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# Relativistic momentum ## Classical Mechanics $\vec{p} = m\vec{v}$ ## Relativistic Mechanics ### Thought experiment Imagine two spaceships moving at relativistic speeds. |...

# Relativistic momentum ## Classical Mechanics $\vec{p} = m\vec{v}$ ## Relativistic Mechanics ### Thought experiment Imagine two spaceships moving at relativistic speeds. | | | :-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | | Mary is in spaceship $S'$, which moves to the right relative to Stan in spaceship $S$ at velocity $v$. Both spaceships are identical. Mary is at the front of her spaceship and Stan is at the front of his, such that at $t=t'=0$, $x=x'=0$. Mary launches a ball with velocity $u'$ up in her frame, and Stan launches an identical ball with velocity -u down in his frame. | ### Classically Velocity Transformation $u_y = u_y' + v_y = u_y' + 0 = u_y'$ Therefore, the y-component of the velocity of the ball, as measured by Stan, would be the same as the y-component of the ball as measured by Mary. Since the masses are the same, the y-component of the momentum would be the same. $p_y = p_y'$ ### Relativistically Velocity Transformation $u_y = \frac{u_y'}{\gamma(1-\frac{vu_x'}{c^2})} = \frac{u_y'}{\gamma}$ Therefore, the y-component of the velocity of the ball, as measured by Stan, would be less than the y-component of the ball as measured by Mary. Since the masses are the same, the y-component of the momentum would be less. $p_y < p_y'$ ### But If momentum is conserved in Stan's frame it cannot be conserved in Mary's frame (or any other frame) **Therefore, classical momentum must be incorrect!** ### Relativistic Momentum $\vec{p} = \gamma m \vec{v}$ where $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ ### Proof $\vec{p} = \frac{m \vec{u}}{\sqrt{1 - \frac{u^2}{c^2}}}$ $p_x = \frac{mu_x}{\sqrt{1-\frac{u^2}{c^2}}} = \frac{m(u_x' + v)}{\sqrt{1-\frac{u^2}{c^2}}}$ $p_y = \frac{mu_y}{\sqrt{1-\frac{u^2}{c^2}}} = \frac{mu_y'}{\sqrt{1-\frac{u^2}{c^2}}}$ $u^2 = u_x^2 + u_y^2 = (\frac{u_x'+v}{1+\frac{vu_x'}{c^2}})^2 + (\frac{u_y'}{\gamma(1+\frac{vu_x'}{c^2})})^2$ $p_y = \frac{mu_y'}{\sqrt{1-\frac{u^2}{c^2}}} = \gamma(1 + \frac{vu_x'}{c^2}) \cdot \frac{mu_y'}{\sqrt{1 - \frac{u'^2}{c^2}}}$ $K = \gamma(1 + \frac{vu_x'}{c^2})$ $p_y = K \cdot p_y'$ $p_x = K \cdot p_x'$ Therefore, relativistic momentum is conserved in all frames! ### Note As $v \to c$, $\gamma \to \infty$ Therefore, it would take an infinite amount of energy to accelerate an object with mass to the speed of light.