Relativistic Inelastic Collision of Protons

[AnniB]

Homework Statement

Let us consider a perfectly inelastic collision between two protons: an incident proton with mass m, kinetic energy K, and momentum magnitude p joins with an originally stationary target proton to form a single product particle of mass M. Due to conservation of momentum, there must be some kinetic energy after the collision. Show that the energy available to create a product particle is

Mc² = 2mc²(sqrt(1 + K/2mc²))

Homework Equations

E₀ = E_f
This is the only equation I can think of. Applying conservation of momentum in just seems like it would add a miscellaneous variable. The others are just figured out from the problem and a couple hints the teacher provided.
E₁ = K + mc²
E₂ = mc²
E_f = Mc² + K

The Attempt at a Solution

E₁ + E₂ = E_f
(E₁ + E₂)² = E_f²
(Mc² + K)² = (K + mc²)² + 2(K + mc²)(mc²) + (mc²)²
Solving this gave me:
(Mc²)² + 2KMc² = 4Kmc² + 4(mc²)²

I can't get this to look like the equation I need. I feel like I either have one term too many, or I'm missing one.

 

[fzero]

Homework Statement

Let us consider a perfectly inelastic collision between two protons: an incident proton with mass m, kinetic energy K, and momentum magnitude p joins with an originally stationary target proton to form a single product particle of mass M. Due to conservation of momentum, there must be some kinetic energy after the collision. Show that the energy available to create a product particle is

Mc² = 2mc²(sqrt(1 + K/2mc²))

Homework Equations

E₀ = E_f
This is the only equation I can think of. Applying conservation of momentum in just seems like it would add a miscellaneous variable. The others are just figured out from the problem and a couple hints the teacher provided.
E₁ = K + mc²
E₂ = mc²
E_f = Mc² + K

It's very unlikely that the product particle will have the same kinetic energy as the incident proton, so your E_f is incorrect.

As a hint, note the factor of 2 in the formula for Mc² is highly indicative that there's a frame in which this energy can be attributed in equal parts to the two reactant photons. What frame might this be?

 

[AnniB]

Would it be one that is moving at half the speed of the approaching proton, so in the reference frame they are at rest after the collision?

What would that do to the energy, though? I tried calculating it how I thought it should be and still ended up with one term too many.

 

[AnniB]

Okay, so I've *almost* got the right equation. There's just a extra term of K²/m²c⁴ that is keeping me from just writing one more step and being done.

I don't know if it's because the energy I used isn't right, but I've used 3 different values for it so far and this is the closest I've gotten.

 

[fzero]

It would help if you showed some work. What energy did you use? We can easily compute the available energy in the frame where the total momentum of the initial protons is zero.

 

[AnniB]

After changing into the frame where both protons are moving towards each other so that there was only rest energy after the collision and so the total momentum should be zero, I have E₀ = 2(K/4 + mc²), which I based off of 1/2mv², but it probably is not good to assume that even applies in this case. I've used E₀ = 2(K/2 + mc²) also, just to check. The first E₀ I used was 2(K + mc²), but I know that can't be right because that's the kinetic energy from the frame of reference set up in the problem itself.

The hints I got from the professor were to set E₁ = K + mc² and to square the expression E₁ + E₂ = E_f, using the initial frames of the problem. When I tried to do that, though, I came up with answer that was even farther off than what I'm getting now.

 

[fzero]

First of all, we're dealing with relativistic equations, so if you wanted to relate kinetic energy to velocity, you need to use

$$K = \frac{p^2}{2m}, ~ p = \gamma m v.$$

However we don't really need that. In the lab frame, the total energy is 2mc² + K. The momentum of the moving proton can be found by solving (K+mc²)² +(pc)² = (mc²)². To find the total energy in the c.o.m. frame, we need to use energy-momentum conservation.

 

[AnniB]

Ah, thank you! I don't know why I didn't think to use that formula earlier.

 

[sweetreason]

why do we solve (K+mc2)2 +(pc)2 = (mc2)2. for the momentum of the moving proton? I don't understand why this equation is true, since we have in general E^2 =p^2c^2+(mc^2)^2?

 

[vela]

That was a typo. It should be (K+mc²)²-(pc)²=(mc²)².