WEBVTT
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hello
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everybody so today we will be talking of another
consequence of special relativity
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so we will do some topics in relativistic
kinematics and if one recalls we had to redefine
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mass we had this famous mass energy equivalence
special relativity gives us that and we also
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have to redefine mass so as to preserve the
conservation of momentum conservation of momentum
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in relativity and talking of both energy and
momentum how are they related
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we know that e = mc square and then this m
or the total mass of body that is it is actually
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related with speed at which this body is moving
there ok
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so if m0 which you see here is the rest mass
of a body then if it is moving at a certain
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velocity v its mass becomes m divided by root
over of 1 - v square by c square ok fine and
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consequently we also have the momentum defined
as the mass multiplied with velocity but then
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the mass again here is velocity dependent
okay so how about the relation between them
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why do not we square them up
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and you know if we square the energy and then
multiply a c square to the momentum do a bit
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of algebra it turns out and then we and then
we subtract p square c square from the square
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of the energy it turns out that we get something
which is which is lorentz invariant or in
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other words it does not change when you change
frames or it does not change in frames which
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are moving with a certain constant velocity
with respect to each other
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so you actually get m0 c square on the m0
being the response of a body and then c square
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being the velocity or the c being the velocity
of light yeah okay that is pretty interesting
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so if we now write it in a certain way actually
if we write all these things in the dimension
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of momentum square
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we get a isolation like you get an e square
that is the total energy square divided by
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the velocity of light square that gets the
and that is get the dimension of that gets
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the dimension of momentum square that is and
that is the linear momentum square that we
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are talking of here and and and then you subtract
the momentum that which i denote by small
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p okay you get something which is invariant
okay
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now this i denote by something like the capital
p squared okay and just make a comment that
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still this is momentum square that is all
let us see is there something behind all these
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things okay now to check that what is there
behind we if you recall what what was a 3
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momentum squared okay so if you know a 3 momentum
square think of the thing in the cartesian
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system or in any is the cartesian okay the
cartesian system
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so what you see that 3 momentum squared you
get the x component square the y component
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square and the z component square or component
1 2 3 squared each we add them up wherein
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invariant quantity and you all know that this
momentum is a vector it has 3 components similarly
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the thing that we talk of when we talk of
a 4 momentum we have we can also talk of something
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called a 4 momentum squared
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and then what do we get and if you define
the dot product of we we define a dot product
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for the case of 4 momentum as you know the
initial the 0th component we call one of the
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complaint the first component was 0th component
and the other 3 components which are equivalent
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to the usual the vector in the 3 dimension
that we all know of and then we define the
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4 momentum the square of the 4 momentum with
the help of the special dot product
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that is why when i say a dot product i do
not put a simple dot but a rather a bigger
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dot okay so that is equal to actually p0 square
- and the 3 dot product of the the momentum
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the linear momentum that is here what we have
here okay now why do we call it as the 4 momentum
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more of a little later but here what we see
is that if we identify this p0 of the 0th
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component of the 4 momentum as e by c
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and then the other 3 components as the linear
momentum itself then what we get is the dot
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product of this on this this 4 momentum dot
product and shown by p capital p dot capital
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p is still an invariant quantity ok now the
notation that i will be using is for a small
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p it will always be the 3 momentum if i put
a vector you also know that it is a this momentum
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vector but for a 4 momentum vector not put
any any vectors on top of it and not only
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that i will always use capital letters for
it
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so so capital p is a for momentum and in the
small p is the 3 momentum and that we talk
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of here where it is also interesting that
we know that the 3 momentum square that is
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an invariant quantity okay you know the dot
product or when you take the square of of
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a vector you get the length of a vector which
is a scalar so that is an invariant quantity
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similarly what we see here is that when we
take the dot product of the 4 momentum itself
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we also get an invariant quantity okay that
is something to think about okay
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now let us let us recall a few more things
about the 4 momentum which we may have covered
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earlier but given todays topic of relativistic
kinematics i thought it will be useful for
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us to do a bit of the 4 momentum a formalism
again a bit ok ok
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so just to recall we can also think of the
relativistic phenomenon in in in minkowski
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space time ok where an event is given by by
by by an event or an world point is specified
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by 4 quantities 3 space quantities one time
coordinate okay of course you can multiply
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the constant velocity of light to time so
that you have the dimension of length for
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all the for all the coordinates okay and then
that is what we do we can simply multiply
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c to the time coordinate and then we have
x y z
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let us say so the 4 space 3 space coordinates
and then we have the same dimension for all
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and then we can simply write x0 to be t0 at
component the other 3 components for the other
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3 the two 3 vector here okay now it is also
interesting when i take the squared norm or
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then the 4 dot product of x with itself
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i will get c square t square - x dot x like
the 4 dot product we had done earlier fundamental
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and how does the components change when we
go from one system to another let us say we
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go from the s frame where you have the coordinates
ct at a certain time and then x y z and then
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the same event if it is measured with the
help of from s prime frame and then you think
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of that frame as and and you measure it at
a certain time t prime and x prime y prime
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z prime
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of course you now know that these are related
by lorentz transformations okay so the components
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of this 4 vector is actually related by lorentz
transformations which we have covered earlier
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and it is act okay just to recall let us let
us read the first line so it is the how is
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the x component in the s prime frame related
with the x component with the s frame
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you see that let us express x prime is actually
equal to gamma times x - beta ct ok so where
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beta is the velocity with which the s prime
frame is moving divided by the speed of light
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okay and then the gamma what you see here
is nothing but the 1 by the square root of
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1 - beta squared okay okay so what is also
interesting is that if one takes the full
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dot product ok the 4 dot product in the s
frame and the s prime frame can easily check
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that it is norm is equal okay
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that is again something we keep in mind it
is that the norm of a full vector is is becoming
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you know what what we check is or for this
in is becoming invariant when when is actually
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invariant when you go from one frame to another
ok now just as an aside we can also recall
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that the norm of a 3 vector is also invariant
when you change frames okay okay
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so to make things a little bit more general
we we call all quantities or other all quantities
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which has 4 sets of sets of numbers here now
and if they transform and if they transform
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in the same way as as ladins lorentz transformations
and not only that that square norm of such
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a quantity a as becoming a become same whether
you are in s frame or the s prime frame then
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such a thing are actually 4 vectors okay
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and to summarize what we have is that we have
a 4 vector in s frame then in the s prime
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frame the same 4 vector will have of course
the same norm but its components will change
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okay but liquid components will change in
accordance with what with accordance with
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deal a lorentz transformations okay so that
is what we have the components of a 4 vector
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will change you know the components are related
by the lorentz transformations and then the
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norm is invariant fine
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so just to put things in perspective again
if we talk of a 4 vector the momentum 4 vector
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here okay so the 0th component is e by c that
is the energy by c and then px py pz are the
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other 3 components and in the s prime frame
which is moving with a certain constant velocity
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v with respect to the s frame the the components
will be different okay of course it will be
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different but the norms of the 4 vectors would
be the same okay
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so now you know just as an aside why do we
call it as a 4 vector okay recall what we
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have in a 3 dimensional space when we have
a vector so what is a vector so if we recall
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what a vector is it is actually that mathematical
physical entity which has 3 components here
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as a 3 vector that is why it is called a 3
vector now a vector what is the special property
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of it it is that if one changes moves from
one coordinate system to another
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now the transformation matrix which one has
to use to go from one frame to another okay
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that same transformation matrix is going to
change the same vector which is in the old
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system to the new system
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for example here in this example if x is a
vector in o system and x prime is a vector
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in the o prime system see the transformation
matrix which is transforming o to o prime
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will be the same which transforms x to x prime
and not only that the length of this vector
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is actually the same as a scalar okay so that
does not change when you change your coordinate
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systems and if you recall what we have just
talked of this quote unquote 4 vector see
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some amazing properties which are quite similar
with the 3 vector first of all it is got 4
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components
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and then the the the components of the you
know when you go from one system to another
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the what is the transformation matrix here
it is the lorentz transformations and it is
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the same lorentz transformations which is
transforming the components of of this quantity
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x of this 4 vector from one system from one
frame to another and not only that it is norm
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is fixed okay okay so that of course justifies
the word the vector here what you have used
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and and also it has 4 component that is how
you call it a 4 vector okay
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so with this let us with this background now
i think we can go over to an application of
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relativistic kinematics okay where let us
take a simple example let us study two body
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collisions okay
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now as in classical mechanics we can study
this in the lab system and also in the center
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of mass system okay so what is a lab system
when we consider the collision of two bodies
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okay so the lab system is one in which one
of the bodies it may be in rest and the other
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comes and hits it okay to keep things simple
i mean we just take the direction of motion
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of one of these bodies of one which is moving
to be along the x axis okay and it is on a
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plane let us say okay anyway
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we also take one of the bodies at rest so
let us just take the body b at rest and the
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body a which is coming with a certain linear
momentum pa and if you see we have also put
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a subscript l and also a vector so that is
that is an indication pa l so that is the
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indication the p has a vector so that is a
3 vector and it is related with the first
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body a and then it is in the lab system so
that is the quantity l that is that is the
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reason i put this quantity here
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now what would be the 4 vector for such a
quant for for these bodies okay if recall
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that the 0th component of the 4 vector relates
with the energy okay so we have the energy
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of particle a in all the body a in lab system
divided by c that is the speed of light and
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then the other 3 components are the 3 vectors
okay or the 3 components of a 3 vector okay
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and what is eal so that is nothing but ma
into c square okay that is the mass of e
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and then what about pbl that is the that is
the 4 vector pbl so recall that again it sits
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at rest so the 3 momentum part of it is 0
okay but it is interesting that it is energy
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is not 0 why because of course it has a rest
mass and then you have rest mass energy so
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you have mb into c square that will be the
rest mass energy okay now if you take mvc
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square divided by c you will get mvc and so
you get the 4 vector pbl that is capital pbl
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to be mbc
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and then the since the 3 vector part is 0
so you have it is 0 okay so there is of course
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another system in which we go over to the
center of mass of the system so all the measurements
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are being made from the center of mass of
the system here
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and in this particular case we have the case
in which the bodies are moving in a set such
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a way that the total linear momentum turns
out to be 0 here okay so the same collision
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process you are looking at the frame you are
looking in a way in which the both are moving
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towards each other in such a way that the
sum of the linear momentum are c okay and
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it is easy how to write the momentum for vectors
here for pa
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notice that i have not used any subscript
l here so except maybe when i only the subscript
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for the bodies or the particles have been
used here so the other subscript the cm is
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not required when superfluous thing so if
i do not write it it means that it is the
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center of mass system otherwise we assume
that i which in the lab system when i write
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it by this letter l
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so the momentum pa that is the 0th component
we know that ea by c again all quantities
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in the center of mass system and then the
3 vector that is the pa and then for the capital
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pb that is the 4 momentum for the particle
b that is eb by c and then it is got a 3 momentum
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pb okay
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so how are they related with each other member
so the center of mass system can be thought
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to be moving with a certain constant velocity
v with respect to the lab system and then
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if our incident direction of collision is
the x axis you know so we just have we just
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written the s frame on the s prime frame we
just rename them as l and l frame that is
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the left frame and the center of mass frame
respectively here
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so in the lab frame notice that the particle
b or the body b is at rest and the particle
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a is coming in coming towards it with a certain
momentum velocity and certain energy here
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okay kinetic energy element of course for
particle b here in the lab system although
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it does not you might recall you might recall
that it has some energy which is which is
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associated with the rest mass energy and then
in the center of mass frame what we have
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we have the particles moving with mentor towards
each other equal momentum so that they are
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opposite oppositely directed and and when
you add them up they cancel out okay so how
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are the things related with each other and
he called that the coordinates in the lab
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and the center of mass frame they are they
are lorentz frames here so they are they are
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related by the lorentz transformations okay
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and what about the components of the 4 vector
of the 4 for momentum so since it is also
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a 4 vector momentum we are we are talking
of the components of momentum 4 vector the
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components will transform according to lorentz
transformation itself okay so it is interesting
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that it is very easy you can you can find
out if the momentum of any particle is you
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know one of them in the left frame and in
the other frame
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and that is in the center of mass frame which
is moving with a constant relative velocity
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with respect to the lab frame you can simply
find it out by by by the help of lorentz transformation
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that is the relation between the lab and the
center of mass moment and of course the energy
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the energy being the 0th component here okay
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it is simple here for example the and if px
is the momentum in the small px that is the
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linear momentum along the x axis in the center
of mass system and then e is just the energy
22:48.710 --> 22:56.150
in the center of mass system then the px l
that is the x component of the momentum in
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the lab system is related with the help of
lorentz transformation that is gamma times
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px + beta e by c ok so it is just the and
then since it is we have taken the motion
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to be along x axis
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so the x and the y and the z components are
the same here ok and what about the energy
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part remember this is the for the remember
this this this transforms like the time like
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components or the ct component when when we
are doing the the position 4 vector okay so
23:31.980 --> 23:37.870
what is the energy in the lab system so that
is simply related with the energy in the center
23:37.870 --> 23:44.419
of mass system by if el is the energy in the
lab system el by c that is actually equal
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to gamma times e by c
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so e being the energy of the center of mass
system + beta times px px being the x component
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of the 3 momentum in the center of mass system
ok fine
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so what is this velocity then with which the
center of mass is moving with respect to the
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lab ok what is this velocity how can we estimate
it with the help of known components with
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the help of known quantities like the mass
of the the particles in the lab system and
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in the linear momentum of the particles in
the lab system ok so let us find it out and
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to do that what we can do is that we can simply
apply the conservation of linear momentum
24:37.390 --> 24:41.330
and the conservation of momentum and also
conservation of energy
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and we will simply take them out from there
okay by adding the corresponding quantities
24:47.820 --> 24:55.030
for for these two particles okay okay so first
let us do the conservation momentum conservation
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first for the for the mentum project for for
both these particles okay and what we find
25:04.580 --> 25:10.309
is that if we add these two quantities in
the lab system and then of course we relate
25:10.309 --> 25:14.880
them to what these quantities are in the center
of mass system
25:14.880 --> 25:21.230
so what we see is that we add pax that is
the x component in the lab system for particle
25:21.230 --> 25:32.549
a and pb x l that is the that is the x component
of the of t of the second particle b in the
25:32.549 --> 25:40.230
lab system by the way remember that it is
actually actually 0 that the x component that
25:40.230 --> 25:49.610
the curve that the momentum of the momentum
of particle b is 0 so pb x l that is actually
25:49.610 --> 25:59.549
0 okay so that is x = gamma times pa + pb
+ beta times ea + eb by c
25:59.549 --> 26:06.550
now pa + pb so that is 0 in the center of
mass frame and and then but the energy is
26:06.550 --> 26:11.020
ea + eb how are they related let us simply
call them as a total energy in the center
26:11.020 --> 26:19.549
of mass frame okay in that case we get a rather
simple looking relation which relates the
26:19.549 --> 26:28.090
component of the component the momentum of
a that is pa xl in in in the lab system with
26:28.090 --> 26:31.179
the total energy in the center of mass system
26:31.179 --> 26:36.910
and remember what is this beta beta is nothing
but velocity with which the center of mass
26:36.910 --> 26:42.419
frame is moving divided by the velocity of
light and then gamma is another kind of mathitical
26:42.419 --> 26:48.350
quantity related with beta and then the c
here is the velocity of light okay or the
26:48.350 --> 26:51.140
speed of light if you wish
26:51.140 --> 26:57.659
again from energy conservation we simply add
up the energies in the lab system and then
26:57.659 --> 27:02.090
corresponding and then you equate that with
the corresponding quantities with the corresponding
27:02.090 --> 27:11.010
patience in the center of mass system and
we simply also use the also use the also use
27:11.010 --> 27:16.480
the quantity that the total momentum the linear
momentum in the center of mass system is 0
27:16.480 --> 27:21.299
then we also get another interesting quantity
okay
27:21.299 --> 27:28.600
what is it that the total energy in the in
the center of mass system and the total energy
27:28.600 --> 27:33.990
in the lab system how are they related you
simply take it out from this solution but
27:33.990 --> 27:41.080
again from the conservation of momentum we
have seen that the total energy in the center
27:41.080 --> 27:47.330
of mass system can be related with the total
momentum in the lab system now we can use
27:47.330 --> 27:53.289
that relation again to to drive out a little
some other quantities here
27:53.289 --> 27:58.650
and then from from these two equations the
one which relates the total center of mass
27:58.650 --> 28:05.250
and the lab energies and then the one which
relates the the total center of mass energies
28:05.250 --> 28:13.000
with the lab momentum of particle a we can
find out what the velocity with which the
28:13.000 --> 28:18.390
center of mass frame is moving with respect
to the lab system you can simply work out
28:18.390 --> 28:26.390
that it is nothing but 3 squared times i mean
times thr times the momentum of the particle
28:26.390 --> 28:34.100
a divided by the total energy with which the
and the the particle a is moving
28:34.100 --> 28:44.029
plus the rest mass less mass energy of particle
b okay so you have everything with respect
28:44.029 --> 28:51.480
so we in terms of everything that you started
with in terms of all lab system quantities
28:51.480 --> 28:59.620
okay so what about the collision process we
have not used the 4 vectors as yet okay we
28:59.620 --> 29:05.520
have just used the conservation of linear
momentum and the conservation of a conservation
29:05.520 --> 29:09.419
of energy separately and we found out this
velocity
29:09.419 --> 29:16.250
now what if we use the conservation of the
the 4 momentum in in the lab and the center
29:16.250 --> 29:20.870
of mass system so what we get from there okay
29:20.870 --> 29:28.130
so for that let us simply add up the 4 momentum
of particle a and particle b and take its
29:28.130 --> 29:34.660
norm in the center of mass system and remember
that the sum of this 4 vector is also 4 vectors
29:34.660 --> 29:39.159
then you could take a norm of that it is going
to be that is going to be invariant whether
29:39.159 --> 29:44.490
you are in the lab system in center of mass
system let us simply call this norm as small
29:44.490 --> 29:51.190
s and given that it is still a 4 vector you
add two 4 vectors you get a you get a greater
29:51.190 --> 29:52.640
4 vector i think it is norm
29:52.640 --> 29:59.279
it is lorentz invariant so s here is a lorentz
invariant quantity then what is it it is the
29:59.279 --> 30:08.150
norm of p a square it is it is the norm of
pa and norm of pb these are the 4 vector norms
30:08.150 --> 30:15.771
plus the 4 vector dot product twice the 4
vector dot product of pa and pb now we all
30:15.771 --> 30:20.049
know how to do these 4 vector dot products
and calculate these norms
30:20.049 --> 30:25.750
and for example what is pa squared that is
the 4 vector of a square and center of mass
30:25.750 --> 30:32.800
system we know that it is also it is ea square
by c square that is the energy of particle
30:32.800 --> 30:41.049
a in the center of mass system minus the 3
vector norm of of particle a and then the
30:41.049 --> 30:48.240
norm of the 4 vector is nothing but ma square
into csquare that is the rest mass energy
30:48.240 --> 30:55.950
square of that that is the rest mass square
of particle e multiplied by c square
30:55.950 --> 31:05.330
similarly you can find the pb square that
is the norm of the vector the 4 vector for
31:05.330 --> 31:13.110
particle b and then you can also find the
4 vector dot product of pa and pb that is
31:13.110 --> 31:19.610
nothing but the product of the energies minus
the 3 vector dot product of pa and pb that
31:19.610 --> 31:25.940
is a bit of an algebra now when you arrange
all these things put it plug it back into
31:25.940 --> 31:28.309
the expression for s okay
31:28.309 --> 31:36.640
what you get is that the sum of the total
center of mass energy squared divided by c
31:36.640 --> 31:49.529
square minus the minus the pa + pb whole squared
okay now if you recall that the total moment
31:49.529 --> 31:54.539
at the total the total momentum at the 3 momentum
when you add these two up in the in the center
31:54.539 --> 32:03.500
of mass system is 0 okay so the total energy
in the center of mass system which is ea +
32:03.500 --> 32:06.130
eb is a lorentz invariant quantity here
32:06.130 --> 32:12.960
you can it is and if you put it in a put in
you know remove the square here and you see
32:12.960 --> 32:20.990
that it is actually equal to c times root
s okay so s being a lorentz invariant quantity
32:20.990 --> 32:26.000
okay so that is a rather important relation
we have got here it is that the total center
32:26.000 --> 32:32.049
of mass energy for this two body collision
that we see here is actually a lorentz invariant
32:32.049 --> 32:39.470
quantity okay fine now s again is a lorentz
invariant quantity
32:39.470 --> 32:48.149
so we can also find the thing in the lab system
okay so we do the thing in the lab system
32:48.149 --> 32:54.110
we take the sum of the 4 vector in the lab
system of particle a and particle b and take
32:54.110 --> 32:55.799
its norm okay
32:55.799 --> 33:03.010
then do the algebra again so we have p al
stands for the subscript l is telling us that
33:03.010 --> 33:12.120
it is a lab system pal squared + pbl square
and then twice pal dot p bl okay so that is
33:12.120 --> 33:23.270
the 4 vector dot product now if we use the
norm of the 4 vectors being invariant in being
33:23.270 --> 33:29.850
invariant here so we know that the norms and
we also use the algebraic expression for the
33:29.850 --> 33:33.539
3 a 4 vector dot products
33:33.539 --> 33:39.590
also remember that the any using the and and
and in d and in writing the 4 vector dot product
33:39.590 --> 33:49.020
we just remember that the of the second particle
here b was 0 so we are not going to have the
33:49.020 --> 33:56.591
3 momentum part here in the 4 momentum dot
product and then the energy of the particle
33:56.591 --> 34:00.830
b in the lab system is just the rest mass
energy
34:00.830 --> 34:10.220
so we plug in all these things into the expression
for the for the invariant s what we get is
34:10.220 --> 34:20.559
is is a quantity which is in terms of the
masses and the energies of and then the energy
34:20.559 --> 34:28.790
the initial energy of particle a in the lab
system okay just as an aside what is the kinetic
34:28.790 --> 34:35.869
energy of particle a in the lab system it
is just the you you you subtract the rest
34:35.869 --> 34:42.639
subtract the rest mass energy from ea l that
is the total energyand you can you can write
34:42.639 --> 34:50.599
the the invariant quantity s in terms of the
kinetic energy also okay
34:50.599 --> 34:57.839
also we can use a approximation when you are
at rather very high energies things called
34:57.839 --> 35:05.920
the ultra relativistic cases when the momentum
of is actually much much greater than ma times
35:05.920 --> 35:12.099
c that is the that is the rest mass times
the velocity of light okay if it is rather
35:12.099 --> 35:21.410
very big then how does this expression for
the invariant s turn out to be and in that
35:21.410 --> 35:27.049
case you know that the energy of the total
energy of particle a in the lab system is
35:27.049 --> 35:36.279
nothing but p al that is the momentum in the
lab system times t times the velocity of light
35:36.279 --> 35:41.200
it can be approximated by that because the
other quantity that is other quantity which
35:41.200 --> 35:47.540
depends on the rest mass will be much more
less compared to this okay we might do an
35:47.540 --> 35:54.979
example for this too to check it out whether
that is or not okay so in that case we can
35:54.979 --> 36:06.329
simply write that the total totally invariant
s in terms of just the lab system in the ultra
36:06.329 --> 36:14.489
relativistic case okay we also know that the
invariant s that we have defined here that
36:14.489 --> 36:18.299
we have got here can be written in terms of
the center of mass quantities
36:18.299 --> 36:23.829
remember the total energy squared divided
by c square that is also an invariant quantity
36:23.829 --> 36:30.099
actually that was also s okay so in this way
we can relate the quantities in the center
36:30.099 --> 36:36.150
of mass system the energy the total energy
in the center of mass system with the total
36:36.150 --> 36:44.359
momentum in the in the total 3 momentum in
the in the in the lab system so in a way it
36:44.359 --> 36:50.660
is a it is a method for us to check if you
are giving this much amount of momentum in
36:50.660 --> 36:51.660
the lab system
36:51.660 --> 36:59.400
how much of it it goes to the total energy
in the center of mass system okay okay so
36:59.400 --> 37:04.209
as i said let us do a small example okay so
here
37:04.209 --> 37:11.209
let us consider the collision of two subatomic
particle let us say we consider the collision
37:11.209 --> 37:16.240
of two particles and then rest masses you
know the rest mass of a proton is something
37:16.240 --> 37:26.410
like 938 940 g ev by c square but it is approximately
let us take it as 1 gev by c square now and
37:26.410 --> 37:32.049
and then we consider the collision of this
in the center of mass system with 3 momentum
37:32.049 --> 37:37.349
of magnitude 30 gev by c okay
37:37.349 --> 37:43.440
this actually rather very picky okay so here
you can look at this figure for exam we have
37:43.440 --> 37:49.789
these two protons a and b they are coming
towards each other with with moment at 330
37:49.789 --> 37:53.219
g v by c
37:53.219 --> 38:01.479
now the question is what is the what is the
linear momentum what is the momentum in the
38:01.479 --> 38:06.089
lab system which will be necessary so that
you have this much amount of momentum in the
38:06.089 --> 38:09.969
center of mass system okay okay
38:09.969 --> 38:15.309
so we simply let us let us find the center
of mass energy of the protons first okay now
38:15.309 --> 38:20.729
it will be of course the same with the center
of mass energies of the move both these protons
38:20.729 --> 38:21.729
can be found
38:21.729 --> 38:27.160
by simply you know be a square and c square
plus mp square c power 4 and then you take
38:27.160 --> 38:36.299
every take the whole thing to the power half
okay so what is that so pa square c square
38:36.299 --> 38:44.719
what is that that is that is 930 squared okay
now to 900 if you add 1 that is the rest mass
38:44.719 --> 38:52.099
rest mass energy is squared okay it is nothing
900 becomes simply 901 so if you take the
38:52.099 --> 38:58.579
square root of 900 or 901 it is approximately
30 okay so it is actually in the ultra relativistic
38:58.579 --> 38:59.619
keys that we are talking here
38:59.619 --> 39:07.079
so we can neglect the rest mass compared to
the momentum here okay or rather rest mass
39:07.079 --> 39:15.799
times t the velocity of light compared to
the moment here okay now we also know we need
39:15.799 --> 39:20.969
to calculate what is the mentor in the in
the lab system okay
39:20.969 --> 39:24.900
so for that let us calculate the invariant
lorentz invariant quantity s we know that
39:24.900 --> 39:32.619
that is related with the total center of mass
energy squared rights s will be 30 + 30 here
39:32.619 --> 39:38.920
because each of these protons have total energy
is 30 in the center of mass system so s is
39:38.920 --> 39:48.930
nothing but 3600 gev square by c square fine
we can also relate this with we will we have
39:48.930 --> 39:54.829
seen how it is related with the lab system
went up okay
39:54.829 --> 40:00.930
that is actually equal to twice the mass of
the protons times the velocity of light times
40:00.930 --> 40:09.029
the linear momentum okay now if you plug in
all these quantities you will get the linear
40:09.029 --> 40:18.009
momentum of in the lab system to be right
1800 gev by c which is huge okay which is
40:18.009 --> 40:26.749
rather huge so you see to get a energy of
30 gv per c you need such a huge momentum
40:26.749 --> 40:33.079
in the lab system okay and this also and then
this number actually justifies the approximations
40:33.079 --> 40:38.219
that we have used that of our ultra relativistic
case
40:38.219 --> 40:46.309
in which the momentum was considered to be
the was much much larger than the rest mass
40:46.309 --> 40:56.959
times the velocity of light here okay so i
hope i have convinced i have you know shown
40:56.959 --> 41:03.449
you some examples of relativistic kinematics
more specifically relativistic collisions
41:03.449 --> 41:17.690
and the use of the 4 momentum of the power
of the use of 4 momentum in relativistic kinematics
41:17.690 --> 41:27.369
thank you very much