From the gamma factor to the velocity (proton)

Problem

From the $\gamma$ factor to the velocity. A proton in an accelerator has $\gamma=1000$. Find its velocity starting from $\beta^2 = 1 - 1/\gamma^2$. How many decimal places separate $v$ from $c$?

Solution

Starting point. From the definition $\gamma = 1/\sqrt{1-\beta^2}$ we get

$\beta^2 = 1 - \frac{1}{\gamma^2}.$

Substitution. With $\gamma = 1000$, $\gamma^2 = 10^6$:

$\beta^2 = 1 - \frac{1}{10^6} = 1 - 10^{-6} = 0{,}999999.$

Square root. Using the expansion $\sqrt{1-x}\approx 1-\tfrac{x}{2}$ for $x = 10^{-6}$:

$\beta = \sqrt{1-10^{-6}} \approx 1 - \frac{10^{-6}}{2} = 1 - 5\cdot 10^{-7} = 0{,}9999995.$

Result.

$\ev{v \approx 0{,}9999995\,c}$

How many decimal places from $c$. The difference is

$1 - \beta \approx 5\cdot 10^{-7},$

that is, about half a part per million: the velocity coincides with $c$ down to the sixth-seventh decimal digit. Even with $\gamma = 1000$ the proton does not reach $c$: there is always a little bit missing.

Links

Topics: Special relativity Concepts: Time dilation Skills: Symbolic setup