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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?

Qmechanic
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physicsrev
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    Here is a related thread https://physics.stackexchange.com/questions/597575/how-to-prove-that-kinetic-energy-equals-k-fracp2-gamma1m – robphy Nov 16 '21 at 19:11

1 Answers1

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From $(3)$ we have

$$(pc)^2 = (E-mc^2)(E+mc^2) = K (E+mc^2)$$

so that

$$K = \frac{p^2 c^2}{E+mc^2}$$

and if we use $E = \gamma m c^2$, the result follows.

Javier
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