WEBVTT
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In our last lecture we had worked out a long
example.
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The idea of working out that example, was
just to give you some feeling of how multiple
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events problems have to be solved. And we
had discussed specifically, the importance
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of identifying the events and filling the
event table. So, in the most of the relativity
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problems, involving kinematics it is important
to realize what are the events and then write
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the co ordinates of these events including
time; time of course, will be treating as
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co ordinates as been telling.
And then fill the event table of course, event
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table can be filled with whatever information
is been given, in a frame. If the information
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is not available in that frame but, in a different
frame that information has to be trans translated
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back to the given frame, by using an appropriate
task formation. So, that is what is the basic
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trick of working out the problems of relativity,
look events, fill the event tables find out
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the information available in the frame. Whatever
are not available in the frame but, available
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in different frame bring them back to your
frame by using an appropriate task formation,
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that is the way to eventually work out these
problems.
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Now, in today lecture we will starts with
giving one more example, involving the speed
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of light before we going little bit ahead
in the course. So, example is comparatively
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simple, we have been working out these type
examples, when there is a train in which particular
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observer sitting and that particular observer
emits light. But, in all earlier problems,
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the direction of the first speed of the light
or direction of the velocity of the light
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was always along the motion of the train or
opposite to the motion of the train which
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we have called is x axis.
So, c was always pointed out was always pointing
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out either in pulse x direction or in minus
x direction. Now, we will work out the different
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examples, in which the speed of li in which
the light will be pointing out in the y direction
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that is the direction which is perpendicular
to x direction and towards the top. So, this
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is what is the example, which I am just describing.
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An observer in a train which we are calling
frame as S prime, flashes the light in Y prime
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direction, from the floor which gets reflected
back from the roof at height of H prime. So,
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that roof at the top which at the height of
H prime, I will show just a figure just now,
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and is received back the light is reflected
back from the top and received back at the
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floor of the compartment, of the train.
Now, what we have to do is to describe the
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motion in S frame which I am calling as an
ground frame or the frame, in which the particular
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train is moving. And view of course, the train
is moving with the constant velocity because,
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we are always in the realm special theory
of relativity. So, in our case, the velocity
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is always assumed to be constant, so this
the problem which really essential problem
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but, a way of explaining various concepts
ones more, with a slightly different type
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of example.
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So, this is the figure which I have drawn
here of course, there is one particular person
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with a light source and then, let us assume
that the light source is right here, at the
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floor of the compartment, this is the train
compartment, which is moving to the right
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with the velocity V as seen by an observer
S. And as you known that this direction we
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always call as x direction, let us assume
the y direction is the long the top. So, this
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particular light is been flashed from the
floor, the light goes up is S seen by the
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observer, vertically up and then comes back.
This is a situation, which is S observed by
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observer S prime in this particular train.
Now, question is that, there is an observer
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setting on the ground, in this ground frame
which I am calling as S frame this train is
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moving with the velocity V. How, this particular
person would perceive this particular lights
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motion that is what we. In fact, we worked
out many type similar types of problem classical
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mechanics earlier, the professor who ever
familiar with classical mechanism was be familiar
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that, a particular person sitting in the train
the ball vertically, upwards and the ball
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comes back and reflected back.
And same of the events or same experiment
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is been seen or visualized by the observer
sitting on the ground, how the perception
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will come different. Now, everything assumption
ally similar except that the instead of ball,
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we have light and because, we are talking
of light. So, we are talking special theory
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of relativity and use the relativistic task
formations, otherwise the problem is somewhat
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similar. So, let us define our events we have
emphases quite a bit on the importance of
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identifying the events.
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So, here event 1, event 1 is light emitted
from the floor of the compartment of the train
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towards roof. So, light is thrown upwards,
so this is event number 1, the light been
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emitted from the roof from the floor of the
compartment and train towards the roof. Then
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the light goes up comes back and receive back
at the ground, so when it is received back
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we called that as event 2. So, my event 2
which have written here, E 2 is light reaching
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the floor after reflection from the roof.
So, light been thrown, even light been thrown
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event number 1, light goes up comes back received
at the ground that is event number 2. Now,
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let us first look how an observer S prime
could observer because, the experiment has
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described with respect the to an observer
S prime, which is sitting in the train. Observer
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in S prime, will always feel the light is
always going vertically upwards.
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So, initially when it goes upwards then the
direction of the velocity is along the pulse
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y direction. When it first goes up and it
gets reflected back, the light starts moving
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downwards and the velocity is now, downwards
is minus y direction that that is what will
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be the perception of the observer sitting
in the train.
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So, this is what I have shown in this particular
figure, this is with the reference to the
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observer in the train, which I am calling
as S prime. Has been given in the example
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that this particular height is H prime, this
arrow shows the direction of the velocity
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initially, the velocity of light initially
which goes upwards, this arrow shows when
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veloc when the light is reflected back and
tries to come back towards the floor.
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So, initially the light moves upwards which
is the plus y direction plus y prime direction
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to be more precise. And then it comes back
to the floor, which we call in minus y prime
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direction.
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So, now, let us tries to write the events
in this particular for these two events in
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S prime frame of reference. So, let us assume
that light is emitted from the origin, so
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at the point when the light was emitted that
point was the origin of S prime, in this particular
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case z axis is not involved because, we are
looking only at the x direction and the y
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direction. So, I am not writing the z co ordinate
but, we got the origin, so I can write that
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x prime equal to 0 and y prime equal to 0.
Let us, also assume the time according to
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the S prime observer was also 0 when light
was emitted. So, it is easy to write the co
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ordinates of event one, which is x prime is
equal to 0 y prime is equal to 0 and t prime
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is equal to 0. So, the light was emitted from
the origin of S prime and when light was omitted
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emitted at that time the work of the observer
S prime was showing a time equal to 0. Now,
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event 2 was discussed is that light goes upwards
and comes back.
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So, light goes upwards this distance is H
prime then light comes backwards, again it
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travels at the distance H prime, remember
the every information has been given in S
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prime frame of different. So, the total distance
that light would travel, will be H prime going
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up and H prime while coming down. So, total
distance is 2 H prime and we know that the
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light travels with the speed of light therefore,
if I have to find out the time taken by the
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light to go up and come back, will be the
total distance 2 H prime that the light has
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traveled divided by the c the speed of light.
So, it means the second event must have occurred
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at a time, which is 2 H prime divided by c.
When the light is received back at this particular
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point remember that point is again at the
origin because, if light goes vertically upwards
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and comes down it is received back exactly
at the same point. Therefore, x co ordinates
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and y co ordinates of the second event has
not changed because, both these events occurred
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exactly at the same point.
The light is emitted from this point, goes
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up goes up and comes back in the particular
point, both these events light been emitted
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and light been received, occur at the same
point which is the origin of S prime frame
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of reference. But, the second event occurred
later then the first event because, light
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to be finite among time to go up and come
down. So, the second event must ever occurred
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at a time 2 H prime divided by c.
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So, this is what I had written here this particular
transparency. That for the event number 2
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x prime is equal to 0, y prime is equal to
0 because, the event does occurred at the
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origin and it occurs at time t prime is equal
to 2 H prime divided by c. So, this is these
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are the co ordinates of the event as seen
in S prime frame of distance.
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Now, let us look at the velocity components.
Again with reference to an observer in S prime
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frame of reference, when the light is moving
upwards we had all the direction pulse y direction.
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So, the light had only a y component, upwards
when the light was returning down again it
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has only the y component because, light has
as seen in S prime frame of reference, is
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only moving either in pulse y direction or
in minus y direction, it has no component
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of velocity in x direction.
That is what an observer sitting on the train,
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sees that light goes vertically upwards, vertically
downwards. So, light never moves in horizontal
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direction or any other direction, so; obviously,
it means that u x prime must be equal to 0
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because, it has no component, no x component
of velocity and it has no z component of velocity.
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So, u z prime is also be equal to 0, it is
only u y prime which is not equal to 0 and
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this is pulse c while going upward and it
is minus c while going downwards, while coming
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downwards.
So, I have written u y prime is equal to pulse
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minus c where, the signs have to be interpreted
appropriately, by choosing pulse sign wherever
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I am referring to the motion of light going
upwards and I have to choose minus sign, when
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the light is coming downwards. So, these are
the velocity components of the light has seen
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in S prime frame of difference. Now, my next
question is that, what will the velocity component
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as seen by an observer on the ground which
I am calling is S frame.
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Remember, it is a disk ground frame the disk
train is moving with the velocity v. Now,
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remember all the information has been given
in the S prime frame of reference and if I
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have to find out the information S frame,
we have to use inverse transformation. So,
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what I do, I would use an inverse velocity
transformation to find out what will be the
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velocity components of the light has seen
is S prime frame and I am sorry seen in S
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frame, which the ground frame of difference.
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So, let us write the velocity transformation
equations, velocity transformation equations
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are u x prime. In fact, I will write u x is
equal to u x prime pulse v 1 pulse u x prime
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v by c square, this is as for as the transformation
of x component is concerned. Remember, we
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have used plus sign here and I had used plus
sign here because, this is an inverse transformation.
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Similarly, I had u y is equal to u y prime
divided by gamma 1 pulse u x prime v divided
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by c square I am not writing for z component
because, as for the z component is concern
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this is sort of irrelevant as for this particular
example is concern. So, let us use this particular
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example, this particular equation transformation
equation and try to find out u x and u y and
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to see it what are those values, as it will
seen in S frame of reference.
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So, let us look at the equation, we have u
x we had u x prime which was equal to 0 remember,
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in S prime of different there is no x component
of the velocity. So, this is 0 v is the relativity
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velocity between the frame, so it is anyway
present here, divide by 1 pulse u x which
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is 0 multiplied by v divided by c square.
So, this particular thing gives you 0, so
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in denominator you are left with 1 and this
becomes 0 pulse v divided by 1 which is just
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equal to v.
So, u x turns out to be equal to v which is
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the relative velocity between the frame. Let
us look at u y this pulse minus c again, we
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said that the sign have to appropriate interpreted
pulse will be when it going upward, minus
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when it is going downwards. So, I just putting
one equation pulse minus c divided by gamma
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multiplied by the same factor here, which
here any we discussed one. So, what you get
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is pulse minus c divided by gamma.
So, the y component of the velocity of the
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light as seen in S frame of reference is pulse
minus c by gamma of course, u z is anyway
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0 because, the u z prime becomes 0. Remember,
the transformation equation of y and z component
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look appear very, very similar. So, what we
have done is no found out the velocity component,
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as seen by an observer in S frame.
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We just note that according to the observer
in S frame, the light is also travelling in
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x direction it also a component in x direction.
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So, give similar to what we had observed classically
that if a ball is thrown vertically upward
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as seen by S prime this ball does not really
move upward as seen by the observer here.
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But, also develop x component of the velocity,
which is same as the velocity of the train
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that is what we seen in classical mechanics,
also that is what we are seeing here. Also
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that this particular light now, has developed
the x component and this x component is same
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as the velocity of S prime as seen in S which
is the same velocity of the train.
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But, now in contradiction to the classical
mechanics we have found even the y component
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has changed. And the y component is now, pulse
minus c by gamma remember, this has to be
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done because, as seen in S frame also the
speed of the light has to be same which is
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c. So, if it has picked up the velocity component
in the x direction, in the y component the
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velocity component must come down in order
to maintain the speed of light to be see,
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also in S frame of reference that is, what
this transformations have to do because, that
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is the first postulate of that is the postulate
of special theory of relativity.
18:05.770 --> 18:12.770
Now, let me just first check whether I am
really getting the velocity of like to be
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c as seen in the S frame of reference we let
us first to a quick exercise because, that
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was I expect y component has come down, x
component has picked up from 0 it gone to
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v, y component has come down but, I expect
that still the speed light must remain c in
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S frame of reference let us verify it.
18:34.540 --> 18:41.540
So, as I have said first check, if the speed
of light is still c as seen in S as expected
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as we expected as per the postulate of special
theory of relativity.
18:47.110 --> 18:52.950
So, when I calculate the magnitude of the
speed of the light, I will use this particular
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standard formula u is equal to under root
of u x square pulse u y square pulse u z square
18:59.570 --> 19:06.260
u z is anyway 0. So, let us note particular
thing, we have seen that u x is just equal
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to v relative velocity of with frame for u
x square I put v square, for u y square we
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just now, seen that u y was pulse minus c
by gamma. There these signs have to be appropriately
19:20.310 --> 19:26.450
interpreted but, ones I square it whether
it pulse sign or minus sign both will make
19:26.450 --> 19:33.140
it pulse. So, this particular equation would
just reduce to v square pulse c square by
19:33.140 --> 19:40.140
gamma square. Now, what I do is just substitute
the value of gamma square.
19:40.330 --> 19:47.330
As you remember, gamma square will be equal
to 1 upon 1 minus v square by c square because,
19:51.050 --> 19:57.160
this gamma square in denominator. So, this
1 minus v square by c square will come into
19:57.160 --> 20:01.180
the numerator that is what I written in this
particular transparency.
20:01.180 --> 20:08.180
This is v square pulse c square by gamma square
and this particular gamma square has been
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written as the c square is still here and
this gamma square has written in written as
20:16.790 --> 20:23.790
1 minus v square by c square. If you expand
here, you will get v square pulse c square
20:24.930 --> 20:30.320
and this c square gets multiplied to this
particular factor, this become minus v square.
20:30.320 --> 20:37.080
So, you will have v square minus v square
that will cancel giving you just C Square
20:37.080 --> 20:44.080
and u becomes equal to c.
In fact, becomes pulse or minus c and depending
20:44.200 --> 20:48.920
upon the way, you look out and I looking only
the magnitude therefore, I am putting only
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the pulse sign. So, what I have shown, is
that the speed of light in the turns out to
20:55.320 --> 21:01.860
c as seen in observer s as expected from the
second postulate from the postulate of special
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theory of relativity.
21:04.030 --> 21:11.030
So, this is what I written now, let us find
now let us find the way the observer in S
21:11.530 --> 21:18.530
would find these events. Now, let look and
how the observer S would find these events
21:19.040 --> 21:26.040
now, according to the observer S the light
has picked up x component of velocity, it
21:26.900 --> 21:31.540
means the light is not going vertically upwards
but, it is going at an angle. And the angle
21:31.540 --> 21:38.270
will depend on the value of gamma of course,
because, u y depends on the value of gamma.
21:38.270 --> 21:44.930
So, if the observer in the S frame wants to
look up the motion of the light.
21:44.930 --> 21:51.930
This is what the he would probably visualize,
he would see that the light has an x component
21:52.080 --> 21:58.220
and also has a y component and is not emitted
vertically. What I have written according
21:58.220 --> 22:03.610
to S the light was not emitted vertically.
Hence would not fall back at the same co ordinate
22:03.610 --> 22:08.410
where it was emitted because, the light does
not go upward goes like this, then it falls
22:08.410 --> 22:13.340
back it have to fall in different point in
x is according to him the x co ordinate must
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have changed for the second event.
22:16.690 --> 22:23.690
So, the interpretation of this particular
observer in S would be that the second event
22:25.750 --> 22:32.750
has occurred not at the same value of x but,
has shifted and shifted by how much, it depends
22:34.660 --> 22:41.660
on how much time light took to go up and come
back, as seen by observer in S frame. Then
22:44.090 --> 22:50.870
whatever was the x component velocity, in
that frame multiplied by that particular time,
22:50.870 --> 22:54.590
let me just explain the point little more.
22:54.590 --> 23:01.590
So, according to the S observer the light
is moving at an angle like this, it comes
23:03.640 --> 23:10.640
back and does not hit at the same point, same
co ordinate as it has started. So, a light
23:13.020 --> 23:20.020
will go like this and come back like here,
if I to find out what is the displacement,
23:20.380 --> 23:27.110
this displacement would depend on the x component
of the velocity light which is v. And the
23:27.110 --> 23:34.050
time taken by the light go up here and come
back here, as seen by the S observer if the
23:34.050 --> 23:39.570
time interval is delta t then this distance
must be equal to v delta t.
23:39.570 --> 23:46.020
Because, v is the x component of the velocity
and delta t is actual time taken for the light
23:46.020 --> 23:47.720
to go up and come down.
23:47.720 --> 23:53.600
So, it is not written the final displacement
of the light in x direction, would be given
23:53.600 --> 24:00.600
by v delta t, where delta t is the time taken
by the light to go back in s frame. And this
24:01.330 --> 24:06.220
displacement happens to be same as the displacement
of that particular point of the train.
24:06.220 --> 24:12.780
Because, the noise observer S at this point,
this particular point also is moving with
24:12.780 --> 24:19.780
the same speed, which is relative speed of
the train. So, this point also moves exactly
24:19.810 --> 24:26.810
by the same distance v delta t and the second
event occurs also at the same shifted by the
24:29.060 --> 24:33.730
exactly the same x co ordinate. So, according
to the observer S prime this point also moves
24:33.730 --> 24:40.020
by the distance v delta t and the shift in
the x co ordinate of the two event is also
24:40.020 --> 24:47.020
equal is to v delta t, that is was the observer
in s frame would see. So, I have put this
24:48.530 --> 24:50.760
particular thing in the picture.
24:50.760 --> 24:56.780
Remember, this height H is going to be same
in the two frames as in S frame because, as
24:56.780 --> 25:01.310
we discussed in one of the earlier example,
there is no change in the y component. The
25:01.310 --> 25:05.720
length contraction occurs on the x component,
as per the y component is concerned because,
25:05.720 --> 25:12.620
y prime is always equal to y is not a transformation.
So, as for as going up and going down concern
25:12.620 --> 25:19.620
the heights are seen to be same. So, though
I am writing H to be to make it the size to
25:19.990 --> 25:26.090
be seen by an observer in S frame.
But, we know that H has to same value as H
25:26.090 --> 25:31.730
prime because, this distance is along the
y direction. So, an observer in an S frame
25:31.730 --> 25:38.730
sees, that light goes this way comes back
like this and the x difference in the x co
25:40.430 --> 25:45.830
ordinate of the two events, will be v delta
t it does not occur at the same point or same
25:45.830 --> 25:51.970
co ordinate from which it was turn. The second
event occurs at the different co ordinate,
25:51.970 --> 25:56.830
this is what will be the perception of observer
in S frame.
25:56.830 --> 26:01.820
Of course, this particular point from which
the light was thrown that also moves by the
26:01.820 --> 26:06.820
second one same amount. So, as far as the
observer S is concerned he will also feel
26:06.820 --> 26:11.860
that, the same point from the same point on
the train from which the light was emitted,
26:11.860 --> 26:18.510
it is received back also by the same, at the
same point, but this point as well as the
26:18.510 --> 26:25.510
light both are moved along the x direction.
So, that was I said move the H equal to H
26:25.510 --> 26:30.620
prime there is no contraction in the y direction.
Let us calculate the delta t to find how much
26:30.620 --> 26:35.670
have been the displacement again I can use
a inverse transformation.
26:35.670 --> 26:42.670
Delta t is equal to gamma delta t prime, this
is pulse sign because, we talking of inverse
26:43.170 --> 26:50.170
transformation v delta x prime divided by
c square. This is the inverse velocity transformation
26:52.470 --> 26:57.960
delta x prime was 0 because, as for as the
x prime is concern the co ordinate of the
26:57.960 --> 27:03.750
both the events are same therefore, x prime
was 0. So, this will becomes 0 and delta t
27:03.750 --> 27:10.270
prime was equal to 2 H prime by c and H prime
equal to H. So, I have written as 2 H by c
27:10.270 --> 27:17.270
this becomes gamma 2 H by c.
So, according to observer in S frame, the
27:17.310 --> 27:23.930
time interval between these two events is
not 2 H by c but, gamma times 2 H by c you
27:23.930 --> 27:28.810
will realize that this is the same as time
dilation formula, which has to happen because,
27:28.810 --> 27:35.810
S prime frame of reference both events occur
at the same point. So, the time interval in
27:35.860 --> 27:40.950
S prime frame of reference was a proper time
interval between these two events therefore,
27:40.950 --> 27:46.140
in S frame this time interval must turn out
to be dilated and that is what we saying,
27:46.140 --> 27:51.360
that this time interval turns out to gamma
multiplied by 2 H by c.
27:51.360 --> 27:58.360
So, this is was I written, note the delta
t prime was proper time interval. Hence, one
27:58.970 --> 28:04.200
could have used time dilation formula directly
for obtaining this result but, we need not
28:04.200 --> 28:09.820
really written, the lots of transformation
if your fast, we can immediately realize that
28:09.820 --> 28:15.050
this is the proper time interval as in S prime
frame of reference. So, in S frame you just
28:15.050 --> 28:22.050
multiply by gamma I will get back the time
interval and delta t of course, it can also
28:23.130 --> 28:27.700
be observed in different fashion, we can also
find out delta t prime slightly different
28:27.700 --> 28:34.680
fashion. If you look from the perception of
the observer in S frame of reference.
28:34.680 --> 28:41.680
This particular light traveled in vertical
distance of H. So, vertical component of the
28:43.690 --> 28:50.690
velocity ver vertical distance travel by the
light was H and the vertical component velocity
28:50.960 --> 28:57.960
was pulse minus c by gamma. So, find out the
distance if I know the speed, the vertical
29:01.580 --> 29:07.000
component of speed I can find out how much
time will be taken. So, the total distance
29:07.000 --> 29:12.760
traveled by the light total vertical distance
travel by the light, as seen in the frame
29:12.760 --> 29:19.760
of reference was H going up, H coming down
and this is achieved by a velocity y component
29:21.570 --> 29:28.330
of the velocity, which pulse minus c by gamma.
Therefore, I can always find out that total
29:28.330 --> 29:34.550
time will be 2 H divided by the y component
of the velocity which is pulse minus c by
29:34.550 --> 29:41.550
gamma. So, this gamma goes upwards and this
becomes 2 H by c to gamma because, I am only
29:43.980 --> 29:49.040
looking only at the time interval pulse minus
does not make any difference. So, I can also
29:49.040 --> 29:56.040
find out this particular time interval delta
t by using this particular way of doing it,
29:56.980 --> 30:03.980
I could have also done a direct inverse transformation
on delta x and obtained, exactly the same
30:05.650 --> 30:09.840
result which I have shown in this particular
transparency.
30:09.840 --> 30:15.260
Delta x can also be found directly by using
inverse Lorentz transformation, delta x is
30:15.260 --> 30:22.260
turned out equal to be gamma delta x prime
pulse v delta t prime delta x prime equal
30:23.190 --> 30:30.190
to be 0 delta t prime equal to 2 H by c. Again
I remain that is x prime is equal to H therefore,
30:31.309 --> 30:38.190
this is equal to gamma times v multiplied
by 2 H by c we get the same result. So, if
30:38.190 --> 30:45.190
you know the delta x value, we can always
find out the v delta t is the same result.
30:45.410 --> 30:52.250
So, we thus see that in S light travels with
the same vertical distance as seen in S prime
30:52.250 --> 30:55.770
but, with a reduced component of velocity.
30:55.770 --> 31:02.470
In addition the light has horizontal component
of velocity same as that of frame. This causes
31:02.470 --> 31:08.660
the x co ordinate of the event 2 to be displaced
from that of event 1, from that event 1 as
31:08.660 --> 31:12.740
in as frame. This is similar to classical
mechanics because, the classical mechanics
31:12.740 --> 31:17.330
also this event will appear to be a displaced
value of x, the difference here of course,
31:17.330 --> 31:23.740
is that in relativity they component of the
velocity has changed as seen in S frame of
31:23.740 --> 31:28.150
reference to make the speed of light same.
31:28.150 --> 31:33.650
In all the speed of light remained c though
it is direction of propagation was different
31:33.650 --> 31:40.650
in S. Now, let us look into the slightly different
issue, generally it is effective almost everyone
31:43.610 --> 31:50.610
sort of knows that one of the outcome are
special theory of relativity, was that speed
31:51.160 --> 31:57.400
is greater than the speed of light were prohibited.
We always said that information does not or
31:57.400 --> 32:02.140
should not travel or could not travel, with
the speed greater than speed of the light
32:02.140 --> 32:07.870
this was consider as the one of the outcome
the special theory of relativity. We are not
32:07.870 --> 32:13.610
discussed this point, so far. So, this one
particular point which was mathematical which
32:13.610 --> 32:19.220
has we see it very easily that almost in every
expression that we are using transformation
32:19.220 --> 32:26.220
the gamma appears which is equal to 1 upon
the root 1 minus v square by c square.
32:26.679 --> 32:33.100
If v really happens to be larger than c then
this particular factor, would turned out to
32:33.100 --> 32:39.050
be larger than 1 therefore, this when I take
under root the gamma would turn to be imaginary.
32:39.050 --> 32:44.210
When gamma turn to be imaginary almost all
the transformation that I am talking about
32:44.210 --> 32:50.710
x transformation, t transformation everything
there may be an imaginary number and do not
32:50.710 --> 32:53.780
really know, how to handle all these imaginary
number or how to physically interpret. So,
32:53.780 --> 32:57.950
there is an mathematical reason that if we
had really v is less than c, probably will
32:57.950 --> 33:03.230
comfortable because, we would not bother about
any thing one of the simplest way of looking
33:03.230 --> 33:07.510
into this particular thing.
But, there is an another important physical
33:07.510 --> 33:13.610
reason of why we think that is speed is greater
than speed of light should not be possible.
33:13.610 --> 33:19.710
And let me some spend time now, to explain
this particular aspect and this I will explain
33:19.710 --> 33:26.480
by using what I call as the order of events.
So, let me first explain by basic things before
33:26.480 --> 33:31.600
I come to this particular concept of speed
greater than or speed being greater than or
33:31.600 --> 33:33.820
speed smaller than speed of light.
33:33.820 --> 33:40.820
So, let us consider a two events, any two
events and these two event let us suppose
33:40.820 --> 33:46.980
now I am writing the full expression let us
call event number 1, E 1 which appears at
33:46.980 --> 33:53.980
a co ordinates x given by the x 1, y co ordinate
is 0, z co ordinate is 0 and it occurs at
33:54.929 --> 34:01.929
a time t 1 any arbitrary event. Let us look
at the second event, second event occurred
34:03.860 --> 34:10.860
at co ordinate x 2 y and z I am taking 0 to
make things simple and occurs at a time t
34:13.940 --> 34:18.960
2.
Now, for simplicity let us assume that x 2
34:18.960 --> 34:25.829
minus x 1 is positive which I am calling as
delta x. So, delta x I am defining x 2 minus
34:25.829 --> 34:32.829
x 1 similarly, let me assume that t 2 minus
t 1 is positive t 2 minus t 1 I am deciding,
34:33.759 --> 34:40.759
defining as delta t. So, I am assuming that
the delta t and delta x both are positive
34:40.960 --> 34:47.960
remember, both the events are seen by the
particular observer in a frame S and two events
34:48.499 --> 34:54.139
have occurred such a fashion that event delta
x is positive, delta t is positive, delta
34:54.139 --> 34:58.970
y delta z is equal to 0 that I never bother.
Because, y co-ordinates and z co ordinates
34:58.970 --> 35:05.970
are assumed to be 0 or to be same or to presides
I have written to 0, so x takes 0. Now, my
35:06.279 --> 35:13.279
question is that if delta x is positive and
delta t is positive, is there a frame in which
35:14.359 --> 35:18.450
delta x can be communicative if of course,
if I go to different frame call it as delta
35:18.450 --> 35:23.980
x prime. Delta x prime has been communicative
and delta t prime has been communicative one
35:23.980 --> 35:30.029
or both of them communicate, is it possible.
So, let us discuss this particular aspect
35:30.029 --> 35:35.809
little bit more in detail, that if delta x
is positive and delta t is positive is it
35:35.809 --> 35:42.809
possible to have a frame, in which delta x
can change sign delta x prime, becomes negative
35:43.049 --> 35:48.180
or delta t can change the sign becomes t prime
becomes negative or both can be communicative,
35:48.180 --> 35:55.180
is it possible to have any such frame. So,
this the question that I am trying to understand
35:55.900 --> 35:59.079
and trying to concentrate on this particular
thing.
35:59.079 --> 36:04.999
So, let us look at the Lorenz transformation.
The Lorenz transformation have applied in
36:04.999 --> 36:11.539
deferential form, this we have discussed even
earlier at rather I writing x I can use x
36:11.539 --> 36:16.049
2 prime, x 1 prime taking the difference but,
I have taking and writing in a differential
36:16.049 --> 36:21.569
form. So, I am writing delta x prime, delta
t prime directly I am transforming and this
36:21.569 --> 36:27.690
is given by Lorenz transformation, which is
delta x prime is equal to gamma delta x minus
36:27.690 --> 36:31.940
v delta t.
Similarly, delta t prime is equal to gamma
36:31.940 --> 36:38.940
times delta t minus v delta x divided by c
square, exactly the same as the Lorentz transformation
36:39.029 --> 36:46.029
I repeat the question, can delta x prime and
or delta t prime be negative. Now, let me
36:49.049 --> 36:54.759
divide this particular problem into two different
cases, one I will call case 1, another I will
36:54.759 --> 37:00.319
call case 2 and I separately look into the
case 1 and case 2 and trying to answer this
37:00.319 --> 37:05.789
question is that, if delta x to be positive
delta t is positive can delta x prime delta
37:05.789 --> 37:09.099
x prime be negative.
37:09.099 --> 37:16.089
So, let us look at the case 1, which is delta
x is less than c delta t. So, we specifically
37:16.089 --> 37:23.089
at looking at one particular case where delta
x is less than c delta t that is my case 1.
37:23.249 --> 37:29.339
I have rewritten this equation delta t prime
equation have slightly rewritten, I have taken
37:29.339 --> 37:35.499
c out of that particular bracket. So, this
becomes gamma bar c because, I have taken
37:35.499 --> 37:40.839
c out of the bracket, so there is a c appears
here. So, this becomes c delta t remember
37:40.839 --> 37:46.140
if I expand then it becomes c delta t and
there was c square here, I have taken one
37:46.140 --> 37:49.579
c out here this becomes v delta x divided
by c.
37:49.579 --> 37:54.869
So, same equation I have written just to make
things little simple interprets I have taken
37:54.869 --> 38:00.789
the c out of the this particular bracket.
So, this is the same equation which I have
38:00.789 --> 38:07.710
written slightly different fashion, let us
take specific example let us assume that delta
38:07.710 --> 38:13.130
x is equal to 0.9 times c delta t. So, the
delta x is less than the c delta t because,
38:13.130 --> 38:20.130
delta x is only 0.9 times delta t.
Now, let us assume that the v is equal to
38:21.109 --> 38:28.109
0.91 c which is slightly larger than this
particular factor of 0.9. So, I have chosen
38:29.460 --> 38:36.400
them for particular value of v which is 0.91
c slightly larger than this factor 0.9.
38:36.400 --> 38:43.400
Let us, look and substitute at this particular
equation. Let me write this particular equation
38:43.829 --> 38:44.579
again.
38:44.579 --> 38:51.579
Delta x prime was is equal to gamma delta
x minus v delta t.
38:53.380 --> 39:00.380
Now, delta x have taken as 0.9 c times delta
t, we have taken as 0.91 times c, this c delta
39:03.470 --> 39:04.839
t.
39:04.839 --> 39:11.839
Let me substitute for delta t prime gamma
divided by c into c delta t minus v delta
39:20.619 --> 39:22.779
x by c.
39:22.779 --> 39:29.779
I put it here in the transparency gamma by
c, c delta t minus v which we have taken as
39:33.269 --> 39:40.269
0.91 c, that c cancels the down c. And this
particular delta x was 0.9 times c delta t
39:41.380 --> 39:48.380
are you can very easily see that is if I take
c delta t out here, out of the bracket and
39:48.450 --> 39:55.450
this becomes 0.9 minus 0.91. So, delta x prime
has became negative, look at delta t prime
39:56.470 --> 40:03.470
if I take c delta t common here, this becomes
1 minus 0.91 multiplied by 0.9 when I multiply
40:06.989 --> 40:10.789
0.91 by 0.9 I will get the factor which is
less than 1.
40:10.789 --> 40:17.789
Therefore, this delta t prime is still not
negative is positive. Because, here you have
40:18.980 --> 40:25.980
1 minus something which is less than 1, instead
of 0.91 I could have used even a larger value
40:26.460 --> 40:30.479
but, still I can always find one particular
factor here, which is slightly larger than
40:30.479 --> 40:35.489
this to make delta x prime negative. But,
because, of this factor which I have used
40:35.489 --> 40:42.489
here, is less than 1 whatever value of v I
used here, whatever this particular factor
40:43.130 --> 40:49.769
I use here I will always get this product
to be less than 1, until this factor what
40:49.769 --> 40:53.619
I am using here, the ratio between v and c
exceeds 1.
40:53.619 --> 40:58.289
So, remember here we have something which
has less than 1, in order that if I want to
40:58.289 --> 41:04.650
make this more than 1 this factor should definitely
exceeding one. Otherwise if I am multiply
41:04.650 --> 41:10.160
this by factor still less than 1 this factor
will always be turning out to be less than
41:10.160 --> 41:15.150
1. And remember, delta t prime can became
negative, only if this factor becomes greater
41:15.150 --> 41:23.410
than 1 and all to that this factor becomes
greater than 1 and this factor here, is less
41:23.410 --> 41:31.950
than 1, this factor here must be greater than
1. So, let us try to consolidate I have said,
41:32.210 --> 41:37.499
delta x prime is negative but, delta t prime
not.
41:37.499 --> 41:42.890
In general for case 1, it is always possible
to find a frame in which delta x prime is
41:42.890 --> 41:49.890
negative, even it was 0.9 or it was 0.92 then
I would have used v is equal to 0.93. But,
41:50.150 --> 41:55.670
I will never able to make delta t prime negative
unless v becomes larger than c because, only
41:55.670 --> 42:01.599
that particular case the ratio between v and
c will exceed one. And then is multiplied
42:01.599 --> 42:06.489
by a factor which is less than 1, can yield
a value which is greater than 1, two factors
42:06.489 --> 42:10.849
both are which are less than 1, when they
are multiplied they will never give a value
42:10.849 --> 42:15.979
greater than 1.
Therefore, in this case delta t prime will
42:15.979 --> 42:22.220
never be about to be negative, irrespective
of what we use, what we you use unless v is
42:22.220 --> 42:29.220
larger than speed of light. But, as far as
delta x prime is concerned you could always
42:29.729 --> 42:36.729
used in appropriate value v which is less
than c and can it still make delta x prime
42:37.930 --> 42:42.549
negative.
42:42.549 --> 42:49.390
Such type of events are called time like separated
events these events are separated as time
42:49.390 --> 42:56.390
like time is, so strong is a pure time interval
cannot be reversed, the position can be reversed
42:56.739 --> 43:00.739
between the two two events I can always find
a frame of reference. In which the position
43:00.739 --> 43:05.849
of the event can be re re reversed delta x
prime can become negative but, I never able
43:05.849 --> 43:12.809
to find out a frame in the delta t prime changes
the sign unless of course, the velocity of
43:12.809 --> 43:15.960
frame difference exceeds speed of light.
43:15.960 --> 43:22.960
Now, let us look out the case 2. Case 2 is
a case when suppose delta x is greater than
43:23.710 --> 43:30.710
c delta t, I write exactly the same equation
delta x prime is equal to gamma times delta
43:32.989 --> 43:39.989
x minus v delta t. I write delta t prime is
equal to gamma by c, c delta t minus v delta
43:41.559 --> 43:47.910
x by c, exactly the same equations which I
have used for case one. But, now because,
43:47.910 --> 43:54.910
delta x have assume to be larger than c delta
t. So, let us take one example, delta x assume
43:55.269 --> 44:01.460
to be 1.2 times c delta t.
So, delta x has becomes larger than c delta
44:01.460 --> 44:07.880
t, we still take v is equal to 0.91 c and
let us see what happens. We substitute the
44:07.880 --> 44:14.880
value delta x is equal to 1.2 c delta t here
and for v I substitute 0.91 c. Similarly,
44:16.089 --> 44:23.089
here for delta x I choose 1.2 c delta t and
for v I choose 0.91 c
44:23.140 --> 44:30.140
I put it in this equation. Here, if we take
c delta t common out of the bracket you will
44:33.690 --> 44:40.690
find that here 1.2 and here we have 0.91 this
of course, is positive if you look at the
44:44.099 --> 44:51.099
delta t prime here c delta t here 0.91 multiplied
by 1.2 multiplied by c delta t. If just multiplied
44:53.690 --> 45:00.690
by 1.2 by 1.91, you get 1.092 this factor
is larger than one if I take c delta t common
45:05.119 --> 45:12.079
outside the bracket, you will get 1 minus
1.092 which makes it negative. So, delta t
45:12.079 --> 45:17.089
prime will turn out to be negative.
So, my conclusion is that at in this particular
45:17.089 --> 45:23.869
case delta t prime is negative but, delta
x prime is not. Now, when I say delta x greater
45:23.869 --> 45:30.549
than delta t could be I could have be anything
but, remember because, this factor is larger
45:30.549 --> 45:37.549
than one, unless we itself becomes larger
than c this will always remain positive. Therefore,
45:40.499 --> 45:47.499
delta x prime will always be positive here,
instead of 1.2 if this particular factor was
45:49.499 --> 45:54.109
not 1.2 but, what something less I could have
still chosen some particular value because,
45:54.109 --> 45:58.369
some particular value still greater than 1.
I could have chosen appropriated value v,
45:58.369 --> 46:02.390
to make this particular factor greater than
1.
46:02.390 --> 46:08.529
So, even if it not 1.2 but, 1.1 I make this
factor slightly larger. So, I can still find
46:08.529 --> 46:12.140
one particular frame, some particular frame
of reference with a very large value of v
46:12.140 --> 46:18.499
in which delta t prime will not be negative.
So, my conclusion is that, in such type of
46:18.499 --> 46:25.499
cases you will never be find a frame of reference,
in which delta x prime turns out to be negative,
46:26.829 --> 46:33.829
unless that frame moves with the speed greater
than speed of light. But, delta t prime you
46:33.920 --> 46:38.769
can always find out some frame of reference
which still moves with speed less than speed
46:38.769 --> 46:45.769
of light but, for which delta t prime will
turn out to be negative.
46:46.049 --> 46:51.079
So, this is what I have written here. In general
for case 2 it is always possible to find a
46:51.079 --> 46:57.859
frame in which delta t prime is negative;
however, it is not possible to find one, when
46:57.859 --> 47:04.859
delta x prime is negative unless v is greater
than c.
47:06.089 --> 47:10.670
Such type of events are called space like
separated events because, this interval is
47:10.670 --> 47:15.259
pure space the, space cannot be reversed,
the space is strong this cannot be reversed,
47:15.259 --> 47:19.650
time interval can be reversed, their sign
can be reversed but, the space sign cannot
47:19.650 --> 47:24.049
be reversed. So, these are called space like
separated events.
47:24.049 --> 47:28.559
Now, we have talked about sign changing this
and that you know what is that mean, you know
47:28.559 --> 47:32.329
what is the physical significance of that.
So, let us come back to the particular significance
47:32.329 --> 47:37.950
of what mean by delta x prime changing the
sign, delta t prime changing the sign become
47:37.950 --> 47:39.190
negative.
47:39.190 --> 47:45.509
So, what is the significance of delta x and
delta t changing the sign, when we go to from
47:45.509 --> 47:47.609
one frame to another frame.
47:47.609 --> 47:54.609
Let us, first look at delta x, delta x prime
be becoming negative in another frame. Essentially
47:55.509 --> 48:02.180
means that the second event occurred at a
value of x which is smaller than the first
48:02.180 --> 48:02.900
one.
48:02.900 --> 48:09.900
See, remember delta x was equal to x 2 minus
x 1. So, if x 2 becomes smaller than x 1 then
48:13.630 --> 48:19.309
delta x turned out to be negative, so all
that it means, that another frame of reference
48:19.309 --> 48:25.259
x 2 turns out to be smaller than x 1 that
is not very surprising. Let us, just look
48:25.259 --> 48:29.160
out an example, which is simple classical
example let us forget relativity let us look
48:29.160 --> 48:34.650
classically, even look classically even classically
this is possible to see that delta x could
48:34.650 --> 48:36.609
change sign.
48:36.609 --> 48:42.019
This figure looks some what complicated but,
let us try to understand it, let us suppose
48:42.019 --> 48:49.019
this is an event which occur here. This is
the frame S in which is sees that events occurred
48:51.670 --> 48:57.849
at the distance of x 1 from him or her and
let us assuming that an lighting striking
48:57.849 --> 49:04.289
here, at a given time. And it occurs at the
distance of x 1 format it is a second lighting
49:04.289 --> 49:10.089
striking and further away from him, at a time
later than this.
49:10.089 --> 49:17.089
So, x 2 is larger than x 1, so delta x was
positive because, this event occurred later
49:17.859 --> 49:24.859
than this event. So, t 2 was larger than t
1 therefore, delta t is positive, so as for
49:25.630 --> 49:31.420
as S is concerned delta x is also positive,
delta t is also positive, Remember I am only
49:31.420 --> 49:36.900
talking classical thing, nothing of relativity
of here, there is an another person train
49:36.900 --> 49:43.749
which is moving with respect to S.
When this is moving then event number 1 occurred
49:43.749 --> 49:50.710
this particular person was close to this observer
S therefore, he finds that this event occurs
49:50.710 --> 49:57.710
at x 1 prime. Of course, if there in classical
mechanics if there exact a line then x 1 should
49:58.059 --> 50:01.779
be equal to x 1 prime is let us forget about
that particular issue, we see that this will
50:01.779 --> 50:08.779
occur at x 1 prime. Now, second event occurs
little later, during this particular time
50:08.890 --> 50:11.869
this compartment was moving towards the right
hand side.
50:11.869 --> 50:16.630
Should depending upon the delta t x is possible
that this event let us occur, let us 10 minutes
50:16.630 --> 50:22.950
afterwards and during 10 minutes this particular
train has moves from here to here, will has
50:22.950 --> 50:27.950
from move here to here. The second event occur
only at this one distance because, he is also
50:27.950 --> 50:34.140
moves respect to us. So, I call it as S prime
observer, the coordinate of the second event
50:34.140 --> 50:39.190
is x 2 prime, while the coordinate of the
first prime was x 1 prime.
50:39.190 --> 50:44.009
So; obviously, x 2 prime smaller than x 1
prime, could depends on what is the speed
50:44.009 --> 50:50.329
and what is the time difference between this
two events, which always possible. That he
50:50.329 --> 50:56.599
finds that this particular event occurs, closer
to him then when the first event occurred,
50:56.599 --> 51:02.619
so according to the observer S prime delta
x has been assigned. So, even classical mechanics
51:02.619 --> 51:09.160
possible delta x was positive in S frame delta
x prime will be communicative in S prime frame
51:09.160 --> 51:14.289
of reference. So, there is nothing surprising
in delta x change in sign as I change my frame
51:14.289 --> 51:15.049
of reference.
51:15.049 --> 51:22.049
But, what does changing sign of time interval
means, it mean the second event occurred earlier
51:23.680 --> 51:29.829
in a different frame of reference then the
first event. So, as far as two lighting comes,
51:29.829 --> 51:34.910
if the second observer observes that the second
lighting of the occurred first and the second
51:34.910 --> 51:41.619
lighting occurred second, that is we mean
delta t changing sign, does it make sense.
51:41.619 --> 51:48.619
Now, if we are talking of two events, which
are totally uncorrelated, which has no bearing
51:50.839 --> 51:56.219
with each other.
Then if it changes I do not bother for example,
51:56.219 --> 52:03.219
I say event number 1, in particular train
departing from New Delhi railway station.
52:03.410 --> 52:10.410
And event number 2 a plane flying off from
New York airport after some moment of time,
52:11.119 --> 52:17.640
this event 1 and event 2 are, so totally uncorrelated.
That particular plane and this particular
52:17.640 --> 52:23.569
train departing has no relation with respect
to each other, the departure of that particular
52:23.569 --> 52:27.160
plane does not depend on the departure of
the train.
52:27.160 --> 52:33.299
But, let us consider some correlated events,
in which there is a cause and effect event
52:33.299 --> 52:40.299
number 1, train is starting from New Delhi
railway station, event number 2 train reaching
52:40.529 --> 52:47.529
at Mumbai railway station this event number
2. Now, unless the train would have started
52:48.190 --> 52:53.059
from New Delhi railway station, it would have
never reached the Mumbai station. So, these
52:53.059 --> 53:00.059
are related events, occur as the second event
depends on event number 1, take another example.
53:01.059 --> 53:08.059
I shoot somebody, with the gun and that person
dies. Now, unless I would have shot that person
53:09.650 --> 53:16.019
would have not died. So, his death which I
call as event number 2 and if I call my shooting
53:16.019 --> 53:19.999
the gun as event number 1, event number 2
has to occur because, of event number 1 had
53:19.999 --> 53:25.509
occurred, if event number 1 had not occurred
then event number 2 will not occur. Now, if
53:25.509 --> 53:31.329
delta t prime changes sign, it means these
orders can also change sign.
53:31.329 --> 53:38.329
Let us just see I am calling this event as
caused events. Reversal of time interval would
53:38.719 --> 53:43.859
mean that it is possible to find a frame in
which causality can be violated, let me explain
53:43.859 --> 53:47.140
this particular point little bit.
53:47.140 --> 53:54.140
If delta t prime changes sign it means, it
may be possible that is some other frame of
53:55.680 --> 54:02.680
reference, one could see a train arriving
before it starts or in some other frame one
54:05.469 --> 54:12.469
could see that a person dies before somebody
has shooting. That is what changing the sign
54:13.420 --> 54:20.420
delta t prime, this what call as causality
because, event number 2 is the effect the
54:21.529 --> 54:26.400
cause is the event number 1 because, of the
event number 2 occurred.
54:26.400 --> 54:32.420
On normally we never see, in our daily life
that causality is violated I should not be
54:32.420 --> 54:36.950
able to physics physical idea says that, should
not be possible for me to find out the frame
54:36.950 --> 54:43.219
of reference. In which these two events which
are one which the cause of another, in which
54:43.219 --> 54:48.289
the time interval could change because, would
mean that in some other frame of reference
54:48.289 --> 54:55.289
a person found to be dies before, he was shot
which we will not be accept.
54:58.619 --> 55:05.619
Now, if the events, the two events I am talking
related events as space like events only that
55:07.440 --> 55:12.910
particular case is possible to involve the
time, delta t prime make delta t prime to
55:12.910 --> 55:18.390
be negative. But, order the events are space
like delta x must be greater than c delta
55:18.390 --> 55:25.390
t, it means the distance between the two these
events must be larger than the c delta t.
55:26.579 --> 55:31.549
It means, if I am shooting the gun here and
the person is dying at the little distance
55:31.549 --> 55:38.549
let us a 2 kilometer away then this distance
should be larger than the c delta t. It means
55:38.630 --> 55:43.390
the bullet must have travelled with speed
greater than c delta t, then only it is possible
55:43.390 --> 55:48.700
that the events the correlate events become
space like.
55:48.700 --> 55:55.700
And if the events are time like where delta
x is less than c delta t any way it is possible
55:59.229 --> 56:04.869
to invert time interval unless than c delta
t anyway it is not possible to involve the
56:04.869 --> 56:11.869
time interval unless v is larger than c. So,
if we say that if v larger than c are not
56:13.749 --> 56:20.359
correlated, it will sustain, it will make
causality in that it will not possible for
56:20.359 --> 56:27.359
us to see that the cause comes later and effect
comes earlier in any other frame of reference.
56:29.529 --> 56:36.499
Therefore, if I say that if I restrict my
speed to speed less than speed of light, that
56:36.499 --> 56:43.499
the most equal to the speed of light I maintain
the causality. That is what ley let to the
56:45.499 --> 56:50.759
conclusion that speed greater than speed of
light should not be acceptable in special
56:50.759 --> 56:53.880
theory of relativity.
56:53.880 --> 57:00.099
So, this what I had say as conclusion, if
speed greater than c are prohibited causality
57:00.099 --> 57:03.650
cannot be violated.
57:03.650 --> 57:08.359
And then I will give the summary of whatever
we discussed today, we of course, discuses
57:08.420 --> 57:16.039
one example. In which we had included the
speed of light and taken the speed of light
57:16.039 --> 57:25.499
and in two different frame of reference. And
then we have said in order to maintain causality,
57:25.599 --> 57:28.229
the speed is greater than c should not be
allowed.
57:28.229 --> 57:29.689
Thank you.