# Homework Help: Baryon decay - maximum energy

1. Nov 19, 2016

### ElectricEel1

1. The problem statement, all variables and given/known data
A Σ0 baryon, travelling 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]

2. Relevant equations
E^2 = m^2 + p^2

3. The attempt at a solution
in the rest frame of the sigma baryon, E_sigma = m_sigma

p_lambda = - p_photon
E^2_lambda = m^2_lambda + p^2 lambda

E_photon = modulus of p photon.

im trying to start by solving for the energy of the lambda particle in the rest frame of the sigma baryon but im stuck. so far i have

E_lam = E_sig - E_photon

2. Nov 19, 2016

### Staff: Mentor

In the rest frame, you have energy and momentum of lambda and photon as unknowns, but you have conservation of energy, conservation of momentum and the energy/mass/momentum relation (separately for photon and lambda) - four equations for four unknowns, and you wrote down the four equations already. Plug the equations into each other until you can solve for a single variable.

3. Nov 19, 2016

### ElectricEel1

thanks. so my four equations are

E_sig = m_sig

E^2_photon = modulus p^2_photon

E^2_lamb = m^2_lamb + p^2_lamb

p_lamb = - p_photon

?

4. Nov 19, 2016

### Staff: Mentor

The first one should be replaced by conservation of energy. The decay products have the same total energy as the original particle.

5. Nov 20, 2016

### ElectricEel1

so if E_sig^2 = E_lam^2 + p_photon^2 = E_lam^2 + p_lamb^2 then since p_lamb^2 = E^2_lamb - m^2_lamb then i can plug that into the equation to get rid of the momentum which gives me

E^2_lambda = (E^2_sig + m^2_lamb)/2 = 1.155 Gev?

is this the condition of maximum energy where they go in opposite directions

6. Nov 20, 2016

### Staff: Mentor

This is the energy of the lambda in the sigma rest frame. In that frame the decay products always go in opposite directions. What leads to the maximal energy of the lambda in the lab frame, where the sigma is moving?

7. Nov 20, 2016

### ElectricEel1

in the frame where the sigma is moving could I have it so the lambda takes all the momentum and the photon is at rest i.e doesnt exist?

8. Nov 20, 2016

### Staff: Mentor

No, that doesn't work, as you calculated already: the lambda has an energy of 1.155 GeV, which is not identical to 1.193 GeV.

Not having a photon would violate energy-momentum-conservation.

9. Nov 20, 2016

### ElectricEel1

im still a little lost with it but in the lab frame the energy of the sigma particle is 2.0Gev so i'll need to use this i guess. could they go the same way? it hasnt yet made sense what makes the sigma energy maximum. The photon is massless so the energy of that will just be the momentum.

10. Nov 20, 2016

Staff Emeritus

11. Nov 20, 2016

### Staff: Mentor

Think about the directions the particles can go to.

This is similar to a classical mechanics problem with a rocket: a rocket (sigma) is flying to the right. It fires its engines (it emits a photon). In which direction should it fire to maximize its speed afterwards?

12. Nov 20, 2016

### ElectricEel1

to be honest there was another thread on this subject and they said to start from there so thats where i started.

it should fire the photon to the right

13. Nov 20, 2016

### Staff: Mentor

"Its speed" refers to the rocket (sigma->lambda). If you fire something to the right you slow down the rocket.

14. Nov 20, 2016

### ElectricEel1

oh i get you now. so the photon would fly out to the left and the lambda would continue right? but in this case they have different momenta?

15. Nov 20, 2016

### ElectricEel1

so to switch to the moving frame i found the lambda velocity to be 0.034c and doing the velocity equation i calculated that that means the lab frame is moving at 0.034c. im not sure if this was the right step to make

16. Nov 20, 2016

Staff Emeritus

17. Nov 20, 2016

### Staff: Mentor

The velocity didn't change when you switched between frames?

@Vanadium 50: Where is your point? Calculating the velocity in the sigma rest frame is a perfectly valid intermediate step.

18. Nov 20, 2016

Staff Emeritus
He hasn't yet answered the question of the configuration resulting in the maximum energy. Until's he's got this, he shouldn't go calculating things willy-nilly. Yes, they might be useful - but they might not.

19. Nov 20, 2016

### ElectricEel1

they asked for the configuration which gives the lambda maximum energy and to calculate the energy.

no, i got the velocity using the relativistic momentum equation then tried to use u' = (u-v)/(1-(uv/c^2)) to switch to the other frame but since its the rest frame i used u=0 so it just cancelled to v

20. Nov 20, 2016

### Staff: Mentor

The other frame is the lab, where the sigma is moving.

@V 50: I think your questions are more distracting than helpful in this case.

21. Nov 20, 2016

### ElectricEel1

realised i did the above part wrong. I found the velocity of the sigma in the lab frame using the 2Gev energy given in the question then used that to transform from rest frame to that frame and got lambda velocity as 0.7c then found momentum as 1.085gev/c and energy 1.56Gev but maybe vanadium is right. Have I got the right idea about the configuration and have I gone about this the right way?

22. Nov 20, 2016

### Staff: Mentor

If you understood the rocket part, you can add the velocities to get the speed of the lambda.

23. Nov 20, 2016

### ElectricEel1

i understood the idea of the rocket part but i've been trying to use a transform to get from the speed in the sigma rest frame and got 0.7c for the speed of the lambda in the lab frame then got the energy from the momentum. i think thats the point im missing. which velocities am i adding together?

24. Nov 20, 2016

### Staff: Mentor

The sigma moves relative to the lab, the lambda moves relative to the sigma. And you get the largest lambda velocity in the lab if ...

25. Nov 21, 2016

### ElectricEel1

It the lambda keeps moving forward relative to the sigma and the lambda moves backwards? The speeds of the photon and sigma are c and 0.69c