00:01
In this problem we have the lambda particle decaying into a proton and a pion.
00:16
And we observed that the proton is at rest at the end of this production, let's say.
00:25
So there are two parts in this problem.
00:35
We are going to compute the energy of the pion, and the initial energy of this original lambda.
00:45
That's it.
00:46
We are going to, of course, employ the relativistic equations for the energy and momentum conservation.
00:53
Well, actually, they are the same energy and momentum conservation equations as in the non -relativistic physics, but the energy expression will be the relativistic energy.
01:05
Okay, so let's get started with the first part.
01:09
What is the energy of the pylon.
01:13
So we write down the conservation of energy and momentum separately.
01:22
Okay, one to entry.
01:23
My convention is like this.
01:26
We have the first particle going into the second one and the third one.
01:32
So one is lambda, two is proton, three is pyon.
01:40
As for the energy, we have e1 equal to e2 plus.
01:44
E3 now we are given that p2 is equal to zero so this gives us p1 equal to p3 let's call them collective the p2 equal to zero also gives us e2 equal to okay now the relativistic energy p2 sucard c squared plus m2 squared c to power 4 so this guy will become m2c squared there's no momentum here in this energy expression now let us focus on the energy equation again this guy we have e1 equal to m2 c squared plus e3 let us take the score of both sides e1 squared equal to m2 squared c to power 4 plus e3 plus 2m2c squared e3 the reason is as follows in e1 the e1 is given by p squared c squared plus m1 squared c2 power 4.
03:36
Okay, let me write it down very quickly.
03:41
So here p squared c squared plus m1 square square c squared 4.
03:49
And e3 is given by the same momentum, p squared, c squared, plus the mass is different.
03:58
M3 squared c to power 4 so i want to take the square of both sides so that we will get rid of these square roots and i'm going to solve this equation for e3 because i want to consider the difference between e1 and e3 e1 squared and e3 squared so that these momentum squared terms will cancel out so that this momentum squared terms will cancel out so so that's why i am taking the square at this very step.
04:33
So let me just raise this because i will demonstrate this shortly in detail.
04:45
Okay, now let's solve this equation for e3.
04:52
This is equal to e1, squared minus m2, squared c to power 4 minus e3, squared divide by 2, 2, m2 squared, c, m2, c squared.
05:12
Okay, now let us insert the old energy expressions.
05:24
E3 is equal to p squared, c squared, plus m3 squared, c2 power 4 under the square root.
05:33
This is equal to p squared, c squared, plus m1, squared, c2 power 4 this is e1 squared minus m2 squared c2 4 minus okay let's write down a 3 squared we have p squared c squared plus m3 squared so this is divided by 2m2 c squared so we have this nice cancellation between the momentum so that there are only masses in the right -hand side so this the beauty of this trick or trick of taking the square of this energy expression and it works only for very special cases like one of the momentum is zero definitely so it's a nice case otherwise we would have to use the most general formalism which is a bit tricky it takes time sold it okay now let's simplify the right -and -side with this cancellation we have m1 squared minus m2 squared minus m3 squared c squared over 2 m2 let us take the square of both sides so we have m1 squared minus m3 minus m3 squared squared squared squared c2 per 4 xx4 xx4 m2 squared and let us solve this equation for this unknown momentum v yet 1 over c squared this guy minus m3 squared s 2 4 square root.
08:08
Now it is, there is a very trivial cancellation here if you expand this square of three terms, three masses, i mean this term.
08:22
And if you make the subtraction after equating the denominators, it will take just one step, but it will have length.
08:32
So let me just simplify the result for you.
08:35
Which is in fact a well -known expression in particle physics.
08:41
And the resultant function, okay, let me write this way.
08:46
So we will have the following.
08:50
There will be this lambda function which i will define very shortly times c squared after canceling out these powers...