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Paraphrasing an extract of the Feynman lectures on special relativity:

Newton's second law can be expressed by the equation: \begin{equation} F=\frac{ \mathrm{d} (mv)}{ \mathrm{d} t} \end{equation} It was stated with the assumption that $m$ is a constant.

However it turns out that the mass of a body increases with velocity. The correct formula for mass can be given by: \begin{equation} m=\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}} \end{equation}

Where the "rest mass" $m_0$ represents the mass of a body that is not moving and $c$ is the speed of light.

Therefore one can change Newton's laws by introducing this correction factor to the mass.

My question: what would be this new expression for $F$ in Newton's second law?

Is the $v$ constant in our formula for $m$ or is it the same $v$ present in the first equation?

I assumed that it was the same $v$, and I found that:

\begin{equation} F=m_0 a\frac{c^3}{\left(c-v)(c+v)\right)^{\frac{3}{2}}} \end{equation}

when I consider $v\ll c$ then I recover $F=m_0 a$. But how can I be sure I have the correct formula?

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It is the same $v$ in your two equations, so it is not constant. However I would like to answer more fully by using the standard notation and concepts in modern presentations of relativity.

The momentum of a body of rest mass $m_0$ is related to its rest mass and its velocity by the formula $$ {\bf p} = \frac{ m_0 {\bf v} }{\sqrt{1 - v^2/c^2}} $$ It is convenient to write this $$ {\bf p} = \gamma m_0 {\bf v} $$ where $$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}. $$ The relationship between force and momentum is the same as in Newtonian physics, namely $$ {\bf f} = \frac{d {\bf p}}{dt}. $$ Using the expression for momentum, the result in terms of rest mass and velocity is $$ {\bf f} = \gamma m_0 \frac{d {\bf v}}{dt} + \frac{d\gamma}{dt} m_0 {\bf v} $$ where I have assumed the rest mass is constant. (This is correct for things like an electric field accelerating an electron, but more generally for a composite system the rest mass might change). Thus the relation between force and change of momentum is simple, but the relation between force and acceleration is not. To calculate $d\gamma/dt$ you can use $$ \frac{d \gamma}{dt} = \frac{d\gamma}{dv} \frac{dv}{dt} $$ where (careful) $v$ is speed not velocity, so $$ \frac{dv}{dt} = \frac{d}{dt} ({\bf v} \cdot {\bf v})^{1/2} = \frac{{\bf v} \cdot {\bf a}}{v} $$

That answers the question, but I will finish by pointing out that you don't have to introduce the notion of a changing mass. It is now thought better to use the term "mass" to mean the rest mass $m_0$. That way, each particle has just one mass which never changes, because we mean the rest mass. Then when you want to consider momentum and energy you use ${\bf p} = \gamma m_0 {\bf v}$ and $E = \gamma m_0 c^2$. This way the stuff which depends on $v$ is contained in $\gamma$ and the stuff which does not depend on $v$ is contained in $m_0$. Since we are always using $m_0$ (the rest mass), we can then do without the subscript zero.

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The relativistic version of the second law is $$F=\frac{\mathrm d(mv)}{\mathrm dτ}$$ where $F$ and $v$ are four-vectors (the four-force and four-velocity), $m$ is rest mass (which needn't be constant), and $τ$ is proper time.

If you try to define a relativistic three-force as a function of three-velocity, you may get something like what you wrote down—though I believe it depends on whether the force is parallel or perpendicular to the velocity. But there's no point in doing it. Just learn four-vectors; it will make your life much easier. It's unclear to me why, 110+ years later, students are still taught the complicated math from 1905 and not the much more elegant approach introduced by Minkowski in 1908.

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  • $\begingroup$ When comparing with experiments, eventually you need to know not just how different 4-vectors relate to one another, but also the actual components, relative to some laboratory apparatus, of a given 4-vector. So then you need to know the 3-vector as well. For some calculations the mathematical task is easier with 4-vectors and proper time. For others it is easier with 3-vectors and laboratory time. $\endgroup$ Oct 13 '20 at 22:49

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