Homework Help: Two protons collding

1. Oct 31, 2007

Niles

1. The problem statement, all variables and given/known data
I have to protons coliding - they have same magnitude of velocity and momentum and they collide head-on:

p + p -> p + (p + p_) (the "p_" is an anti-proton).

I have to find the minimum combined kinetic energy of the two protons for this process to run.

3. The attempt at a solution

I use P_total = P_1 + P_2 <=>

(all the masses squared ...) * c^2 = (P_1 + P_2)^2.

But this is where I encounter my problem. The 4-vector for e.g. P_1 is [E/c ; gamma*m*v] - I don't know v?

Is there another way of doing this?

- another thing: When I have found the total energy, do I have to subtract it with the rest mass of a proton x 2 (2*m_p*c^2) to find combined kinetic energy ?

Last edited: Oct 31, 2007
2. Oct 31, 2007

Avodyne

You don't know v because that's what you're trying to find: if you knew v, you would know the total energy (and yes you have to subtract the rest energy to get the kinetic energy).

You've written P_1 in components; what is P_2 ? Then, what is P_1 + P_2 ?

3. Oct 31, 2007

Niles

I want to find E - finding E and v is overkill, isn't it?

1) P_2 = [E/c ; -gamma*m*v]

2) P_1 + P_2 = 2E/c?

Last edited: Oct 31, 2007
4. Oct 31, 2007

Avodyne

Yes, finding v is overkill. And more precisely, P_1 + P_2 = [2E/c; 0].

5. Nov 1, 2007

clem

"p + p -> p + (p + p_) (the "p_" is an anti-proton)"

You must mean -->p+p+(p+p_)
You need a cm energy W=4M.
The initial cm energy is W=(T+2M).
Solve for T.

6. Nov 1, 2007

Niles

So I get that T = 2*M?

I have another question, related to the first post. If one of the protons are at rest, I have to find the kinetic energy for the process to begin:

I would use that P_1 + P_2 = P_f (of the 4 protons) - so

2*m^2*c^2 + 2*E*m = 4m^2 * c^2 (I have squred it)

Two questions for that equation:

1) Is it correct?

2) When I have 4m^2 * c^2, how do I do this? (938 MeV/c^2)^2 * c^2 or what? So 4m^2 * c^2 = 938 ^2?

Last edited: Nov 1, 2007
7. Nov 1, 2007

Niles

If I do what Clem did in his post for the second scenario (where one proton is in rest, other is moving), I get that:

W = 4M

W_cm = (T+M), so T = 3M? Does that make sence?

8. Nov 1, 2007

Niles

Ok, the two questions in my post are solved (#1 and #2). Now the original question - where the 2 protons have equal magnitude momentum - still remains.

This is what I have so far:

(P_1 + P_2)^2 = P_f^2 <=>

(2E/c;0)^2 = 16*m^2*c^2 <=>

4E^2/c2^2 = 16*m^2*c^2, where I find the total energy E and use: E - 2*m*c^2 = E_kinetic.

Is this correct? I can use this method for all particles as long as I change the masses, right?

9. Nov 1, 2007

Niles

Guys, I solved the problem.. I forgot that I found the total energy for EACH proton, not the total total-energy for both particles.. that's why my result didn't add up.

Sorry for posting so many times, but all my problems are solved now - thanks.

10. Nov 1, 2007

clem

"I have another question, related to the first post. If one of the protons are at rest, I have to find the kinetic energy for the process to begin:"

If p_2=0 (at rest), then
W^2=(E_1+M)^2-(p_1)^2=2M^2+2ME_1.
This equals (4M)^2.
Solve for E_1. Then T_1=E_1-M=6M.