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

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?

• You mean "when", as opposed to "why"? Commented Dec 14, 2022 at 16:58
• Do you know about four-fermion interactions?
– J.G.
Commented Dec 14, 2022 at 17:15
• No, I mean "why". Commented Dec 14, 2022 at 17:21
• I'm not sure if I understand your question. Of course there are choices for $a$, $b$, $c$, and $d$ where $a+b\rightarrow c+d$ is a valid reaction. Are you asking for a set of conditions where this type of reaction is not possible? Commented Dec 14, 2022 at 18:29
• Sorry for the confusion, I've just edited my question. I've been told by my professor that due to energy balance the reaction can never happen but he didn't go any further. I was wondering if that reaction could occur in vaccuum, and in case it couldn't, if there was some kind of restriction like the decay one I've said. Thanks! Commented Dec 14, 2022 at 18:49

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.

• Thanks! It's all clear now. Commented Dec 15, 2022 at 13:31