WEBVTT
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hello everybody so today we come over to the
third part of lecture on on special relativity
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and having done the postulates of relativity
and seen consequences like length contraction
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let us move over to another interesting consequence
that is called a time dilation okay so just
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to set up the thing a little bit more let
us let us just have the results which we had
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in our last lecture and it is mainly on line
contraction
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so what did you see there so what we saw there
was if you have borrowed you know if you have
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some some object at rest in in a certain frame
you would not measure its length okay find
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that the length of this rod let us say it
is it is l0 okay now if you go and measure
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this in a in a different frame from which
this is this this rod appears to be moving
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okay with a uniform velocity the length of
the moving rod appears to be shortened in
01:51.800 --> 01:56.060
the direction of its motion okay
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of course so so that the result that we got
so if you see the length of this object is
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rod from s frame okay remember that the rod
is at rest in the s prime frame which is moving
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with a certain uniform velocity v then what
is the length that we are going to see in
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the s frame its l0 times the root over of
1 - v square by c square so obviously l is
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less than l0 okay so the length of a moving
rod or a moving object appears to be shortened
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as compared to the to the measured length
in its rest time okay
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so well so having seen something to do with
the space coordinates i mean that is the length
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contraction what we do i mean what what is
it that we have with the time component in
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special relativity does something happen there
okay remember in galilean transformation of
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course time in both these frames in s and
s prime frames are the same and they are they
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are moving with a certain uniform velocity
with each other
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but is it so in in in special relativity if
you if you remember your lorentz transformations
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so there is space component even in the time
vision okay and that in a sense leads to a
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another interesting concept in a consequence
in in special relativity it is called time
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dilation okay
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so what is it way in what is the in long and
short of it it is just moving clocks appear
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to run slow okay so it is nothing to do with
how good the machine inside the clock was
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it is it is it is it is the concept of physics
here okay or in other words the time expands
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for a moving body okay right so let us take
an example by short one what how do you how
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do you how do you measure such a thing how
do you get the feel of such a thing of things
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like time dilation okay
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so if you consider a person who is inside
a train and the train is moving with a certain
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velocity v okay with respect to the station
of course station or stations in this case
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okay then for the person who is sitting in
this train he matches his own watch i mean
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this he has synchronized his watch with one
of the station clocks earlier okay let us
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say then in the next station when the train
passes he looks at the station clock and then
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he will see that the times in his watch and
the station clock will not match okay
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so that is again what is meant by you know
an example of a time dilation but let us have
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a more more pictorial example of this of this
concept okay so which will clarify it a little
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bit more
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see that you have two stations and then you
have a synchronized clocks in these two stations
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okay you synchronize in a sense that so they
give the same they same that the the time
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so synchronized okay so so well i should say
that if you have two stations you need two
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clocks there okay so so we have our two clocks
one on the on the left of your screen and
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one of the right of your screen as written
as station a and station b
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and then their clock stay there keeping fine
then what happens suppose let us say a person
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moves with with a certain velocity you know
the uniform velocity v okay you know in a
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train coach and then he has a watch on his
wrist let us say so he is talking of a wristwatch
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and then he checks his time in distress what
only adjusted according to the to the time
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in in station a so that he synchronizes his
clock with station a okay
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now when he passes station b okay what he
will observe is that the the the time or the
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whatever time it is in his wristwatch and
whatever time it is in station b and in station
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clock they will not match okay so this is
a more toriel way of saying the thing okay
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well why is it so now for that we have to
be a little bit more quantitative
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and so let us consider two events explain
what these events are let us let us just define
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what is the space time coordinates of these
events first okay so so we consider two events
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we call it event 1 and event 2 seen from two
different frames of reference s and s prime
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okay and as always we take the s prime frame
that is moving with a certain uniform velocity
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with respect to s okay
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if the the space time coordinates of these
two events in in either of these two events
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in s frame okay that is x1 y1 z1 and t1 so
that is measured at t1 and then the second
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event occurs at position x2 y2 z2 and at time
t2 ok and correspondingly these two events
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occurs in the in the primed frame with x1
prime y1 prime and z1 prime at this position
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and time t1 prime and then at the second event
occurs at at the positions x2 prime y2 prime
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z2 prime at time t2 prime
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then what we have or if you want to be a little
bit more specific now if you wish to define
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these events in terms of our train example
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we take the frame in which the so first we
define what our s frame and this s prime frames
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are okay so the frame in which the person
with his clock is at rest is the s prime frame
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okay now this frame so it is moving with a
certain uniform velocity relative to stations
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a and b okay and the these are fixed in frame
s okay so now you define what our frames are
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so the stations are so you somebodys watching
from the station so that is s frame okay
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and then you and then the person in the train
sitting sitting in the train looking at his
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wristwatch okay so that is the s prime frame
so so so when the person passes station a
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let us call that as event one okay so it is
got a certain position at certain time okay
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and similarly when the person passes station
b let us call that as event two seen from
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two different frames of course the positions
are different and the times are also different
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okay
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so so what we have remember in the s prime
frame in which the person with his clock is
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at rest the events e1 and e2 has occurred
at the same space coordinate when the different
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times why is that you see he is at rest within
the within the within the train carriage let
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us say and then he just looks at his watch
at the same position so his position coordinates
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are the same okay but then he from his point
of view he has looked at it at different times
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and that is why he gets different times okay
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and just to keep the mathematics a little
bit simpler we we also take you know so moving
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in the common x and x prime direction so that
the y primes are also y’s and zs are also
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the same here but in terms of coordinates
in the s prime frame we have x1 prime = x2
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prime and then y2 prime = y1 prime and z2
prime = z2 prime z prime = z2 prime that is
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it but the times are different okay so so
what are these times quote unquote the times
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in in s frame that is seen from from from
the station okay
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so from the unprimed frame if we call this
time t1 ok so we just have the lorentz contraction
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equations or lorentz lorentz transformation
equations to be more precise here
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so so t1 that is t1 prime + vx prime by c
square divided by root over of 1 - v square
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by c square and t2 is t2 prime + vx2 by c
square into root over of 1 - v square by c
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square ok so the time interval between these
two events according to the observer in s
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ok so according to the measurement i made
in s frame so that is delta t let us say so
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what is delta t now
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so delta t is just t2 - t1 okay and that is
t2 prime -t1 prime + v by v square into x2
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- x1 prime divided by the usual the denominator
root over 1 - v square by c square ok now
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what is the time interval according so that
is the time interval of these two events according
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to the observer in s frame ok but in the s
prime frame remember the positions of these
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two events were the same ok so we have
x2 prime and x prime prime are the same so
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x2 prime - x1 prime see you okay so that will
give us what
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that will give us that this delta t is delta
t prime so we have defined delta t prime as
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the time difference as measured by the observer
in the s prime frame remember the s prime
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frame is the one in which the person sitting
in this train carried is looking at his watch
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okay so what we get so what we get is delta
t prime is delta t into root over of 1 - v
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square by c square okay so just if you get
a feeling that you use immediately see that
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delta t prime is actually less than this delta
t here okay
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now what does it mean physically it means
that the time interval measured by the moving
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clock relative to s will be smaller than the
time interval measured by the clock stationary
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s okay so in other words what we have is it
is confirmation of the statement with which
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we started with that moving clocks run slower
however i should remind you that if the velocity
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is much much less than the speed of light
then you will immediately see that these two
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time intervals are the same okay
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so they are approximately the same so next
time you come from let us say delhi to roorkee
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and then you look at the station clocks supposedly
i mean you are moving with a uniform velocity
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and all these things with respect to the station
in all cases you should not you you well you
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should not and you would not see this distinct
this time dilation okay okay so so having
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seen this this thing let us have a look at
what proper time is okay
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so what is the proper frame was a proper frame
of reference so a reference reference frame
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in which body is at rest it is called its
proper frame we know that and we have also
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measured we know what a proper length is so
the length of the body measured in this proper
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frame is called its proper length so the time
measured by a clock at rest in the proper
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frame is the proper time okay so just like
the proper length is the proper time invariant
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well i think it is but let us just have a
look at it once again okay so so what is so
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the time interval so what is the proper time
interval to be more precise
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so it is the time interval measured between
two events via single oclock at rest at the
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same place is the proper time interval between
two events we just had a example of a person
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sitting in a train looking at his watch at
two different stations so that is an example
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of you know position is not changing according
to him in he is sitting in the same place
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in a carriage so his position is not changing
and then he looks at his own watch
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so that is a single clock and he is sitting
at rest in the train so according to him he
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is measuring the proper time okay by a single
clock however i mean at the same time interval
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you know am i going to measure it from another
frame having a relative velocity with respect
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to the proper frame so you need two clocks
at different places so that is exactly what
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we had we had two stations and two clocks
and then they had to be synchronized a beginning
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well in the sense this time sometimes is also
this kind of time is also time interval rather
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is also called non proper okay
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so so if if the velocity of the second frame
we another frame be v relative to the proper
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frame with proper time interval delta tau
between these events then the non proper time
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interval delta t measured from the second
frame of course by subtracting the time measured
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from two different clocks will be delta tau
divided by root over of 1 -v square by c square
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okay
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and so what is this delta tau that is the
proper time so it is delta t into root over
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of 1 minus v square by c square okay
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now if in another frame in which it is moving
with the velocity v prime its respect to the
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proper frame if the time interval measured
between the same two events is delta t prime
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then obviously delta t prime is again delta
tau divided by root over now v prime squared
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by v squared so you immediately see that if
you just write out the delta tau here
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the delta tau is nothing but delta t into
root over of 1 - v square by c square that
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is the time corresponding to this in a measured
in the frame the time interval measured in
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the frame it is moving with a certain velocity
v then t prime this time interval measured
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in the frame which it is moving with a certain
speed v prime so that is equal to delta tau
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okay and so delta tau is invariant quantity
okay
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so the proper time just like the proper length
is an invariant quantity okay so let us look
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at an example okay let us look at some examples
here on on these on these consequences of
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time dilation proper time and see what else
we can do from here okay let us take a simple
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but interesting sample so you know a muon
is that is a subatomic particle okay
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and a muon is observed to move approximately
something like 800 meters on an average during
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its lifetime in a laboratory okay so if one
looks up the mean lifetime of a muon from
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one of these particle physics books are these
data tables you see that it is something like
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2 into 10 power - 6 seconds okay so if we
just go on write the velocity as the distance
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covered and then its mean lifetime it is okay
800 meters by 2 into10 power -6 seconds we
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end up with 4 into 10 to the power 8 meters
per second okay
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this is much bigger than the speed of light
okay now this is this is harrison in in special
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relativity and what do we do i mean obviously
we have not considered something so what is
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that
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well one needs to consider and that what is
what given as mean lifetime of this muon is
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actually its proper lifetime okay so when
you are measuring the the length in the laboratory
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okay so you need to divide this by the this
quote unquote improper lifetime so the measurement
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of the time and the measurement of the the
distance this muon has travelled should be
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done in the same frame that is it okay
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so so this quote unquote improper lifetime
of this mueon has measured in a frame moving
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with this velocity v with respect to this
proper frame so what is that so if you remember
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this the few slides back we talked of the
proper time interval that is divided by root
23:17.010 --> 23:24.440
over 1 - v square by c square so the proper
time here is 2 into 10 to the power -6 seconds
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so we divide that by root over 1 - v square
by c square okay
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and that is what we get the the speed of the
muon is now 800 meters divided by delta t
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that is this improper lifetime and if you
put in this this expression which we obtained
23:48.230 --> 23:55.750
earlier you land up with the speed of this
muon to be a 4 5th of at the speed of light
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so this is still a less than the speed of
light okay and it does not you know not more
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than speed of light as we had done as we had
seen it would be if if you do it you know
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in a wrong sense of course okay
24:17.260 --> 24:29.260
so let me now move over to a another interesting
concept it is the relativity of simultaneity
24:29.260 --> 24:37.920
ok so and then i shall later on talk of things
called earlier you know the concepts of earlier
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a little later okay so the concepts like if
you if a certain event is you know following
24:45.320 --> 24:50.880
a sequence i mean some some event has occurred
earlier and then later and then does it appeared
24:50.880 --> 24:53.970
to be in the same sequence in some other frame
or not okay
24:53.970 --> 25:03.180
but before that we do a little you know more
basic thing it is called this relativity of
25:03.180 --> 25:09.160
simultaneity so so what is it that is what
that we are going to say here
25:09.160 --> 25:15.670
what we are seeing is that two events are
said to be simultaneous if they occur at the
25:15.670 --> 25:21.300
same instant of time of course i mean that
is the that is the basic definition of what
25:21.300 --> 25:27.980
you would see for a simultaneous event okay
but the question is if these events are to
25:27.980 --> 25:32.820
be simultaneous in one inertial frame so is
it going to be simultaneous in another frame
25:32.820 --> 25:36.780
of reference which is you know i mean talk
of inertial frames so we the frames here im
25:36.780 --> 25:41.130
moving with uniform velocity with respect
to each other okay
25:41.130 --> 25:47.820
so if it is simultaneous in one frame is it
going to be simultaneous in another frame
25:47.820 --> 25:56.550
okay so for that i think it will be good if
you see there are more pictorial fashion yes
25:56.550 --> 26:01.280
we talk of event one okay
26:01.280 --> 26:10.840
so we say event one so now it is it is it
is occurred at a certain position so the x1
26:10.840 --> 26:23.060
y1 z1 say at certain time in in in in s frame
let us say okay and then the same event is
26:23.060 --> 26:32.140
seen to be at positions at position x1 prime
y1 prime z1 prime and then the time measured
26:32.140 --> 26:42.590
is t prime ok so that is the event seen from
the the frames s and s prime so remember s
26:42.590 --> 26:51.880
prime is also moving with a certain uniform
velocity with respect to the s frame ok
26:51.880 --> 27:03.030
now if you talk of another event okay seen
by the observer in s and s prime frame so
27:03.030 --> 27:12.480
it is occurred at a different position and
time bolting it again so ok so what is it
27:12.480 --> 27:17.200
what is the relation between this between
these time components so we know it we know
27:17.200 --> 27:19.710
it is t1 prime that is
27:19.710 --> 27:32.160
equal to t1 - vx1 by c squared divided by
root over 1 - v square by c square ok so and
27:32.160 --> 27:40.190
same thing for the for the second measured
time in in the frame s prime
27:40.190 --> 27:51.660
so that t2 prime is just t2 – so -v x2 by
c square divided by 1 - v square by c square
27:51.660 --> 27:58.360
remember this unprimed things are for the
s frame and then this prime things are for
27:58.360 --> 28:12.510
the s prime frame ok so what is the difference
between these two times ok as seen well the
28:12.510 --> 28:14.470
difference between these two primes is obvious
28:14.470 --> 28:23.810
so it is t2 prime - t1 prime so we find that
the same as t2 -t1 so the things on the right
28:23.810 --> 28:35.590
hand side are the the coordinates time that
we have in the s frame okay so that is t2
28:35.590 --> 28:42.480
– t1 - v of x2 -x1 by c square and divided
by root over 1 - v square by c square okay
28:42.480 --> 28:52.590
so if this events are simultaneous in s frame
ok then of course it is occurred you know
28:52.590 --> 29:05.280
at the same time so we take t1 = t2 okay so
that would mean that we have from this equation
29:05.280 --> 29:10.980
of difference of times in the prime frame
so we have on the right hand side we have
29:10.980 --> 29:13.880
taken out t1 - t2 okay
29:13.880 --> 29:21.240
however so the position so that is x1 - x2
so that well it is not 0 why because if it
29:21.240 --> 29:27.900
is 0 so you would not be able to distinguish
between these events okay so so that is not
29:27.900 --> 29:36.750
0 okay so that would mean that the difference
of measured times of these two events in the
29:36.750 --> 29:48.500
prime frame is not 0 because none of the quantities
here are 0 you can see here okay so what you
29:48.500 --> 29:56.350
get so what we get is the measured in in the
s prime frame need not be the same
29:56.350 --> 30:04.620
even if the times are the same in you know
the measured times are the same in the s frame
30:04.620 --> 30:12.780
okay so in other words to be more precise
what we have is that the events in e1 and
30:12.780 --> 30:13.850
e 2 okay
30:13.850 --> 30:21.040
although they may be simultaneous in one frame
of reference they need not be simultaneous
30:21.040 --> 30:35.240
in another inertial frame ok ok so we come
then to another interesting concept it is
30:35.240 --> 30:45.020
the this thing called the sequence of events
okay the concept of earlier and later as we
30:45.020 --> 30:51.320
said i mean what we saw is that if something
is happening you know earlier in a certain
30:51.320 --> 30:57.840
frame and then another event is happening
a liter in a certain frame
30:57.840 --> 31:04.190
is this order of events preserved okay so
why do we say that because we have just seen
31:04.190 --> 31:09.700
that you know this simultaneous itself is
a is a relative concept here so it depends
31:09.700 --> 31:16.760
on i mean it need not be a cement something
which is simultaneous in one frame of reference
31:16.760 --> 31:24.400
need not be simultaneous in a another frame
of reference so will this walker here too
31:24.400 --> 31:27.900
i mean in the sequence of events
31:27.900 --> 31:36.220
so we consider two events you e1 and e2 again
and we see these events from two different
31:36.220 --> 31:44.590
frames s and s prime okay so this s prime
frame it is moving with a certain uniform
31:44.590 --> 31:54.840
velocity with respect to s okay so we written
down what the space time coordinates of these
31:54.840 --> 32:05.090
two events are in s frame and the s prime
frame okay so the question that we now ask
32:05.090 --> 32:13.520
ourselves is that if these events occur in
a certain sequence in one frame
32:13.520 --> 32:22.070
so can it occur in the reverse sequence in
another frame so for that we again have to
32:22.070 --> 32:32.460
look at the the difference of times as measured
in the s and this the s prime frame okay
32:32.460 --> 32:38.810
so what is the difference of time measured
in the s prime frame so that is t2 – t1
32:38.810 --> 32:47.010
that is the time measured in the difference
of time measured in the s frame minus of the
32:47.010 --> 32:55.350
it is the velocity v into x2 - x1 that is
the those are the positions as measured in
32:55.350 --> 33:02.070
the s frame by c squared then the usual factor
to power of 1 - v square in the denominator
33:02.070 --> 33:12.370
okay now if let us say event 2 has the second
event has occurred later than event 1 in the
33:12.370 --> 33:19.761
frame s so obviously t2 - t1 is positive okay
33:19.761 --> 33:32.190
now that means that for the reverse to happen
in s frame okay what we have we have you know
33:32.190 --> 33:44.350
t2 - t1 was positive in s frame but t2 prime
- t1 prime so that is not positive okay so
33:44.350 --> 33:54.000
in that case what it means is that we have
nothing in the numerator so that is t2 - t1
33:54.000 --> 34:03.410
- t into x2 - x1 divided by c square that
has to be negative okay that is obvious and
34:03.410 --> 34:05.490
denominator here is positive ok
34:05.490 --> 34:21.710
so now if that is so we would require x2 - x1
divided by t2 - t1 to be greater than c square
34:21.710 --> 34:37.149
by v okay now now if these events are causally
connected that is in a sense that event e2
34:37.149 --> 34:47.510
is happening as a consequence of event 1 happening
okay it is more like you know you in a cricket
34:47.510 --> 34:53.659
field and then think of your favorite cricketer
and then what he does is he autographs a ball
34:53.659 --> 34:56.879
and it throws it at you okay and so you catch
the ball
34:56.879 --> 35:02.470
so he of the cricket of your favorite cricketer
autographing a ball and then that is called
35:02.470 --> 35:10.000
an event one and then you catching you know
the throw that he throws that event 2 so this
35:10.000 --> 35:17.080
has to be done at a certain you know so certain
that the message has to be passed at a certain
35:17.080 --> 35:27.830
speed here okay so you catching an autograph
ball is in a sense dependent on the your cricketer
35:27.830 --> 35:30.240
signing the ball okay
35:30.240 --> 35:36.710
so unless the cricketer signs the ball and
throws it to you you cannot catch the ball
35:36.710 --> 35:44.280
okay so in a sense some messages been passed
in this case the example of two events which
35:44.280 --> 35:47.070
are causally connected okay
35:47.070 --> 35:53.780
so in a sense what we have is that some message
which is originating from the first event
35:53.780 --> 36:01.870
reaches the second place the position of the
second event traveling with some speed let
36:01.870 --> 36:10.100
us say is some speed u okay then the minimum
speed that this you should have is actually
36:10.100 --> 36:15.760
x2 - x1 divided by t2 - t1 difference of these
positions and the difference of this measure
36:15.760 --> 36:22.550
time has a minimum thing that value that you
should have okay u that is the the alphabet
36:22.550 --> 36:24.700
u i mean
36:24.700 --> 36:34.190
now what does it mean now this u in this case
would have c squared would be more than c
36:34.190 --> 36:42.010
square by v okay why because x2 - x1 divided
by t2 - t1 that is more than c square by v
36:42.010 --> 36:51.120
so that is the condition one has to satisfy
remember if the sequence of event is is reversed
36:51.120 --> 36:57.640
in in another frame and the events are causally
connected okay then we will end up with a
36:57.640 --> 36:59.050
problem
36:59.050 --> 37:08.730
the problem is that v here is distance c okay
or at most it is c so at most its c but so
37:08.730 --> 37:14.160
you cannot you cannot go more than the speed
of light well then what about u the alphabet
37:14.160 --> 37:26.080
with u that is the speed u see that it is
more than c ok so this again is it is not
37:26.080 --> 37:33.550
allowed so because why i said again because
v is the speed s prime frame is moving uniformly
37:33.550 --> 37:42.770
with respect to a frame s ok so in a sense
it would mean that the message between the
37:42.770 --> 37:47.590
events e1 and e2 is travelling with a speed
greater than the speed of light
37:47.590 --> 37:55.470
so this is not allowed ok so this condition
is not fulfilled this condition is the wrong
37:55.470 --> 37:57.560
condition okay
37:57.560 --> 38:06.770
now since it is not possible to fulfill this
condition if the events e1 and e2 are causally
38:06.770 --> 38:18.580
connected then it means that the concept of
earlier and later between events will be preserved
38:18.580 --> 38:25.210
in all inertial frames okay so that is the
point here so if the events are causally connected
38:25.210 --> 38:34.950
then it is going to be a pisser but if the
events are not causally connected ok so event
38:34.950 --> 38:41.330
the second event let us see is happening independent
of whether event occurs not ok
38:41.330 --> 38:46.510
in that case the opposite can be true okay
so that is the case when they are not causally
38:46.510 --> 38:57.420
connected ok so what we saw is that all the
these things called the simultaneity of the
38:57.420 --> 39:05.580
relativity of simultaneity tells us that if
two events are simultaneous in certain inertial
39:05.580 --> 39:14.920
frame so it did not be simultaneous in the
another inertial frame ok but the concepts
39:14.920 --> 39:23.680
of earlier and later so that is preserved
if the events that we are talking of they
39:23.680 --> 39:29.430
are causally connected okay
39:29.430 --> 39:39.521
so as usual let us have a few examples to
clarify our concepts okay so let us take two
39:39.521 --> 39:49.960
events here e1 and e2 and in s frame and we
give the position coordinates of these these
39:49.960 --> 39:57.700
events here in the s frame and also write
down what is the times here
39:57.700 --> 40:05.240
so here let us say e1 is happening at position
x1 is x0 and then the you know the y and zs
40:05.240 --> 40:14.640
are 0 here so ok and the x2 is happening at
a position 2 x0 and then the the time t1 that
40:14.640 --> 40:27.210
is x0 by c so c being the speed of light and
t2 is x0 by 2c okay now if the this event
40:27.210 --> 40:35.130
is simultaneous so see that in in s frame
this this event is not simultaneous it is
40:35.130 --> 40:42.950
happening at different times okay but if this
event is simultaneous in another frame so
40:42.950 --> 40:50.200
which is moving with a certain velocity v
let us say along the common x x prime axis
40:50.200 --> 40:54.330
ok relative to the s frame
40:54.330 --> 41:04.990
what is this velocity and what is the time
that what is the simultaneous time where that
41:04.990 --> 41:13.720
is being measured in s prime frame okay so
how do you go about doing these things so
41:13.720 --> 41:22.770
let us look at the time intervals had seen
from the s frame and the s prime frame so
41:22.770 --> 41:27.320
since the events are simultaneous in the s
prime frame
41:27.320 --> 41:37.190
so t2 - t1 so that is 0 so if you look at
the second step of the second equation so
41:37.190 --> 41:44.460
on this numerator you put in the times that
we had for the event in for the events other
41:44.460 --> 41:53.119
that we had in the s frame and then also the
difference of their positions and divided
41:53.119 --> 41:58.750
by the root over of v square minus 1 - v square
by c square v being the velocity with which
41:58.750 --> 42:09.840
s prime frame is moving with respect to the
s frame so we immediately get what we get
42:09.840 --> 42:12.140
the left hand side to be 0 of course
42:12.140 --> 42:23.470
and we can immediately find what v is and
you see that v is actually -c by 2 okay so
42:23.470 --> 42:29.850
how do you find the time in the prime frame
remember in the prime frame the events are
42:29.850 --> 42:43.110
simultaneous so so t1 prime that is nothing
but t1 - vx by by c squared ok divided by
42:43.110 --> 42:48.170
root over 1 - v square by c square and then
what you are going to do is they are going
42:48.170 --> 42:55.520
to substitute t1 by the measured time you
know for a certain position
42:55.520 --> 43:03.110
so if x is x0 we look this problem that the
this was measured at time t1 = x0 by c ok
43:03.110 --> 43:10.180
and we know the velocity now with which this
prime frame is moving so that is -c by 2 so
43:10.180 --> 43:20.620
immediately get what this time is in the prime
frame of reference so t1 prime is root 3 x0
43:20.620 --> 43:30.300
by c ok so let us go to a another problem
ok well let us look at this problem first
43:30.300 --> 43:36.840
okay so in a certain inertial frame let us
say so we have two events which occur at the
43:36.840 --> 43:38.250
same place
43:38.250 --> 43:46.710
and they are separated by a time interval
of four seconds okay now in another inertial
43:46.710 --> 43:53.820
frame the question is what is the spatial
separation what is the spatial separation
43:53.820 --> 44:00.970
in these coordinates here between these two
coordinates if these are separated by a time
44:00.970 --> 44:14.510
interval of 6 seconds okay so that is what
we do we first try to find what is the velocity
44:14.510 --> 44:19.730
with which these frames are moving with respect
to each other okay
44:19.730 --> 44:30.369
so call the prime frame that is the you know
so the frame in which the time separation
44:30.369 --> 44:36.250
between these two events are 6 seconds so
that is the prime frame see it colleague like
44:36.250 --> 44:46.440
that so what is the corresponding you know
portion of this time difference in in the
44:46.440 --> 44:48.470
s frame how do you do it
44:48.470 --> 45:01.500
so it is t1 - t2 t2 - t1 rather -v x2 - x1
divided by c square root over of 1 - v square
45:01.500 --> 45:10.510
by c square okay so here we have so you put
instead of t2 - t1 you put that to be 6 and
45:10.510 --> 45:19.670
t2 - t1 you put that to be 4 and you find
what b is and this v in this particular case
45:19.670 --> 45:30.100
would be root over 5 by 3 into the speed of
light okay now having found that it is very
45:30.100 --> 45:38.310
easy to find what the spatial separation is
so we just find what the spatial separation
45:38.310 --> 45:39.310
is
45:39.310 --> 45:51.250
so that is x2 prime - x1 prime that is x2
- x1 -v into t2 - t1 to root over of 1 - v
45:51.250 --> 45:59.680
square by c square we know the you know the
spatial separations in the s frame that is
45:59.680 --> 46:09.690
0 it is happened at the same place so what
is the spatial separation in the prime frame
46:09.690 --> 46:16.900
so that actually is you put in the proper
times here so i mean the proper times not
46:16.900 --> 46:22.400
in the sense of proper time proper length
i mean you did put in the right times here
46:22.400 --> 46:36.010
so that is t2 - t1 and you get the difference
to be - 2 into root over of 5c okay so the
46:36.010 --> 46:40.510
proper length the proper separation is the
modulus of this of course so that is twice
46:40.510 --> 46:56.410
of 5c okay so we have seen so far it is another
consequence of time dilation we have talked
46:56.410 --> 47:08.550
of you know we have talked how or why moving
blocks of run slow you know things are its
47:08.550 --> 47:17.780
we derived it from we derived this consequence
from the lorentz transformation
47:17.780 --> 47:25.170
then we talked of the relativity of simultaneity
so in which we showed that two events which
47:25.170 --> 47:31.530
are simultaneous in an inertial frame need
not be a simultaneous in another frame okay
47:31.530 --> 47:41.230
and then in the third part we saw that the
sequence of events they are preserved in all
47:41.230 --> 47:49.450
inertial frames if the events are if this
events are causally connected okay
47:49.450 --> 47:59.880
so in the next lecture what you want to see
is talk of doppler effect in in light and
47:59.880 --> 48:05.800
see what next we can do from there okay doppler
effect i think we yeah bit familiar with doppler
48:05.800 --> 48:16.750
effect in sound where you know if the you
have a source moving you know use your standing
48:16.750 --> 48:23.040
some you standing at a distance and then from
a distance a train is coming at you okay and
48:23.040 --> 48:26.520
the train is flowing it is on or a giving
a whistle
48:26.520 --> 48:33.760
then you see that as the train moves towards
you the frequency that changes okay so that
48:33.760 --> 48:41.210
is the doppler effect so the source is moving
here so or if the source is at rest and the
48:41.210 --> 48:50.320
observer is moving the observed frequency
of sound is also different so we want to look
48:50.320 --> 48:56.840
at whether doppler effect is there in light
or not well that is a thing for the next lecture
48:56.840 --> 49:10.160
thank you very much