# Lambda decay, momentum of the pion and proton

## Homework Statement

I have a lambda decaying into a pion and a proton. The lambda is moving with velocity 0.9c and I know the mass of the lambda as well as the pion and proton (these are known constants). I need to find the momentum of the pion and the proton after the decay happens.

## Homework Equations  Conservation of momentum and conservation of energy

## The Attempt at a Solution I don't know at this point which momentum to choose out of the two. I solved the math using a software so I would assume it is right.

## Answers and Replies

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Orodruin
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The question you need to ask yourself is why there are two solutions. Once you have figured that out it should help you answer your question.

The question you need to ask yourself is why there are two solutions. Once you have figured that out it should help you answer your question.
There are two solutions because I assume you solve kind of the quadratic equation (so + or - the square root)
But I am having trouble realizing which answer is "wrong" since none of them give anything negative

Orodruin
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Why would either be wrong? What are the possible decays in the Lambda rest frame.

Edit: I suggest you do the computation in the Lambda rest frame first and then interpret your solutions there.

Edit: I suggest you do the computation in the Lambda rest frame first and then interpret your solutions there.
That's kind of the first problem I am working on, the lambda is at rest and I need to find the momentums of the other particles.
I end up getting that the momentum is equal to the mass, which seems wrong to me. They don't have the same mass and if the momentum of particle A (mother) is 0, then Pb=-Pc ... so this can't be right.

Orodruin
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Please show your computations in the Lambda rest frame. No, I am not referring to the possible decay channels, I am referring to the possible momenta of this particular decay.

Please show your computations in the Lambda rest frame. No, I am not referring to the possible decay channels, I am referring to the possible momenta of this particular decay.
https://imgur.com/rbiJNbT

In this page I tried kind of two different things and at one point got Pc=mc and then the other got a factor of sqrt2/2

Orodruin
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I am sorry, but it is impossible to tell what you have tried to do from that photo. Please write it out here along with argumentation describing what you are doing and why.

I am sorry, but it is impossible to tell what you have tried to do from that photo. Please write it out here along with argumentation describing what you are doing and why.
Basically I was trying to apply this logic Where I can write expressions for the momentum because it will be conserved, it will be a sum. Then I wanted to solve for an expression for the energy which I could equate to the more commonly known And solve for the momentum, substituting as necessary since I can find an expression for Pa, and it is equal to 0 anyways

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Orodruin
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I am sorry, it is still unclear. It is not clear from your handwriting what ##p##s represent 4-momenta and what ##p##s represent 3-momenta. Perhaps you are even mixing them up. This is why I suggest that you write down your entire reasoning again here, taking care to explain what you are doing in every step.

I am sorry, it is still unclear. It is not clear from your handwriting what ##p##s represent 4-momenta and what ##p##s represent 3-momenta. Perhaps you are even mixing them up. This is why I suggest that you write down your entire reasoning again here, taking care to explain what you are doing in every step.
This is the best I can do by myself. I haven't learned this stuff formally so trying to learn as I go, it's for a project and not even homework it will not be graded.  It's just a lot of algebra and I thought maybe there was an easier way, there must be some concepts I just don't understand properly like the four vectors that I could apply here.

If I fill this in with the values of the lambda decay, I get that Pc=0.1485 Gev/c which is roughly the mass of the pion.

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Orodruin
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If I fill this in with the values of the lambda decay, I get that Pc=0.1485 which is roughly the mass of the pion.
No, you get ##\pm## that.

No, you get ##\pm## that.
Yes, and Pb=-Pc.

I'm not sure what I'm missing out on here

Orodruin
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How do you interpret the ##\pm##?

How do you interpret the ##\pm#?
Not sure, depending on the direction of travel along the x axis

Orodruin
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Not sure, depending on the direction of travel along the x axis
So, are both solutions physical?

So, are both solutions physical?
Well I would think so. I assume that it can decay in any direction so the pion & proton are not forced to go "forwards"

Orodruin
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Well I would think so. I assume that it can decay in any direction so the pion & proton are not forced to go "forwards"
So what does this tell you about the two solutions you found in the case of a moving ##\Lambda##?

So what does this tell you about the two solutions you found in the case of a moving ##\Lambda##?
then they are both possible. But I don't understand really, is my answer right? That the momentum would be + or - this whole expression (0.14)

So what does this tell you about the two solutions you found in the case of a moving ##\Lambda##?
The thing is, ultimately I am writing code to plot 10000 random lambda decays. I will plot the momentum and kinetic energy of the pion. So there aren't just 2 possibilities for the momentum. It will always depend on the angle at which it decays right ? Because 10 000 random decays could produce 10 000 different momentum right ?

Orodruin
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The thing is, ultimately I am writing code to plot 10000 random lambda decays. I will plot the momentum and kinetic energy of the pion. So there aren't just 2 possibilities for the momentum. It will always depend on the angle at which it decays right ? Because 10 000 random decays could produce 10 000 different momentum right ?
Right, if you are looking at it in more than 1 space dimension you have a large distribution of momenta and energies depending on the angle. You do not need to write code to do this. What you need to do is to consider the decay isotropic in the rest frame of the ##\Lambda## and get the distribution on energy and momenta from there. You will find that the resulting decay products are emitted preferentially in the forward direction. This effect is called beaming.

Right, if you are looking at it in more than 1 space dimension you have a large distribution of momenta and energies depending on the angle. You do not need to write code to do this. What you need to do is to consider the decay isotropic in the rest frame of the ##\Lambda## and get the distribution on energy and momenta from there. You will find that the resulting decay products are emitted preferentially in the forward direction. This effect is called beaming.
Okay, I will look into this beaming effect.

The point of my code wasn't to do the math, I was supposed to make a histogram. I was just trying to understand what's going on here in this situation.

Orodruin
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The point of my code wasn't to do the math, I was supposed to make a histogram. I was just trying to understand what's going on here in this situation.
I would suggest that doing the maths is a better way to understand what is going on as you will learn why things behave as they do rather than just looking at the result on a histogram. This plot is from my SR lecture notes and shows the beaming of the neutrino resulting from a decay ##\pi^+ \to \mu^+ + \nu_\mu##, where the original pion is moving at ##v = 0.5c## (energy is in units of MeV): The blue curve is the energy as a function of the angle ##\theta## to the forward direction and the orange curve is the distribution in the cosine of the angle.

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I would suggest that doing the maths is a better way to understand what is going on as you will learn why things behave as they do rather than just looking at the result on a histogram. This plot is from my SR lecture notes and shows the beaming of the neutrino resulting from a decay ##\pi^+ \to \mu^+ + \nu_\mu##, where the original pion is moving at ##v = 0.5c## (energy is in units of MeV):
View attachment 226393
The blue curve is the energy as a function of the angle ##\theta## to the forward direction and the orange curve is the distribution in the cosine of the angle.
Yes, I agree there is no point in doing a code & histogram if I don't understand at all what's happening in the situation.

Maybe it is out of context or not relevant to this, but just out of curiosity, do you know what the point of making a momentum and kinetic energy plot would be ? I understand in your plot making the energy with the angle, because then you can really see the change depending on the direction of the decay and it seems to show some pertinent information. I'm going to somehow have to include the angle in my plot anyways I assume, even though it was never explicitly asked of me....