Question

Show that${\mathit{E}}^{\mathbf{2}}\mathbf{=}{\mathit{p}}^{\mathbf{2}}{\mathit{c}}^{\mathbf{2}}\mathbf{+}{\mathit{m}}^{\mathbf{2}}{\mathit{c}}^{\mathbf{4}}$follows from expressions (2-22) and (2-24) for momentum and energy in terms of m and u.

Short answer

The relation of relativistic energy in terms of mass and momentum is derived bysquaring the relativistic energy relation in terms of mass and velocity andexpanding it using the binomial identity.

Step-by-step solution

Square the relativistic energy relation and expand the expression.

The relativistic energy and momentum expressions are given below,

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

Squaring the energy expression and solving further,

$\begin{array}{c}{E}^2=\frac{m^2{c}^4}{(1-\raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.)}\\ =m^2{c}^4{(1-\frac{u^2}{c^2})}^{-1}\end{array}$

Using the binomial expression:${(1-x)}^{-1}=1+x+x^2+x^3+\mathrm{...}$

$\begin{array}{c}{E}^2=m^2{c}^4[1+\frac{u^2}{c^2}+\frac{u^4}{c^4}+\mathrm{...}]\\ =m^2{c}^4[1+\frac{u^2}{c^2}(1+\frac{u^2}{c^2}+\frac{u^4}{c^4}+\mathrm{...})]\\ =m^2{c}^4[1+\frac{u^2}{c^2}{(1-\frac{u^2}{c^2})}^{-1}]\\ =m^2{c}^4+\frac{m^2{u}^2{c}^2}{(1-\frac{u^2}{c^2})}\end{array}$ … (1)

Express the above equation in terms of mass and momentum

The relativistic momentum expression is given below,

$\begin{array}{l}p={\gamma }_{u}mu\\ =\frac{mu}{\sqrt{1-\frac{u^2}{c^2}}}\end{array}$

Therefore, the equation (1) becomes,

$E^2=m^2{c}^4+{\left(\frac{mu}{\sqrt{1-\frac{u^2}{c^2}}}\right)}^2{c}^2$

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