Special Relativity -- Kinetic Energy

[Barry Melby]

Homework Statement

An electron e− and positron e+ moving at the same speed in the Earth reference frame collide head-on and produce a proton p and an antiproton p¯. The electron and positron have the same mass. The proton and antiproton also have the same mass. The mass of the proton is 1836.15 times the mass of the electron.

Calculate, in the Earth reference frame, the minimum value possible for the ratio of the electron's kinetic energy to its internal energy in order to have the reaction e− + e+ →p + p¯ take place.

Want: K/E_internal = ?

Homework Equations

K = (y-1)mc^2 --- y = lorentz factor
E_internal = mc^2

The Attempt at a Solution

I'm not even remotely sure what this question is asking. Can someone offer me some guidance?

 

[nrqed]

Homework Statement

An electron e− and positron e+ moving at the same speed in the Earth reference frame collide head-on and produce a proton p and an antiproton p¯. The electron and positron have the same mass. The proton and antiproton also have the same mass. The mass of the proton is 1836.15 times the mass of the electron.

Calculate, in the Earth reference frame, the minimum value possible for the ratio of the electron's kinetic energy to its internal energy in order to have the reaction e− + e+ →p + p¯ take place.

Want: K/E_internal = ?

Homework Equations

K = (y-1)mc^2 --- y = lorentz factor
E_internal = mc^2

The Attempt at a Solution

I'm not even remotely sure what this question is asking. Can someone offer me some guidance?

You have the correct equations, so you need to figure out the ratio which is simply $\gamma-1$.
To find that, the key point is that you want the minimum energies for the e+ and e- to have the reaction take place. This is a special case. What can you say about the proton and antiproton produced then?

 

[Barry Melby]

You have the correct equations, so you need to figure out the ratio which is simply $\gamma-1$.
To find that, the key point is that you want the minimum energies for the e+ and e- to have the reaction take place. This is a special case. What can you say about the proton and antiproton produced then?

Well i know that the mass of the proton and the antiproton are 1836.15 times the mass of the electron and positron. Would $\gamma$ = 1/(1-1836.15^2)?

 

[Barry Melby]

Well i know that the mass of the proton and the antiproton are 1836.15 times the mass of the electron and positron. Would $\gamma$ = 1/(1-1836.15^2)?

No no, never mind. That wouldn't make sense.

Where do i go?

 

[nrqed]

No no, never mind. That wouldn't make sense.

Where do i go?

As I mentioned earlier, the key point is that you want the *minimum* energy for the e+ and e- to have the reaction take place. This is a special case. What can you say about the proton and antiproton produced then?

 

[Barry Melby]

I don't really understand what you're saying.

Do i have to do something with momentum?

 

[nrqed]

I don't really understand what you're saying.

Do i have to do something with momentum?

In a general problem, you would have to apply conservation of momentum. Here, you are luckier because that won't be necessary. What is the speed of the proton and antiproton produced, in your question?

 

[Barry Melby]

it doesn't appear that the proton or antiproton are moving at all.

 

[nrqed]

it doesn't appear that the proton or antiproton are moving at all.

Exactly! Now, what is the total energy of the system after the reaction?

 

[Barry Melby]

Exactly! Now, what is the total energy of the system after the reaction?

The total energy would be zero i would suspect because they have no velocity. How does this relate to $\gamma$?

 

[nrqed]

The total energy would be zero i would suspect because they have no velocity. How does this relate to $\gamma$?

Not quite. The total energy is not just kinetic energy, in relativity. It includes rest mass energy. So what is the total energy, taking this into account? (Once you find that, you can write an expression for the total energy before. Using conservation of total energy, you will get an equation for gamma that you can then solve for.

 

[Ray Vickson]

The total energy would be zero i would suspect because they have no velocity. How does this relate to $\gamma$?

$E = M c^2 = 0$ only if $M = 0$.