00:01
In this problem, we're looking at a proton collision.
00:04
So it's two protons, the masses are the same.
00:08
One proton is stationary.
00:09
The other is coming in at some momentum p1.
00:12
The problem asks if the kinetic energy 110 megachron volts is enough to produce a pion from that collision.
00:21
The rest mass of the pion is 139 .6 million electron volts per c squared.
00:29
So just by spec, the answer to that is going to be no.
00:33
The most efficient scenario is if the equal masses are moving equal and opposite directions and collide head -on and then come to a dead stop because a total momentum of the system is zero in that scenario.
00:50
Then all of the kinetic energy is available to be converted into the rest mass of a new particle.
00:58
But the important thing is, in this problem, is you might like, in the same thing we want to say, oh, well, 139 .6 is the kind of energy you need to produce the pion.
01:11
But that's not true because if one proton is incoming with a high speed and the other is stationary, then when they collide, the, if there's enough energy, to produce a pion, then the total momentum in the system is not zero.
01:35
This situation, e1 plus p2 equals p total.
01:44
That means the composite system that results, the two protons in the pion, have the momentum of the incoming particle.
02:00
And so what happens is you're losing kinetic rna and g, to the momentum post -collision.
02:11
So if they were equal and opposite, the new rest mass that you can get increases linearly or proportionally with k, the kinetic energy.
02:22
But if one of the protons is stationary, then the mass increases only going to go as a square root of the kinetic energy.
02:33
So you're losing energy, essentially to the fact that the system has momentum after the collision.
02:43
So this problem, as a result, is significantly harder than the previous problems that were similar in nature.
02:55
So there's two ways you might go about doing this.
02:58
One way is to convert this collision into the center of mass between the two protons so so that the protons are coming in at equal and opposite speeds, which would mean that they have the same gamma.
03:16
If you were to write the momentum's out explicitly, gamma, m speed one, gamma, m speed two.
03:26
The speeds are equal in magnitude.
03:30
The gamas are the same.
03:32
Then you would set the total energy to include the rest mass energy of the pion on the right -hand side.
03:39
Then you would calculate the speed of the protons out of the gamma term, and then you would switch back to the stationary proton frame, and then use relativistic velocity addition to find what the speed is of the incoming particle under that frame conversion, because that encodes that extra energy for the pie and then you calculate the kinetic energy from the speed.
04:16
I don't like dealing with gammas and speeds and relativistic velocity addition, so i'm not going to do it that way.
04:25
I'm going to keep the original frame of reference.
04:27
I'm going to make a total inelastic collision where there's some new combined mass m and that whole system has the total momentum of, it's the same as the momentum and the incoming particle.
04:53
I'm going to then, you know, apply that conservation momentum to simplify the equations.
04:59
They're all going to be energy equations.
05:01
There's going to be no speed terms.
05:03
I'm going to rearrange the fine k in terms of the rest mass energies, where this new composite mass includes the proton and pion masses.
05:12
So all that being said, we have to write energy conservation equations.
05:22
So the energy of the incoming particles, it's kinetic energy.
05:30
I'm not going to put a subscript on because it's the only kinetic energy in the problem.
05:37
So e2, the second proton, is stationary, so it has no kinetic energy.
05:44
So it has just the rest mass energy.
05:48
Now, in general, the energy momentum relationship is this.
06:07
You can calculate, you can derive that from the total energy gamma -mc -squared and the momentum gamma -muse.
06:20
You can just rearrange stuff and eventually derive that.
06:26
So that means for the first particle, you have the incoming term.
06:45
Since that, e2 squared, this is zero because the momentum is zero for the second particle.
06:59
And then ef, this is the composite post -collision energy...