Maria Huey

2021-12-26

In the theory of relativity, the mass of a particle with velocity v is
$m=\frac{{m}_{0}}{\sqrt{1}}-\frac{{v}^{2}}{{c}^{2}}$
where ${m}_{0}$ is the mass of the particle at rest and c is the speed of light. What happens as $v\to {c}^{t}$?

William Appel

Expert

$\underset{v\to {c}^{-}}{lim}\frac{{m}_{0}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}=\mathrm{\infty }$
Since v is close to c and $v
The direction of the inequality does not change because v and c are the speed of the velocity of the particle and the velocity of light, respectively. The magnitude of a vector is always a positive number.
$-\frac{{v}^{2}}{{c}^{2}}\succ 1$ (multiplying both sides by -1)
$1-\frac{{v}^{2}}{{c}^{2}}>0\to$ the denominator is a small and positive number.
The numerator ${m}_{0}$ is a positive number
The division is then a huge positive number.

Do you have a similar question?