# Centre-of-mass energy of a head-on proton collision in the SI units

I am confused about the relation between the natural and SI units in the special relativity. For example, I have found that I am not able to solve a classical problem of finding the center-of-mass energy of a head on collision of two protons in the SI units. I defined the four vector of proton 1 and 2 to be: $$\vec{P_1} = (E_p/c, \vec{p})$$ and $$\vec{P_2} = (E_p/c, \vec{-p})$$

Then I defined the square of the center-of-mass energy to be:

$$s = (\vec{P_1} + \vec{P_2})^2 = |\vec{P_1}|^2 + |\vec{P_2}|^2 + 2 \vec{P_1} \vec{P_2} = (E_p^2/c^2 - |\vec{p}|^2) + (E_p^2/c^2 - |\vec{p}|^2) + 2 (E_p^2/c^2 - \vec{p}\vec{p} \cos(90 ^{\circ}))$$

$$s = 4E_p^2/c^2$$

$$\sqrt{s} = 2E_p/c$$ in the SI units ($$\sqrt{s} = 2 E_p$$ in the natural units)

However, the equation derived in the SI units seems to be wrong. For example, assuming $$E_p$$ to be 1 GeV = $$10^9 \times 1.602 \times 10^{-19} = 1.602 \times 10^{-10} J$$. Then $$\sqrt{s} = 2E_p/c = \frac{2 \times 1.602 \times 10^{-10}}{3 \times 10^8} = 1.068 \times 10^{-18}$$ J $$= 6.7 eV$$ which is obviously wrong.

$$s$$ is energy squared, so your definition of $$s$$ should have $$c^2P^2$$.