What is the maximum energy of the Λ particle after the Σ0 baryon decay?

[ElectricEel1]

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

E^2 = m^2 + p^2

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 I am stuck. so far i have

E_lam = E_sig - E_photon

 

[mfb]

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.

 

[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

?

 

[mfb]

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

 

[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

 

[mfb]

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?

 

[ElectricEel1]

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?

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 doesn't exist?

 

[mfb]

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.

 

[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.

 

[Vanadium 50]

im trying to start by solving for the energy of the lambda particle in the rest frame of the sigma baryon

Why? They didn't ask this. What do they ask?

 

[mfb]

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?

 

[ElectricEel1]

Why? They didn't ask this. What do they ask?

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

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?

it should fire the photon to the right

 

[mfb]

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

 

[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?

 

[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. I am not sure if this was the right step to make

 

[Vanadium 50]

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. I am not sure if this was the right step to make

Did they ask for this? What did they ask for?

 

[mfb]

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. I am not sure if this was the right step to make

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.

 

[Vanadium 50]

Where is your point?

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.

 

[ElectricEel1]

Did they ask for this? What did they ask for?

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

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.

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 canceled to v

 

[mfb]

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.

 

[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?

 

[mfb]

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

 

[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 that's the point I am missing. which velocities am i adding together?

 

[mfb]

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 ...

 

[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

 

[mfb]

Why do you expect the lambda to move backwards now?

 

[ElectricEel1]

The lambda forwards and photon backward sorry

 

[mfb]

Right.

 

[ElectricEel1]

Awesome. So I already calculated the energy and now I have the configuration too? Thanks

 

[mfb]

I don't see where you got the right final answer. It got scattered across posts a bit.

 

[ElectricEel1]

i had 0.034c as the velocity of the lambda in the sigma rest frame. then using 2gev as the energy of the sigma in the lab frame, I calculated the momentum and used the relativistic momentum equation to find the sigma velocity. Then I used u' = (u-v)/(1-uv/c^2) to where u = 0.034c and v = 0.69c (sigma velocity in lab frame) to get the lambda velocity and then the energy. Or should I have used u = (u' + v)/(1 + u'v/c^2) since I am transforming from a rest frame to the actual lab frame

 

[mfb]

One will lead to a value that doesn't make sense, the other one is the right answer.