Particle A (Energy E) hits particle B (at rest), producing particles C

_{1}, C

_{2}, …:

A + B → C

_{1}, C_{2},… C_{n}Calculate the threshold energy (i.e., minimum E) for this reaction in terms of the various particle masses.

Now one could try to use conservation of four-momentum: p

_{μ}

^{i}= p

_{μ}

^{f}. In that case the four-momenta would have to be calculated in the same frame. The problem is that the

*lab*frame is convenient for the left hand (initial) side of the reaction, while the center of momentum frame is convenient for the right hand side. That is because in the center of momentum frame, the threshold occurs when the particles C

_{1}, C

_{2},… C

_{n}are created with no kinetic energy. All their energy, in that case, is in their mass.

Here comes the utility of invariants. The four-momentum squared is also conserved, i.e., the same before and after, but has the added bonus that it is the same in all frames. Therefore we can calculate it in the lab frame for the left hand side, and in the center of momentum frame for the right hand side, and those two must be equal.

For the left hand side (where particle B is at rest)—and taking the speed of light to be unity,

p

_{μ}p

^{μ}= (E + M

_{B})

^{2}- P

_{A}

^{2}, where P

_{A}is the square of the three-momentum for the incoming particle A.

expanding that out, and using P

_{A}

^{2}= E

^{2}- M

_{A}

^{2}gives

p

_{μ}p^{μ}= 2 E M_{B}+ M_{A}^{2}+ M_{B}^{2}(1)Now we need to calculate the same invariant for the right hand side, which is trivial in the threshold case in the center of momentum where all the final particles are at rest.:

p

_{μ}p^{μ}= M^{2}(2)where M is just the total mass of the final particles, M = M

_{1}, M

_{2},… M

_{n}

Even though (1) and (2) were worked out in different frames, invariance means we can equate their answers, leading to the result:

E = (M

^{2}- M_{A}^{2}- M_{B}^{2}) / 2 M_{B}
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