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 [itex]E_{\gamma}=|p_{\gamma}|[/itex] 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.