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Why is $$\Sigma^0 \rightarrow \Lambda +\pi^0$$ not a possible process?

Charge and baryon number both are conserved. There's no issue with strangeness that I can tell. The masses in $\frac{Mev}{c^2}$ are approximately $$1193 \rightarrow 1116 + 135.$$ This gives me a red flag... It doesn't seem energetically favorable, but then again, couldn't the $\Sigma^0$ have a lot of kinetic energy prior to decay?

A different example, $e^- + e^+ \rightarrow \mu^-+ \mu^+$, clearly has less mass beforehand than after but is said to occur. I'm a bit confused here about how and when to apply energy conservation.

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  • $\begingroup$ Also, there is no information with regards to the particle velocities. $\endgroup$
    – zahbaz
    May 11 '15 at 8:17
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You should always apply energy conservation, and it ought to hold in all reference frames, including the frame in which the sigma is at rest. In the sigma's rest-frame, $$ E_{\text{initial}} = E_\Sigma = m_\Sigma $$ and $$ E_{\text{final}} = E_\Lambda + E_\pi \ge m_\Lambda + m_\pi $$ Thus we have that, $$ E_{\text{initial}} < E_{\text{final}} $$ The process is forbidden by energy conservation (it must be the case that $E_{\text{initial}} = E_{\text{final}}$).

In the case of $ee\to\mu\mu$, there is no frame in which both the electrons are at rest - in every reference frame, the initial state has more energy than the sum of the electron masses, $$ E_{\text{initial}} > 2m_e. $$ Thus, $ee\to\mu\mu$ may conserve energy, if the pair of electrons has sufficient energy.

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Energy conservation always applies.

Your mistake is in thinking that adding masses will solve the problem instead of clarifying some aspects.

In the case of the sigma-zero the decay at rest allows to see that the sum of the constituent masses is larger than the mass of the sigma-zero. For a decay to happen there should be energy left over to go to the kinetic energy of the decay products, thus, correctly it is a no go from energy conservation process.

In the case of the e+e- one is talking of scattering, i.e. the electron and the positron have kinetic energy and are defined by a four vector each. The addition of the two four vectors gives a large invariant mass, as happened in the colliders at SLAC and LEP. This large invariant mass leaves a lot of energy to appear as kinetic energy in the interaction products.

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  • $\begingroup$ I can see why e+e- with considerable kinetic energy can decay, but why isn't that the case for the sigma as well? Why is the sigma at rest but the e+e- not? $\endgroup$
    – zahbaz
    May 11 '15 at 8:15
  • $\begingroup$ Wait. It's the fact that e+e- is an interaction of two particles that changes things, isn't it? Given enough KE, there could be enough energy in the interaction to produce the muons. In the sigma-zero's case, though, its velocity doesn't really come into the picture, as it doesn't interact with anything. Instead, the question is whether or not it spontaneously decays... and this does not depends on its velocity. The decay happens in the rest frame. Is this correct? $\endgroup$
    – zahbaz
    May 11 '15 at 8:31
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    $\begingroup$ Because of special relativity you know that the products of your decay/reaction/scattering doesn't depend on the reference frame you choose. For a sigma, there is a simple frame, the rest frame. For colliding electrons there is no such frame where both electrons would be at rest. You could put either one of them at rest, but in practice we work in the center of momentum frame, ie the frame in which the momenta cancels out. $\endgroup$ May 11 '15 at 8:37
  • $\begingroup$ as @MichaëlUghetto said . The simplest frame is the rest frame for decays and for any total four vector sum of particles. A single particle in its rest frame just has it mass. More than one will add up to a four vector that will have an invariant mass, larger than the sum of the masses of the constituents, according to their energy and momentum. $\endgroup$
    – anna v
    May 11 '15 at 9:03

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