left(\frac{c^2}{1-\frac{v^2}{c^2}}\right)\\ =m_0^2{c}^2\left(\frac{v^2+\left(c^2-v^2\right)}{1-\frac{v^2}{c^2}}\right)\\ =m_0^2{c}^2\left(\frac{v^2}{1-\frac{v^2}{c^2}}+\frac{c^2-v^2}{1-\frac{v^2}{c^2}}\right)\end{array}$

Solve further

  $\begin{array}{c}{E}^2=m_0^2{c}^2\left(\frac{v^2}{1-\frac{v^2}{c^2}}+c^2\frac{1-\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)\\ =m_0^2{c}^2\left(\frac{v^2}{1-\frac{v^2}{c^2}}+c^2\right)\\ =\frac{m_0^2{v}^2{c}^2}{1-\frac{v^2}{c^2}}+m_o^2{c}^4\\ =\left(\frac{m_0^2{v}^2}{1-\frac{v^2}{c^2}}\right)c^2+m_0^2{c}^4\end{array}$   ....... (VI)

  $$ Square both sides of the equation $\left(\text{VI}\right)$

  $p^2=\frac{m_0^2{v}^2}{1-\frac{v^2}{c^2}}$   ........ (VII)

By equation (VI) and (VII),

  $\begin{array}{c}{E}^2=\left(\frac{m_0^2{v}^2}{1-\frac{v^2}{c^2}}\right)c^2+m_0^2{c}^4\\ E^2=p^2{c}^2+m_0^2{c}^4\\ E=\sqrt{\left(p^2{c}^2+m_0^2{c}^4\right)}\end{array}$

Conclusion:

The Relativistic energy of a particle in terms of momentum is

  $E=\sqrt{\left(p^2{c}^2+m_0^2{c}^4\right)}$.

Thus, the correct option is $\left(\text{a}\right).$