# Error of relativistic kinetic energy

I have recently begun working on the special relativity theory. I have then made the taylor series for the gamma factor to show that we get the classic formula for kinetic energy:

$$E _ { k i n } = m c ^ { 2 } \left( 1 + \frac { 1 } { 2 } x ^ { 2 } + \frac { 3 } { 8 } x ^ { 4 } - 1 \right) = m c ^ { 2 } \left( 1 + \frac { 1 } { 2 } \frac { v ^ { 2 } } { c ^ { 2 } } + \frac { 3 } { 8 } \frac { v ^ { 4 } } { c ^ { 4 } } - 1 \right) = \frac { 1 } { 2 } m v ^ { 2 } + m c ^ { 2 } \cdot \left( \frac { 3 } { 8 } \frac { v ^ { 4 } } { c ^ { 4 } } \right)$$

Question: Now I have been trying to figure out how some people claim, that an error at 1% of energy appears at 11% lightspeed.

I can, however, not reach that conclusion. I get 14% instead:

$$\frac { 1 } { 2 } \frac { v ^ { 2 } } { c ^ { 2 } } = \frac { 1 } { 100 } \Rightarrow v ^ { 2 } = \frac { 2 c ^ { 2 } } { 100 } \Rightarrow | v | = \sqrt { \frac { 2 c ^ { 2 } } { 100 } } = 0,14 c$$

Can any of you help me?

## 2 Answers

The classical kinetic energy is $$\frac{1}{2}mv^2$$. Dividing the relativistic kinetic energy by this yields

\begin{align} \frac{(\gamma - 1)mc^2}{\frac{1}{2}mv^2} &\approx \frac{\frac{1}{2}mv^2 + \frac{3}{8}mv^4/c^2}{\frac{1}{2}mv^2} \\ &= 1 + \frac{3}{4}\frac{v^2}{c^2} \end{align}

Thus, the relativistic and classical kinetic energies differ by 1% when $$\frac{3}{4}\frac{v^2}{c^2} \approx 0.01$$, i.e. when $$v \approx 0.115c$$.

Your error was using the first term of the Taylor series (the $$\frac{1}{2}\frac{v^2}{c^2}$$), which is actually just the classical kinetic energy, rather than the second one, which is the first-order relativistic correction.

You actually want to make the second term have 1% the value of the first: $$\frac{3}{8}mc^2 \beta^4 = 0.01 \left ( \frac 1 2 mc^2 \beta^2 \right )$$ where I'm using $$\beta = v/c$$. Solving for $$\beta$$ $$\beta^2 = \frac{4}{300} \quad\Rightarrow\quad \beta =0.115$$ which is the 11% you were looking for.