# KE at speeds close to c

julianwitkowski
I'm curious as to the reasons why KE = ½ m ⋅ v2 only works at speeds much less than c?

Also, how does the equation change?

Thank you!

Gold Member
julianwitkowski
I prefer not to use relativistic mass at all and in fact its not needed. Physicists don't use it too. Its just that in both Newtonian and Relativistic mechanics, linear momentum and kinetic energy depend on the reference frame, but with different forms. In relativity we have $KE=(\gamma-1)mc^2$ and $\vec p=\gamma m \vec v$. By m, I mean rest mass and this is the only concept of mass I use.