According to the theory of relativity, the energy, E, of a body of mass m is given as a function of its speed, v, by $E=m c^{2}\left(\frac{1}{\sqrt{1-v^{2} / c^{2}}}-1\right)$ , where c is a constant, the speed of light.Assuming v

Question

According to the theory of relativity, the energy, E, of a body of mass m is given as a function of its speed, v, by $E=m c^{2}\left(\frac{1}{\sqrt{1-v^{2} / c^{2}}}-1\right)$ , where c is a constant, the speed of light.Assuming v < c, expand E as a series in v/c, as far as the second non-zero term.
A) $m c^{2}\left[1+\frac{1}{2} \frac{v^{2}}{c^{2}}+\cdots\right]$
B) $m c^{2}\left[\frac{1}{2} \frac{v}{c}+\frac{3}{4} \frac{v^{2}}{c^{2}}+\cdots\right]$
C) $m c^{2}\left[1+\frac{1}{2} \frac{v}{c}+\cdots\right]$
D) $m c^{2}\left[\frac{1}{2} \frac{v^{2}}{c^{2}}+\frac{3}{8} \frac{v^{4}}{c^{4}}+\cdots\right]$

Quizplus · Accepted Answer

The correct expansion of $E$ in terms of $v/c$ involves a binomial expansion of the term $\frac{1}{\sqrt{1-v^{2} / c^{2}}}$. The first non-zero term after the constant term is proportional to $v^2/c^2$, and the second non-zero term is proportional to $v^4/c^4$, which matches option D.

According to the Theory of Relativity, the Energy, E, of a Body