WEBVTT
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hello everybody so but today we are going
to talk of the consequences of special relativity
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i am sure this portion is not new to you
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but we will talk of the doppler effect in
light the the prerequisite for that is of
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course you know a bit off from time dilation
things that we have covered earlier well i
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am sure you have heard of doppler effect in
sound let us see what it is there in light
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when we talk of electromagnetic waves example
light okay and after that we will check or
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rather we will see whether the mass of a body
actually way better in values with velocity
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we may have heard of such things but let us
see where it comes from okay
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so let us first of all consider a you know
we consider a light source and then it is
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emitting foot or by emitting light of a certain
frequency let us just let us call that frequency
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v 0 okay and you have an observer here we
will show it by oh here on to towards the
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right of your screen i mentioned that of the
doppler effect in sound i am sure where instead
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of this light source you will be having someone
some source which is emitting sound okay
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which is making some noise and then emitting
certain frequencies and then it is all it
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is all fine when they are at both of them
both the observer and the source is at rest
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at each other but what happens if all of a
sudden one of them starts moving okay so let
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us say the one of the source start moving
and then the observer starts moving well if
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you have the observer starts moving towards
these the source of the sound
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do not you feel that you know the frequency
of the sound that you are hearing has increased
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it is more like a car coming towards you you
know and it blows the car blows its horn then
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the more it comes towards you you find the
pitch of the sound increasing well so what
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is it with light do we observe certain the
same things or is it a little bit different
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let us check that out of course need to mention
the very beginning so for propagation of light
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here you do not need a material medium
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because it can propagate to vacuum but for
sound of course you need a material medium
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okay so that is what we do we consider a light
source which is emitting light of a certain
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frequency that is nu0 and then we have an
observer okay now what we can also do you
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can also consider this a light source you
know as a clock that is ticking like this
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tick tick tick and then added each tick each
each one of these ticks okay and it is emitting
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a light wave ok
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so it is so what you are considering you are
considering a light source as a clock in a
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sense that is ticking nu0 times per second
and so on each tick its emitting a wave
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ok so it is now why do we put all these clocks
and things here then because if you put this
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clocks you know time dilation you really you
can make the correlation and then we can use
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the all the principles of relativity that
you have learnt earlier
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and now let us see what happens if you have
the source here you have the light source
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which is emitting a certain frequency and
the observer starts moving now we can have
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three situations here either the observer
is moving transversely or perpendicular to
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the direction of these light waves or it is
moving towards a light source or it is moving
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away so let us let us consider each of them
one by one ok ok
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so case number one so in this what we have
is that the observer that is we denote by
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o here and it is moving uniformly with a certain
velocity v okay transverse to the direction
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of the light waves ok now if you are you know
in the frame of the light source itself where
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the things were at rest okay and what is the
what is the proper time in the sense between
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the ticks you know you consider the light
source to be a clock which is ticking and
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each of these ticks emitting a it is a emitting
a light wave okay so what is the proper time
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here it is call it t0 then how is it related
with the frequency well you know you know
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frequency is in hertz and time it is in second
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so it is 1 by nu0 okay so that is the proper
time interval between the tick in the light
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source frame okay now what about how would
they observer then find this time to be okay
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now in the frame of the observer and the time
elapsed between two ticks would be t now t
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will not be same as t0 which was the proper
time remember where t will be t0 divided by
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root over of 1 - v square by c square okay
it is this very simple time dilation formula
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that you used it okay now what would then
be what would then be our the frequency of
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the light as observed by the observer well
you take just the inverse of the time here
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remember this time is the one measured by
the observer in in his or her frame ok so
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let us call this frequency as nu t it is going
to be 1 by t and then you know what 1 by t
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is you are going to you just see in the previous
paragraph so you will see that its root over
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1 - v square by c square whole divided by
t0 now t0 is also related with the frequency
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at which the light is being emitted ok so
you see that so nu t that is the that is the
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frequency as measured by the observer
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that will be nu0 times root over of 1 - v
square by c square now since v is always less
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than c here okay it is going to be vt or the
since the velocity of the of the observer
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is always going to be less than and the velocity
of light okay so what you find as nu t that
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is the frequency as measured by the observer
will always be less than the proper frequency
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okay proper time interval and then you know
the proper frequency if i can put it that
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way
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so nu t here is less than u0 so that is the
frequency that is that will be measured by
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an observer who is moving transversely to
the to the source of light okay so that leaves
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us with two other cases one in which the observer
is moving towards the towards the light source
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and in the other when the observer is moving
away from the light source ok so let us consider
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the other one first
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either namely the one in which the observer
is moving away from the light source okay
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so here to you know what is the proper time
between ticks effect it is going to be 1 by
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where t0 which is like 1 by nu0 that is when
nu0 is the frequency at which the light is
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being emitted at this source okay and remember
here the observer is moving with a velocity
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v away from the light source so in the frame
of the observer if t is the time that you
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know small t is the time that the observer
will measure between each of the ticks so
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as to say
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then for in time t remember the observer is
also moving with a certain uniform velocity
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v so the observer has already traveled vt
that is that is the unit of distance here
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away from the source between the ticks ok
so this implies that the time interval between
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two successive ticks as measured by the observer
ok will be what will we be vt by how much
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will it increase it will increase by vt by
c where t is the time as measured by you know
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by the observer here right
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and and vt by c and c is the velocity of light
and that has the same you know that so the
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value of c is going to be the same in in both
frames okay
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now what will be the total time between the
arrival of successive waves or the successive
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light waves in the frame of the observer it
will be first the time between two ticks as
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measured by you know by the observer plus
the amount of time you know the observer has
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moved away in a sense by which time i saw
was already moved by a certain distance between
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these two ticks ok so the time lapse between
the arrival of successive waves will be capital
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t which is small t + vt by c ok
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now you you simplify this as t plus and then
take a bracket or 1 + v by c ok now how is
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t related with the proper time small t that
is small t is related with the proper time
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t0 by small t being equal to t0 by root over
of 1 - v square by c square and then you have
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the bracket 1 + v by c so that simply simplifies
to the one given at the bottom of your screen
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that is t0 into root over of so 1 + v by c
divided by 1 - v by c
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now how would you find the you know the frequency
here is that all you need to take is the inverse
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of the time time interval in this case you
see here i put it as v by nu - so that is
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1 by capital t now that you just you just
invert the expression that you see in the
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top of screen and then you relate t0 with
the frequency that is the frequency with which
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the light is being emitted by the light source
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and then you immediately find what is a nu-
okay and then you can relate nu - to2 nu0
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and then you see that nu- is actually less
than nu 0 here ok so in every case you see
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that the change in frequency so far we have
seen that it is depending on the velocity
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of the observer ok the velocity with which
the observer is moving away from the the light
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source here ok so what about the last case
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so in this case let us say the observer is
moving towards the light source ok so so by
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the same logic which we had applied earlier
the observer travels a distance which is v
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times small t towards the light source between
successive ticks successive ticks of the imagine
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the clock that we have considered here ok
so the total time interval between the arrival
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of successive waves here light waves here
would be that is the capital t
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would be small t and then remember it is -vt
by c ok y - because the observer is moving
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towards the it is moving towards the light
source and then we have kept the velocity
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of light is the same in both frames i mean
way whether you are in the moving frame uniformly
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moving frame or in a rest frame ok so that
is why it is the same c so it is it is natural
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now so you do the simplification you you know
you convert the small t into t0 t0 being the
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proper time ok
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and here what we will find is that when you
take the observed frequency when you when
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you find the observed frequency it is going
to be put v nu plus here that is nu + is 1
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by capital t and we are going to see that
nu + here is more than nu0 okay it is going
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to be more than the frequency of and the the
emitted light okay so we have studied the
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three cases in which the first case the observer
was moving perpendicular to the light source
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okay
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then the one was moving away from the light
source and then towards the light source however
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i should also mention here that in all cases
you have seen that enough in all these cases
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the shift in the frequency as observed by
they the the by the observer okay depends
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on the velocity of the observer here okay
so you do see it in all these in all these
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formulae here another important point that
i should mention here is that in each of the
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formula that you have seen so far i have always
considered to light source to be at rest
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why because had i not for example the light
source was moving and then the observer was
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was at rest okay that would have been a similar
situation because it is the relative velocity
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which is important here unlike the case of
doppler effect in sound okay so it is not
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only in the doppler effect in sound it is
not the relative velocity you have to take
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into consideration the velocity of the speed
of the source and speed of the observer here
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but here what we have it is the relative velocity
of between the observer and the source which
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is going to be important here so it is if
the source is moving or the observer is moving
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it does not matter as long because the velocity
that we are considering here is actually the
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velocity v that you see in all the formula
will be will always be the relative velocity
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okay so having done this let us spend a couple
of minutes on on where possible applications
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of of this derivation would be okay
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well as i told you you see in the the shift
in frequency is related with the relative
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velocity with which the object is moving okay
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so that immediately told people that maybe
it could be used to to measure the speeds
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of distance taylor objects okay like stars
so what do what do people do what do astronomers
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do in that case what they do is that they
take a photograph of you know the spectra
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of elements which is in the star okay so from
the light which is emitted from the star the
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light spectra is analyzed so and you know
where the frequencies are people find out
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okay
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and then they compare it with the known specter
of elements present in the star ok so these
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known spectra of elements you can find the
specter of elements by a some experiment in
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the laboratory here on earth and then they
compare that spectra with the spectra that
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was obtained from this star now if the star
is moving away or you know there is a relative
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velocity it is there is a relative motion
between earth and the star there
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and then the star we are going to have a shift
in spectral frequency okay now from the shift
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in spectral frequency you can then find the
speed of the star okay so it is it is got
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lots of applications in astronomy of course
then many other you know some other discoveries
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were also based on this okay another application
actually a very interesting application of
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this would be to measure the the temperature
of hot plasma in nuclear fusion experiments
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or ok the temperature of very hot gases okay
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now what they do is that well the principle
is almost similar you know to what had what
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i have outlined earlier it is that they take
the spectra of you know when things are moving
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at such high speeds so they are emitting radiation
and and this this radiation is analyzed and
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then if if it is a known gas so the from the
known spectra you can find out what is the
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shift in the spectral frequency okay
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and then you and then you can relate that
with the speed of the molecule or the gas
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so which is there inside your experiment and
then if you relate if you if you remember
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your kinetic theory of gases so from this
velocity or you know the mean velocity of
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the root mean square velocity you can always
relate that to the temperature okay that that
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can always be done from the kinetic theory
okay so that will give an estimate of the
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temperature involved here okay
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so so we have studied the doppler effect in
in light as against it you know that now as
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against doppler effect in sound ok and then
we have also talked a bit about few applications
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fine so now we are going to shift gear a little
bit and do something a little bit different
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and go to another consequence of special relativity
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and study if the mass of a body actually varies
with the velocity with which it is moving
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when i know that this is one of the things
that many of many of us are rather aware of
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when when we study physics or any of the elementary
courses of science fatigue mass increasing
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with velocity and e equal to mc square ok
but all of it in due time ok now to do this
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do this problem of where the mass of a body
varies with velocity
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let us start with up with an example ok and
then we will come to a conclusion ok so concentrate
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on the on the on on the thing written down
as s prime frame ok
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so what we have here is that we are considering
the inelastic head on collision of two identical
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particles a and b in in s prime frame let
us see so what is this s prime frame well
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it is a frame which is let us say it is moving
with a certain velocity certain uniform velocity
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v with respect to another frame s let us say
okay now in it we have the inelastic collision
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of two bodies here named as a and b so they
are of identical masses m prime both are of
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m prime and they are moving with the same
speed you know opposite to each other
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and then they are heading towards and head
on collisions mean speed is u prime here the
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wall put primes here because you know to keep
in to keep you know in to make since we are
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talking of s prime frame so we are dealing
with prime coordinates here let us put it
21:20.040 --> 21:27.760
that way okay and then all these velocities
u prime you know they are moving parallel
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to the x axis okay fine on the way are identical
masses also
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now how is the same event observed in s frame
okay well you have different below you have
21:46.630 --> 21:54.360
a you know in the s frame you goingt to see
the same thing as some you know the velocities
21:54.360 --> 22:01.240
may not be the same the velocity is although
they are moving the speeds are all the same
22:01.240 --> 22:08.530
but if you if you take the modulus of of the
speeds in s prime frame they are the same
22:08.530 --> 22:11.590
the directions may be different but the speeds
are the same
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but then the velocities of the speeds are
not going to be the same in s frame a simple
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reason it is that you need to invoke einsteins
velocity addition formula to find what is
22:23.290 --> 22:28.930
the velocity in in s frame okay and you go
to see immediately that it is not going to
22:28.930 --> 22:36.640
be the same it is this written in you know
in for your convenience at towards the end
22:36.640 --> 22:45.240
of the slide and for example if you just consider
you one of the mass of the velocity of mass
22:45.240 --> 22:48.740
okay one mass a rather okay
22:48.740 --> 22:55.180
you are going to see in terms of the ss prime
quantities and then the velocity v its u prime
22:55.180 --> 23:02.420
+ v divided by 1 + u prime v by c square however
if it was minus you since was moving in the
23:02.420 --> 23:09.960
other direction okay u2 that is the other
you know the other quantity that is -u prime
23:09.960 --> 23:16.500
+ v1 - u prime v by c square so i am going
to be the same okay so you might be a little
23:16.500 --> 23:22.130
bit confused why you have written a different
masses for identical masses ok
23:22.130 --> 23:27.780
so in the ss prime frame you had identical
masses both of mp okay but why is it that
23:27.780 --> 23:33.190
they are different in in the s frame well
at this stage i do not know why they are different
23:33.190 --> 23:38.200
rather i just take them to be different so
just to keep myself in conjunction with the
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idea that i have different velocities okay
i do not know what is going to the masses
23:42.130 --> 23:48.930
so if it is if my if my relief from my relativity
from whatever i do it turns out from all this
23:48.930 --> 23:53.429
conservation laws a marine work on if it turns
out that these masses are the same they will
23:53.429 --> 23:58.660
turn out to be same m 1 will be equal to m2
but right now i am going to take m1 and m2
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different ok
23:59.660 --> 24:05.050
let us just see what happens okay so that
is what we have we have the s prime frame
24:05.050 --> 24:11.360
and then in the s frame we are going to see
what it is going to be of the same even the
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same collision how is going to how how one
is going to see in s frame and remember s
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prime frame is moving with a certain uniform
velocity v with respect to the s frame ok
24:24.100 --> 24:30.720
so on top of that of course on the back of
our minds we we have the conservation of linear
24:30.720 --> 24:32.950
momentum in our in the collisional process
24:32.950 --> 24:39.090
well this is rather something very sacrosanct
i mean we are not going to violate all these
24:39.090 --> 24:44.900
conservation principles ok whether it is non
relativity or relativity of classical mechanics
24:44.900 --> 24:49.080
i mean the conservation laws a conservation
laws you are supposed to you are supposed
24:49.080 --> 24:55.360
to con you supposed to maintain them okay
and and then on top of that you also have
24:55.360 --> 25:00.440
the total mass to be conserved that the total
variable mass you know the total mass
25:00.440 --> 25:09.510
for example if it is considered s prime frame
you have m prime a and then the other mass
25:09.510 --> 25:11.679
b also of m prime
25:11.679 --> 25:17.520
so after collision the mass is going to be
m prime + m prime that is 2 m prime that is
25:17.520 --> 25:21.050
what i mean the total mass remains conserved
in the collision process in each of these
25:21.050 --> 25:32.890
frames okay okay so so what does it mean here
well from the linear momentum conservation
25:32.890 --> 25:36.790
what we have in this conditional process if
you look at the s prime frame
25:36.790 --> 25:41.010
it is very simple it is going to be zero is
not a terminates they are moving that you
25:41.010 --> 25:45.550
have an identical masses the speeds are identical
and opposite to each other moving parallel
25:45.550 --> 25:54.340
to the x and x prime axis so u prime m prime
u prime + m prime - of u prime that is going
25:54.340 --> 26:04.140
to be 0 so momentum so even if the if so that
tells you that you are going to have an inelastic
26:04.140 --> 26:09.830
collision in the end so you know the mass
is going to whatever it is the velocity or
26:09.830 --> 26:16.350
the it has to be at rest inn in the s prime
frame after the collision something of that
26:16.350 --> 26:17.350
later o
26:17.350 --> 26:22.260
but what about this the to the final mass
of the combination there is final mass of
26:22.260 --> 26:26.100
the combination m prime + m prime it is going
to be 2m prime there is nothing rocket science
26:26.100 --> 26:30.200
in that but the only thing is that it is going
to be maintained after collision that is what
26:30.200 --> 26:40.990
i want to say ok so that is what we have the
thing ok so just one look one more slide to
26:40.990 --> 26:46.660
to recapitulate on this collisional process
before the collision okay
26:46.660 --> 26:55.030
so we have the s prime frame where we have
identical masses a and b okay they have similar
26:55.030 --> 26:58.820
masses they have the same mass m prime and
they are moving opposite to each other parallel
26:58.820 --> 27:05.540
to the x prime axis and then one is observing
the same thing from the s frame and then the
27:05.540 --> 27:11.040
s prime frame is moving in a certain velocity
v with respect to the s frame and from the
27:11.040 --> 27:14.700
s frame we are going to see the same collisional
process
27:14.700 --> 27:19.510
but then the velocities are not going to be
different and then we have taken different
27:19.510 --> 27:23.120
masses here we do not know what we do not
know whether the masses will be same or different
27:23.120 --> 27:30.480
but let us see if from all these conservation
laws what turns out to be what whether we
27:30.480 --> 27:41.450
have indeed different masses whether m1 a
separate is different from m prime ok ok
27:41.450 --> 27:50.830
so what happens after collision remember the
collision is perfectly inelastic so what it
27:50.830 --> 27:55.620
means is a combined particle the combined
particles will be at rest and i will stick
27:55.620 --> 28:00.140
together in the s prime frame so that is what
they are they will stick together in s prime
28:00.140 --> 28:07.840
frame the total mass is 2m prime and they
are at rest now what will be the velocity
28:07.840 --> 28:15.160
with which these thing will be seen to move
from s frame ok so is it v we are going to
28:15.160 --> 28:17.320
find that out ok
28:17.320 --> 28:24.510
but remember here what will be the total mass
of this ab system here at in s frame it is
28:24.510 --> 28:37.010
going to be m1 + m2 ok his total mass is conserved
in this coalitional process right so whatever
28:37.010 --> 28:41.419
the velocity as we said so remember after
collision will still after collision
28:41.419 --> 28:46.220
we have we are still considering s prime frame
to be uniformly moving with a certain speed
28:46.220 --> 28:55.520
v with respect to the air spray now in s prime
frame a and b is at rest right is at rest
28:55.520 --> 29:01.500
so what does it tell you and then the s prime
frame is moving with velocity v with respect
29:01.500 --> 29:08.059
to s so what is the velocity with which a
and b will be seen to be moving from s frame
29:08.059 --> 29:14.809
well it is b that is what we have so in s
frame we have these two particles they are
29:14.809 --> 29:16.210
sticking together okay
29:16.210 --> 29:19.870
since it is sticking together prime frame
that is sticking together is going to have
29:19.870 --> 29:25.430
mass m1 + m2 and it is going to move with
a certain velocity v which is same as the
29:25.430 --> 29:34.490
relative velocity between s and the s prime
okay so just to recapitulate the entire collision
29:34.490 --> 29:36.390
process
29:36.390 --> 29:42.660
what you see in the top of the slide is the
situation before the collision okay both in
29:42.660 --> 29:50.450
s prime frame and the s frame just have a
nice look on the situation what is happening
29:50.450 --> 29:57.160
in the s prime frame and the situation that
we have in the s frame okay that is the that
29:57.160 --> 30:02.080
is the situation before collision and then
after collision it is perfectly inelastic
30:02.080 --> 30:07.570
collision so it is sticking together in s
prime frame and total mass is 2m prime and
30:07.570 --> 30:08.960
it is at rest
30:08.960 --> 30:13.980
and then from s frame we are going to have
the total mass to be m1 + m2 that total mass
30:13.980 --> 30:21.660
is conserved here and is moving with velocity
v okay so next what we are going to do is
30:21.660 --> 30:27.059
that we are going to invoke the conservation
of momentum in the s frame remember if you
30:27.059 --> 30:30.679
invoke the conservation of momentum in the
s prime frame it is going to be 0 is not it
30:30.679 --> 30:36.690
i mean the total momentum total linear momentum
in the s prime frame is 0 okay
30:36.690 --> 30:42.140
and what is it if you invoke that same conservation
of momentum in the s frame what is it going
30:42.140 --> 30:45.070
to be okay
30:45.070 --> 30:54.080
it is going to be m1 u1 + m2 u2 that is the
total momentum in one in the s frame before
30:54.080 --> 31:00.929
the collision and then after the collision
it is going to be m1 + m2 whole times v that
31:00.929 --> 31:06.430
is the velocity with which this combination
if a + b or the same m1 + m2 is moving ok
31:06.430 --> 31:14.120
rearrange the term do as little simplifications
immediately get what m1 + m2 is m1 + m2 is
31:14.120 --> 31:21.660
nothing but v - u2 divided by 1 - v well that
is very simple you just take all the m1s you
31:21.660 --> 31:27.620
know all the terms involving em once or one
side all the terms involving m2 and once and
31:27.620 --> 31:29.460
then do simplifications okay
31:29.460 --> 31:38.340
and then what you do is that you substitute
for u1 and u2 the expressions from einsteins
31:38.340 --> 31:44.510
velocity addition formula so you know that
very well so u 1 is 2 prime + v divided by
31:44.510 --> 31:52.340
1 + u prime v by c square and what is u 2
u 2 is - u prime + v divided by 1 -u prime
31:52.340 --> 32:00.190
v by c square so you want to substitute these
two velocities u1 and u2 okay in the expression
32:00.190 --> 32:04.610
given in the middle of your screen then given
by this m1 by m2
32:04.610 --> 32:11.669
well i am doing a little bit of the algebra
a bit rather little bit of the algebra which
32:11.669 --> 32:17.179
only but if you miss this slide other next
not going to miss the most of the physics
32:17.179 --> 32:22.580
okay i am just doing it so that you will have
the expression you can you can derive all
32:22.580 --> 32:30.430
these things by yourselves okay so what i
have done is that now what will be m1 + m2
32:30.430 --> 32:36.550
in terms of all the quantities in the prime
frame okay
32:36.550 --> 32:44.530
so it is 1 + u prime v by c square divided
by 1 - u prime v by c square again what we
32:44.530 --> 32:52.900
do is that ok so we have derived up to the
the expression the top of the slide again
32:52.900 --> 32:57.710
what we see is that a little more simplification
actually you can derive this if you invoke
32:57.710 --> 33:06.809
again the the the einsteins velocity addition
formula 1 - u1 square divided by c square
33:06.809 --> 33:13.070
and what is 1 - u2 square divided by c square
so you put in what is u1 and u2 in terms of
33:13.070 --> 33:15.110
u prime vs and all these things
33:15.110 --> 33:19.429
you see rather very interesting expressions
you can you can you can actually check that
33:19.429 --> 33:25.770
out okay it is 1 - u prime v by c square whole
square divided by 1 + u prime v by c square
33:25.770 --> 33:33.900
whole square now from the top of the slide
you can see that this is nothing but m2 by
33:33.900 --> 33:44.289
m1 whole squared ok now now check what have
check the one on the left hand side okay so
33:44.289 --> 33:49.820
it is 1 - u1 square by c square divided by
1 - u2 square by c square and then on the
33:49.820 --> 33:54.900
extreme right hand side you have this m2 by
m1 and whole of square
33:54.900 --> 33:58.480
is very interesting these are the two expression
of the left hand side and the right hand side
33:58.480 --> 34:02.680
disregard the one in the middle okay what
do you see you see the masses you see the
34:02.680 --> 34:06.880
velocities okay what do you again see
34:06.880 --> 34:16.549
what you see is m1 into 1 - 1 square by c
square and you take a square root of that
34:16.549 --> 34:23.619
is equal to m2 into 1 – u2 square by c square
take u square root of that and that is it
34:23.619 --> 34:30.460
variant and that is not changing okay so what
you see is that what you have here so m1 and
34:30.460 --> 34:40.290
m2 are the masses of identical particles okay
when the velocities are u1 and u2 respectively
34:40.290 --> 34:44.090
i mean when i say m1 and m2 are masses of
identical particles
34:44.090 --> 34:49.220
and you know does not carry much of a meaning
and then you could say that identical particles
34:49.220 --> 34:53.810
how come they have different masses okay but
you see that they are moving at different
34:53.810 --> 35:00.700
speeds okay so this combination some rather
interesting combinations of m1 and u1 of these
35:00.700 --> 35:08.720
speeds with which they are moving and then
use that combination given by m1 times 1 - u1
35:08.720 --> 35:12.310
square by c square that or that square root
of that
35:12.310 --> 35:17.750
that is the thing that is going to be invariant
change the velocity the expression for the
35:17.750 --> 35:24.400
mass changes okay that is becoming invariant
okay so what is invariant is this combination
35:24.400 --> 35:31.550
but in whatever you know whatever mass and
then its corresponding velocity that is possible
35:31.550 --> 35:32.550
here
35:32.550 --> 35:37.070
so what we have so in a frame if v is the
velocity of the particles v is the velocity
35:37.070 --> 35:45.710
of the particle and m is its mass then the
quantity that will have m into 1 - v square
35:45.710 --> 35:52.770
by c square root over of that that is going
to be invariant okay now this is very interesting
35:52.770 --> 36:00.200
why now watch this well rather what what should
be the invariant quantity let us check that
36:00.200 --> 36:07.800
out now if you take the velocity with which
the mass is moving to be 0
36:07.800 --> 36:12.690
so so it is basically the mass is at rest
so you consider you considered a frame in
36:12.690 --> 36:16.830
which the particle is at rest and then you
measure the mass there
36:16.830 --> 36:23.090
now if the mass is m0 there okay so you plug
in those quantities into the formula so m
36:23.090 --> 36:29.940
into 1 - v square by c square so the mass
is m0 and the velocity is 0 see the one in
36:29.940 --> 36:35.740
the bracket 1 - 0 square by c square root
of that so that is going to be 1 no problem
36:35.740 --> 36:41.150
and then the right hand side you have m0 so
that is going to be invariant quantity m0
36:41.150 --> 36:46.980
is a mass of the same particle in a frame
in which it is at rest
36:46.980 --> 36:54.990
so that is the formula we are looking for
see that if you know that the invariant quantity
36:54.990 --> 37:00.320
is actually m0 here you can now relate the
mass of a body which is moving with a certain
37:00.320 --> 37:01.760
velocity v
37:01.760 --> 37:09.690
as m = m0 which is the mass of the body in
in the inner frame in which is at rest and
37:09.690 --> 37:16.070
then divided by root over of 1 - v square
by c square so so what we have is the mathematical
37:16.070 --> 37:21.570
formula and then this is very interesting
of what the mass of the body will be if it
37:21.570 --> 37:33.690
is moving with a certain velocity v ok now
let us see how this is related so let us have
37:33.690 --> 37:37.261
a graphical illustration of this variation
of mass and velocity
37:37.261 --> 37:49.490
it is if you if you look at at the graph on
your screens what i have plotted here on the
37:49.490 --> 37:59.030
y axis is the mass and on the x axis i have
v by c ok but is basically it is it is the
37:59.030 --> 38:05.310
velocity in the units of velocity of light
so it is someone you know it is starting from
38:05.310 --> 38:13.660
0 to the speed of light so when v by c is
one it is the speed of light so okay and when
38:13.660 --> 38:20.760
when v by c is 0 that you know that the body
is at rest okay
38:20.760 --> 38:32.540
now if we consider a mass of you know of which
is of at rest which is m0 to be off let us
38:32.540 --> 38:42.820
say 10 kg okay now that is the green line
okay so that is the green line and so whether
38:42.820 --> 38:48.140
if you increase the velocity or not so that
is m0 it is the it is the rest mass so that
38:48.140 --> 38:56.740
is 10 kg but as you increase the velocity
okay see the red see the red curve see that
38:56.740 --> 39:02.140
at small speeds when the velocity of when
the velocity of the body is not too high or
39:02.140 --> 39:06.300
it is too high not too high compared to the
speed of light
39:06.300 --> 39:11.970
you see that it is it is increasing its increasing
and then all and then when it when it reaches
39:11.970 --> 39:15.770
towards the speed of light you see all of
a sudden it increases very fast i mean is
39:15.770 --> 39:22.070
it is almost explodes okay so you do see that
the mass of the body is increasing in a certain
39:22.070 --> 39:31.510
way when when when the velocity is increased
okay now i also wish to remind you one more
39:31.510 --> 39:38.400
thing is that when you are at very small speeds
when on basically when you are non relativistic
39:38.400 --> 39:40.109
speeds okay
39:40.109 --> 39:47.260
when v is much much less than c you can immediately
figure out look at the formula for example
39:47.260 --> 39:48.570
let us first look at the formula
39:48.570 --> 39:56.520
so m = m0 divided by root over of 1 - v square
by c square so when is much much when v is
39:56.520 --> 40:04.600
much much lesser than c c by c almost tends
to 0 so the denominator in this fraction is
40:04.600 --> 40:12.021
going to be 1 okay so m will roughly be equal
to m0 when the velocity is much much less
40:12.021 --> 40:17.670
than the velocity of light and that is exactly
what you see in the graph ok so when velocity
40:17.670 --> 40:21.170
is much much less than the velocity of light
40:21.170 --> 40:25.660
you have your you are in the non relativistic
limit in the classical limit the classical
40:25.660 --> 40:31.640
world that we all live in okay you see that
the red line actually converges with the green
40:31.640 --> 40:41.680
line so it actually fits with our experience
okay okay so now let us see so so let us let
40:41.680 --> 40:50.210
us see what have another example of what the
consequences of this of this formula would
40:50.210 --> 40:56.349
be let us just consider the you know the rest
mass of an electron okay
40:56.349 --> 41:02.390
the rest mass of an electron if you look at
the tables it is something like 911 into 10
41:02.390 --> 41:09.910
power minus 31 kilograms now suppose this
is moving with the it is a pretty high velocity
41:09.910 --> 41:18.040
it is eighty percent the speed of light ok
so will its mass be equal to you know how
41:18.040 --> 41:24.580
much will it what we must be will it be a
very near to 911 and 10 power minus 31 kilograms
41:24.580 --> 41:33.970
let us see so that is the rest mass and the
velocity v here is 80% of the speed of light
41:33.970 --> 41:41.820
so what you see here will be m so that is
the measured mass will be equal to m0 divided
41:41.820 --> 41:50.670
by root over of 1 - v square by c squared
okay now m0 you know what that is 911 into
41:50.670 --> 41:58.859
10 power minus 31 kilograms okay and then
1 - 08 square and c square divided by c square
41:58.859 --> 42:04.831
ok now this turns out to be something like
15 into 10 power minus 31 kilograms so you
42:04.831 --> 42:11.920
do see an increase in mass okay when you when
when the velocity is increasing
42:11.920 --> 42:17.780
and that too when the velocity is much much
well which it is actually 80o% the speed of
42:17.780 --> 42:28.900
light is quite a huge amount of which is quite
a large velocity okay so by hope today i have
42:28.900 --> 42:35.210
you know you be able to cover some part of
the consequences of special activity we talked
42:35.210 --> 42:45.359
of doppler effect in light and then the the
interesting phenomenon of the mass of a body
42:45.359 --> 42:47.010
varying with its velocity
42:47.010 --> 42:56.200
in the next section what we will talk off
is the relation between mass and energy okay
42:56.200 --> 43:02.660
and i will heavily draw upon what we have
done so far about the consequences of relativity
43:02.660 --> 43:23.830
specially about the variation of mass with
its velocity well thank you very much