# Photon striking a proton

1. Oct 28, 2015

### Sophia Su

1. The problem statement, all variables and given/known data
A high energy photon moving to the right strikes a proton that is moving to the right at a speed u/c= 0.40 in the laboratory frame, creating a proton and a neutron: γ + p → p + n For the purposes of this question, assume that protons and neutrons have an equal mass m.
a) What is the minimum energy that the photon must have in order for this nuclear reaction to occur? (answer in units of mc2 )
b) For the conditions in part a), what speed will the proton and neutron have in the laboratory frame?
c) Suppose the photon was moving to the right, and the proton was moving to the left at 0.40c before the collision. Will this change the answer to a)? Why or why not?

2. Relevant equations

E=Gamma*m*c^2

3. The attempt at a solution

I am so confused. I simply found the gamma of the proton before collision.

2. Oct 28, 2015

### Orodruin

Staff Emeritus
This reaction violates the standard model of particle physics (baryon number violation) as well as the conservation of angular momentum. It will simply not occur.

Apart from that, you will need to provide more of an attempt for us to be able to help you. I suggest you think about what conservation laws that are required to be fulfilled (apart from the ones which the problem seems to ignore.

3. Oct 28, 2015

### SolusVir

Ok so I have the same question and don't really know how to tackle it either. I'm assuming we are going to be using conservation of momentum and conservation of energy, and my initial thought was that I would have to move to the frame of reference where the proton is at rest. I'm going to take a bit of a stab here, but i"m assuming this is the center of mass frame. At first I thought that would make the the photons relative velocity in this new frame 0.6c, but im beginning to feel as if i need to perform some kind of lorentz transformation to convert this velocity to the new frame of reference. Since the addition of the energy from the photon causes the proton to become two particles, or break into two particles-
(honestly i couldn't fully grasp how a proton plus energy made a neutron. Is the photon turning into a neutron, is the energy from the photon breaking the proton into a smaller proton and a neutron? Aren't all protons in this question assumed to have the same mass, in which case how would i get a smaller proton? idk, if someone could shed light on what the conceptual idea is there that would be great even tho Orodruin has mentioned this is not a physical possability.)
-im assuming it's like the binding energy, which i tink is basically what we're solving for, in which case the total energy of the system i think is suppose to equal

E(total) = (Kinetic energy + rest mass of photon) + (rest mass of proton) - B(Binding Energy)

Since a photon is massless its total energy can be represented by E=pc, where p is momentum, so

E(tot) = p(photon)*c + M*c^2 -B.

When i get to this point I realize I've achieved pretty much nothing cuz the solution doesn't seem any closer, i dont know how to find the momentum when it doesnt have a mass if i can't equate it to energy, and im not even sure if im solving for the correct variable.

If anyone could maybe give a conceptual idea of whats going on in the question and lay a skelaton down of the things we need to find in order to work our way to the solution, that would awesome. Maybe even the first equation we should use, because i have seen a number of different forms of the same relationship between momentum and energy used in different ways. Im pretty sure I'm somewhere far in left field atm, but I've missed a number of classes due to conflicting work shifts and this is what I'm piecing together from my book and online sources. If anyone could grab the proverbial chalk from my hand and lead me, sophia, and whoever else has missed too many lectures in our class on some kind of path to finding an answer, that would sweat.

4. Oct 29, 2015

### Staff: Mentor

A Lorentz transformation always transforms the speed of a photon from c in one frame to c in any other frame. The speed of light is invariant. What do change from one frame to another are the energy and momentum of the photon.

5. Oct 29, 2015

### Staff: Mentor

No, these two frames are not the same. "Center of mass frame" is actually poor and confusing terminology here, because the photon has zero mass. What we really mean by "center of mass frame" is the frame where the total momentum is zero. In this frame, the proton is not at rest, but is moving to the left, towards the photon that is moving to the right.

It's actually not necessary to do the math for this problem in the "center of mass" (zero-momentum) frame. I did it in the lab frame, using the numbers given in the problem statement. Nevertheless, it is very useful to consider what this process looks like, qualitatively, in the "center of mass" frame, because it tells you what the "minimum energy" condition means in terms of the outcome, both in this frame and in the lab frame.

In the "center of mass" frame, what does this process look like, specifically in terms of the directions of motion of the two incoming particles, and of the two outgoing particles? What condition on the two outgoing particles gives you the minimum total energy in this frame, and therefore the minimum energy for the incoming photon (and proton)?

Don't worry about the conceptual details of the process here, just think of it as a "black box". As Orodruin noted, this specific process is impossible because it violates certain conservation laws even though you can make it satisfy conservation of energy and momentum. If it makes you feel better, rephrase the problem as "photon plus particle 1 goes to particle 2 plus particle 3", where particles 1, 2 and 3 are different particles that happen to have the same mass. OK?

The whole point of this problem is conservation of relativistic energy and relativistic momentum. Other details of the process are simply distractions.

There's no binding energy here. The photon and proton before the collision are not bound. The proton and neutron after the collision are not assumed to be bound. (We're not talking about deuterium here.)

Last edited: Oct 29, 2015