# Special relativity - Center of mass frames vs Laboratory frames

In the reaction where particle A hits particle B, which is at rest. This produces particles C and D. For the minimum energy of particle A for the reaction to take place I have two calculations. I don't know if I'm confusing the center of mass frame with the laboratory frame, but first let me present the two approaches:

$$Approach~1$$

By squaring the initial and final 4-momentum vectors we have:

$$P_i^2 = (P_A+P_B)^2 = \left(\frac{E_A+m_Bc^2}{c},~\vec{p_A}\right)^2 = \left(\frac{E_A+m_Bc^2}{c}\right)^2~-~\vec{p_A^2}$$

Combining this with $$\vec{p_A^2}=\frac{E_A}{c}-m_A^2c^2$$ gives $$(P_A+P_B)^2 = 2E_Am_B+m_A^2c^2+m_B^2c^2$$

For the minimum energy of particle A, particles C and D have only rest mass energies and no 3-momentum. Thus the final 4-momentum vector squared is$$P_f^2 = (P_C+P_D)^2 = (m_Cc^2+m_Dc^2)^2$$

Equating the initial and final 4-momentum gives the minimum energy of particle A:$$E_{min} = \frac{[(m_C+m_D)^2-m_B^2-m_A^2]c^2}{2m_B}$$

This is the correct minimum energy but the following approach yields something close to it

$$\\$$$$Approach 2$$

If we just use plain conservation of energy $$E_i = E_f$$ and square both sides: $$(E_{min}+m_Bc^2)^2=(m_Cc^2+m_Dc^2)^2$$$$E_{min}^2+2E_{min}m_Bc^2=(m_Cc^2+m_Dc^2)^2-m_B^2c^4$$

If we use $$E_{min}^2=m_A^2c^4+p^2c^2$$, then:$$p^2+2E_{min}m_B=[(m_C+m_D)^2-m_B^2-m_A^2]c^2$$

If $$p=0~$$ we get the same result as in the first approach. But this should be the momentum of particle A which was not zero. However could we argue that the second approach operates in the center of mass frame. From what I understand, A and B have a combined total 3-momentum of zero in their center of mass frame and therefore you could set $$p=0$$ in the second approach.

Setting the 3-momentum to zero doesn't feel right, but why is the result of the 2nd approach so close to the 1st? Is it correct to assume the 1st approach operates in the laboratory frame but the 2nd approach in the mass-center frame?

(What about particles C and D? Since they have no 3-momentum, their 4-momentum is the same in the laboratory and mass-center frames?)

If we notice that the energy conforms to the Lorentz transformation, the second approach is also feasible. Writing the Lorentz transformation of 4-momentum vector from one frame to another with relative velocity $$v$$ in the $$x$$-direction, we have:

$$E=\frac{E'+vp'_x}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$p'_x=\frac{p_x-\frac{vE}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$$

In this problem,

$$E=E_m+m_Bc^2$$

$$E'=m_Cc^2+m_Dc^2$$

$$E_m^2=p_x^2c^2+m_A^2c^4$$

We take $$v$$ to be the velocity of CM frame, that is to say

$$p'_x=\frac{p_x-\frac{v(E_m+m_Bc^2)}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}=0$$

Which yields $$v=\frac{p_xc^2}{E_m+m_Bc^2}$$. Solving these equations gives

$$E_m+m_Bc^2=\frac{m_Cc^2+m_Dc^2}{\sqrt{1-\frac{p_x^2c^2}{(E_m+m_Bc^2)^2}}}=\frac{m_Cc^2+m_Dc^2}{\sqrt{1-\frac{E_m^2-m_A^2c^4}{(E_m+m_Bc^2)^2}}}$$

$$E_m=\frac{(m_C+m_D)^2-m_A^2-m_B^2}{2m_B}c^2$$

This result is exactly the answer.