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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?

Qmechanic
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Hury H
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3 Answers3

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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.

innisfree
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  • Is $F^\mu \equiv m a^\mu$ true if $m$ varies with $\tau$? – md2perpe Nov 14 '18 at 09:54
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    @md2perpe the $m$ in this case is the rest mass. Any professional relativist will only ever use the rest mass. The relativistic mass is a concept now abandoned to the dustbin and subpar books on relativity. – John Rennie Nov 14 '18 at 10:01
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    I know. But what if the rest mass for some reason would vary with $\tau$? Wouldn't the formula then be $F^\mu \equiv \frac{d}{d\tau}(m u^\mu),$ where $u$ is the four-velocity? – md2perpe Nov 14 '18 at 10:12
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    @md2perpe, if rest mass varies with time, it can be because the body is ejecting smaller particles in various directions. These smaller particles will exert forces on the original body but direction of these forces depends on the details of motion of those smaller particles. Thus there is no unique equation of motion that could capture all cases of variable mass. For simple case like isotropic outflow of particles (so there is no net force due to them), an equation of motion could be derived. – Ján Lalinský Nov 14 '18 at 13:08
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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.

J.G.
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    This is incorrect, systems with time-dependent mass do not obey single general equation like that, because the motion and value of force depends on the manner in which the mass is lost. See my answer here https://physics.stackexchange.com/a/142773/31895 – Ján Lalinský Nov 14 '18 at 12:35
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$$\sum F=\frac{dp}{dt}=\frac{d(mv)}{dt}=^*m\frac{dv}{dt}=ma$$

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

Steeven
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    Constant mass cannot even be assumed in special relativity, where the relativistic mass, which is what combines with velocity to make momentum, depends on the speed. – md2perpe Nov 14 '18 at 09:51