Question

Define and plot the percentage error in the kinetic energy computed from the classic mechanics expression a compared to the full relativistic formula as a function of the fraction of the speed of light (v/c).

Answer 1

Relativistic:

$E_{kr}=mc^{2}-m_{0}c^{2}=m_{0}c^{2}\left ( \frac{1}{\sqrt{1-\left ( \frac{v}{c} \right )^{2}}}-1 \right )$

Classic: $E_{kc}=\frac{mv^{2}}{2}$

Consider E_kr=100%, and the difference between E_k=requested percent.

$\frac{E_{kr}-E_{kc}}{E_{kr}}=\frac{c^{2}\left ( \frac{1}{\sqrt{1-\left ( \frac{v}{c} \right )^{2}}} \right )-\frac{v^{2}}{2}}{c^{2}\left ( \frac{1}{\sqrt{1-\left ( \frac{v}{c} \right )^{2}}} \right )}$

Not very sure if this is the percent asked in the text (such a parameter may be calculated in many ways).

I choose to work for E_kc/E_kr, which looks like:

$\frac{E_{kc}}{E_{kr}}=\frac{x^{2}}{2\left ( \frac{1}{\sqrt{1-x^{2}}}-1 \right )}$

x=v/c

The plot of E_kc/E_kr=fc(x) looks like (I might take too many points, but just to be sure the graph looks as it should):

0.4 0.6 0.8

I think the plot of $\frac{\Delta E}{E_{kr}}=fc\left ( \frac{v}{c} \right )$ looks similar.