WEBVTT
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Hello, in our last lecture we had described
the concept of force, in classical mechanics
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force is well known. We discussed how we redefine
it when we come to the special theory of relativity
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then we realized that the way we define force
in the special theory of relativity the acceleration
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And force need not be in the same direction,
while in classical mechanics, if I apply force
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in a particular direction the acceleration
is also in the same direction, but in relativity
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it is not like that it need not be like that.
Then we consider two special cases; one when
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we are talking of only one dimensional motion.
It means that the force is being applied in
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the same direction as the velocity of the
particle, and a situation in which the force
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is applied perpendicular to the direction
of velocity. In these two cases we found out
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that the force and the acceleration will turn
out to be the same, will turn out to be in
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the same direction, sorry will turn out to
be in the same direction; these are the things
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that we discussed last time.
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So, we recapitulate, we discussed the definition
of force and relativity, the acceleration
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in general is not in the same direction as
force. We considered two special cases when
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force and acceleration are in the same direction
and this is what I have just now told these
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two are the two special cases; one where the
motion is along the line, and the force is
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applied along the direction of velocity.
And the second condition, when the force is
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applied perpendicular to the direction of
velocity of the particle. In the last lecture,
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we had sort of said that we should we will
find out the equivalent of what we call is
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parametric equations using these definition
of force in relativity. So, we will spend
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some time today to discuss that particular
aspect then of course we will talk about the
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force transformation.
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So, this is the equation that we have derived
earlier force multiplied by time. Let us just
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recapitulate the nomenclatures u naught is
the initial velocity of the particle. If force
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is applied in the same direction of velocity
of the particle was the force is applied in
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the same direction as the velocity of the
particle and the motion is only in one direction.
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We are not talking about the vector signs,
so we have remove the vector signs. So, u
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naught is the initial velocity of the particle
or we now call initial speed of the particle
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and a constant force is applied for a time
t and after time t, u t is the final velocity
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of the particle. This u here, there is a u
t here there is a u naught here, there is
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a u naught here and m naught of course, the
rest mass of the particle.
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So, this is the equation which we had derived
last time which relates force to the final
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velocity of the particle and the initial velocity
of the particle of course, it assumes that
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force is constant as a function of time. So,
my next step would be to find out this u t
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as a function of time, so that I can directly
get one equation from which I can find out
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the speed of the particle this is the way
the traditional kinematic equations in classical
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mechanics are derived. So, let us attempt
to do that.
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This is what I have written this equation
can be rewritten in a different form to enable
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us to calculate the speed after a given time
t, when a particle is subjected to a force
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F in one dimension motion which essentially
means, reorganization of these terms because
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the equation has already been obtained. It
is only sort of reorganizing the term, so
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that by direct substitution I can find out
the velocity at a particular time when a constant
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force is applied to the particle. So, let
us try to reorganize these terms.
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So, this was my equation which we have just
now written and explained. I define a quantity
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A is not a very standard symbol, we just force
divided by m naught rest mass of the particle
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this I am calling as A which is some sort
of acceleration. You know in some sense because
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it has the dimension of acceleration, but
I am writing it as capital A. If I take this
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define f by m naught as capital A then this
m naught which is appearing in both these
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equations can be brought here. We brought
it here this becomes F of one m naught and
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this quantity will become A this 1 upon under
root 1 minus u naught upon u naught square
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upon c square.
I can write as gamma u naught which is the
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standard nomenclature which we have been always
using it if this quantity can be written A
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gamma u naught. This will become gamma u naught
multiplied by u naught this quantity I take
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on their right hand side because this I had
divided by m naught. This becomes this quantity
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becomes A t m naught disappears from the right
hand side of these equations.
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This equation I take it this side and write
in the form of gamma u naught, it is the left
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hand side become A multiplied by t plus gamma
u naught u naught. So, this term remains exactly
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as becomes except for m naught which we have
already taken by replacing f by A from the
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right hand side, we just have u t upon under
root 1 minus u t square by c square.
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My idea is to write u t in terms of everything,
so what I will do I will square this quantity
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and try to solve for u t. It is a simple very
simple algebra, but just arrive at the equation
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which is comparative simplified equation top
determine final velocity in most of the kinematic
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problems that we are used to the classical
mechanics. Essentially, you are given force
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and given acceleration and you have to find
out the velocity of the particle at a particular
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time.
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This is what I am doing in the next transparent
sheet; this is the same equation what I had
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written in the last transparent sheet exactly
the same I have squared it. I have squared
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it this becomes this square A t plus gamma
u naught u naught A t plus gamma u naught
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u naught squared, when I square the right
hand side this will become u t square and
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in the denominator the under root will go
away. So, it will become just 1 minus u t
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square by c square this I take on the left
hand side this will get multiplied to this
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particular quantity c squared this is what
is happening here.
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This quantity after squaring and bringing
to the left hand side becomes 1 minus u t
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square upon c square this quantity squared
is put here becomes equal to u t square. Now,
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it is simple we have tried to organize the
terms with u t square collect all the terms
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that have u t square and those terms, which
do not have u t square and solve for u t square
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this is what I am doing in the next transparent
sheet. So, this u t square term has just been
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taken on the other hand side.
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Let me just work it out we had 1 minus u t
square by c square and here we have this quantity
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A t plus gamma u naught u naught square, whole
square is equal to u t square. If I take one
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and multiply this, this becomes a constant
quantity and this in retain on the left hand
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side this becomes A t plus gamma u naught
u naught square, this contains u t square.
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So, I take it on the other hand side, so this
becomes u t square this term will give you
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one and because this I am changing to right
hand side. So, this negative will become plus
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this whole quantity divided by c square this
is what this equation would look like this
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is what I am writing in the next transparent
sheet precisely.
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If you look here at the transparent sheet,
A t plus gamma u gamma u naught u naught square
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and this u t square 1 plus A t plus gamma
u naught u naught square divided by c square,
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after that simple I divide this quantity on
the both both the side. So, left hand side
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becomes this divided by this square, and then
I take another root once I do that this is
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the value of u t which I get u t once I take
the under root this just becomes A t plus
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gamma u naught u naught this is on the numerator
this gets divided.
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So, this becomes 1 plus A t plus gamma u naught
u naught square by c square and of course,
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because I am taking another root, so this
becomes another root. So, this equation would
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tell me if I know what is the value of A that
is f upon m naught, if I know what is the
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force and if I know what is the initial velocity
u naught. Then if you tell me what is going
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to be the speed of the particle after time
t of course it assumes very clearly that motion
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has to be only along a single line, because
if the motion is not along the single line
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these equations will not work unlike kinematic
equations where you could take the components
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and write those equations here.
It is very clear in this case that force and
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acceleration would not be in the same direction.
So, in this case one has to be careful it
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is only for the one dimensional motion, that
I can write this type of equation. And as
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you have seen this equation is much more complicated
than the usual kinematic equation that we
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have been used to writing v is equal to u
plus a t. Suppose, we are opting this equation
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of velocity as a function of time and also
like to find out the distance travelled by
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the particle in the same time t and at the
same conditions.
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This is an equivalent of the kinematic equation
that we write s is equal to u t plus half
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a t square, which tells you that if you know
the acceleration then you know what is the
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distance travelled by the particle in a given
time t. Of course, knowing the initial velocity
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u equivalent of that equation can be d found
out by integrating this particular equation
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and that is what I will do, next to find out
the distance travelled by the particle in
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a given time t.
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This is what I have written here the equation
gives the speed of the particle with initial
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speed u naught after time t under a constant
force F. Force is very clear that may repeat
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I assume that the motion is in one dimension,
force is applied in the same direction in
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which the velocity of the particle exist.
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We can integrate the above equation further
to find the distance travelled by the particle
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that is what I said and going to do next.
For that matter, let us just assume for simplicity
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that this is along the x direction, because
we are talking only of one particle frame,
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there is no question of transformation as
of now. So, you have just a frame and you
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would like to know you have a particular particle
which is moving in a particular direction.
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Let us assume that this particular dimension
direction is x direction just for simplicity.
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Therefore, once I try to integrate this equation
this u t can be written as d x d t, because
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the motion is along the x direction.
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Therefore, if I write this u t as d x d t
this equation becomes d x and I take d t on
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the right hand side, so this whole thing to
d t after that I can integrate it, this is
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what I have written. This particular transparent
sheet 0 to x d x assuming that the particle
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starts at time t is equal to 0 from x is equal
to 0. Of course, we assume the time starts
13:56.420 --> 14:02.100
from the time when we are trying to measure
the distance and goes after time t here, x
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is total distance travelled by the particle.
I have to integrate this particular equation
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once I integrate this equation then I can
find out what will value of distance that
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the particle travels at this particular time
t, like before I would like to take the indefinite
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form of this integral and solve. Of course,
the left hand side is very simple because
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this just gives you the value of x because
once I integrate, I will get x and if you
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put the limit this will just be x. So, the
left hand side is obvious trivial only the
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right hand side has to be integrated though
the integral looks complicated, but you can
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see that this integral as you will see this
integral can be evaluated. So, let us go to
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your next transparent sheet, where I write
indefinite form.
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Let us solve right hand side integral in indefinite
form by using a substitution, I take a substitution
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and using that particular substitution I will
able to solve this integral.
15:04.620 --> 15:07.930
So, this right hand side integral which I
have written, I have removed the limits because
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I want to solve this in indefinite form just
to make things easy. You could have done the
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other way also if you like it and I out the
substitution this particular quantity which
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is under the root 1 plus A t plus gamma u
naught u naught divided by c square. Of course,
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this quantity is also squared this whole quantity
within the bracket. I write as sub variable
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that I am just taking this variable to be
p to sub variable which is put here, which
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is put inside the bracket.
Suppose, I take this for substitution I have
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to change this by variables integration also
from d t to d p, for that I have to differentiate
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this quantity. Once I differentiate this quantity
on the right hand side I will get d p, but
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let me try to first differentiate this particular
quantity. If I differentiate 1 I will get
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one times m, so I will get 0, so that does
not contribute to the differential.
16:06.030 --> 16:13.030
Once, I start differentiating this, this whole
square is multiplied by constant 1 upon c
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square or rather multiplied by a constant
c square, this constant c square will remain
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as it is in fact that is a reason this c square
have taken on the right hand side. Then I
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have to differentiate this quantity which
is on the numerator, if I differentiate that
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particular quantity this is f x square or
rather f t square. So, first I have to differentiate
16:37.810 --> 16:43.540
assuming f t to be one single variable this
becomes two times f t, so this becomes 2 multiplied
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by this quantity, which is in the bracket
which is a multiplied by t plus gamma u naught
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u naught.
Then I have to differentiate this quantity
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in the bracket itself, once I differentiate
this quantity in the bracket you know I have
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to find out d f d t and for that this quantity
is constant, initial velocity is constant
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then gamma u naught is also constant. Then
this quantity when I differentiate I just
17:09.260 --> 17:16.260
get A, so this A this 2 is because of this
differential this A is because of this differential
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and divided by c square.
I have taken on the right hand side, so what
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I get 2 A t plus gamma u naught u naught A
d t is c square d p what I will do here. I
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have A t plus gamma u naught u naught this
particular quantity I will replace by c square
17:37.500 --> 17:44.500
d p. Of course, this multiplied by d p, I
will replace by c square d p, d by 2 A this
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what I am doing in the next transparent sheet.
17:51.850 --> 17:58.850
This was my original integral this quantity
I have written as p, so this becomes under
17:58.940 --> 18:05.940
p. And as it is discussed this multiplied
by d p, can be written as c square divided
18:06.610 --> 18:13.610
by 2 A multiplied by d p this is what I have
written c square upon 2 A is constant.
18:13.780 --> 18:20.780
So, I can take it out and what remains here
is the integral of 1 upon under root p, which
18:31.130 --> 18:38.130
becomes integral of p to the power minus half
d p. This will become n plus 1 which becomes
18:44.370 --> 18:51.370
3 by sorry minus half this becomes p to the
power half must be divided by half. So, this
18:55.870 --> 19:02.460
becomes 2 p to the power half, so this is
p to the power half is at the root p, this
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what I have written here, in this particular
transparent sheet c square divided by 2 A
19:08.350 --> 19:15.350
multiplied by 2 p. This two of course, cancels
out and then what I get is c square by 2 A
19:15.780 --> 19:21.620
root p. So, this rather easy integral to solve
not all that difficult I substitute the value
19:21.620 --> 19:27.430
of p and then put that into the integral,
then I will get x as a function of p in fact
19:27.430 --> 19:29.500
p contains time.
19:29.500 --> 19:36.500
So, now the integral as I said becomes c square
by a under root p, so this quantity was p
19:38.440 --> 19:43.910
which we had defined it of course, depend
on time and depends on A which is f upon m
19:43.910 --> 19:50.910
naught. We can substitute that into the integral
and we can write the full integral and the
19:53.010 --> 19:59.410
solution of the full integral by putting limits,
if I put the limit I have to I mean the left
19:59.410 --> 20:03.880
hand side was very clear because it was integral
of d x. So, first you put the limits you just
20:03.880 --> 20:09.650
get x, but here when you put the limit you
have to put the limit t and subtract minus
20:09.650 --> 20:14.120
put t is equal to 0 and then get the answer.
So, this is what I will be doing in the next
20:14.120 --> 20:18.410
transparent sheet taking this write it.
20:18.410 --> 20:25.410
So, that it makes little clear quantity written
here was just c square by A 1 plus A t plus
20:29.750 --> 20:36.750
gamma u naught u naught square divided by
c square, first quantity when I put the limit
20:40.440 --> 20:47.440
t this quantity remains identical. So, when
I put 0 then have to subtract whatever I get
20:50.800 --> 20:57.800
substitute u 0, if I substitute 0 you will
get c square by A and the root 1 plus gamma
20:59.280 --> 21:06.280
u naught u naught, from there is no square
here gamma naught u naught divided by c square.
21:18.510 --> 21:23.890
This is what I am writing in the next transparent
sheet here, that x is equal to c square by
21:23.890 --> 21:24.840
A.
21:24.840 --> 21:31.840
This quantity remains as it is, minus c square
upon A under root 1 plus gamma u naught u
21:32.420 --> 21:39.420
naught upon c whole square. As you can see
that this equation is also much more complicated
21:41.090 --> 21:46.040
than simple equation s is equal to u t plus
half a t square, which we have been using
21:46.040 --> 21:52.180
for kinematics in the traditional classical
mechanics. Now, what this equation can give
21:52.180 --> 21:59.180
me x the distance travelled by the particle
in time t, if the particle starts with a speed
21:59.770 --> 22:05.710
u naught and is under the influence of a force
F and it has a rest mass m naught.
22:05.710 --> 22:12.650
So, a is equal to F upon m naught of force
assuming that the motion is purely in one
22:12.650 --> 22:18.620
dimension this is where the equivalent of
the kinematic equations, we are used to in
22:18.620 --> 22:24.700
kinematics in classical mechanics. Now, let
us go to the next step which is crucial for
22:24.700 --> 22:30.690
us to arrive at the transformation of electric
field and magnetic field that we are going
22:30.690 --> 22:37.690
to discuss, as the last topic for this course.
So, question is that if I know the force of
22:39.380 --> 22:46.380
the particle in a given plane what will happen
to the force as we viewed by another frame.
22:48.130 --> 22:54.540
This what we call then transformation we have
been talking about momentum transformation,
22:54.540 --> 22:57.200
we have been talking about momentum transformation,
but it means that I know the velocity components
22:57.200 --> 23:01.920
in a given frame and I want to find out the
velocity components of the same particles
23:01.920 --> 23:08.920
in a different frame. Similarly, I know the
momentum components in the given frame and
23:09.240 --> 23:13.680
I want to find out the momentum components
in a different frame.
23:13.680 --> 23:20.100
Now, my question is if I know what are the
force components F x, F y, F z, I would like
23:20.100 --> 23:25.850
to find out what are the F x, F y, F z in
different frame. In classical mechanics they
23:25.850 --> 23:30.770
are expected to be same, but as we will be
seeing in relative mechanics they do not turn
23:30.770 --> 23:37.770
out to be same the force components all four
are framed. Remember, F is equal to d v d
23:38.960 --> 23:45.960
t, so once you are talking of transformation
of momentum how we are going to transform
23:47.750 --> 23:48.980
the force.
23:48.980 --> 23:53.809
This in have written as transformation force
with the new definition of force in place.
23:53.809 --> 23:58.730
Let us try to see if it is transformation
to seek its transformation as we change the
23:58.730 --> 24:03.970
frame, I know the components of force in a
given direction what are the components of
24:03.970 --> 24:09.980
a force in a different frame on that, what
we will do? We will go back to the concept
24:09.980 --> 24:15.770
of fierce vector, which we have been using
earlier we had defined the Lorentz transformation
24:15.770 --> 24:19.340
in terms of four vector.
We have defined velocity transformation in
24:19.340 --> 24:25.720
terms of force vector and we defined energy
momentum in terms of force vector. So, let
24:25.720 --> 24:32.720
us try to construct a force four vector and
if I am able to construct a force, four vector
24:34.040 --> 24:39.309
obviously that will give me the transformation
of force, so that is the way I will look into
24:39.309 --> 24:46.309
it, I will first try to define the force four
vector.
24:49.450 --> 24:56.450
As you have said earlier define the momentum
four vector, as this the notation that we
24:58.140 --> 25:03.990
have used if you remember is that these are
the four components of the four vector, first
25:03.990 --> 25:09.850
three components are x, y and z. Components
of momentum of the particle, first component
25:09.850 --> 25:16.720
is p x second component is p y third component
is p z and the fourth component of momentum
25:16.720 --> 25:23.350
called momentum or normally call momentum
energy. Four vectors are dependent on the
25:23.350 --> 25:27.890
total relativistic energy of the particle
and in the notation that we have used.
25:27.890 --> 25:34.890
We write it as i E by C, where E is the total
energy, total relativistic energy of the particle.
25:37.030 --> 25:43.140
Obviously, if I have to construct a momentum
for force, I have to differentiate with respect
25:43.140 --> 25:48.470
to time or I have to differentiate differentiate
something; I have to differentiate this with
25:48.470 --> 25:54.480
respect to something that has a dimension
of time. If I differentiate this with respect
25:54.480 --> 26:00.410
to d t again we land into the same problem,
where we are trying to construct from a displacement
26:00.410 --> 26:06.690
four vector to a velocity four vector because
d t is a frame dependent quantity, time interval.
26:06.690 --> 26:13.690
If we take a particular time a moment of the
particle and at a little later time the moment
26:14.010 --> 26:21.010
of the particle is the difference this delta
t time is a frame dependent quantity. If I
26:21.760 --> 26:27.640
differentiate this particular thing with d
t, then I will not be able to get the four
26:27.640 --> 26:34.309
vectors because in that case d t will be frame
dependent quantity. If I have to construct
26:34.309 --> 26:38.690
another four vectors out of this I have to
multiply or divide with something which is
26:38.690 --> 26:44.620
a four scalar which does not depend on the
frame, this we have discussed earlier.
26:44.620 --> 26:51.620
If you realize if you remember it is a proper
time interval which is frame independent quantity.
26:51.710 --> 26:57.080
Therefore, if I have to construct a force
four vector from a moment four vector I cannot
26:57.080 --> 27:02.260
differentiate with respect to d t, which is
different in different frame I must differentiate
27:02.260 --> 27:07.730
with respect to d tau. The proper time interval
because d tau is a frame independent quantity,
27:07.730 --> 27:14.130
if you change your frame d tau will not change.
Therefore, if I have to take a force four
27:14.130 --> 27:16.050
vector I prefer to differentiate this with
respect to d tau.
27:16.050 --> 27:23.050
That is what I have written to generate a
force four vector, we have to differentiate
27:23.980 --> 27:30.980
with proper time which is four scalar therefore,
I define force vector, four vector force four
27:31.429 --> 27:38.429
vector as d p d tau where tau is the frame,
tau is a frame independent quantity. If you
27:42.299 --> 27:47.850
remember whatever we have done at the time
at velocity transformation eventually, I will
27:47.850 --> 27:53.360
write to write in terms of the force component
in my own frame. If I have a particular frame
27:53.360 --> 27:59.340
and in my frame I would prefer to write this
in terms of d t and as that is my own frame,
27:59.340 --> 28:06.030
but my definition gives the differential in
terms of d tau. I can always convert this
28:06.030 --> 28:13.030
d tau into d t, so if I know the velocity
of the particle at that instant of time then
28:13.450 --> 28:20.450
I can calculate gamma u.
Then I know d tau is equal to d p divided
28:20.750 --> 28:26.480
by gamma u this also wee have earlier used
in the case of force four, I am sorry velocity
28:26.480 --> 28:33.480
four vector. So, this d tau can be written
as d p divided by gamma u, this quantity force
28:34.670 --> 28:41.670
can be written as gamma u d p d t the advantage
is that now I write this in a particular frame.
28:42.500 --> 28:47.240
If you are in a particular frame you would
like to know things in your frame to construct
28:47.240 --> 28:53.150
a force four vector, so it is better if I
know things in terms of my own frame.
28:53.150 --> 29:00.070
So, my prescription is that I have to calculate
the moment four vector in my frame then differentiate
29:00.070 --> 29:05.309
with respect to time with my own frame, because
this is going to a frame dependent quantity.
29:05.309 --> 29:10.549
So, what I have to do I multiply this by gamma
u, where gamma is 1 upon under root 1 minus
29:10.549 --> 29:14.929
c square by c square, where u is this speed
of the particle measured in my own frame.
29:14.929 --> 29:21.929
If I know the speed of the particle the moment
of four vector, I know gamma u and using that
29:22.090 --> 29:28.510
particular d t I can find out what is force
four vector.
29:28.510 --> 29:32.490
This what I have said the four components
of the force four vector, vector defined in
29:32.490 --> 29:39.070
this way would then given me like this I calculate
with my own frame, what is the x component
29:39.070 --> 29:44.040
of momentum differentiate with respect to
time as measured with my frame. This is this
29:44.040 --> 29:48.140
quantity take the y component of momentum
differentiate with respect to time as measured
29:48.140 --> 29:51.990
in my own frame, this will give me the second
component.
29:51.990 --> 29:57.160
I take the z component of the momentum differentiate
with respect to time in my own frame I will
29:57.160 --> 30:02.040
calculate and get the third component. I know
the energy of the part in my frame, I differentiate
30:02.040 --> 30:07.650
with respect to time in my own frame all these
quantities are given in my own frame then
30:07.650 --> 30:11.720
I do what is the speed of the particle, I
calculate gamma u which is the under root
30:11.720 --> 30:17.120
1 minus u square by c square. I know gamma
u, I have multiplied this gamma u by these
30:17.120 --> 30:21.799
quantities then these four vectors are now
written in with respect to everything that
30:21.799 --> 30:27.429
is known in my own frame.
So, I know how to write it, so the force four
30:27.429 --> 30:33.150
vectors will be given by this particular quantity
all these quantities multiplied by gamma u
30:33.150 --> 30:38.620
will generate the force four vectors. If you
remember exactly the way we had calculated
30:38.620 --> 30:44.510
or we have we have constructed a velocity
four vector, now once we know that this is
30:44.510 --> 30:50.670
a velocity four vector, I am sorry force four
vector then I know how it is transformed.
30:50.670 --> 30:56.100
Once I go from one frame to another frame
because this is the way force transformation
30:56.100 --> 31:00.429
four vector class formation takes place through
a standard equation.
31:00.429 --> 31:05.480
Let me just first read this thing the first
three quantities in the bracket are equal
31:05.480 --> 31:11.960
to the components of the force once I said
d p x d t, we have already defined as the
31:11.960 --> 31:18.960
force d p y by d t. I have defined as f y
d p z by d t, I have defined as f z and the
31:20.179 --> 31:24.570
last quantity depends on the rate of change
of the total energy which we can write in
31:24.570 --> 31:29.240
terms of power, which we have done earlier
in the last lecture. So, before going to the
31:29.240 --> 31:34.630
transformation let me just rewrite this thing
this components in terms of the force and
31:34.630 --> 31:37.590
d e d t.
31:37.590 --> 31:43.760
Therefore, the components of force four vector
can be written as gamma u, first component
31:43.760 --> 31:50.760
was d p x d t which by new definition is f
x d p by d t is the new definition is f y
31:51.910 --> 31:58.870
d p z d t. If I call this by new definition
is f z. If you remember we had calculated
31:58.870 --> 32:05.870
d p of E in the last lecture and that we have
found out be power F dot u. So, d p d t I
32:06.890 --> 32:13.890
have written F dot u divided by c, so energy
happens to be the forth component of energy
32:14.280 --> 32:19.630
moment four vector time happens to be the
fourth component of the displacement four
32:19.630 --> 32:26.630
vector for the force four vector, the force
component turns out to be power.
32:26.910 --> 32:33.910
So, once we have generated everything in terms
of forces, if I go to different frame in general
32:34.210 --> 32:40.640
f x is likely to change f y will change f
z will change. Of course, u will change and
32:40.640 --> 32:44.090
of course, gamma u will change because the
velocity of the particle would change. Once
32:44.090 --> 32:50.870
I change my frame of preference, I can write
the transformation equation as given in this
32:50.870 --> 32:52.280
particular transparent sheet.
32:52.280 --> 32:59.280
These are my four components of the force
four vectors in the particular frame s prime
33:10.610 --> 33:16.559
this is the transformation matrix which depends
on gamma, gamma depends on v the relative
33:16.559 --> 33:21.220
velocity between the frames. Remember, we
have three different speeds, so we should
33:21.220 --> 33:23.809
not get confused these are the components
that I want to find out in a frame s this
33:23.809 --> 33:30.059
a particle which is moving in s its velocity
is being measured by an observer sitting in
33:30.059 --> 33:37.059
frame s. That is what I am calling as u the
same particle is being observed by an observer
33:38.280 --> 33:45.280
in s prime and that observer finds the speed
of the same particle to be u prime, so that
33:45.309 --> 33:49.419
is second speed.
The third speed is that the relative velocity
33:49.419 --> 33:54.250
between the force, the two frame which always
assumed to be along the x direction and all
33:54.250 --> 34:00.700
those things which are the conditions of Lorentz
transformation that is v. We have three speeds
34:00.700 --> 34:06.809
and correspondingly we have three gamma’s,
we have gamma u we have gamma which depend
34:06.809 --> 34:13.190
on v, we have gamma u prime which depend on
u prime. This is my transformation matrix
34:13.190 --> 34:19.339
which I have to open up if I have to find
out what will be f x prime relating to other
34:19.339 --> 34:26.339
components in s frame that is what we see
next.
34:26.770 --> 34:33.770
Let us expand it within this particular transformation
matrix here just to take some time to make
34:34.629 --> 34:41.629
it clear, if I have write gamma prime f x
prime this will be gamma multiplied by this
34:41.940 --> 34:48.940
component 0 multiplied by this component plus
0 multiplied by this component plus i beta
34:51.499 --> 34:58.499
gamma multiplied by this component. So, that
will be the first term, let us just see first
35:00.410 --> 35:05.640
term once I multiplied here gamma u, gamma
u will come out and f x there will be gamma
35:05.640 --> 35:11.710
and gamma u which will also be similar i square
will make it minus 1. So, second term will
35:11.710 --> 35:15.380
be F dot u by c into beta.
35:15.380 --> 35:22.380
This is what I have written here that is the
first term gamma u prime f x prime, which
35:22.380 --> 35:29.380
is the first terms on the matrix. The column
matrix on the left hand side is equal to gamma
35:29.869 --> 35:36.869
gamma u which we have agreed will come out
multiplied by f x minus beta F dot u divided
35:37.940 --> 35:44.940
by c. Let us look at the second component,
second equation here in the paper it is gamma
35:47.190 --> 35:54.190
u prime f y prime is equal to 0 multiplied
by this one multiplied by this plus 0 multiplied
35:55.279 --> 36:02.279
by this plus 0 multiplied by this which gives
gamma u prime f y prime is equal to gamma
36:02.999 --> 36:09.999
u f y gamma u prime f z prime.
Similarly, will be gamma u f z and the fourth
36:11.479 --> 36:18.479
equation will be in gamma u prime this quantity
minus i beta gamma multiplied by gamma u f
36:19.670 --> 36:26.380
x plus 0 multiplied by this plus 0 multiplied
by this plus gamma multiplied by this. Here
36:26.380 --> 36:31.549
also you will see that there is a gamma and
gamma u there is a gamma and gamma u which
36:31.549 --> 36:38.549
will come out of the matrix. So, these are
my equations gamma u dot f y prime is equal
36:39.489 --> 36:45.190
to gamma u f y gamma u prime f z prime is
equal to gamma u f z and this is the fourth
36:45.190 --> 36:52.190
equation gamma u prime i F prime dot u prime
by c because the fourth term in the column
36:52.829 --> 36:57.349
matrix of the left.
This i beta beta with this i gamma gamma u
36:57.349 --> 37:04.349
and i comes out and you will get minus beta
f x plus f dot u by c, these are expanded
37:07.930 --> 37:14.309
this particular matrix equation has been expanded
into the four equations. Now, we will try
37:14.309 --> 37:21.049
to simplify these equations by using a notation
using a equation, which we had opted at the
37:21.049 --> 37:27.539
time of velocity transformation to write in
a compact form which is a usual form and which
37:27.539 --> 37:34.539
probably is much more useful as far as our
futures endeavors are concerned.
37:34.769 --> 37:40.380
So, this is what I have written, we shall
use a previous result to simplify the force
37:40.380 --> 37:47.380
transformation equation this result was obtained
while writing the velocity four vectors.
37:49.380 --> 37:54.150
This was my velocity four vector transformation,
see remember this transformation matrix is
37:54.150 --> 37:59.390
exactly same in that case. If you remember
what was forming the four vector was gamma
37:59.390 --> 38:06.390
u u x gamma u u y gamma u u z and gamma u
i c and when we change the frame it became
38:08.220 --> 38:13.450
gamma u prime u x prime gamma u prime u y
prime gamma u prime u z prime and gamma u
38:13.450 --> 38:19.319
prime i c prime.
Let us look at only the fourth equation which
38:19.319 --> 38:25.329
is this equation, which is gamma u prime i
c which is equal to gamma multiplied by gamma
38:25.329 --> 38:32.299
u I am sorry minus i beta gamma multiplied
by gamma u x plus gamma u multiplied by gamma
38:32.299 --> 38:37.539
u i c. I am looking only the fourth equation
first three equations will give me standard
38:37.539 --> 38:38.720
velocity transformation.
38:38.720 --> 38:45.720
Now, you just look at this particular fourth
equation which is here gamma u prime i c is
38:46.489 --> 38:50.029
equal to this particular quantity this we
have done earlier, discussed earlier in great
38:50.029 --> 38:55.390
detail when we are discussing the velocity
transformation. And by reorganizing the term
38:55.390 --> 39:01.960
on which I would not like to occupy time here
is what we this what we get gamma u prime
39:01.960 --> 39:06.989
is equal to gamma gamma u 1 minus v u x upon
c square.
39:06.989 --> 39:13.569
So, on the left hand side the force transformation
equation the four equations that we have got
39:13.569 --> 39:20.059
where ever gamma u prime comes I will replace
it by gamma gamma u 1 minus v u x upon c square.
39:20.059 --> 39:25.309
Therefore, I will be able to get a clean equation
writing f x prime in terms of the components
39:25.309 --> 39:30.210
this is what I am doing in the next transparent
sheet.
39:30.210 --> 39:37.210
So, I have said we use this equation in the
expansion of the force transformation equation.
39:40.200 --> 39:46.819
This was my first term which we have opted
by expanding that particular matrix equation
39:46.819 --> 39:53.819
gamma u prime f x prime is equal to gamma
gamma u f x minus beta F dot u by c just now
39:56.349 --> 40:00.440
we have obtained by expanding that particular
matrix. This is the equation which I have
40:00.440 --> 40:04.489
written in earlier transparent sheet, which
we had obtained from the velocity transformation
40:04.489 --> 40:11.029
which related gamma u prime in terms of gamma
gamma u so gamma u prime is equal to gamma
40:11.029 --> 40:16.180
gamma u multiplied by 1 minus v u x upon c
square.
40:16.180 --> 40:21.910
So, this is gamma u prime here I will substitute
by this particular quantity gamma gamma u
40:21.910 --> 40:28.089
multiplied by 1 minus v u x by c square once
I do here. Let us put it here gamma gamma
40:28.089 --> 40:34.869
u cancel from here, what I will be getting
1 minus v u x upon c square which will get
40:34.869 --> 40:41.869
multiplied to f x prime and right hand side
will just become f x minus beta f dot u upon
40:42.440 --> 40:49.440
c. Now, it is rather simple I can just divide
this quantity both sides and I will get f
40:51.150 --> 40:57.410
x prime is equal to a quantity that I am doing
in the next transparent sheet.
40:57.410 --> 41:04.410
I have divided 1 minus u x upon c square and
remember there was a beta and there was a
41:05.700 --> 41:10.489
c here, there was beta was v upon c there
was also a c here. So, this become v upon
41:10.489 --> 41:17.489
c square this becomes v upon c square F dot
u divided by 1 minus v u x upon c square may
41:19.190 --> 41:25.130
be you will see the resemblance that you will
get from the equation. That was the time when
41:25.130 --> 41:29.779
we were doing Lorentz transformation, we were
doing the time transformation we were getting
41:29.779 --> 41:36.630
something like this, and this is something
similar to that except the fourth term to
41:36.630 --> 41:42.029
be F dot u.
This will be the transformation of the x component
41:42.029 --> 41:49.029
of the force, let us look at the transformation
of the y component of the force of course,
41:52.969 --> 41:57.460
before first let me first discuss it because
this equation sometime is written in different
41:57.460 --> 42:01.329
form. Let us first work out that particular
form then we will do their vital component
42:01.329 --> 42:06.609
transformation and transformation just wanted
to make one more point.
42:06.609 --> 42:13.609
That member f x prime will also depend on
force f x f y and f z see like in the case
42:15.119 --> 42:22.119
of moment of transformation x prime also dependent
on the energy which depend on E square should
42:22.410 --> 42:28.029
depend also on the x y and z component of
moment of not just the x component of the
42:28.029 --> 42:33.200
momentum. Similarly, here it is also dependent
on f y and f z sometimes the equation is written
42:33.200 --> 42:38.640
in a different form avoiding to write this
F naught. So, let us just try to write that
42:38.640 --> 42:42.229
equation in that form before you go to the
transformation of the y component and the
42:42.229 --> 42:42.390
z component.
42:42.390 --> 42:49.390
So, what we do we just expand this F dot u
f x is v upon c square remains like this and
42:55.049 --> 42:58.779
like simple dot product of two vectors. If
we have dot product of E and v we have e x
42:58.779 --> 43:05.779
dx plus e y d y plus e z d z and if you have
F dot u you will get f x u x plus f y u y
43:05.969 --> 43:12.969
plus f z u z. This is what I have written
minus v upon c square F dot u is f x u x plus
43:13.739 --> 43:20.739
f y u y plus f z u z just simply expanding
and writing what I will do, I will take this
43:23.140 --> 43:27.359
f x common here.
If I take f x common I will get 1 minus v
43:27.359 --> 43:34.359
b u c square I am sorry we will get f x 1
minus v u x by c square. So, there is a u
43:36.279 --> 43:41.039
x here there is a u here, so I will get a
v u x by c square and that quantity will cancel
43:41.039 --> 43:48.039
with this particular quantity only as far
as the x components are concerned as far as
43:48.279 --> 43:54.950
these components are concerned there is no
cancellation, so I can rewrite slightly this
43:54.950 --> 44:01.950
This was my equation, so as I said f x 1 minus
v upon c square f x 1 minus v u x upon c square
44:04.049 --> 44:11.049
this cancels and you get this simple f x these
two quantities, there is no cancellation.
44:11.259 --> 44:18.259
If you get f x minus v upon c square which
is here f y u y which is here plus f z u z
44:19.079 --> 44:25.119
which is here the of course, divided by 1
minus v u x upon c square. So, this tells
44:25.119 --> 44:30.279
you that f x prime is equal to f x minus all
these quantities.
44:30.279 --> 44:36.390
Remember in classical mechanics we expect
v f x prime to be equal to f x. So, we can
44:36.390 --> 44:41.289
see very easily that when v is going to be
close to c then all these terms will cancel
44:41.289 --> 44:45.289
and just f x prime and u will be negligible
rather and this will give you f x prime is
44:45.289 --> 44:52.289
equal to f x. Now, let us go to the transformation
of the y component of the force.
44:54.819 --> 44:59.369
For second and third both are similar, so
let us look for simplifying the second term
44:59.369 --> 45:06.059
this was the equation gamma u f y prime is
equal to gamma u f y. This was my second equation
45:06.059 --> 45:13.059
which we had obtained from opening the matrix
I use again gamma u prime is equal to gamma
45:13.319 --> 45:20.319
gamma u 1 minus v u x upon c square exactly
similar way here. There will be gamma this
45:21.509 --> 45:26.099
gamma u will cancel from this gamma u, on
the right hand side multiply by 1 minus v
45:26.099 --> 45:32.089
u x upon c square. So, this gamma will not
cancel here, we will just have a f y prime
45:32.089 --> 45:39.089
into gamma into the whole of this quantity
1 minus v x upon c square divided by equal
45:40.719 --> 45:42.479
to y f y.
45:42.479 --> 45:49.479
Therefore, f y prime is equal to f y divided
by gamma 1 minus v u x upon c square. If you
45:52.279 --> 45:57.719
remember this was similar to the velocity
transformation of the y component and because
45:57.719 --> 46:04.719
the z component equations are exactly identical
their f z prime is equal to f z gamma 1 minus
46:05.400 --> 46:08.999
v u x upon c square. Remember in the x component
there is no gamma there because the gamma
46:08.999 --> 46:14.769
has cancelled out here gamma has not cancelled
out gamma u has cancelled out. So, gamma is
46:14.769 --> 46:21.769
written in the denominator just like the velocity
transformation that we had got obtained there.
46:22.599 --> 46:27.289
You can see that in non relativistic limit
that f x prime is equal to f x, that we have
46:27.289 --> 46:33.349
discussed. You can also see in non relativistic
limit this will become one this when v and
46:33.349 --> 46:39.450
u x are much smaller than c this quantity
is negligible in comparison to one f y prime
46:39.450 --> 46:43.289
will be equal to f y f z prime will be equal
to f z. So, normally in non relativistic limit
46:43.289 --> 46:48.529
we do get f x prime is equal to f x f y prime
is equal to f y and f z prime is equal to
46:48.529 --> 46:55.529
f z, but in relativity forces also transformed
in a different frame they appear to be different.
46:56.170 --> 47:00.460
Now, let us look at the fourth component which
component which is related to transformation
47:00.460 --> 47:06.690
of power, so power also transforms. So, once
I go from one particular frame to another
47:06.690 --> 47:11.309
frame the power will also transform. Now,
let us look at the transformation of the fourth
47:11.309 --> 47:18.309
component of force for vector which is related
to the transformation of power.
47:18.509 --> 47:25.099
This was my equation which we have obtained
by expanding the matrix, which was the fourth
47:25.099 --> 47:31.769
component equation again I write for gamma
u prime gamma gamma u 1 minus v u x divided
47:31.769 --> 47:38.769
by c square here. There is a gamma gamma u
which will cancel, it will also cancel which
47:39.190 --> 47:46.190
is being to somewhat simpler equation it is
F prime dot u prime is equal to f dot u minus
47:49.799 --> 47:56.799
v f x divided by 1 minus v u x upon c square.
So, this gives me the transformation of power
48:00.200 --> 48:05.700
once I move from frame s to s prime how my
power changes.
48:05.700 --> 48:12.700
This is the transformation equation that governs
that particular transformation equation. Now,
48:14.259 --> 48:19.650
let us look at the inverse transformation
this tradition whenever we write a transformation,
48:19.650 --> 48:26.170
we also write an inverse transformation. It
means if my quantities are given in s prime
48:26.170 --> 48:30.559
frame of reference, we have to find put the
equivalent quantities in s frame of reference
48:30.559 --> 48:36.170
how do I transform like for instance in velocity
transformation.
48:36.170 --> 48:41.749
My velocity components are given in s prime
frame of reference and I want to find those
48:41.749 --> 48:47.309
velocity components in s frame of reference
that is what we call as inverse transformation
48:47.309 --> 48:53.150
descriptions are simple. As we have discussed
number of times that all prime quantities
48:53.150 --> 48:59.749
have to be changed to unprime quantities and
wherever there is a v replace it by minus
48:59.749 --> 49:06.660
v this is as simple as that, so that is what
we are doing. We are now writing the inverse
49:06.660 --> 49:12.749
transformation equation just by interchanging
the prime and unprime and replacing v by minus
49:12.749 --> 49:13.160
v.
49:13.160 --> 49:20.160
So, this is my force component, so on the
left hand side we have f x is equal to f x
49:27.499 --> 49:33.700
here there was f x. So, this becomes f x prime
there was a minus v because we have replaced
49:33.700 --> 49:40.700
a v by minus v, so this becomes plus this
becomes v upon c square F dot u. Now, F prime
49:41.009 --> 49:48.009
dot u prime should be prime divided by 1 minus
v u x prime by c square. Similarly, f y will
49:51.329 --> 49:58.329
now become f y prime gamma 1 plus v u x prime
upon c square because v has been replaced
50:00.219 --> 50:07.219
by minus v. Similarly, here f z will be equal
to f z prime divided by gamma 1 plus v u x
50:08.289 --> 50:14.140
prime by c square, so this becomes inverse
transformation by force.
50:14.140 --> 50:21.140
Similarly, we can write the inverse power
transformation also exactly identical things
50:21.390 --> 50:28.390
here become F dot, I am sorry become F prime
dot u prime plus v x prime divided by 1 plus
50:29.599 --> 50:35.219
v u x prime by c square. For the left hand
side we have F dot u prescription is simple
50:35.219 --> 50:39.209
we do not spend too much of time in describing
this, we just interchange all quantities as
50:39.209 --> 50:46.209
I said make all prime quantities all unprime
quantities make them prime replace v by minus
50:46.410 --> 50:53.410
v. This case we get inverse transformation
before we close the lecture, let us look at
50:54.729 --> 51:01.729
a special case which is quite useful and illustrative
and we will also be using in our next lecture,
51:03.200 --> 51:08.479
we consider a special case of a force transformation.
51:08.479 --> 51:15.390
Let us assume that there is a frame s in which
this particular particle is instantaneously
51:15.390 --> 51:20.930
at rest of course, it is at rest at that given
point of time, but a later time because it
51:20.930 --> 51:26.170
is under the influence of force it is speed
will change. So, no longer it will at rest
51:26.170 --> 51:31.769
at that particular frame assume that there
is a frame in which at a given instant in
51:31.769 --> 51:37.109
which I am interested. In this particular
particle is at rest is whatever was the velocity
51:37.109 --> 51:43.979
of that particular particle at that particular
frame that particular the frame seems to go
51:43.979 --> 51:50.979
with the same velocity. Therefore, this particular
particle is at rest in that particular frame
51:51.890 --> 51:58.890
in that case u would be 0 u will be 0 and
the force equations become somewhat simple
52:00.200 --> 52:00.799
equations.
52:00.799 --> 52:07.099
So, this is the f x prime this is the transformation
of the x component because this particle is
52:07.099 --> 52:14.099
instantaneously at rest in s frame, so u becomes
0 once u becomes 0 this whole quantity becomes
52:14.849 --> 52:21.009
0. Of course, if u is 0 x is also 0, so this
quantity becomes 0 and denominator you have
52:21.009 --> 52:28.009
only one which gives me f x prime is equal
to f x. So, if we measure the force on the
52:29.739 --> 52:35.009
particle in the s frame at that point of time
when the particle was instantaneously at rest
52:35.009 --> 52:41.170
in s prime frame also it will appear to be
at rest the x component of the force will
52:41.170 --> 52:43.559
turn out to be the same.
52:43.559 --> 52:50.559
Let us look at y f y prime is equal to f y
divided by gamma one minus v u x upon c square
52:51.799 --> 52:57.479
u and x being 0 in s frame, you will have
this quantity equal to 0. So, you will just
52:57.479 --> 53:04.479
have f y prime is equal to f y divided by
gamma gamma be always greater than 1. Your
53:04.880 --> 53:11.809
f y prime will be smaller than f y and of
course, because these transformation equations
53:11.809 --> 53:18.479
are similar for component I will also get
a z prime is equal to z divided by gamma.
53:18.479 --> 53:25.009
So, what get in this particular case an f
x prime is equal to f x by gamma f y prime
53:25.009 --> 53:31.150
is equal to f y by gamma and f z prime is
equal to f z by gamma.
53:31.150 --> 53:37.499
We thus see the force is smaller because the
f y components have become smaller in a frame
53:37.499 --> 53:44.499
in which the particle is not at rest. So,
the force is large in a frame in which the
53:44.569 --> 53:49.989
particle is at rest which you can call as
a proper frame.
53:49.989 --> 53:56.989
So, to end the lecture we just summarize whatever
we have done we derived the equations similar
53:57.289 --> 54:03.599
to the equations in classical mechanics. We
also defined the force four vector and discussed
54:03.599 --> 54:06.619
the transformation of force.
Thank you.