Can the particle reaction $a+b \rightarrow c+d$ occur in vacuum?

Question

In particle physics, can the reaction $a+b \rightarrow c+d$ occur in vaccum from the point of view of energetic balance?

Both $a,b,c,d$ are generic and different particles.

It is true, for example, that in vacuum, if particle $a$, let's say, decays into $b,c$ and $d$, its rest mass must be bigger than the one of the other three particles, but in the case above, in case it couldn't occur, would the same reason apply?

Answer 1

If all the particle of the reaction have a non-zero mass in the point of view of energetic balance this reaction is always posible in vaccum.

We have two cases:

• $m_a$ + $m_b$ > $m_c$ + $m_d$. Here there isn´t any problem. The excess of energy is translate in a non-zero kinetic energy of c and d.

$E_a + E_b + K_{a+b} = E_c + E_d + K_{c+d}$

$K_{c+d} = E_{a+b}-E_{c+d} + K_{a+b}$ will never be negative.

• $m_a$ + $m_b$ < $m_c$ + $m_d$. Then we have the opposite but you can compensate the lack of energy if you give kinetic energy to a and b. This is how particle accelerators obtain heavy particles from protons for example.

$E_a + E_b + K_{a+b} = E_c + E_d + K_{c+d}$

If we make $K_{c+d} = 0$ (the less energetic case)

$K_{a+b} = E_{c+d} - E_{a+b}$ You can always find a Kinetic energy that make this expresion true.

You don´t have to worry about the momentum conservation since you can solve the problem in the mass center inertial system. So $p_a = -p_b$ and $p_c = -p_d$. That is why this don´t work with photons or other particles with no mass.