# Relation between kinetic energy and momentum in relativity

In a recent discussion a friend of mine claimed that kinetic energy ($$K$$) an momentum ($$p$$) in relativity can me expressed by

$$K=\frac{p^2}{(1+\gamma)m} \tag{1}$$

This equation if holds, has some cool significance for me, because it may show a smoother connection between classical and relativistic mechanics, as it is extremely easy to see that if $$v \rightarrow 0$$, $$K=\frac{p^2}{2m}$$ without the necessity for the expansion of the square root.

However, from what I know

$$K=E-mc^2=(\gamma-1)mc^2 \tag{2}$$ $$(pc)^2=E^2-(mc^2)^2 \tag{3}$$

I am having some hard time trying to deduce equation (1) from equation (2) and (3), so I believe that equation (1) may be incorrect. Therefore my question is:

Does equation (1) holds for a relativistic particle?

From $$(3)$$ we have
$$(pc)^2 = (E-mc^2)(E+mc^2) = K (E+mc^2)$$
$$K = \frac{p^2 c^2}{E+mc^2}$$
and if we use $$E = \gamma m c^2$$, the result follows.