Why can't n -> p + minus-pi-meson?

[nonequilibrium]

Hello,

In my book it says that
$$n \to p^+ + \pi^-$$
is not possible due to energy conservation issues: the mass of the rhs is larger than that of the lhs.

Now I was wondering, is it not possible to give the neutron e.g. some vibrational energy (like a drop of liquid continuously changing from a pancake to a dumbell-shape) which could be used to create the extra energy/mass required for the rhs? If not, what principle is stopping us? (Is there perhaps some conservation of "vibration" like there is a conservation of angular momentum?)

Thank you

EDIT: extra question: and if we put the meson on the other side: $$p^+ \to n + \pi^+$$, why is this not possible?

 

[Vanadium 50]

Let's start with the extra question, because until you solve that, there's no point in going further. What is the mass of the right hand side? What is the mass of the left hand side?

 

[nonequilibrium]

So for
$$p^+ \to n + \pi^+$$
we have (in MeV/c²):
LHS: 938.3
RHS: 939.6 + 139.6
so the mass of the LHS is smaller than that of the RHS, so the question why this decay is impossible comes down to original question: why can't I give the proton enough vibrational energy to pay for the extra mass on the RHS.

 

[Vanadium 50]

Insofar as we are considering the proton as elementary, if it vibrates, the energy is kinetic, and at the instant of decay there is a frame in which it is at rest. The problem then reduces to the one well already solved.

 

[nonequilibrium]

But the frame in which it is as rest at, is a non-inertial frame of reference; so we're not allowed to switch over to it.

Or what about a photon hitting the proton?

Thank you for your time btw.

EDIT: okay apparently the decay is possible when a photon hits the proton, but then it's a different situation, namely
$$p^+ + \gamma \to n + \pi^+$$ (it was an exercise in my book)

Weird thing is, another exercise was the question if
$$\Delta^{++} \to \Delta^+ + \pi^+^$$
is possible, and it said "yes", even though the mass of the particle on the LHS is already less than the first particle on the right side! Very confused now...

 

[Vanadium 50]

"at the instant of decay there is a frame in which it is at rest"

As far as the Delta decays, look at the width of the Deltas.

 

[nonequilibrium]

I'm a bit puzzled by your last post. How does that quote tie together with your Delta comment? I probably don't get the connection because I don't get your Delta comment itself, because I don't know what you mean with "the width of the Deltas"? (my knowledge of particle physics doesn't go beyond an introductory modern physics course)