# Is the rate of the reaction $a+b\rightarrow c\rightarrow e+f$ the same as $e+f\rightarrow c\rightarrow a+b$

In particle physics, is the rate of the reaction $$a+b\rightarrow c\rightarrow e+f$$ the same as the one of $$e+f\rightarrow c\rightarrow a+b$$ ?

If so, then what does prevent that the final states goes back to the initial state, so that we never observe the final state ?

Kinematics, mostly.

More specifically, the final-state particles generally are moving away from each other, and so are unlikely to collide, whereas the initial-state particles generally are moving toward each other, so they are likely to collide.

Note that there are, in fact, situations in which you do see the final-state particles return to the initial state. One great example is positronium production and annihilation. A photon of energy greater than $$2m_e$$ can produce an electron-positron pair, and another photon is required to interact with the electron or position to obey energy and momentum conservation. So the reaction here is $$\gamma+\gamma\to e^++e^-$$. If the photon's energy is only slightly greater than $$2m_e$$, the electron-positron pair will become bound in a positronium atom. Eventually, this positronium atom annihilates, turning back into two photons (sometimes three, but for our purposes we only care about the two-photon decay), so we have $$\gamma+\gamma\to e^++e^-\to\gamma+\gamma$$. If you wait more than a few microseconds after observing the initial state, then you'll never see the two-electron final state. In this instance, the kinematics are compatible with a return to the initial state, since the "final-state" electrons don't move away from each other, but instead remain bound near each other.

• would you say that the plasma phase is obtained once reaction kinematics can have equal probabilities? – anna v Aug 6 at 5:33

Yes, these reactions occur at the same probability, given the same kinematics.

But even with interactions occurring at the same rate, we still expect to see some of the final state. If the interactions $$\rm A\to B$$ and $$\rm B\to A$$ occur with the same probability, but you start out with 100 of particle A and 0 of particle B, then obviously the interaction $$\rm B\to A$$ cannot occur at all. The interaction $$\rm B\to A$$ will be more rare than $$\rm A\to B$$ until a balance is reached between the populations of A and B, 50/50 in this case. This is the principle of detailed balance.

Additionally, there are kinematic concerns, but @probably_someone's answer covers those pretty well.