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peguerosdc

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- 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

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:

Thanks!

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

Thanks!