# Relativistic version of Newton's 2nd law

Paraphrasing an extract of the Feynman lectures on special relativity:

Newton's second law can be expressed by the equation: $$$$F=\frac{ \mathrm{d} (mv)}{ \mathrm{d} t}$$$$ It was stated with the assumption that $$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: $$$$m=\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}}$$$$

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:

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

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

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.

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.

• 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. Oct 13, 2020 at 22:49

i don't know if my solution is correct but i guess this preety much looks like the solution

• Welcome to Physics! Please use MathJaX when posting equations instead of taking pictures of your notes. You can find a tutorial on how to do so on our sister site. Jan 3 at 13:18
• Ok kind sir,thank you, I wasn't aware of that Jan 4 at 14:31