# Understanding crossing symmetry: inverse beta decay

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• peguerosdc
peguerosdc
TL;DR Summary
When applying cross symmetry to ##\nu + n \rightarrow e^- + p##, why in the result ##\bar{\nu} + p \rightarrow e^+ + n## the arrow is pointing to the right?
Hi! This is a very very noob question, but I am starting to get into particle physics and I don't understand the application of crossing symmetry in the inverse beta decay.

Crossing symmetry says (from Griffiths) that, in a reaction "any of these particles can be 'crossed' over to the other side of the equation, provided it is turned into its antiparticle, and the resulting interaction will also be allowed". And the Compton scattering is mentioned as an example.

Then, why another common example is the inverse beta decay used to detect the neutrino? The beta decay reaction is:

$$\nu + n \rightarrow e^- + p$$

And the inverse beta decay reaction is:

$$\bar{\nu} + p \rightarrow e^+ + n$$

But if you just do crossing symmetry to the beta decay equation, I understand that you should get the reverse reaction (with the arrow pointing to the other side):

$$e^+ + n \rightarrow \bar{\nu} + p$$

That is, I am:
• crossing ##e^-## from the right to the left as ##e^+##
• crossing ##\nu## from the left to the right as ##\bar{\nu}##
• leaving ##n## where it is
• leaving ##p## where it is
So, why is the arrow not pointing to the left in the correct inverse beta decay reaction?

Thanks!

Mentor
If ##a \to b## is a possible reaction then ##b \to a## is a possible reaction, too (assuming you have enough energy). They are closely linked. Similarly, ##\overline a \to \overline b## is a possible reaction, which you can convert to ##b \to a##. Here a and b can be any set of particles.

peguerosdc
peguerosdc
If ##a \to b## is a possible reaction then ##b \to a## is a possible reaction, too (assuming you have enough energy).

@mfb thanks for your reply! From what I read in Griffiths', due to energy conservation if ##m_a > m_b##, (where ##m## stands for mass) then I would need to supply enough energy to make it up in the reaction ##b \rightarrow a##. Is this what you mean?

And, starting from a given reaction (assuming it is possible i.e. it complies with conservation laws), is this the only requirement to see it going in the opposite direction?

Mentor
If the sum of masses is larger in a then the reverse reaction will need to have the particles colliding with some kinetic energy, yes. If it's smaller then you need that kinetic energy for a->b.

Conservation laws always say "x stays the same". That's true for both directions, which means both directions always satisfy the conservation laws (which means they are possible). The only catch is the energy. In e.g. ##e^+ e^- \to \gamma \gamma## the photons will always have at least 511 keV in the center of mass frame, but you don't write it down explicitly. In ##\gamma \gamma \to e^+ e^-## the photons need at least 511 keV in the center of mass frame to make the reaction possible. If you try to collide two visible light photons you are not reversing the reaction. You are starting with a condition that cannot be the final state of the reverse direction.

vanhees71 and peguerosdc
peguerosdc
Got it! Your example helped me a lot. Thank you!