Dp/dt in the relativistic domain

[Repetit]

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?

 

[jtbell]

It still works, so long as you take into account the time-dependence completely. Either use the product rule:

\frac{dp}{dt} = \frac {d}{dt} (m_r v) = \frac{dm_r}{dt} v + m_r \frac{dv}{dt}

or write it in terms of invariant mass:

\frac{dp}{dt} = \frac {d}{dt} \left( \frac {m_0 v} {\sqrt {1 - v^2 / c^2} } \right) = m_0 \frac {d}{dt} \left( \frac {v} {\sqrt {1 - v^2 / c^2} } \right)

 

[Meir Achuz]

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?

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 mean
dp/dt, just say dp/dt.