# Homework Help: Relativistic collision of two protons to produce a proton-antiproton

1. May 3, 2014

### Wavefunction

1. The problem statement, all variables and given/known data

What is the minimum proton energy needed in an accelerator to produce antiprotons by the
reaction:

$P+P \rightarrow P+P+P+\bar{P}$

The mass of both protons and antiprotons is $m_p$. Assume first that the initial protons have equal
energy (the lab frame is the center-of-momentum frame). What energy is required if protons are
collided with a fixed target (one of the initial protons is at rest). This is one of the reasons
modern particle accelerators collide two beams.

2. Relevant equations

$P^{\mu} = \begin{pmatrix} \frac{E}{c}\\\vec{p}\end{pmatrix}$

3. The attempt at a solution

Setup: I think what I need to do is to look at the zeroth component of the momentum 4-vectors since it contains the relativistic energy of the particle. In the first case, since the protons have equal energy the center of momentum frame is already the lab frame so I don't need to boost to a different frame or anything like that. In the second case I will need to boost to a frame where the second particle at rest in the lab frame (K -frame) is moving at the same velocity $v$ as the first particle (K'-frame). Then I need to compare the zeroth component of the momentum 4-vectors in the K'-frame. Is my line of thinking correct, along the right direction, or flat out wrong? Thanks for your guidance guys

Last edited: May 3, 2014
2. May 3, 2014

### dauto

Your logic is correct but that's not the easiest way to solve the problem. The easiest way to solve the problem is to write all the 4-momenta for all the particles and use energy and momentum conservation combined with the particles' energy-momentum relations. So for the incoming proton you have the 4-momenta Pin = (E/c,p), the target 4-momentum is Ptarget = (mc,0), and all the outgoing particles have identical 4-momenta Pout = (E'/c,p')

Last edited: May 3, 2014
3. May 3, 2014

### Wavefunction

So in the first case I would have the following:

$p^{\mu}_{1i}+p^{\mu}_{2i}= p^{\mu}_{1f}+p^{\mu}_{2f}+p^{\mu}+\bar{p}^{\mu}$

$\begin{pmatrix} \frac{E}{c}\\ \vec{p} \end{pmatrix} + \begin{pmatrix} m_pc \\ \vec{0} \end{pmatrix} = 4\begin{pmatrix} \frac{E'}{c}\\ \vec{p}' \end{pmatrix}$

Which would give me 2 equations:

Eqn 1 $E+m_pc^2=4E'$ and Eqn 2 $\vec{p}=\vec{p}'$

I'm not exactly sure what to do here, but if I use $E'=\sqrt{(p'c)^2+(m_pc^2)^2}= \sqrt{(pc)^2+(m_pc^2)^2}$ then I get:

Eqn 1 $E=4\sqrt{(pc)^2+(m_pc^2)^2}-m_pc^2$

4. May 3, 2014

### dauto

Eqn 2 should be p = 4 p'

5. May 3, 2014

### fayled

Sorry to interrupt here - I've been using this method a lot to solve problems recently without understanding it too well.

I was wondering what in this method actually constrains the energy required to be a minimum. Obviously all the products stick together which is necessary, but why does this method yield only the minimum energy solution and not infinitely many others for which say the incident proton has a larger energy/moves faster and thus the products have larger energies/move faster. I just can't see the step in my working (despite getting correct answers) that does this.

It's obvious when working in the CM frame because we constrain the products not to move. In the lab though the products can have any energy/momentum E' and p'.

6. May 3, 2014

### dauto

If the proton had higher energy, the particles wouldn't stick together. As you said it yourself, that fact is obvious when working in the CM frame.

7. May 3, 2014

### Wavefunction

Okay I corrected the factor of 4: how does it look now?

8. May 3, 2014

### Wavefunction

Its okay I'd like to learn why the method works as well :)

9. May 3, 2014

### fayled

So if the products all stick together, there is only one mathematical solution for this to happen, which by judging it from the CM frame has to be the minimum energy one?

10. May 3, 2014

### Wavefunction

This might be a really dumb question, but why would the particles not stick together if the proton had a higher energy. Also is the in-elasticity of the collision required for the proton anti-proton production to occur?

11. May 3, 2014

### dauto

Now use the relation between E, p, and m along with the equation you found and solve for E

12. May 3, 2014

### dauto

Yes.

13. May 3, 2014

### dauto

If there was extra energy that energy would end up as kinetic energy of the products in the CM frame. That means the products would fly away from each other (they can't fly together because that would give them a total momentum different from zero, but that's impossible because the CM is defined as the referential where the total momentum is zero.

14. May 3, 2014

### Wavefunction

Haven't I already solved for $E$? I used $E' = \sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}$ to get and equation for $E$ in terms of $p$ an $m_p$

15. May 3, 2014

### dauto

You solved for E but p is unknown. Use E2 = (cp)2 + (mc2)2 to eliminate p

EDIT: the equation had a typo. I fixed it

Last edited: May 3, 2014
16. May 3, 2014

### Wavefunction

ah ok I see thank you! I'll post my result when I finish the calculation(:

17. May 3, 2014

### Wavefunction

Now I'll let $E^2 = (pc)^2+(m_pc^2)^2$ then

Eqn 1 $(pc)^2+(m_pc^2)^2 = [4\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}-m_pc^2]^2$

$(pc)^2+(m_pc^2)^2 = 16(\frac{1}{4}pc)^2+16(m_pc^2)^2-8\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}m_pc^2+(m_pc^2)^2$

$(m_pc^2)^2=16(m_pc^2)^2-8\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}m_pc^2+(m_pc^2)^2$

$1= 16-\frac{8\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}}{m_pc^2}+1$

$2=\frac{\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}}{m_pc^2}$

$(2m_pc^2)^2-(m_pc^2)^2=(\frac{1}{4}pc)^2$

$\frac{4\sqrt{3(m_pc^2)^2}}{c}= p$

$p = 4\sqrt{3}m_pc$ plugging back into Eqn 1 I get:

$E = 4\sqrt{(\frac{1}{4}[4\sqrt{3}m_pc]c)^2+(m_pc^2)^2}-m_pc^2$

$E = [4\sqrt{13}-1]m_pc^2$

18. May 3, 2014

### Wavefunction

Okay now my next question is: is this the answer to part 1 or part 2 because it seems like this is the answer to part 2.

19. May 3, 2014

### dauto

Yeah, that's part 2. Part 1 is easier.

20. May 3, 2014

### Wavefunction

Ah ok yea I see: for part 1 all I have to do is set all the energies on the RHS of the reaction equal to the rest energy for a proton and the 3-momentum equal to 0 such that

$2E=4E'$ and $p'=0$ Using $E'^2 = (m_pc^2)^2$ I get

$E=2m_pc^2$ since $2 < 4\sqrt{13}-1$ thats why they use two proton beams instead of one proton beam and a "stationary" proton.

21. Mar 17, 2016

### Laurafb92

Imagine that a supernova explosions occurs 100 pc from Earth and accelerates a proton (rest mass: 1.67×10-27 kg) to a speed of 99.999999% of the speed of light. Answer the following questions:

(e) If, before reaching the Earth, this cosmic ray proton had collided with a stationary proton in the interstellar medium, what is the maximum number of anti-protons which might be produced by the collision. State briefly why this is a maximum number? [6]

22. Apr 6, 2016

### Robert Lasater

I am not a physics person, but having worked (successfully, I think) the original problem on this thread, I believe I have some insights for you.

First, that proton accelerated by the supernova has a humongous amount of kinetic energy relative to the interstellar medium (effectively same as the reference frame of the earth), many, many times the energy of its rest mass from E=mc^2. That means when it collides with some proton at rest in the reference frame of the interstellar medium, if the collision is inelastic, there is enough energy available to create many new particles. Conservation laws require these new particles to be created as matter, anti-matter pairs.

Now the way I read your problem, the assumption is all available kinetic energy gets turned into proton, antiproton pairs. The question thus becomes how many such pairs can be created. Although that proton from the supernova has a humongous amount of kinetic energy, it does have an infinite amount of kinetic energy. So the number of proton, antiproton pairs is limited.

And I write "available kinetic energy", because that supernova proton also has a tremendous amount of momentum. And that momentum is conserved. The particles that result from that collision will have some momentum - and thus some kinetic energy. Not all the kinetic energy of the supernova proton will be available to create pairs of protons and antiprotons.

The way I would approach this problem is to transform the collision into the relativistic frame where the two protons have the same momentum (and kinetic energy), but since the two protons are approaching each other, the net momentum is 0 - the Center of Momentum frame. In this COM frame the supernova proton is still moving quite fast - a minute fraction of the speed of light - but now the interstellar proton is moving at the same speed.

And when the two collide, the results of that collision, the two original protons, plus the proton-antiproton pairs, are all at rest in this COM frame. Notice the particles present after the collision are all moving very fast relative to the interstellar medium.

Starting with the formula for adding relativistic velocities, it's not too hard to get a formula for the size of that boost required to be in the COM frame. But because the supernova proton has a velocity (relative to the interstellar medium) that is so very, very, VERY close to the speed of light, you may either need a very high precision calculator, or you will need to apply the binomial formula. For the formula I derived includes the square root of a number that here will be very, very close to 0.

With the magnitude of the boost, you should be able to immediately derive the kinetic energy of the two protons as observed in the COM frame. Remember now all the kinetic energy is available to make proton, antiproton pairs. And remember in the COM frame, both protons have the same kinetic energy, I have not worked it out in detail, but I am confident the kinetic energy will be many times the rest mass of the protons. You should find quite a few antiprotons could be created in the collision between the supernova proton and the interstellar medium proton.