## Conceptual Physics (12th Edition)

Unlike in the classical formula, p = mv, the increase in a particle's relativistic momentum, $p = \gamma mv$, does not depend in a linear fashion upon its speed.
The relativistic formula for momentum, $p = \gamma mv$, is greater than the classical formula $p = mv$. This shows that the relativistic momentum increases with $\gamma$, not linearly with speed. Near v = c, the momentum increases at a faster rate (percentage-wise) than speed so a 5 percent increase in momentum for a 1 percent increase in speed is quite possible. Here is the numerical proof. Let v = 0.907c. Find $\gamma$ at v = 0.907c. It is $\frac{1}{\sqrt{1-v^{2}/c^{2}}}$, or $\frac{1}{\sqrt{1-0.907^{2}}}$, which is 2.37. Now let the speed v increase by 1 percent. v = 0.916c Find $\gamma$ at v = 0.916c. It is $\frac{1}{\sqrt{1-v^{2}/c^{2}}}$, or $\frac{1}{\sqrt{1-0.916^{2}}}$, which is 2.49. With a 1 percent increase in speed, the factor $\gamma$ has increased by 5 percent, and so has the momentum.