WEBVTT
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okay so we will be talking of the consequences
of special relativity today having been introduced
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to this subject in a previous lecture okay
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so we will be talking mainly on length contraction
but of course before that we will spend
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some time in in in learning how velocities
are added in special relativity okay as opposed
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to human you know this the galilean velocity
addition formula that we have so well used
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to right so you know if you want to come to
the subject we just summarized a little bit
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of what we did in last time
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so we were seeing we were talking of newtonian
mechanics being invariant under galilean transformations
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electrodynamics i mean the laws of electrodynamics
specially maxwells equations so we saw that
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they were invariant under lorentz transformations
so it was a peculiar situation i mean a branch
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of physics another branch of physics not being
been variant on the same transformation okay
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and it is actually this reconciliation of
transformations of transformation laws of
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mechanics and electrodynamics from a theoretical
sense which led einstein to the special theory
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of relativity of course having mentioned this
i mean i should also mention the the michelson
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morley experiment because had that not being
there it would have been very difficult to
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see the they really need for for for such
a for such a theory okay right
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so we recapitulate the postulates of relativity
once again the principle of relativity
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the states of course that it is a very very
basic principle that the laws of physics are
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to be same in all in jail system so there
is no preferred inertial system okay so what
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does it mean it means that if you are in a
system and you do an experiment there and
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then there is another system which is moving
with uniform velocity with your system will
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not be able to differentiate that
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so basically systems which move with uniform
velocity relative to each other they are absolutely
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similar in the sense that the laws of physics
are going to be the same in all these systems
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okay
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the the second postulate and it is it is it
is very important theory it is the principle
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of the constancy of the speed of light okay
so well i mean in in simpler words it says
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that the speed of light in free space it has
to be the say it has to have the same value
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i mean we can see here and if you know that
it is 3 into 10 to power 8 meter per second
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i mean which is almost 3 into 10 power 8 meter
per second we know that it is right 299 something
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in all inertial systems
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now this in the first instance may sound very
strange because because if you if you think
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of everyday experience suppose you you shine
a light say you light a torch and then in
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the direction of this light let us say you
are moving in a vehicle with there an imaginary
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vehicle of course in a very fast way so let
us say you are moving with 2 into 10 to power
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8 meter per second so what happens so what
speed will you see the light will see this
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light from your normal experience of galilean
relativity okay
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so galilean velocity addition formula and
all these things so the light is moving forward
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3 in 10 power 8 you are moving in the same
direction in 2 into 10 power 8 so you are
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supposed to see the light in 3 - 2 that is
1 into 10 power 8 meter per second well relativity
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tells you that no you ever see it 3 in 10
power 8 meter per second and so where we will
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see how this is possible because if you if
you believe in this galilean transformations
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if you will be if you do the thing in gradient
transformation of course this does not come
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about
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so there is another transformation where such
thing is possible and that is we thought what
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that was that was actually lorentz transformations
and if you i am just summarizing the transformations
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once again for you if you have let us say
a frame s prime which is moving with a certain
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velocity v that is certain uniform velocity
i should had with respect to another frame
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s okay so they have let us say the common
x x prime x axis here
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and then a point a general point is xyz is
there and then you put in the fourth component
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that time here and then at the same point
from the s prime frame that is x prime y prime
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z prime and t prime and then those relations
are given by the transformation equations
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which you see here so it is x how does the
the quantities in the primed frame relate
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the quantities in the unprimed frame okay
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so that is so that is the that is given by
the lorentz transformations so here you see
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x = gamma times x prime + beta ct prime now
i have defined what this beta and the gammas
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are so beta is also it is v by c so it is
dimensionless and gamma is 1 by root over
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of 1 - beta square okay if you write it in
this form you will see that when you write
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the time component it has almost similar looking
expression
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so of course instead of time i multiplied
that by c that is the velocity of light so
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that it has the dimension of length okay so
i got ct so you see you see compare this equation
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for x and compare this equation for ct okay
you see that it contains in the equation x
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you have both x prime and t prime and then
in the equation for ct you have both t prime
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and x prime of course in this opposite fashion
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so it looks very symmetric so it is quite
useful to right it in this way but for the
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sake of simplicity and because we are doing
this for the first time i will use all these
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expansions whenever we come across it okay
so so what are the limits i mean does this
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in some limit go back to the galilean transformation
well actually it does where does it do so
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add small speeds that is if one is moving
with speed much much lesser than the speed
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of light here
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so you see v is much much lesser than v so
of course v square by c square that tends
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to 0 here now you can immediately figure out
that x prime it becomes you know this the
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the the quantity below the denominator for
x prime x prime = x -vt divided by root over
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of 1 - v square by c square so this this turns
out to be just the galilean equivalent that
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is x prime = x - vt and since you are moving
along the x x prime direction
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so the x y prime for the z prime are the same
okay now what happens for the t primes okay
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now it is interesting when lorentz transformation
the time’s in both these frames are not
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the same they are also dependent on the coordinates
and all the space coordinates of the system
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it is it is evident from this equation itself
okay now add small speeds at small speeds
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smaller much smaller than the speed of light
that is
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you will see that these times t prime and
t they become equal they become approximately
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equal actually if you do this you take v by
c to be much much less than so it is almost
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tending to 0 it is much much less than 1 instead
to 0 that is okay so that is the thing that
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we get lorentz transformations they go to
galilean transformations add small speeds
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so that is the that is the thing that is the
thing
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so at at smaller velocities newtonian mechanics
is still the thing that one can follow and
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that is that is it that we that is what we
do in our daily lives i mean we in our daily
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lives rarely do we go at speeds such high
speeds and speed of light okay sothe newtonian
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mechanics is valid is correct at at small
speeds okay
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but having said that let us now have a look
at how velocities are going to be added in
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these two frames okay so what are the relative
velocities or whats the galilean velocity
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addition formula here okay i am sorry let
us see the the lorentz velocity at the enstien
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velocity addition formula according to lorentz
transformations because the galilean velocity
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transformations are addition laws we have
already done okay
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but how do you do it i mean the process is
similar if you remember i have drawn this
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this this diagram for these frames once again
see x o by the way the x axis is always there
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it is protruding out of the screen i am just
not drawn it so that the picture looks a less
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clumsy that is all okay so so how do you find
the velocities in each of these frames you
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are going to do you are going to take this
differential this dx by dt
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now remember the coordinates are going to
take for each frame and then the time which
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is suitable for that frame itself okay now
so in the s frame it is dx by dt and in the
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prime frame it is obviously dxprime by dt
prime remember in in when we were doing this
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galilean velocity addition formula this t
and t prime were the same but here they are
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not okay i mean just for the for the sake
of simplicity and just for your remembrance
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i have listed it on the left
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so you see that they are not the same so if
you if you take dx prime by dt it is not the
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same as dx prime by dt prime okay but you
need to take this properly now if you do this
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if you do dx by dt and du prime dx prime and
u prime = dx prime by dt prime we are going
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to land up with the velocity addition formula
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and so how does it look like so so what is
the velocity in s frame having known the velocity
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of a particle or some velocity in the prime
frame so that is the one given on the left
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hand side of your screen you will see that
if the s prime frame is moving with a certain
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uniform velocity v and then in this frame
a velocity is measured as u prime this velocity
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in the unprimed frame is going to be measured
as u prime + v divided by 1 + u prime v by
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c square okay
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so this is the einsteinian velocity addition
formula okay yeah just just to remind you
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had this been the normal or the more familiar
a galilean velocity addition you would not
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have the denominator here you just have u
would be just u prime + v and so on okay but
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then in some limit as we said i mean these
transformations go at small speeds the lorentz
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transformations become the galilean transformation
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so to here it must be so here these velocity
addition formula must go to the the einsteinian
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formula must go to the galilean formula small
speed so we will check that we will check
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that okay so similarly of course can do the
reverse transformations and find u prime to
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u - v divided by 1 - uv by c square so that
you find so if you know the velocity in the
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sv so how is it so what is the velocity u
prime you are going to measure in a frame
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s prime that is which is moving away with
a uniform velocity v ok
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so as we are talking of the limits so at small
limits when u is small v small you immediately
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see what you immediately see that this denominator
so u v by c square this becomes negligibly
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small so called limit speed equal to 0 you
might you might say so that u prime = u - v
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or if you feel added your arms you are fastidious
about this partition so u = u prime + v that
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is all so that gives us the galilean limit
at small speeds right
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how about some examples let us see what happens
if in with this einsteinian velocity addition
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formula where you have a frame which is moving
let us say your s prime frame is moving away
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with the velocity 09 c that is cs being the
speed of light okay and in this s prime frame
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you measure some speed speed of some object
you know things can have such large speeds
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that is relativistic speed some subatomic
particles do have such relativistic speeds
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to be like 09c right
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so if you directly go and apply the galilean
formula you know it is fine 9 + 09 you want
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to get more than c of course you know that
that is forbidden relativity so how is it
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how is it that comes what is it that comes
here so for u prime you are going to substitute
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9c and for v 09c and then in a denominator
you see 1 + 081 c square by c square and then
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value you get something like 99c okay
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so it is still still less than c right so
we have checked that even if individual velocities
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we have in this particular case is almost
touching c the added velocity is the from
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the einsteinian formula it still gives you
less than c i mean has it should okay
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yeah but the interesting case is you might
say is what happens if one of these speeds
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is c itself okay that is if u prime is c okay
in s prime frame which is moving with with
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the velocity some v with respect to the s
frame so what is the speed that is seen in
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u so in other words so this is the case you
have let us say your light foot so you have
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a photon or speed light which is moving with
speed or the cost of the speed of light c
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in one frame
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what are you going to observe in another frame
okay with respect to which your frame is moving
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with a certain velocity v okay so this is
a direct consequence you want to test that
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is a direct consequence of the second postulate
of relativity so u prime you are going to
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substitute by c v is of course the speed with
which the s prime frame is moving with respect
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to the s frame and then the denominator you
have 1 + cv by c square and what does it give
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you mm-hmm
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it gives you just c okay so that is the speed
of light is the same in all inertial frames
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i mean this we had started with postulates
no wonder that is so but then at least you
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are verifying here that the einsteinian velocity
addition formula that we just use right now
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is consistent with this postulates okay so
just one final example perhaps to find the
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relative velocities of particles
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so let us say you have two particles a and
b which are moving in opposite directions
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okay one moving towards left with point with
–c by 2 right well this plus and minus is
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just tells you the direction the direction
is different that is on the vector direction
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is different so they are moving with speeds
half the speed of light okay in opposite directions
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okay the question is what is the relative
velocity of 1 with respect to the other
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what is the relative velocity of a with respect
to b and what is the relative velocity of
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b with respect to a yeah it is very common
so i mean we all have travel in train so you
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see when when you travel in a train in a certain
direction so you if you can you can always
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feel you can always feel that a train which
is moving in the other direction to you is
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somehow moving what is the what is the relative
velocity of that train it moves it moves quite
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fast is not it okayand in this so we are going
to wind up you were to find out a similar
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situation here okay
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so here how do you do that so we take particle
b to be at rest in frame s prime so which
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is so which is moving with s with the other
frame s with a uniform velocity c by 2 here
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so again the velocity of a relative to s is
whatever what what is measured in s is c by
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2 okay so what is the relative velocity so
what you are going to find is the velocity
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of a relative to s is c by 2 and therefore
relative to s prime and you see s prime is
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the frame in which b is taken to be at rest
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so that is the frame you if you find this
velocity of a with respect to s prime you
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want to find the relative velocity with b
that is it so you use the velocity this addition
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formula here so that you are using the the
primed coordinates here so u prime is u - v
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1 - uv by c square and then you just put in
the numbers c by 2 the perfect the the the
20:50.230 --> 20:57.179
proper numbers here and what do you get you
get 4c by 5 still less than c
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i mean if it had been the galilean way it
would be c but here it is 4 by c and what
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is the other way now the velocity of b relative
to a it should be exactly the opposite right
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so it is -4c by 5 you could have also used
the velocity addition formula to get this
21:19.230 --> 21:30.129
speed i mean that you can do it as an exercise
of course ok right so having now seen how
21:30.129 --> 21:36.139
velocities are done let us go to some of the
other corollaries of special relativity
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and especially an interesting thing called
length contraction okay so what is it so it
21:44.330 --> 21:50.260
just tells you that the length of a moving
body appears to be shortened in the direction
21:50.260 --> 21:57.690
of its motion okay so for moving bodies appear
to be shorter okay we want to find out how
21:57.690 --> 22:09.000
ok so for this let us consider a rod okay
the end points of this rod are in s prime
22:09.000 --> 22:14.740
let us say so this rod is in s prime and it
is at rest on this x axis ok
22:14.740 --> 22:25.460
so just consider an unidirectional thing here
one dimension thing here and you have the
22:25.460 --> 22:36.000
x the you have the ends of this rod as x1
prime and the the other end as x2 prime okay
22:36.000 --> 22:43.340
sothe length of the rod which is at rest in
this frame is recall that at l0 this is you
22:43.340 --> 22:51.590
just measure this what how do you do so it
is not you you just measure you can take a
22:51.590 --> 22:57.570
measuring tape put it on one end of this sort
and just put the other one at the other end
22:57.570 --> 23:11.870
okay so you how do you what do you find that
is all l0 is x1 x2 prime -x1 prime okay
23:11.870 --> 23:19.470
however one thing is to be noted here is that
we are not bothered about the time at which
23:19.470 --> 23:24.990
these two measurements are done because this
thing is at rest so it does not we are not
23:24.990 --> 23:30.031
bothered about the time okay so you pew that
is what you do you put you will liesurely
23:30.031 --> 23:37.539
put one end of your measuring tape on at x1
and then go and put the other end at x2 and
23:37.539 --> 23:42.120
then see that the length that is all okay
23:42.120 --> 23:52.769
now what happens i mean if you want to measure
this in some other frame s let us say so you
23:52.769 --> 23:59.000
want to measure the length of this rod s now
remember relative to s s prime is moving with
23:59.000 --> 24:04.429
a certain velocity and hence the rod which
is at rest in s prime is also moving with
24:04.429 --> 24:13.350
a certain velocity with respect to s prime
okay so that is it so in this frame s right
24:13.350 --> 24:20.240
remember that the rod is now moving parallel
to the x x prime axis with a certain uniform
24:20.240 --> 24:23.289
velocity v ok
24:23.289 --> 24:31.110
now it is very important that to measure the
length of the door in the s frame it is n
24:31.110 --> 24:35.559
coordinates must be measured at the same instant
of time okay
24:35.559 --> 24:43.490
this is very important now ok now for this
what you need of course you need synchronize
24:43.490 --> 24:47.200
clocks and then they have and you have two
different different observers who are going
24:47.200 --> 24:56.740
to measure the endpoints of the rod here okay
so let us say at this particular time t = t0
24:56.740 --> 25:07.320
in s prime in s prime in s frame that is not
in the s prime frame the two ends are measured
25:07.320 --> 25:14.490
as x1 and x2 okay so what is the length of
this rod in s frame
25:14.490 --> 25:22.659
you want to just subtract these two coordinates
so l = x2 - x1 and say okay so this is my
25:22.659 --> 25:31.889
length in in the s frame okay relative to
which new rod is moving with a certain velocity
25:31.889 --> 25:40.470
but remember there is a relation between the
quantities in the s frame and the prime frame
25:40.470 --> 25:48.370
in the in the frame in the x x prime coordinates
and the unprimed coordinates x which are unprimed
25:48.370 --> 25:53.370
so they are given by the lorentz transformations
okay
25:53.370 --> 25:59.220
so we use the lorentz transformations so what
is it again just to remind you so what is
25:59.220 --> 26:05.090
x x prime so that is x – vt now here at
a certain so on the right hand side of this
26:05.090 --> 26:13.649
equation so all these quantities are in the
quantities measured in the s frame okay so
26:13.649 --> 26:21.100
on the left hand side the quantities which
are measured in the primed frame okay so from
26:21.100 --> 26:30.590
this so you you can find the relations of
x1 prime and x2 prime in terms of the corresponding
26:30.590 --> 26:33.799
quantities in the s frame okay
26:33.799 --> 26:39.059
but remember along with these positions the
time also enters into this picture but then
26:39.059 --> 26:43.509
the point is that the time you measure this
at the same time you have to measure the ends
26:43.509 --> 26:50.059
of the rod in s frame at the same time so
it is going to be the same here
26:50.059 --> 26:58.929
so if you subtract these two equations what
you get so you get x2 prime - x1 prime = x1
26:58.929 --> 27:10.159
x2 -x1 okay what does this give you what is
x2 prime - x1 prime that is l0 ok so that
27:10.159 --> 27:15.360
is the length of this rod in in the prime
frame and then what is x2 minus x1 that is
27:15.360 --> 27:22.799
l that is the length of the rod in the s frame
so and then notice that the denominator is
27:22.799 --> 27:29.769
still there okay so that is root over of 1
- v square by c square so you get this expression
27:29.769 --> 27:36.899
so what is the length of this rod in the s
frame remember it is moving in the s frame
27:36.899 --> 27:42.270
it is as if the rod is moving with a certain
velocity v okay you see that is this l is
27:42.270 --> 27:51.889
nothing but l0 to root over of 1 - v square
by c square clearly you see that this l the
27:51.889 --> 28:01.090
length as measured in the frame in which the
rod is moving is less than l0 okay well to
28:01.090 --> 28:07.279
be very exact it should be less than equal
to l0 you can even have v = 0 in some sense
28:07.279 --> 28:08.279
okay
28:08.279 --> 28:17.619
so if v =to 0 l = n0 okay so this shows that
the length of the rod it appears to be shortened
28:17.619 --> 28:28.749
in a frame relative to which the rod is moving
okay so this is length contraction okay okay
28:28.749 --> 28:38.110
but having said that let us just be familiarized
with another concept of the thing called proper
28:38.110 --> 28:49.049
length well you have just seen that if you
know if if this rod is moving with a certain
28:49.049 --> 28:55.840
velocity if it moves at different velocities
okay it is going to be the measured length
28:55.840 --> 28:57.129
is going to be different
28:57.129 --> 29:03.660
so if a rod has a length l in a frame relative
to which it is moving uniformly with a velocity
29:03.660 --> 29:09.640
v1 along its length this is important it is
moving along its length because there could
29:09.640 --> 29:13.649
be a case where it is not moving along its
length it is making some angle with the x
29:13.649 --> 29:22.669
axis we will try to see if you can do that
too and similarly let us say in another frame
29:22.669 --> 29:29.929
this rod is moving with a certain velocity
v2 and then the measured length is l2
29:29.929 --> 29:40.549
then you must have figured out by now that
l1 divided by root over of 1 - v1 square by
29:40.549 --> 29:52.409
c square = l2 divided by root over of 1 - v
square v 2 square by c square = l0 okay and
29:52.409 --> 29:58.570
what is l0 you well you can also write l0
in the form of the other things that you have
29:58.570 --> 30:08.450
written on the left l0 l0 root over by 1 – 0
square by c square so it is as if moving in
30:08.450 --> 30:17.590
a frame which is it is not which is that in
which the rod is at rest okay
30:17.590 --> 30:26.279
but what do you see y see that the quantity
l by root over of 1 - v square is invariant
30:26.279 --> 30:34.039
it is not changing you change it so if you
change if you change v the length changes
30:34.039 --> 30:40.669
but then the whole quantity of the length
divided by root over of 1 minus velocity square
30:40.669 --> 30:46.710
by the speed of light square that is becoming
invariant in all inertial frames okay and
30:46.710 --> 30:56.779
this is called the proper length okay so the
proper length of a rod is of an object is
30:56.779 --> 31:01.580
a length measured in a frame in which it is
at rest
31:01.580 --> 31:08.539
and obviously figured out that this is also
the largest of all possible length measurements
31:08.539 --> 31:21.450
of this rod okay having done this theory i
think it might be useful if we have a few
31:21.450 --> 31:25.279
examples
31:25.279 --> 31:37.320
so let us again have an observer a and and
let this person find the length of rod to
31:37.320 --> 31:48.899
be one meter and then and also to be moving
with velocity let us say 08 c okay now the
31:48.899 --> 31:55.869
same rod if it appears to have a velocity
06 the speed of light relative to another
31:55.869 --> 32:04.749
observer in another frame okay so see that
we have not specified the length here of this
32:04.749 --> 32:05.749
rod
32:05.749 --> 32:11.519
so the first question that we ask ourselves
is what is the proper length of the rod okay
32:11.519 --> 32:18.919
and secondly what will be the length measured
by the second observer okay delayed relative
32:18.919 --> 32:29.690
to which it is moving with velocity c 06 c
naught point cm so what is the proper length
32:29.690 --> 32:35.879
we just saw in our previous slide that it
is the length divided by the root over of
32:35.879 --> 32:39.869
1 minus the velocity square by the speed of
light square
32:39.869 --> 32:48.140
now you do that you do the length of the rod
was 1 meter so you have put one meter there
32:48.140 --> 32:59.269
and then for the velocity was point eight
so which was like 06 for c square takes square
32:59.269 --> 33:08.360
of that that is and then you see that this
proper length is l0 and 0 is 167 meters ok
33:08.360 --> 33:16.330
so this is the land that would be measured
in a frame in the rest frame of the rod ok
33:16.330 --> 33:21.960
so what would be the length then measured
by the other observer relative to which it
33:21.960 --> 33:25.169
is moving with a certain velocity okay
33:25.169 --> 33:34.190
that is simple the recall that as l subscript
b lb that is n0 0 in the sense that that is
33:34.190 --> 33:43.419
the proper length and l0 into over 1 - v square
by c square and then the v here is 6 okay
33:43.419 --> 33:53.269
6c that is so you find here it is 133 meters
that is all so so when it was moving with
33:53.269 --> 34:03.399
08 c the length was point 18 the length was
1 meters when it was when the when it is moving
34:03.399 --> 34:14.649
with with a lesser speed of 06 c the length
measured appears to be 133 meters okay and
34:14.649 --> 34:25.500
then so you see the higher the speed the shorter
is the length that you want to measure okay
34:25.500 --> 34:26.500
right
34:26.500 --> 34:33.899
so we come over to another interesting case
now in the previous cases we had this rod
34:33.899 --> 34:41.520
to be parallel to the x-axis okay but now
we are having distort let us see it is it
34:41.520 --> 34:45.700
is moving making a certain angle with the
x-axis okay
34:45.700 --> 34:51.620
so it is not moving along its length so there
is a component along its length along the
34:51.620 --> 34:59.910
x-axis and then there is a component along
the y-axis also okay so the y axis and also
34:59.910 --> 35:06.579
i mean the z axis is also there it is protruding
out of the screen again i have not drawn it
35:06.579 --> 35:16.099
so that the figure does not look too clumsy
okay now what we take this rod in the xy plane
35:16.099 --> 35:25.380
itself okay and then this rod it is in the
s prime frame where is it at rest
35:25.380 --> 35:33.819
and then it is making a certain angle theta0
and then this s prime s prime frame is moving
35:33.819 --> 35:39.559
with a uniform speed v with respect to the
s frame
35:39.559 --> 35:49.910
so the question that we asked ourselves is
that so what is the length of the rod as measured
35:49.910 --> 35:58.210
in the s frame and what angle does this rod
make with the x-axis in the s frame okay so
35:58.210 --> 36:04.390
if it is with a certain angle theta0 does
it is it theta0 again in the s frame where
36:04.390 --> 36:13.820
we were to find out okay so remember of course
that this s is moving with a certain velocity
36:13.820 --> 36:22.809
v relative to the common x x prime x axis
here i mean s prime is moving with a uniform
36:22.809 --> 36:26.640
velocity v okay
36:26.640 --> 36:37.079
so what we do in the s prime frame measure
what is the length of the rod along the x
36:37.079 --> 36:47.420
axis its x2 prime - x1 prime and you are going
to find this out to be l0 cos of theta 0 okay
36:47.420 --> 36:57.079
and then if l0 is the proper length of the
rod that is and what will be the component
36:57.079 --> 37:03.830
perpendicular to this direction of motion
so it is y2 prime - y1 prime that is simply
37:03.830 --> 37:13.839
l0 sine of theta 0 okay now when measured
in the s frame okay
37:13.839 --> 37:20.619
now remember in the s frame this rod is moving
with a certain velocity with a certain uniform
37:20.619 --> 37:33.260
velocity v with respect to the s frames the
end coordinates of this rod at time you are
37:33.260 --> 37:38.549
measuring this end coordinates of this rod
at a certain time t = t0 here so you find
37:38.549 --> 37:48.270
them to be x1 y1 and x2 y2 so what is this
difference x2 - x1 here well it is lcos theta
37:48.270 --> 37:56.869
okay so l is the length as measured in the
what that with the length of the rod in the
37:56.869 --> 38:04.500
in the s frame and then the theta angle is
the angle that will make with the x axis in
38:04.500 --> 38:05.869
the s frame okay
38:05.869 --> 38:15.569
well so x2 - x1 that will give you what the
along the length along the x axis that is
38:15.569 --> 38:25.190
l cos theta and then y2 - y1 that will be
l sine of theta okay so how are they related
38:25.190 --> 38:32.780
well we just figured out again that it is
well by the lorentz transformations of course
38:32.780 --> 38:41.270
so you use the transformations equations between
the prime coordinates and the unprimed coordinates
38:41.270 --> 38:48.790
and you find them to be well it is given on
your screen so the things you it is done at
38:48.790 --> 38:54.420
as the prime coordinates on the left hand
side and the unprimed coordinates on the right
38:54.420 --> 39:00.640
hand side okay and then the unprimed coordinates
in the s frame the measurement of the length
39:00.640 --> 39:03.480
is done at a certain instant okay
39:03.480 --> 39:11.549
so you definitely require to observe us with
synchronize clocks to do this and then synchronize
39:11.549 --> 39:19.099
and then measurement is done exactly at t
= t0 here and since it is the rod is moving
39:19.099 --> 39:34.720
parallel to the xx prime axis here my coordinates
are the same yeah so y1 prime = y1 prime is
39:34.720 --> 39:44.819
actually y1 and y2 prime is also equal to
y2 okay
39:44.819 --> 39:56.490
so now if you subtract x2 prime from x1 prime
and y2 prime you subtract from y1 prime of
39:56.490 --> 40:02.819
the other way round so what do you get you
get this you get x 2 prime - x1 prime that
40:02.819 --> 40:13.729
gives you x2 - x1 divided by root over of
1 - v square by c square okay and y2 prime
40:13.729 --> 40:21.980
- y1 prime is y2 - y1 so therefore you get
l cos theta which you know to be l0 cos theta
40:21.980 --> 40:30.510
0 which you know to be the length along the
x prime axis that is x2 prime - x1 prime is
40:30.510 --> 40:36.760
again equal to l cos theta divided by root
over of 1 - v square by c square
40:36.760 --> 40:45.980
it is very interesting of course and then
the in the direction perpendicular to the
40:45.980 --> 40:55.720
direction of motion that is in the y direction
l0 sine theta 0 is same as l sine theta 0
40:55.720 --> 41:04.270
okay so what is the length of this rod in
s frame okay so we come back to our old question
41:04.270 --> 41:13.760
and what angle does it make with the x axis
in the s frame okay so to do that all we have
41:13.760 --> 41:19.099
to do is divide those angles which we found
41:19.099 --> 41:25.860
and we found that we will find that it is
a tan theta 0 = tan theta root over 1 - v
41:25.860 --> 41:33.109
square by c square how do you do that that
is simple you have this relations l0 cos theta
41:33.109 --> 41:39.240
0 is l cos theta divided by root over 1 - v
square by c square and then this l 0 sine
41:39.240 --> 41:45.849
theta 0 is l sine theta so we divide the equation
with the sine with this equation with the
41:45.849 --> 41:49.990
cost and then you get this tan that is all
41:49.990 --> 41:55.859
so this gives you first the angle you know
so the angle is not the same the angle with
41:55.859 --> 42:04.980
which the angle this rod makes with the x-axis
in the s frame okay it is data and then the
42:04.980 --> 42:11.789
angle this rod makes with the s prime with
the x prime axis in the s prime frame that
42:11.789 --> 42:20.380
is theta 0 okay okay so just check that they
are not the same and then this is the relation
42:20.380 --> 42:22.770
you can use to find these angles
42:22.770 --> 42:32.020
what about the length remember l was the length
in the s frame so l square you know this very
42:32.020 --> 42:41.210
you so identity of l cos theta square + l
sine theta squared and we know these what
42:41.210 --> 42:49.520
is l cos theta and l sine theta in terms of
the proper length in the s prime coordinates
42:49.520 --> 42:57.789
in the coordinates of the s prime frame so
we get the thing everything in terms of the
42:57.789 --> 43:03.150
proper length in terms of all the quantities
in the primed frame okay
43:03.150 --> 43:11.069
so interestingly this comes this l that is
the length of the rod comes out to be l right
43:11.069 --> 43:17.849
so l so what is it so in the s frame its mesh
it is moving with a certain velocity uniform
43:17.849 --> 43:27.070
velocity v so that is and the length is measured
to be l and that l = l0 that is the proper
43:27.070 --> 43:33.609
length of this rod to root over of 1 - v square
by c square and now see that you have a cos
43:33.609 --> 43:40.700
it is cos square theta 0 okay so what is theta
0 that is the angle the rod makes with the
43:40.700 --> 43:50.460
x prime axis that is the frame in which it
is at rest okay ok
43:50.460 --> 43:56.589
so let us spend a couple of minutes with this
equation just to see whether it is right or
43:56.589 --> 44:05.260
not and whether remember when we were doing
some land contraction formula earlier
44:05.260 --> 44:12.150
it was l = l0 into root over 1 - v square
by c square and then the rod was moving along
44:12.150 --> 44:21.940
its length so if we put theta 0 = 0 so that
that simulates this the system or the rod
44:21.940 --> 44:28.710
to move along its length we get back our old
line contraction formula okay l = l0 into
44:28.710 --> 44:36.280
root over 1 - v square by c square ok but
what happens we put theta 0 = 0 is equal to
44:36.280 --> 44:39.740
pi by 2 okay that is perpendicular to its
direction
44:39.740 --> 44:51.700
well l = l0 then ok so if it is moving perpendicular
to its direction so the perpendicular component
44:51.700 --> 44:54.770
that is not the contracted
44:54.770 --> 45:00.120
so let us be more specific so the component
of the length parallel to the direction of
45:00.120 --> 45:07.770
motion is length contracted by a certain amount
dependent on the velocity ok that go and the
45:07.770 --> 45:13.329
quantity is root over of 1 - v square by c
square and the component perpendicular to
45:13.329 --> 45:26.490
the direction of motion is unaltered okay
right so that being our idea of llength contraction
45:26.490 --> 45:29.580
in the next segment in the next lecture
45:29.580 --> 45:37.140
we will talk of another consequence of special
relativity and that is called time dilatation
45:37.140 --> 45:38.099
till then goodbye