WEBVTT
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In our last lecture, we had discussed the
Doppler effect in light. As we had said that
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Doppler effect is sound is well-known, we
visualize many times in our daily life. But
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in light, it is quite is like different treatment
and probably not so much apparently known.
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We also discussed a special Doppler effect,
which you call as a transfer Doppler effect,
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which is normally it is not have any classical
law will not be sound for that matter. These
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two aspects we discussed then we gave some
flavor of some experiments in which Doppler
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effect is used.
Just to give an idea that you know the Doppler
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effect that we have dop has certain usefulness
as far as certain part of physics is concerned
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effect. And even in medical science Doppler
effect is used quite effectively. So, I think
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Doppler effect is finding more and more uses,
but what we did in last lecture just to give
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some idea about the Doppler effect. Today
we would like to go to a slightly different
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concept basically it is for problem solving.
We define what we call as a center of frame
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of reference. Normally in classical mechanics
when we are not dealing with relativity, center
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of mass is always described in a fashion,
such that we have a standard definition of
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a center of mass how to find it out. And if
we change our frame of reference to that center
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of mass of a system of particles, then we
call that particular frame of reference as
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center of mass frame of reference. The advantage
of dealing with center of mass frame of reference
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is that, if there is no external force with
the system center of mass becomes an automatic
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inertial frame of reference.
Because if there is no external force to a
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system of particles, the center of mass must
be at rest or must moving with constant velocity
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as seen from any other special frame of reference.
So, if you move our self to the center of
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mass automatically we are landing up ourselves
into an inertial frame of reference. Now,
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we slightly modify this particular definition
as we have discussed earlier.
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And we say that center of mass frame of reference
is that frame of reference in which the sum
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of the momentum of all the particles is 0,
this center of mass frame of reference is
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occasionally referred as C frame. So, what
we have said that center of mass frame is
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a frame in which the sum of momentum of a
system of particle is zero. Normally as we
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have already said that there is a set of particles
which considered is a system in which we are
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interested in. So, there could be a force
inter in between the particles, it means a
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force being applied by one particle on any
other particle that we call as internal force,
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and if f force is applied by a particle, which
is not part of the system that is what we
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call as external force.
So, I repeat that if we find that some of
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the moment, if we frame of reference in which
the sum of momentum of all the particles of
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the system is 0 that is what we call as C
frame or center of mass frame. Similarly,
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we can define or we generally call a L frame
as a laboratory frame of reference. This frame
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of reference is a frame in which you have
describe your problem or you describe your
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experiment or in which you were performed
the experiment.
04:04.590 --> 04:08.790
This frame of reference could be a C frame
or center of mass frame of reference or could
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be any other inertial frame of reference.
So, I may describe a problem my problem in
04:15.080 --> 04:19.690
a frame of reference which does not happen
to be center of mass frame of reference or
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I could choose to describe the problem only
in a center of mass frame of reference. So,
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laboratory frame of reference is generally
a frame of reference in which we have described
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my problem or my experiment.
Now, why we talk specifically of C frame specially
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relativity will see that many problems become
very simple if we work it out in the C frame.
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Though the problem may not have given to you
C frame it might have been given in some other
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frame, but still if you transform the problem
into C frame the problem that is how to be
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simpler. So, in today’s lecture we will
be essentially discussing some problems in
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which we will find the use of C frame to be
convenient frame and problem easy frame.
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So, this is what I have written. In STR which
is special theory of relativity problems can
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be simplified when transformed to C frame.
I would also like to mention that when we
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actually perform many HEP, HEP is stands for
high energy physics, if you perform experiment
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like collision experiments, scatter experiments
in high energy physics. Many times we prefer
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would design this experiment in C frame of
reference in center mass frame of reference.
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So, that overall there is energy which will
be seen little later that you know energy
05:38.310 --> 05:42.840
that is turns to be in smaller C frame of
reference. Basically, because you do not have
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spend energy to accelerated to mass frame
of reference. Effectively means that if you
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have an particular particle and you want to
design an experiment, which another particle
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comes and colloids it. In that case this frame
of reference is not a center of mass frame
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of reference because the net momentum is not
zero.
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So, you rather like to have an experiment
in which both the particle as come together
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and colloids such that center mass. So, that
it become in a center of mass frame of reference
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or the net initial momentum is zero. So, let
us come to one particular problem which is
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actually simple problem, but it illustrates
the fact that how it can become simple if
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it is solved in C frame. And it is rather
difficult to solve in L frame even though
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the problem appears to be very simple.
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Let us look at the problem; there is a particle
of rest mass capital M naught, which has a
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total energy of 3 M naught c square. Now,
we must be clear that when we are talking
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of total relativistic energy, it means it
includes rest mass energy. So, the total energy
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which means kinetic energy plus rest mass
energy is 3 M naught c square. This particular
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particle decays into two identical particles.
So, this particular particle goes away and
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eventually decays into two particles each
one of them have a rest mass of M naught.
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So, their identical particles whatever might
be the particles may not be a relativistic
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sort of experiment just for understanding
the problem.
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So, in decays into two masses two particles
which both of them which have a rest mass
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of M naught. Now, I have to find out what
will be the velocity of these particular particles,
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when they have decayed. Given the fact that
the decay product, it means the both the particles,
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move along the direction of the motion of
the present of the parent particle. Let me
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just read again, a particle of rest mass capital
M naught and total energy 3 M naught c square,
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decays into two identical particles of rest
mass small M naught. Find the velocities of
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the two particles given that the decay products
move along the direction of the motion of
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their parent particle.
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If I want to draw into a small and simple
picture essentially it means that, this is
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a particle which is of mass rest mass capital
M naught. This is moving in a given frame
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and eventually decays into two particles two
small particles. Each one of them has rest
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mass M naught; obviously, because it has to
conserve energy and conserve momentum. Therefore,
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this particular particle also be moving what
has been told that both these particles move
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in this same direction as this particular
particles, it means the this also moving in
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this direction this direction or opposite
direction.
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But the total motion is confined only to one
dimension which we can call for an example
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direction that does not matter what direction.
But everything the problem is just a single
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dimension problem a one dimensional problem
in which everything has been described. On
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a traditional classical mechanics problem
where we have to conserve energy and momentum
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this problem would have been absolutely simple.
But we will see that if we try to work out
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this problem this particular fashion.
You see that there are certain complications
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and it is not easy to work out this particular
problem, until we go to a different frame
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of reference which is the center of mass frame
of reference. Let us look at this particular
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problem, first try to let us attempt to solve
only in the L frame the laboratory frame the
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weight has been described. We will discuss
what are the difficulties which comes across.
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So, as we said the idea is that we need to
apply the conservation of energy and conservation
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of momentum in this particular frame. Energy
has been given because there is only one particular
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particle initially, so energy has been given
s 3 M naught c square. So, I have to find
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out this energy and we have to find out corresponding
what is a momentum of particular particle.
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Once we find the momentum, I know the initial
energy, I know the initial momentum it decays
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into two particles. We have to write initial
energy is equal to the sum of the energy of
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the resultant particles. Similarly, initial
momentum vector is equal to the initial the
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final momentum that is what we have to do.
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So, let us first find out the initial momentum,
we have been given that E is equal to 3 M
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naught c square. And if you remember what
we have said that the energy is given by gamma
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M naught c square, where M naught is a rest
mass of any particular particle.
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And in this particular case because the energy
has been given as 3 M naught c square, it
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is quite clear that gamma is equal to 3, only
thing because this has been described in laboratory
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frame of reference. So, let us write a subscript
L to make it clear that this gamma that we
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are talking or the velocity that I will be
talking on which this particular gamma depends.
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It is actually the velocity in the L frame
or in the laboratory frame.
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Now, once we have this particular gamma L,
I can always find out what is velocity of
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this particular particle. And once I find
out the velocity of the particle I can always
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find out the momentum of the particle. Of
course, I can also use direct energy momentum
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relationship that also I will use just to
show that we get the same result. Let us first
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try to find out the velocity value using this
gamma L.
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If you remember the value of gamma the expression
for gamma is given as one upon under root
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1 minus v square by c square, sometimes we
are using v, sometimes we assume. But I should
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be clear now when we use v when we use u.
Now, I want to solve it for v upon c, I want
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to know if I want to have I know the value
of gamma I want to find out the value of v
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by c. So, let me square it and take inverse
of it, if I square and take inverse of it
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I will get one upon gamma square, square it
under this goes away.
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Take inverse, so 1 minus v square upon c square
comes on the numerator, this becomes 1 minus
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v square by c square. So, I can solve this
particular equation. So, this becomes v square
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by c square is equal to 1 minus 1 by gamma
square, which means v by c square is what
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we call as beta square in our notation.
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So, I can write this as beta square as sorry
there is no do it here, beta square is equal
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to gamma square minus 1 upon gamma square.
So, this is the expression which I have written
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this particular transparency that beta is
equal to under root gamma square minus 1 upon
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gamma square. So, this expression gives me
the value of beta which is v by c, using the
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value of gamma that we find, in the present
case gamma is 3. So, I substitute it here
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this becomes three square minus 1, this becomes
under root 8 divided by gamma square which
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is 9, this 9 can be taken out of the under
root sign. So, this becomes under root 8 by
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3. So, this p this beta is under root 8 by
3, which is v by c.
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Now, once I know the speed I can always find
out what will be the momentum of the particle.
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And the momentum of the particle as we know
is given by the momentum is equal to gamma
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M naught v. We know gamma is equal to 3, M
naught is M naught and v is under root 8 by
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3, c we have just not found out. So, I have
substituted this in this particular expression.
13:55.899 --> 14:01.819
So, p L, L is again because I am describing
this laboratory frame is equal to 3 multiplied
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by M naught multiplied by root 8 by 3 c, which
gives me root 8 M naught c.
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Force as I said I could have found out this
particular expression also by using directly
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the energy momentum relationship, which is
E square is equal to p square c square plus
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M naught square c to the power 4 this is very
commonly used relationship. So, from this
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particular thing I have to evaluate momentum
I know the energy. So, p square c square will
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turn out to equal to be E square minus I take
this on the other side, minus M naught square
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c to the power 4, E being 3 M naught c square,
E square will become 9 M naught square c to
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the power 4.
So, I substitute it here, I have to subtract
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this expression minus M naught square c to
the power 4. So, this becomes 8 M naught square
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c to the power 4, c square I will cancel out
and p I will get under root 8 M naught c.
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So, what we have done so far I know the energy
the initial energy of the system which happens
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to be just one particular particle, which
now eventually decays into two particles.
15:05.419 --> 15:12.419
So, this particular particle has a total energy
of 3 M naught c square and it has a total
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momentum of under root 8 M naught c.
So, once this particular particle decays into
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two particles, the total energy must remain
same. Similarly, the total momentum must remain
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same because this purely one dimensional problem.
So, I have not taken the x axis and y axis
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because just described. I have not described
it in the vector form or I need that this
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particular value of momentum must be conserved.
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So, I will be able to write the following
equations which I have written in the next
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transparency. First we have found out what
is the energy and the momentum of the original
15:47.559 --> 15:52.869
particle, what we have to do is to write the
energy and momentum conservation equations.
15:52.869 --> 15:59.869
We have seen it has been given in the problem
that the initial energy is of is equal to
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3 M naught c square, there is only one particular
particle so there is only one particular particle
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system.
Once they are two particles, which have resulted
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out of the decay of this particular particle
if E 1 L and E 2 L are the energies, then
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3 M naught c square must be equal to E 1 L
plus E 2 L. Now, corresponding to 3 M naught
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c square I have seen that the value of the
momentum is under root 8 M naught c remember
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this is original particle, so I have used
capital M naught here.
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This must be equal to the sum of the momentum
of the two particles we have been told that
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the particles move along the same direction
its original direction. So, I can write just
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in a scalar form, I do not have to use vector
form because as we have said this is purely
16:43.819 --> 16:50.319
one dimensional problem. So, this must be
equal to p 1 L plus p 2 L, where p 1 L is
16:50.319 --> 16:54.309
the momentum of the first particle, p 2 L
is the momentum of the second particle. As
16:54.309 --> 16:59.720
I have said E 1 L is the energy of the first
particle E 2 L is the energy of the second
16:59.720 --> 17:01.869
particle.
In a traditional classical mechanics this
17:01.869 --> 17:07.059
would been a very very simple trivial dissemble
problem, all you have to do is to express
17:07.059 --> 17:13.780
this energy into momentum in the form of momentum
or this momentum get form of energy. And just
17:13.780 --> 17:18.319
to solve this equation you have to one knows
and two equations to solve it, this one dimension
17:18.319 --> 17:24.289
is simple. Mainly because in traditional classical
mechanics in which there are no relativity,
17:24.289 --> 17:29.210
the relationship between E and p are purely
simple. Of course, in that case rest mass
17:29.210 --> 17:36.210
what we try to do is the conserve the kinetic
energy, it not bother about that yes of course.
17:36.850 --> 17:40.809
Now, in this particular case what will find
out that energy and momentum relationships
17:40.809 --> 17:47.759
are little more involved, which has been written
in the next transparency.
17:47.759 --> 17:52.679
So, the energy of the first particle is related
to the momentum of the first particle by this
17:52.679 --> 17:58.960
particular expression, E 1 L square is equal
to p 1 L square c square plus m naught square
17:58.960 --> 18:05.169
c to the power 4. Remember I am using small
m not because the energy and momentum of the
18:05.169 --> 18:11.139
particle that I am going to relate has rest
mass of small m naught. This is not the original
18:11.139 --> 18:14.630
particle, the original particle has decayed\
the resultant particle, which I have got which
18:14.630 --> 18:18.700
has rest mass of M naught. And originality
in this particular problem the two particles
18:18.700 --> 18:23.659
have the same rest mass of this M naught that
is what I have given. So, I am using small
18:23.659 --> 18:29.720
m naught here in both these expressions.
Exactly similar equation I write for as a
18:29.720 --> 18:34.549
relationship between energy and momentum of
the second particle, E 2 L square is equal
18:34.549 --> 18:40.960
to p 2 L square c square plus m naught square
c to the power 4. Remained that this L has
18:40.960 --> 18:47.740
written just to make sure that I am talking
of laboratory frame of reference. As we will
18:47.740 --> 18:52.580
see that this is not easy to solve this equations
because if I try to express this particular
18:52.580 --> 18:58.759
equation in that particular form, let us say
let us try to express energy in terms of the
18:58.759 --> 19:03.559
momentum. Then I have to take under root of
this particular expression to find out what
19:03.559 --> 19:04.250
this E 1 L.
19:04.250 --> 19:11.250
So, I must write E 1 L is equal to under root
of p 1 L square c square plus m naught square
19:18.190 --> 19:25.190
c to the power 4. Similarly, for E 2 L I have
to write under root p 2 L square c square
19:30.419 --> 19:37.419
plus m naught square c to the power 4. When
I write in the conservation of energy expression,
19:38.049 --> 19:45.049
I have to write M naught c square is equal
to this plus this so this becomes this under
19:46.809 --> 19:53.809
root plus this under root. If you can solve
this equation of course, you will get a energy
19:57.809 --> 19:59.259
and momentum in fact, will get p 1 L and p
2 L.
19:59.259 --> 20:02.820
Taking the other two other equation also to
consideration writing p 1 L also in terms
20:02.820 --> 20:07.500
of the initial momentum I will be able to
solve it. But as you can see because of the
20:07.500 --> 20:11.870
presence of these square roots etcetera these
are not very straight forwarding equations
20:11.870 --> 20:16.710
to solve. On the other hand this equation
becomes fairly this problem becomes fairly
20:16.710 --> 20:21.769
simple if I go to the center of mass frame
of reference. So now, let us attempt to solve
20:21.769 --> 20:26.299
this particular problem in center of mass
frame of reference of course, here the problem
20:26.299 --> 20:29.179
fairly simple because there is only originally
one particle.
20:29.179 --> 20:34.120
So, if I have to find out what is the velocity
of the center of mass it is going to the velocity
20:34.120 --> 20:38.759
of the particle the initial particle that
I am talking about. So, I go to a frame of
20:38.759 --> 20:45.029
reference in which the initial particle is
at rest and because in this frame the particle
20:45.029 --> 20:49.750
is at rest the momentum is 0, there is only
one particle. So, that automatically becomes
20:49.750 --> 20:54.250
the C frame or the center of mass frame of
reference. So, let us describe let us go to
20:54.250 --> 20:59.730
a frame of reference in which the original
incident particle or un-decayed particle will
20:59.730 --> 21:03.100
not be incident, but un-decayed particle was
at rest.
21:03.100 --> 21:10.100
So, this what I have said, let us go to C
frame as there is only one single particle
21:11.889 --> 21:15.740
we shall have the following conditions. I
know very clearly because if I have gone to
21:15.740 --> 21:19.929
the frame of reference of that particle itself
the particle if not moving in that frame.
21:19.929 --> 21:24.149
Therefore, its total energy has to only the
rest mass energy that cannot be any other
21:24.149 --> 21:31.149
energy therefore, the total energy C frame
will have to be just m naught c square. And
21:32.659 --> 21:37.850
of course, if the speed of the particle is
0, then momentum is also 0. So, therefore,
21:37.850 --> 21:44.330
the C frame the initial value of momentum
is zero which has to be if it is to be called
21:44.330 --> 21:51.330
as C frame and the total initial energy is
just capital M naught c square.
21:52.370 --> 21:58.419
Now, let us try to see if I have to conserve
energy and momentum in this particular frame
21:58.419 --> 22:05.169
of reference, remember because the rest mass
of the two particles that I am getting as
22:05.169 --> 22:10.799
a result of decay happened to be same. The
only way these particular particles can decay
22:10.799 --> 22:17.799
is when they move back-to-back, it means one
particle goes in this particular fashion,
22:19.799 --> 22:25.019
another particle of course, goes in this particular
fashion. That is only way that they can conserve
22:25.019 --> 22:28.169
and make momentum 0, because remember initial
the momentum is 0.
22:28.169 --> 22:33.740
So, the final momentum also has to be 0, if
both the particles moves in the same direction
22:33.740 --> 22:39.590
there cannot be momentum cannot be 0, we have
been told that a only one line. So, there
22:39.590 --> 22:45.299
is only possibility is that one goes in this
particular fashion, another goes in this particular
22:45.299 --> 22:52.299
fashion. And because their rest mass is happened
to be same therefore, there energy has to
22:52.460 --> 22:58.889
be equally shared because in order to conserve
momentum both must have same momentum obviously.
22:58.889 --> 23:03.429
And energy and momentum relationship depends
on their rest masses, their rest masses being
23:03.429 --> 23:08.620
same in forces that their energy is also same.
Let me just explain this particular particle
23:08.620 --> 23:10.179
little bit more.
23:10.179 --> 23:16.809
So, we have E square is equal to p square
c square plus m naught square c to the power
23:16.809 --> 23:23.809
4. If first particle p’s is same as the
second particle, m naught is same as a second
23:24.470 --> 23:29.899
particle, then E of that particle also has
to be same as second particle. Therefore,
23:29.899 --> 23:35.429
the two particles in the center of mass frame
of reference must have same energy as well
23:35.429 --> 23:39.960
as the same momentum. Same momentum because
the initial momentum was 0, so final momentum
23:39.960 --> 23:45.210
also has to be 0 and energy have to be same
because the rest mass is also happened to
23:45.210 --> 23:47.750
be same.
Therefore, we can see that the problem has
23:47.750 --> 23:54.750
to be simple, it means each one of the particles
must be having the energy of half m naught
23:54.889 --> 24:00.429
c square. So, this is what I have written
here that M naught c square should be equal
24:00.429 --> 24:07.379
to 2, this is I can write as gamma c m naught
c square. So, this m naught c square these
24:07.379 --> 24:10.120
are the total energies have to be equally
shared.
24:10.120 --> 24:17.120
And this energy of each particle can always
be written as we just now have said gamma
24:17.460 --> 24:22.360
M naught c square and because this center
of mass frame of reference, so I am writing
24:22.360 --> 24:28.730
gamma c. So, each particle must be having
the same energy this two must be equal to
24:28.730 --> 24:33.970
the total initial energy in this frame, which
is M naught c square. So, this is what I have
24:33.970 --> 24:39.529
written M naught square M naught c square
is equal to 2 gamma c m, small m naught c
24:39.529 --> 24:45.360
square. It immediately gives me the value
of gamma c, which is c square cancels here
24:45.360 --> 24:52.139
M naught divided by 2 small m naught. Once
we have found out the value of gamma c I can
24:52.139 --> 24:55.840
immediately find out what will be this p’s
of this particular particle, in this particular
24:55.840 --> 25:01.929
frame of reference by using the expression
which you have just written earlier.
25:01.929 --> 25:06.879
So, the two particles will move in a opposite
direction with the following speed, this expression
25:06.879 --> 25:11.710
is same expression which we just now I have
worked out little bit earlier. And if I substitute
25:11.710 --> 25:15.850
the value of gamma c which I have obtained,
I will get that this speeds of the particle
25:15.850 --> 25:21.559
will be like this, by will given by this expression.
One moving in plus direction another moving
25:21.559 --> 25:25.820
in minus direction, you can call it x direction
one moving in plus x direction another moving
25:25.820 --> 25:29.879
in minus x direction. So, these are the two
velocities.
25:29.879 --> 25:35.409
But what we have found out are the velocities
in C frame of reference, if I have to find
25:35.409 --> 25:41.029
out 11 laboratory frame of reference, all
I have to do is simple velocity transformation.
25:41.029 --> 25:47.320
I have to know what is the velocity of the
C frame, which by the way we have found out
25:47.320 --> 25:51.340
earlier. Once we know the velocity of that
particular frame take care of proper science
25:51.340 --> 25:58.340
and you can transform and obtain the values
of speeds or some velocities in laboratory
25:58.480 --> 26:03.799
frame of reference. So, we transform into
C frame then brought it back to L frame, but
26:03.799 --> 26:06.679
remember the problems very simple, we do not
have deal with all those under roots and time
26:06.679 --> 26:11.440
to work out and trained to solve those equations
problems essentially very simple.
26:11.440 --> 26:17.519
So, these I have I have written these speeds
have only to be transformed back to L frame
26:17.519 --> 26:23.749
taking care of appropriate sign. The speed
of C frame in L frame needed to apply transformation
26:23.749 --> 26:30.749
this was already found out earlier, if you
remember earlier, the earlier case which we
26:30.899 --> 26:37.899
have done here. We already found out that
beta is equal to under root 8 by 3, so from
26:38.720 --> 26:42.299
that we can found out p is equal to under
root 8 by 3 c. Once we know the relative velocity
26:42.299 --> 26:47.879
between the frames we know the velocities
of the particles and C frame moves a velocity
26:47.879 --> 26:52.399
transformation bring it back to laboratory
frame. So, will not work it out further I
26:52.399 --> 26:55.460
think this work go out roots simply.
26:55.460 --> 27:02.460
Now, let us go to a slightly more involved
problem it is not really involved it is slightly
27:02.870 --> 27:09.870
more in the sense that instead of one initial
particle we have two particles and things
27:10.269 --> 27:17.269
are not really one dimension equation. So,
let us read the problem, an electron of total
27:18.639 --> 27:25.129
energy 1.4 MeV, again when we say total energy
it means it is root less mass energy is total
27:25.129 --> 27:32.129
energy is 1.4 MeV. M is stands about 6, electron
volt collides with another electron, which
27:32.230 --> 27:35.929
is at rest in L frame. So, in laboratory frame
of reference there is one electron which is
27:35.929 --> 27:38.240
at rest, another electron comes and hits it.
Colloids it or it can scatter or gets scattered
27:38.240 --> 27:45.240
whatever you want to call it that is what
important.
27:48.169 --> 27:51.960
After the collision the target electron, which
means the electron which is originally at
27:51.960 --> 27:58.960
rest is found to get scattered at an angle
of 45 degrees, but not in the laboratory frame,
27:59.210 --> 28:03.799
but in the C frame in the center of mass frame
of reference. So, the problem involves both
28:03.799 --> 28:08.830
laboratory frame and C frame. So, in that
sense the problem is little more involved.
28:08.830 --> 28:15.279
So, after the collision the target electron
which has originally the electron which is
28:15.279 --> 28:18.690
at rest is found to get scattered at an angle
of 45 degree in C frame.
28:18.690 --> 28:23.749
So, from the nature of the problem itself,
particular problem itself is cleared that
28:23.749 --> 28:29.799
we have to give talking about C frame. Find
the energy and the momentum components of
28:29.799 --> 28:36.799
target electron after the scatter in S and
C in C and L frames. So, after this scatter
28:37.450 --> 28:44.450
after the scattering has taken place what
are the momentum and energies of target electron
28:46.999 --> 28:51.730
in the frames of course, rest mass energy
of the electron has been given as 0.51 MeV.
28:51.730 --> 28:56.950
So, I think the problem is clear that this
one particular electron, which is being hit
28:56.950 --> 29:03.730
by an another electron, one electron is at
rest, another electron comes and hits here
29:03.730 --> 29:07.090
hits it.
The electron which was at rest is found to
29:07.090 --> 29:13.429
move it an angle of 45 degrees in center of
mass frame of reference, k of because when
29:13.429 --> 29:16.619
you are talking of 45 degrees. It means it
no longer a one dimension it has to be you
29:16.619 --> 29:23.190
have to work out in two-dimension. And then
you have to find out what will be the energy
29:23.190 --> 29:27.450
and momentum of the particle, work out for
the both the particles necessary or target
29:27.450 --> 29:33.809
electron in laboratory frame of reference
as well as C frame of reference.
29:33.809 --> 29:38.700
What are the issues in this thing; first of
all it is a two particle system from the beginning
29:38.700 --> 29:43.889
as we have said lets no longer a one particle
system. So, like an one particle is very easy
29:43.889 --> 29:46.909
to find out what is center of mass frame of
reference, because you have to just go to
29:46.909 --> 29:49.830
the reference frame of reference of that particle
itself in which this particle has to be at
29:49.830 --> 29:56.830
rest. So, here it is not so obvious, but we
have to work it out.
29:58.139 --> 30:02.999
The first method which is slightly I mean
it is straight forward method, but little
30:02.999 --> 30:08.330
more longer method in fact will give you a
comparatively more straight forward method.
30:08.330 --> 30:14.559
Let us assume that C frame travels in the
laboratory frame of reference with a speed
30:14.559 --> 30:19.460
v. Find the momentum and energy on both the
particles in the C frames. So, assume could
30:19.460 --> 30:25.919
in an arbitrary frame of reference, which
moves relative to L frame with a speed v.
30:25.919 --> 30:30.440
Then find the momentum and energy on both
the particles in the C frame, then take the
30:30.440 --> 30:35.309
sum of the momentum put it equal to 0. Because,
I am looking for that particular frame of
30:35.309 --> 30:38.450
reference in which the sum of momentum is
0.
30:38.450 --> 30:43.159
So, once I put I take the momentum of first
particle and momentum of second particle in
30:43.159 --> 30:48.340
a frame of reference, which is moving with
a speed v relative to L. And equate this particular
30:48.340 --> 30:55.340
momentum sum of the momentum to 0 and solve
for v that will be the velocity of the center
30:55.759 --> 30:58.580
of mass frame. So, this is very straight forward
standard method of finding out even if we
30:58.580 --> 31:03.710
had n particles that is what I am doing principle.
If I want to do in most simple fashion then
31:03.710 --> 31:09.190
find out the momentum go to a particular frame
of reference with moves with a speed v.
31:09.190 --> 31:13.590
Transform all the momentum to that particular
frame of reference, takes summation of momentum
31:13.590 --> 31:20.309
put it equal to zero. That velocity that will
obtain will be the velocity of the center
31:20.309 --> 31:25.210
of mass frame of reference. So, let us first
work it out that that way, before we go to
31:25.210 --> 31:30.799
more involved little more trickier way, but
at simple method.
31:30.799 --> 31:35.450
So, let us first find out the total initial
energy of the two electrons, as we have seen
31:35.450 --> 31:42.450
that one electron is at rest in the laboratory
frame. So, I am doing this particular problem
31:43.239 --> 31:49.489
first in L frame. So, that is 0.15 electron
MeV and the other electron, which is at which
31:49.489 --> 31:55.419
is coming and hitting it that has a total
energy 1.4 MeV. So, total energy in the laboratory
31:55.419 --> 31:58.559
frame is 1.91 MeV. So, this is the total energy
which is available to me in the laboratory
31:58.559 --> 31:58.610
frame of reference.
31:58.610 --> 32:05.610
Let us work out on momentum. Second particle
is at rest and the target particle is at rest
32:09.110 --> 32:13.409
in the laboratory frame of reference. So obviously,
this is momentum is zero or we have to work
32:13.409 --> 32:19.049
out the momentum of the other particle which
is coming and hitting it here, find out what
32:19.049 --> 32:23.179
is the momentum of that particular particle.
I know the energy of that particle which is
32:23.179 --> 32:27.789
1.4 MeV, I use the same standard expression
E square is equal to p square c square plus
32:27.789 --> 32:33.399
m naught square c to the power 4. M naught
square c to the power 4 will just 0.51 square,
32:33.399 --> 32:37.639
0.51 MeV square, so this is what I have written
0.51 MeV square.
32:37.639 --> 32:43.460
So, this is p square c square plus M naught
square c to the power 4 and this was E square
32:43.460 --> 32:48.210
which is the total energy. So, I am solving
for momentum like we have did for earlier
32:48.210 --> 32:54.369
problem. So, this becomes 1.4 square minus
0.51 square, only thing which I have done
32:54.369 --> 33:00.100
I have taken a special units here. So, that
this particular normal issue had been p L
33:00.100 --> 33:06.460
c, but I have taken units of MeV by c. So,
this value of c I did not writing here. So,
33:06.460 --> 33:11.679
these this very conventional unit I would
traditional units sorry, in which we express
33:11.679 --> 33:16.769
momentum in the units of MeV by c energy by
c.
33:16.769 --> 33:23.220
So, energy we express in MeV or g E v whatever
depending upon the type of problem and express
33:23.220 --> 33:28.809
momentum in MeV by c or g V by c. So, the
momentum of this particular particle is one
33:28.809 --> 33:35.809
point we yes work it out sort it 1.304 MeV
by c of course, the momentum the second electron
33:38.629 --> 33:41.919
is 0 that is what we have just now said.
33:41.919 --> 33:45.159
Now, transform it; go to a frame of reference
which is moving with a speed v in the L- frame,
33:45.159 --> 33:52.159
I use momentum transformation. Momentum transformation
standard, gamma, what is the original momentum
33:54.220 --> 34:01.220
p x, p 1 x prime is equal to gamma p x minus
v e by c square. This is the momentum transformation
34:02.129 --> 34:09.129
equation momentum, we have just find out 1.304.
I am writing the unit to make it clear minus
34:11.119 --> 34:17.679
v is the speed relative speed between the
frames the energy is 1.4 MeV divided by c
34:17.679 --> 34:21.260
square.
For the second particle the initial momentum
34:21.260 --> 34:28.260
0. So, gamma multiplied by 0 into 0 minus
0, v into its energy is only is rest mass
34:28.929 --> 34:35.899
energy. So, v multiplied by 0.51 MeV divided
by c square. In fact, you just take the sum
34:35.899 --> 34:41.470
of these two put it equal to 0 solve for v.
That will give me immediately the velocity
34:41.470 --> 34:48.040
of the center of mass frame of reference.
That is what I have written for C frame E
34:48.040 --> 34:51.909
1 x prime plus p 2 x prime is equal to 0.
34:51.909 --> 34:58.909
So, just sum these things and work out for
v you will get v is equal to 0.683 c from
34:59.559 --> 35:04.710
this particular way of working it out. And
then once you know the velocity you can find
35:04.710 --> 35:10.890
out gamma, once you have found out gamma then
you can transform the energy and find out
35:10.890 --> 35:17.540
what is the energy of these particles in the
center of mass frame of reference. Then center
35:17.540 --> 35:20.690
of mass frame of reference initial some momentum
is anyway 0, energy momentum conservation
35:20.690 --> 35:27.690
and you will be able to solve the problem.
But as I told that is slightly quicker method
35:28.760 --> 35:34.430
to work it out by using the convention of
the 4 vectors. So, let us adopt the method
35:34.430 --> 35:38.859
two of working out this particular problem
going to center of mass frame of reference
35:38.859 --> 35:44.510
by using the concept of four vectors. As we
will see that this is slightly tricky, but
35:44.510 --> 35:46.690
it is much more simpler.
35:46.690 --> 35:51.839
So, this is what I have said method two, one
can use the fact that the length of four vector
35:51.839 --> 35:57.299
would be same in all the frames. We have discussed
it earlier, when we describe the concept of
35:57.299 --> 36:02.339
four vectors that the length of four vectors
is what we call as a four scalar. Thus we
36:02.339 --> 36:07.770
change the frame of reference the individual
components of the four vector may change,
36:07.770 --> 36:12.880
but its length is unchanged that is a four
scalar. So, once I go from one frame one initial
36:12.880 --> 36:16.220
frame to another initial frame of reference
length is going to be same.
36:16.220 --> 36:21.309
So, what I can do now calculate the length
of energy momentum four vector in the laboratory
36:21.309 --> 36:27.680
frame of reference for the system of the particles.
And I know that this particular length is
36:27.680 --> 36:32.670
going to be same in the center of mass frame
of reference and I know in the center of mass
36:32.670 --> 36:36.690
frame of reference summation of momentum would
be 0. So, it will have session of only four
36:36.690 --> 36:41.250
terms of energy if you remember the four terms
of the energy momentum four vectors, where
36:41.250 --> 36:48.250
p 1, p 2, p 3 and rather p x, p y and p z
and fourth term is i E upon c. So, fourth
36:48.510 --> 36:54.910
term depends on energy, let us just work it
out and see how similar this particular problem
36:54.910 --> 36:55.299
becomes.
36:55.299 --> 37:02.299
So, first let us write the length of the four
vector for this summation for the system of
37:02.369 --> 37:09.369
the particle. So, first three terms you are
taking the length. So, it will E 1 square
37:10.569 --> 37:15.460
plus E 2 square plus E 3 square plus a 4 square,
first three terms are only the momentum terms.
37:15.460 --> 37:19.819
So, essentially it means the first three terms
will give me a summation of summation of p
37:19.819 --> 37:26.240
L I square where L is the momentum the laboratory
frame of reference. Some over the all particles
37:26.240 --> 37:30.150
this case they are only two particles.
So, what I will do? I will add the momentum
37:30.150 --> 37:37.069
of the two particles and this particular term
will be obtained from that. Similarly, this
37:37.069 --> 37:41.490
particular term depends on the summation of
the energy of two particles because I am applying
37:41.490 --> 37:45.660
this particular I am calculating the length
of energy momentum four vector for the system
37:45.660 --> 37:50.410
of the particles, for two particles together,
not for the individual particles. So, when
37:50.410 --> 37:56.309
I am writing for the two particles together
this summation will be the sum of the energies
37:56.309 --> 38:02.430
of two particles; first particle has the energy
of 1.4 MeV, the second particle is at rest,
38:02.430 --> 38:05.790
it has only rest mass energy which is 0.51
MeV.
38:05.790 --> 38:11.660
So, summation of p L i will be only 1.91.
sum of first energy of first particle plus
38:11.660 --> 38:15.970
sum of energy sum of energy first particle
and the second particle, first particle having
38:15.970 --> 38:22.549
energy of 1.4, second particle having energy
0.51. Similarly, here summation of the momentum
38:22.549 --> 38:28.650
magnitude of the momentum for the first particle,
which is 1.4 and for other particle is 0.
38:28.650 --> 38:34.290
So, this becomes 1. Sorry, I am sorry… The
momentum of the first particle is 1.304 for
38:34.290 --> 38:36.640
the second particle is 0. So, just because
1.304 square.
38:36.640 --> 38:42.690
I calculate this number, this turns out to
be an imaginary number, the length of four
38:42.690 --> 38:47.160
vector can always be an imaginary, we have
discussed this problems earlier. So, this
38:47.160 --> 38:54.160
will be given by 1.396 i MeV by c. Now, I
know if I go to any other frame of reference
38:56.780 --> 39:02.020
this length is for the same system of particle
is going to remain same. But if I go to a
39:02.020 --> 39:06.680
specific frame which is called center of mass
frame of reference in that summation of momentum
39:06.680 --> 39:13.680
will be zero, it means this particular term
would be 0. So, if I want to write in center
39:14.559 --> 39:19.569
of mass frame of reference, this term is 0
and only the energy terms will be present.
39:19.569 --> 39:25.020
So, it is much easier to find out the energy
in the center of mass frame of reference,
39:25.020 --> 39:30.010
let us see how? First let us realize that
because the two particles that I am talking
39:30.010 --> 39:37.010
are 1 electrons. So, what will happen in the
center of mass frame of reference; one particle
39:39.020 --> 39:44.859
comes like this, another particle comes like
this, both are electrons. They have to come
39:44.859 --> 39:48.930
in opposite direction because they have to
make the total momentum 0, so their momentum
39:48.930 --> 39:55.930
must be in opposite direction. Now, they can
scattered one of the electron goes this way,
39:56.700 --> 40:02.010
another electron goes this way.
We have been told that this angle 45 degree,
40:02.010 --> 40:07.849
this has been given in the problem that is
angle is 45 degrees, all right? Now, the only
40:07.849 --> 40:12.710
way because initial momentum same as the final
momentum. So, this is the term goes this way
40:12.710 --> 40:17.299
making an angle of 45 degree, the other electron
has to go opposite back to back just like
40:17.299 --> 40:21.740
in earlier case, except for the fact that
is not in the same line. But now, tilted at
40:21.740 --> 40:27.700
an angle of 45 degree, but otherwise this
electron also to go to back to back because
40:27.700 --> 40:33.980
the rest mass of this electron is same, this
is electron. So, rest mass is same therefore,
40:33.980 --> 40:40.359
the energy has to be same and the momentum
is also same. So, once we realize that this
40:40.359 --> 40:46.740
is what is going to happen, then I realize
that this particular energy, must be same
40:46.740 --> 40:52.150
as this particular energy as we have discussed
earlier.
40:52.150 --> 40:58.510
Now, what would summation of the what will
be length of the four vector in C frame, summation
40:58.510 --> 41:05.510
of p i is 0. It means it must be minus summation
of E c i square by c square, when the symbol
41:06.510 --> 41:13.319
c here, to demonstrate that this is centre
mass frame of reference. So, this particular
41:13.319 --> 41:18.520
thing because the E c is 0 some of the I am
applying again I am reminding I am applying
41:18.520 --> 41:23.130
from the system of the particle and for the
system of the particle is centre of mass frame
41:23.130 --> 41:29.500
of reference summation E is 0.
So, it contains only one term which is this
41:29.500 --> 41:34.720
and this must have exactly the same value
as we have done earlier, because the length
41:34.720 --> 41:39.910
is going to be same in this frame for the
same system of particles. So, I just evaluate
41:39.910 --> 41:45.190
this particular thing and in peaceful I get
summation of E c i and I realize I take under
41:45.190 --> 41:50.410
root. And this is sum of the two energies
and because energies are going to be shared
41:50.410 --> 41:55.329
equally by the two electrons. So, each one
of them will have an energy of 1.396 divided
41:55.329 --> 42:01.950
by 2. Let me repeat this length of the four
vector has only one term which is here which
42:01.950 --> 42:07.339
contains the submission of energy of two electrons.
The submission of the energy of the two electrons
42:07.339 --> 42:14.339
and with the energy of both equal of two electrons,
it means E c i is just… By this particular
42:14.770 --> 42:19.450
summation is just the double of the energy
of the individual electron. So, this one particular
42:19.450 --> 42:25.299
1.396 if I divide divided by 2 and I will
get the energy of the two electrons.
42:25.299 --> 42:31.000
This is what I have written the energy of
the two scattered electrons of the magnitude
42:31.000 --> 42:38.000
of the momentum are same. We therefore, get
E c is just as 1.396 divided by 2 is equal
42:39.240 --> 42:44.950
to 0.698 MeV. So, each of the electron will
have this much energy because of the symmetry
42:44.950 --> 42:49.990
involved in the centre of mass frame of reference.
The two particles have to go back to back
42:49.990 --> 42:54.190
and both being the same particles with same
same rest mass have to have the same energy
42:54.190 --> 42:59.109
and of course, they have to have a same momentum.
I can calculate what will be momentum of the
42:59.109 --> 43:03.680
particle by just using exactly the same relationship.
I will get that the momentum of the individual
43:03.680 --> 43:10.680
particles individual electrons will be 0.477
MeV by c in C frame of reference this centre
43:11.650 --> 43:18.650
of mass frame of reference, I found out energy
and momentum. Now, is C frame will have laboratory
43:19.200 --> 43:24.910
frame of reference make a transformation.
Again you have to find out the speed of C
43:24.910 --> 43:28.890
frame with respect to L frame of course, if
you have adopted the second method we have
43:28.890 --> 43:34.160
so not obtained the speed on the first method
we found out, the speed of the centre of mass
43:34.160 --> 43:36.470
frame of reference.
But in this particular method we have not
43:36.470 --> 43:41.640
yet found out, but we can now find out because
in this particular C frame the energy of the
43:41.640 --> 43:48.309
electron is 6.98 MeV, while its rest mass
energy was 0.51 MeV. So, the gamma will be
43:48.309 --> 43:53.220
0.698 divided by 0.51, once I know gamma I
can calculate what will be the speed.
43:53.220 --> 43:58.740
This what I have done of course, before that
let me let me also tell that we have to calculate
43:58.740 --> 44:03.559
x component and the y component of the momentum.
Because now, make an angle of 45 degrees and
44:03.559 --> 44:07.450
because it is an angle of 45 degrees. So,
x component and y component will be just whatever
44:07.450 --> 44:14.280
is the momentum divide by root 2 cos 45 or
sine 45 both are under 1 upon root 2. So,
44:14.280 --> 44:19.190
these are the x component and the y component
of the momentum and I must you know we should
44:19.190 --> 44:24.010
be clear, that we have to transform not just
one momentum, we have to transform x component
44:24.010 --> 44:30.039
of momentum and y component of momentum and
also the energy of the particle to L frame.
44:30.039 --> 44:36.750
So, as I said relative speed of c can be obtained
by 0.698 divided by 0.51, which gives me the
44:36.750 --> 44:43.470
value of gamma is equal to 1.369. And I can
calculate the beta as we have done using the
44:43.470 --> 44:47.829
expression gamma square minus 1 divided by
gamma square, which gives me same value, which
44:47.829 --> 44:54.829
I obtained by the first method which is 0.683
c. I know the velocity I know p x, I know
44:56.809 --> 45:01.250
p y, I know E the centre of mass frame of
reference, applied transformation equations
45:01.250 --> 45:05.430
go back to laboratory frame of reference.
You know all the energy all the momentum back
45:05.430 --> 45:09.480
to laboratory frame of reference if you are
not very sure you have done mat everything
45:09.480 --> 45:13.220
correctly. Again apply conservation of energy
and momentum in that frame they have to be
45:13.220 --> 45:16.910
obeyed in that frame of reference if you have
done any mistake.
45:16.910 --> 45:22.420
So, that is what I have said the energy and
momentum can be obtained in laboratory frame
45:22.420 --> 45:26.710
using energy momentum transformation standard
transformation, we have been using those equations
45:26.710 --> 45:31.240
earlier. One can also assert in that the energy
and momentum are also conserved in that L
45:31.240 --> 45:36.089
frame which they have to, but that is not
a part of problem to show that. But if one
45:36.089 --> 45:41.400
wants to be show to see that I have done no
mistake, this is the way we can calculate.
45:41.400 --> 45:47.359
I just wanted to point you out one particular
thing that, once we did when described the
45:47.359 --> 45:52.450
experiment on centre of mass frame of reference
the energy that we got was much less, in the
45:52.450 --> 45:58.569
laboratory frame of reference it was 1.4 plus
0.51 MeV. While in the case of the centre
45:58.569 --> 46:01.809
of mass frame of reference is 1.396 MeV.
46:01.809 --> 46:06.319
So, just let us state this particular fact
this if we have to decide this particular
46:06.319 --> 46:11.030
experiment, then in the centre of mass frame
of reference the energy that is being used
46:11.030 --> 46:16.720
only 1.396 MeV, while in the laboratory frame
of reference its 1.91 MeV. So, this is what
46:16.720 --> 46:21.450
sort of illustrates what I have been dealing
earlier, which I told earlier that many times
46:21.450 --> 46:26.089
it is simple it is energetic favorable to
design the experiment centre of mass frame
46:26.089 --> 46:33.089
of reference. So, do not spend energy in accelerating
or letting this centre of mass. Now, let us
46:35.680 --> 46:42.490
take one more problem in which also we are
using the concept of four vectors and the
46:42.490 --> 46:46.910
concept of centre of mass frame of reference,
this is also very simple problem, but involve
46:46.910 --> 46:49.010
many more particles.
46:49.010 --> 46:55.950
So, problem is as follows; that there is a
proton p which has a kinetic energy K, this
46:55.950 --> 47:00.160
problem has been given in terms of kinetic
energy. Most of the time there problems are
47:00.160 --> 47:04.530
given in terms of total energy, in this it
happens to be given in terms of kinetic energy,
47:04.530 --> 47:08.819
no issues we know how to convert it, no problem.
We also know the expression between energy,
47:08.819 --> 47:15.819
kinetic energy and the momentum, so we can
use that expression in. So, we have a proton
47:15.890 --> 47:21.690
it has a kinetic energy K and this is incident
on another proton, which is rest in the laboratory
47:21.690 --> 47:25.530
frame of reference. And even if I would not
accept L frame, I would have realized that
47:25.530 --> 47:30.210
this one is centre of mass frame of reference.
Because, one particle at rest, another particle
47:30.210 --> 47:34.010
is moving, so obviously, the centre of mass
cannot be at rest. So, this obviously, not
47:34.010 --> 47:36.200
a centre of mass frame of reference.
47:36.200 --> 47:43.200
Now, this proton when it incidents on the
another proton results into four different
47:46.740 --> 47:51.440
particles. After the interaction four particles
are formed not two particles as in the earlier
47:51.440 --> 47:57.490
case. The four particles we have found, three
of which them are proton and the fourth one
47:57.490 --> 48:04.490
is an anti-proton. Just to you sure anti-proton
is has a same mass as a proton same rest mass
48:06.180 --> 48:12.289
as a proton. But it has a charge which is
negative to it, it has a charge which is negative,
48:12.289 --> 48:15.130
protons are positively charged particles,
while anti-proton is a negatively charged
48:15.130 --> 48:19.440
particle. So, you get three protons and one
anti-proton.
48:19.440 --> 48:25.430
Now, question is little more tricky; what
should be the least value of kinetic energy
48:25.430 --> 48:31.039
K? What should be least kinetic energy that
you must supply to that incident proton? So,
48:31.039 --> 48:37.910
that this particular reaction becomes possible.
See from two protons you are generating four
48:37.910 --> 48:42.859
particles; three of which are protons and
one of them is anti-proton. So obviously,
48:42.859 --> 48:48.799
if you at the both the protons were at rest
you could have not done it, because the energy
48:48.799 --> 48:53.559
that was available to you is only two M naught
c square, where M naught is a rest mass of
48:53.559 --> 48:56.520
the proton.
You are getting four particles, each one of
48:56.520 --> 49:03.520
which has a rest mass of M naught as mass
of rest mass of proton it was not possible.
49:04.430 --> 49:07.539
Obviously, we require certain amount of energy,
you cannot just make of conservation of energy,
49:07.539 --> 49:13.520
because momentum also has to be conserved.
But if you go back to the center of mass frame
49:13.520 --> 49:18.740
of reference there certain amount of easiness.
Because, in centre of mass frame of reference
49:18.740 --> 49:24.650
I am very clear initial momentum is 0 and
in that particular frame of reference, to
49:24.650 --> 49:29.039
a require least amount of energy is that I
can create all 4 protons at rest.
49:29.039 --> 49:35.299
If I think in terms of centre of mass frame
of reference I have 2 protons, which are coming
49:35.299 --> 49:39.970
back-to-back like this and hitting it. They
have to have equal momentum because this is
49:39.970 --> 49:44.849
centre of mass frame of reference they have
the same rest mass. Now, if you are giving
49:44.849 --> 49:50.819
them too much of energy you have 4 protons,
which may move anywhere else, 4 protons plus
49:50.819 --> 49:55.839
3 protons plus 1 anti-proton, 4 particles.
But if I want this particular thing to have
49:55.839 --> 50:02.680
least amount of energy I can have such a velocity,
such that these 4 particles are all at rest.
50:02.680 --> 50:05.559
And because all the four particles are at
rest then momentum is still 0.
50:05.559 --> 50:12.450
So, this is what I am trying to explain I
have written the transparency the least energy
50:12.450 --> 50:18.750
required would be when all the four particles
are created at rest. But that is only possible
50:18.750 --> 50:22.510
in the C frame of reference, because it is
only in that particular frame of reference
50:22.510 --> 50:27.109
that the momentum 0. In laboratory frame the
two particles, the four particles not be created
50:27.109 --> 50:31.500
at rest, because the momentum will not be
conserved. Because, initial momentum is non-zero,
50:31.500 --> 50:36.980
it is only in C frame that the initial momentum
is 0. So, you can imagine that all the four
50:36.980 --> 50:39.900
particles are created at rest. So, we go to
center of mass frame of reference.
50:39.900 --> 50:46.900
The energy is C frame corresponding to the
case when all four of them are at rest must
50:49.059 --> 50:54.289
be equal to 4 m naught c square, where m naught
is the rest mass of the proton. Because, each
50:54.289 --> 51:00.490
four all the four particles are at the same
rest mass m naught and they are not moving,
51:00.490 --> 51:05.010
because I want the least amount of energy.
So, the total energy which will be generated
51:05.010 --> 51:10.440
in the centre of mass frame of reference has
to be 4 m naught c square. Therefore, the
51:10.440 --> 51:14.990
square of the length of the four vector this
frame, because this momentum 0 is minus E
51:14.990 --> 51:21.720
square by c square. So, it must be minus 16
m naught square c square. So, this will be
51:21.720 --> 51:27.029
the length of the four vector in the centre
of mass frame of reference after the reaction
51:27.029 --> 51:28.160
is over when we have got four particles.
51:28.160 --> 51:35.160
But what will be the length in the laboratory
frame of reference, before the interaction
51:38.210 --> 51:45.210
we have momentum which is not equal to 0,
I can find out what is a momentum, there is
51:46.270 --> 51:49.770
only one particle which is moving. So, I can
find out what will be the total momentum of
51:49.770 --> 51:54.960
that particular particle moving its energy.
I will know the energy of the first particle,
51:54.960 --> 51:58.500
I know the energy of the second particle.
I can find out the length of the four vector
51:58.500 --> 52:03.029
just like I did in the previous problem.
52:03.029 --> 52:07.710
Because, there is only one term I have not
written summation here, here of course, I
52:07.710 --> 52:13.440
have write the totally, not put the summation
sign makes things simple. So, this p square
52:13.440 --> 52:17.200
will be given by K square by c square plus
2 m naught K, this is the standard expression
52:17.200 --> 52:24.200
relating momentum to the kinetic energy, what
we did earlier. The total energy, energy of
52:24.289 --> 52:28.400
the first particle which is coming with the
speed is kinetic energy plus rest mass energy.
52:28.400 --> 52:32.940
So, K plus m naught c square there is another
proton, which is at rest which has energy
52:32.940 --> 52:38.869
of just m naught c square.
So, total energy is K plus 2 m naught c square,
52:38.869 --> 52:42.700
this is summation of E, I have not written
summation sign here, this is the total energy
52:42.700 --> 52:47.770
of the four vector. So, that the length of
the four vector in laboratory frame of reference
52:47.770 --> 52:54.770
is minus 2 m naught K, minus 4 m naught square
c square. Now, this length of four vector
52:56.039 --> 53:00.750
after collision has to be or after the interaction
has to be same in the length frame in the
53:00.750 --> 53:04.220
laboratory frame of reference and has also
be described in the in the center of mass
53:04.220 --> 53:07.630
frame of reference.
So, what I do I take the length of the four
53:07.630 --> 53:14.630
vector of the system of the particles before
interaction in L frame and equate it to the
53:15.279 --> 53:20.559
length of the four vector after reaction in
C frame. So, this is what I am doing I am
53:20.559 --> 53:25.750
taking before interaction the length laboratory
frame and equating it to the length of the
53:25.750 --> 53:31.230
four vector, in after the reaction in C frame,
which I know is minus 16 m naught square c
53:31.230 --> 53:34.630
to the power 4, if I want the least energy.
53:34.630 --> 53:41.630
This is what I have written equating the length
in L before interaction to the length in C
53:41.970 --> 53:46.990
frame after interaction. I must have minus
16 m naught square c to the power m naught
53:46.990 --> 53:51.700
square c square, given by this particular
expression, I solve this K terms out to be
53:51.700 --> 53:57.690
equal to 6 m naught c square. So, this is
the least k needed in laboratory frame. So,
53:57.690 --> 54:02.440
in laboratory frame of reference this particular
particle must have kinetic energy of at least
54:02.440 --> 54:09.440
6 m naught c square in order that reaction
becomes possible. In summary we have discussed
54:11.750 --> 54:17.039
some problems in center of mass frame of reference
and demonstrated how it becomes easy to solve
54:17.039 --> 54:21.990
these problems and we also gave some examples
of the use of four vector concept.
54:21.990 --> 54:22.769
Thank you.