Max Energy Λ in Σ0 Decay: Explained & Calculated

[Matt atkinson]

Homework Statement

A Σ0 baryon, traveling with an energy of 2 GeV, decays electromagnetically into a Λ and a photon.
What condition results in the Λ carrying the maximum possible energy after the decay? Sketch how the decay appears in this case, and calculate this energy. Explain your reasoning.
[Mass of Σ0 is 1.193 GeV/c2; mass of Λ is 1.116 GeV/c2]

Homework Equations

Relativistic kinematic equations;
CoM invariant mass

The Attempt at a Solution

So I am not sure on the condition for maximum energy of the Λ, is it when the photon is traveling with opposite momenta?
because the photon is a mass less particle it couldn't have 0 velocity right?

 

[Orodruin]

What energy will the $\Lambda$ have in the rest frame of the $\Sigma$?

 

[Matt atkinson]

So in the rest frame of the $\Sigma$
The energy would be
$$E_{\Lambda}=E_{\Sigma^o}+p_{photon}c$$
Right?
Because if The $\Sigma$ decays at rest then the momenta of the $\Lambda$ and photon will be equal and opposite.

 

[Orodruin]

What does energy and momentum conservation tell you?

 

[Matt atkinson]

If we are looking at the rest frame of the $\Sigma$ energy conservation gave me tee equation above and doesn't momentum conservation just tell me that;
$$0=p_{photon}-p_{\Lambda}$$
$$p_{photon}=p_{\Lambda}$$

 

[Matt atkinson]

I was wondering should use that $$W^2=(\sum E)^2-(\sum p)^2$$
for both the lab frame of the $\Sigma$ and then the center of mass frame after the decay?

which gives
$$(E_{\Sigma^o})^2-(p_{\Sigma^o})^2=(E_{\Lambda}+E_{photon})^2-(p_{\Lambda}-p_{photon})^2$$

 

[Matt atkinson]

Im not quite sure what the condition is to give the $\Lambda$ max energy.

 

[Orodruin]

What does $(E_{\Sigma^o})^2-(p_{\Sigma^o})^2$ evaluate to? What is the relation between the $\Lambda$ and photon energies and momenta?

 

[Matt atkinson]

So $E_{\Sigma}^2-p_{\Sigma}^2=m_{\Sigma}^2$, as for the second question I'm not sure.

 

[Orodruin]

How about what you just wrote down, but for the $\Lambda$ and for the photon instead?

If we are looking at the rest frame of the Σ\Sigma energy conservation gave me tee equation above

Your formula for energy conservation is a bit off and you should correct it.

 

[Matt atkinson]

So for the photon and $\Lambda$
$$(E_{\Lambda}+E_{photon})^2-(p_{\Lambda}+p_{photon})^2$$
And the Energy conservation should be $E_{\Sigma}=E_{\Lambda}+p_{photon}c$
Right? And expanding the relation before you get;
$$E_{\Lambda}^2+p_{photon}^2+2E_{\Lambda}p_{photon} - p_{\lambda}^2-p_{photon}^2 -2p_{photon}p_{\Lambda}$$?

 

[Orodruin]

I suggest starting from just the conservation equations without squaring. The only interesting thing at the moment being: What is the energy of the $\Lambda$ in the rest frame of the decaying $\Sigma$?

 

[Matt atkinson]

So to consider the problem with the entire problem with everything in the rest frame of $\Sigma^0$?
In which case if the energy of the $\Lambda$ is;
$$E_{\Lambda}=E_{\Sigma^0}-E_{photon}$$
and The momentum would be;
$$p_{\Sigma^0}=p_{\Lambda}+p_{photon}=0$$
Therefore;
$$p_{\Lambda}=-p_{photon}$$
Which would give;
$$E_{\Lambda}=E_{\Sigma^0}-p_{photon}=E_{\Sigma^0}+p_{\Lambda}$$
because photon has E=pc, so;
$$E_{\Lambda}-p_{\Lambda}=E_{\Sigma}$$
is that correct?
And does this mean that the condition for maximum energy is, to consider the $\Lambda$ in the rest frame with the electron traveling in the opposite direction?

 

[Matt atkinson]

I just don't understand where the $2GeV$ energy of the $\Sigma$ comes into it if, we are looking at the rest frame because in that frame it has 0 momentum and $E_{\Sigma}=mc^2$

 

[vela]

So to consider the problem with the entire problem with everything in the rest frame of $\Sigma^0$?
In which case if the energy of the $\Lambda$ is;
$$E_{\Lambda}=E_{\Sigma^0}-E_{photon}$$
and The momentum would be;
$$p_{\Sigma^0}=p_{\Lambda}+p_{photon}=0$$
Therefore;
$$p_{\Lambda}=-p_{photon}$$

This equation tells you that the $\Lambda$ and the photon have equal and opposite momenta.

Which would give;
$$E_{\Lambda}=E_{\Sigma^0}-p_{photon}=E_{\Sigma^0}+p_{\Lambda}$$
because photon has E=pc, so;
$$E_{\Lambda}-p_{\Lambda}=E_{\Sigma}$$
is that correct?

The signs are probably going to mess you up here. You should keep in mind that momentum is a vector, so it's really $E_\gamma = \lvert p_\gamma \rvert$ for the photon.

You need to somehow get rid of $E_\gamma$, $p_\gamma$, and $p_\Lambda$ because they're all unknowns. You want to solve for $E_\Lambda$ in terms of the masses of the $\Lambda^0$ and $\Sigma^0$.

And does this mean that the condition for maximum energy is, to consider the $\Lambda$ in the rest frame with the electron traveling in the opposite direction?

I think you meant photon, not electron. As you found above, conservation of momentum requires that the $\Lambda^0$ and photon go in opposite directions in the rest frame, so no, it's not the condition for maximum energy.

As Orodruin has suggested, first solve for the energy of the $\Lambda^0$ in the rest frame of the $\Sigma^0$. This may help give you insight into what the maximum energy condition is.

 

[Matt atkinson]

Ah sorry yes I did mean photon not electron.
So when I'm using the fact that the photon has no mass, it should be $E_{\gamma}=|p_{\gamma}|$ because $E_{\gamma}^2=p_{\gamma}^2+0^2$, and as you said momentum is a vector so the energy isn't, so the energy is just the magnitude of the momentum vector.
Thank you both so much for your help, I managed to get the solution after a lot of work.