Nuclear physics reaction, minimal energy to make it work

[Coffee_]

1. Consider the reaction that happens when a photon collides with an atomic electron in rest $\gamma+e^{-} <---> K^{-} + v_{e}$. The masses of all particles can be considered as known. What is the minimal energy that the photon needs to have to make this reaction work?


2. Conservation of four momentum


3. Frankly, I am quite lost. At first glance I'd say that $E_{\gamma}+m_{e}=E_{K}+E_{v}$ and then say that the minimal energy required is to create the K and neutrino in rest, and so substituing the masses for the energy in the last expression. This won't work with conservation of momentum though. Hoewever once I consider that it gets harder because of the allowed angles after collisioN.

 

[haruspex]

I'd say that $E_{\gamma}+m_{e}=E_{K}+E_{v}$ and then say that the minimal energy required is to create the K and neutrino in rest, and so substituing the masses for the energy in the last expression. This won't work with conservation of momentum though. Hoewever once I consider that it gets harder because of the allowed angles after collisioN.

There's only one angle to worry about, no? Having set an unknown for the angle the departing neutrino makes to the incoming photon's path, the angle for the K follows by conservation of momentum in the direction orthogonal to the photon's path. Then it's a matter of minimising the energy with respect to the one angle.

 

[Coffee_]

There's only one angle to worry about, no? Having set an unknown for the angle the departing neutrino makes to the incoming photon's path, the angle for the K follows by conservation of momentum in the direction orthogonal to the photon's path. Then it's a matter of minimising the energy with respect to the one angle.

Do you mind elaborating on that a bit? You seem to have seen this easily without calculations so I'm probably overlooking something.

 

[haruspex]

Do you mind elaborating on that a bit? You seem to have seen this easily without calculations so I'm probably overlooking something.

Post your momentum equations, then we'll have a reference point for discussion.

 

[Coffee_]

Post your momentum equations, then we'll have a reference point for discussion.

Here you go: http://imgur.com/CpEDxzi

I just wrote down what I knew, but I wasn't sure where to go from there because there seem to be too much unknown.

 

[haruspex]

Here you go: http://imgur.com/CpEDxzi

I just wrote down what I knew, but I wasn't sure where to go from there because there seem to be too much unknown.

You've written $E_\gamma$ for both the energy and the momentum of the photon. Is there some convention I don't know, or is that a mistake?
Your last equation gives a relationship between the two angles. You can take one angle as the independent variable and the other to be a function of it. Differentiate wrt the independent angle to find the minimum $E_\gamma$.

 

[Coffee_]

You've written $E_\gamma$ for both the energy and the momentum of the photon. Is there some convention I don't know, or is that a mistake?
Your last equation gives a relationship between the two angles. You can take one angle as the independent variable and the other to be a function of it. Differentiate wrt the independent angle to find the minimum $E_\gamma$.

In natural units, since the mass of a photon is zero, the momentum and energy are equal. Thanks for the tip I'll try it out and reply with results. In a previous post you mentioned something being perpendicular, will this lead to it or is it a different approacH?

 

[haruspex]

In natural units, since the mass of a photon is zero, the momentum and energy are equal.

OK, I thought it might be something like that. can't help feeling it's a bit confusing though. Dimensional consistency can be a useful check on the algebra.

In a previous post you mentioned something being perpendicular,

Do you mean where I wrote 'orthogonal'? You already did that bit in your eqn 4.

 

[Coffee_]

OK, I thought it might be something like that. can't help feeling it's a bit confusing though. Dimensional consistency can be a useful check on the algebra.

Do you mean where I wrote 'orthogonal'? You already did that bit in your eqn 4.

Tried playing with it for a bit without much luck. It sucks because basically it's an algebra momentum-energy problem which I was supposed to have learned to solve in classical mechanics intro courses. I guess it's because we always worked with given values, not used to having 4 variables in these problems.

 

[PeroK]

Tried playing with it for a bit without much luck. It sucks because basically it's an algebra momentum-energy problem which I was supposed to have learned to solve in classical mechanics intro courses. I guess it's because we always worked with given values, not used to having 4 variables in these problems.

Can you find a simple argument for both angles being 0?

Hint: if you can't see it in the original frame, try thinking about the collision in the CM frame.

Remember you are trying to minimise the energy.

 

[Coffee_]

Can you find a simple argument for both angles being 0?

Hint: if you can't see it in the original frame, try thinking about the collision in the CM frame.

Remember you are trying to minimise the energy.

Oh right! In the CM frame, they can be created in rest without problems, this means an opening angle of 0 in the lab frame! Right?

 

[PeroK]

Oh right! In the CM frame, they can be created in rest without problems, this means an opening angle of 0 in the lab frame! Right?

Yes, absolutely.

 

[Coffee_]

In the CM frame, what is the total momentum after the collision?

Always 0. This means that as long the momentum vectors of the particles are the same and in opposite directions the momentum is conserved. The only worry now is the total energy after the collision which depends only on the magnitude of those vectors. Well if I pick both magnitudes as 0 , I still have total momentum 0 - check , and minimal incoming energy of both particles.

 

[Coffee_]

Yes, absolutely.

Thanks a lot for your help, so now I have the expression for the photon energy in function of the known masses in CM frame. Do you now recommend finding an expression for $\beta$ of the CM relative to the lab frame and then transforming the photon energy into that frame? Or just work in in lab frame and assume angle 0?

 

[PeroK]

Thanks a lot for your help, so now I have the expression for the photon energy in function of the known masses in CM frame. Do you now recommend finding an expression for $\beta$ of the CM relative to the lab frame and then transforming the photon energy into that frame? Or just work in in lab frame and assume angle 0?

You can do it either way, but if you know how to do energy-momentum transformations, try using the CM frame.

 

[Coffee_]

You can do it either way, but if you know how to do energy-momentum transformations, try using the CM frame.

My attempt here: http://imgur.com/u2izMjJ.

In the right lower corner it's a derivation for $\beta_{CM}$ in function of the photon energy in the lab frame. The expression will be very ugly, and then that expression will have to be plugged in the other one in the left lower corner. I doubt an elimination of $E_{\gamma}$ will be possible.

 

[PeroK]

My attempt here: http://imgur.com/u2izMjJ.

In the right lower corner it's a derivation for $\beta_{CM}$ in function of the photon energy in the lab frame. The expression will be very ugly, and then that expression will have to be plugged in the other one in the left lower corner. I doubt an elimination of $E_{\gamma}$ will be possible.

You're missing a trick. Think about energy-momentum transformations for a system of particles.

 

[Coffee_]

You're missing a trick. Think about energy-momentum transformations for a system of particles.

Alright since you mentioned more particles I tried using the fact that the electrons energy transforms as well, this gives me a much better expression:

http://imgur.com/GvM5u8p

Oh nvm I made a mistake I'll post an edit in a sec.

EDIT: correct version - http://imgur.com/kyVNJjk

 

[PeroK]

Alright since you mentioned more particles I tried using the fact that the electrons energy transforms as well, this gives me a much better expression:

http://imgur.com/GvM5u8p

Oh nvm I made a mistake I'll post an edit in a sec.

The trick you're still missing is that energy-momentum transforms for a system, as it does for individual particles:

$E_1^2 - P_1^2c^2 = E_2^2 - P_2^2c^2$

Where the E's and P's are the total energy and momentum of a system of particles in two frames.

 

[Coffee_]

The trick you're still missing is that energy-momentum transforms for a system, as it does for individual particles:

$E_1^2 - P_1^2c^2 = E_2^2 - P_2^2c^2$

Where the E's and P's are the total energy and momentum of a system of particles in two frames.

Oh! I never heard of this fact before I'm still in an intro course to relativity. Yeah now it is indeed a quick calculation. Is my final link techincally correct as well?

 

[PeroK]

Oh! I never heard of this fact before I'm still in an intro course to relativity. Yeah now it is indeed a quick calculation. Is my final link techincally correct as well?

You can show this easily by showing that the sum of the individual energy-momentum transforms according to the Lorentz Transformation. Hence, the total energy-momentum of a system of particles is also a four-vector with the above frame-invariant.

You should get a very simple answer to this one. Start from what you got in the CM-frame for total energy-momentum.

 

[Coffee_]

You can show this easily by showing that the sum of the individual energy-momentum transforms according to the Lorentz Transformation. Hence, the total energy-momentum of a system of particles is also a four-vector with the above frame-invariant.

You should get a very simple answer to this one. Start from what you got in the CM-frame for total energy-momentum.

Is it alright to do it with just the four vector product invariance then?

$p_{\gamma CM} p_{e CM} = p_{\gamma LAB} p_{e LAB}$ where these are four vector products. According to the four vercot product convention in our class this will give (I've already used the fact that the angle between them is $\pi$ here.:

$2E_{\gamma CM}E_{e CM} + 2p_{\gamma CM}p_{e CM}= 2E_{\gamma LAB} E_{e LAB}$ Now these are all numbers, no vectors anymore. p's represent the magnitudes of the three-vector momentum.

Everything is known/can be reduced except for $E_{\gamma LAB}$ and thus eliminating it yields the correct solution?

 

[PeroK]

I'm not sure that's so useful,here. You don't know the individual energies (or momenta) in the CM frame, but you do know the total energy and momentum. Unless you work with the total energies it's going to get messy.

 

[Coffee_]

I'm not sure that's so useful,here. You don't know the individual energies (or momenta) in the CM frame, but you do know the total energy and momentum. Unless you work with the total energies it's going to get messy.

I do know them in the CM frame assuming the decay happens so that the resulting particles are at rest, in the CM frame in natural units (c=1 , hbar:=1):

$E_{\gamma} + E_{e} = m_{k} + m_{v}$

$E_{\gamma}+\sqrt{p^{2}+m_{e}^2} = m_{k}$ (neglect neutrino mass)

Alright here we know that both particles have momentum $p$, for a photon which hass no mass in my units the energy is equal to the momentum, so $p=E_{\gamma}$

$E_{\gamma} + \sqrt{E_{\gamma}^2 + m_{e}^2} = m_{k}$

Only the energy of the photon in this frame is unknown and can be expressed quite nicely eventually. This is on it's turn equal to p of both particles, and so everything in the CM frame can be expressed.