The equation \(\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}\) is valid in the relativistic domain, provided that the time-dependence of mass is fully accounted for. The discussion emphasizes the use of the product rule for momentum, \(\frac{dp}{dt} = \frac{d}{dt} (m_r v) = \frac{dm_r}{dt} v + m_r \frac{dv}{dt}\), or expressing momentum in terms of invariant mass, \(\frac{dp}{dt} = \frac{d}{dt} \left( \frac{m_0 v}{\sqrt{1 - v^2 / c^2}} \right)\). It is noted that while \(\frac{dp}{dt}\) can define "force" in both special relativity (SR) and Newtonian physics, the term "force" is best avoided in the context of SR to prevent confusion.
PREREQUISITESPhysicists, students of physics, and anyone interested in the applications of Newtonian mechanics in the relativistic domain.
F=dp/dt can be taken as one definition of "force" in either SR or Newtonian physics. However there are other possible, equally reasonable, definitions of "force" in SR. It is best to stop using the word "force" in SR. If you meanRepetit said:Just a quick question which came to my mind when reading a historical article on Newtons second law. Is \overrightarrow{F}=\frac{d\overrightarrow{p}}{dt} correct in the relativistic domain because the mass is not necessarily constant?