A New Look at Energy — Question 3

Question 3

The first correction to classical kinetic energy is quartic in speed

For a massive particle, $K=(\gamma-1)mc^2$, with $\gamma=(1-\beta^2)^{-1/2}$ and $\beta=v/c$.

1. Expand K through order $\beta^4$.

2. At $v=0.60c$, compare the exact, classical and quartic-approximation energies, all in units of $mc^2$.

Question 3 - Solution

The first correction to classical kinetic energy is quartic in speed

(a) Method and calculation

$$\gamma=1+\tfrac12\beta^2+\tfrac38\beta^4+O(\beta^6).$$$$\boxed{K=\tfrac12mv^2+\tfrac38mv^4/c^2+O(mv^6/c^4)}.$$

(b) Interpretation and further calculation

$$\boxed{K_{\mathrm{exact}}/mc^2=.2500},\quad\boxed{K_{\mathrm{cl}}/mc^2=.1800},\quad\boxed{K_4/mc^2=.2286}.$$$$\frac{K_{\mathrm{exact}}-K_{\mathrm{cl}}}{K_{\mathrm{exact}}}=\boxed{.28},\qquad\frac{K_{\mathrm{exact}}-K_4}{K_{\mathrm{exact}}}=\boxed{.0856}.$$

Physical check

The quartic term improves the approximation, but beta 0.60 is large enough that higher terms are still visible.