Why equation $F = m a$ is not valid in relativistic physics but $F = dp / dt$ is valid?

Question

I read that the equation $$F = m a$$ is not valid in relativistic physics. Instead $$F = \frac{d }{dt} \frac{mv}{ \sqrt{1- \frac{v^2}{c^2} } }$$ is valid in relativistic physics but why is that?

Answer 1

I don't understand the question or the answers. The relativistic version of Newtons's second law indeed reads $$ F^\mu \equiv m a^\mu, $$ where $F$ and $a$ are four-vectors. The 3-space vector component of the force is $$ \vec F = m \frac{d^2 \vec x}{d\tau^2} $$ where $\tau$ is proper time.

Answer 2

Even before Einstein, $F=ma$ wasn't considered universally valid; it should be $F=\frac{d}{dt}(mv)$, which includes an extra term for time-dependent $m$, such as in rocket propulsion. If you take that equation as your starting point, the relativistic result is unsurprising because mass scales as $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. The $m$ in your final equation is actually the rest mass.

Answer 3

$$\sum F=\frac{dp}{dt}=\frac{d(mv)}{dt}=^*m\frac{dv}{dt}=ma$$

At the star $^*$ we assume constant mass, which cannot be assumed in general relativity.