WEBVTT
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Hello, in our last lecture we had discussed
some of the issues of classical mechanics.
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We started with a concept of a frame of reference,
and we discussed what we mean by inertial
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frame, we said that when we talk of Newton’s
laws of motion, Newton’s laws of motion
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are equally valid in all inertial fields.
Therefore, one does not have to be a specific
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to a given frame while describing our motion,
so long the frame is inertial. Unfortunately,
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when we came to electromagnetic theory our
conclusions were somewhat different. We realized
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that electromagnetic theory it appeared as
if all the frames of reference or all the
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inertial frames of reference they are not
equivalent. There are some frames in which
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the velocity could be given by c is equal
to 1 upon under root of 7 naught 1 more this
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is what we discussed.
Because of 7 naught plus mu naught appeared
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to be fundamental constants, therefore we
thought it is coming only one frame of reference,
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in which this is speed can be given by this
particular expression, otherwise the fundamental
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constants themselves will become frame independent
which we did not like as an idea.
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So, we thought that probably there is one
preferred frame of reference, in which the
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expression of light is given by this standard
formula involving the fundamental constants
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c is equal to 1 upon under root epsilon naught
of mu naught. We are not very happy with the
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situation, because it appeared that in mechanics
all the inertial frames used to be equivalent,
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but electromagnetic theory they are not is
it so, that different branches of physics
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re replay different type of game probably
we do not expect.
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We expect that the nature would be uniform,
different parts of physics should tell us
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the same thing. So, if in mechanics all the
inertial frames turn out to be equivalent,
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we expected that the same thing should have
happened, but we had lots of problems with
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electromagnetic theory.
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So, this is what I am trying to recapitulate,
we had discussed that from mechanical point
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of view all inertial frames appear to be equivalent,
however electromagnetic theory seems to suggest
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that there is a preferred inertial frame of
reference in which the speed of light is given
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by the fundamental constants.
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We said that we must go back to experiment
to find out whatever we are saying is true.
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So, the issue which has to be tested is that
is there any special frame of reference in
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fact this inertial, this special frame of
reference, we decided to call by Ether or
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decided at that time to call it by Ether,
and we thought that so only in this particular
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frame of reference that the expression of
velocity of light is given by the c is equal
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to 1 upon under root epsilon of mu.
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So, all we have to do is to test now find
out an experimental mean by which we could
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test whether this particular hypothesis is
correct. It means that if you go to any other
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frame, which is different from Ether frame,
then the speed of light should be different,
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and normally should be given by this standard,
velocity addition formula which is very well
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known in classical mechanists. So, official
what I am trying to say, is that if I change
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my frame of reference from Ether even though,
the frame of reference is inertial the speed
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of light would be different and this will
follow the standard additional velocity, additional
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formula which is well known in the classical
bionics.
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What is the implications of the hypothesis?
Let us try to understand, what the Ether hypothesis
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implies let us, take a very simple example
and assume that we have an object and a beam
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of light in Ether medium. Let just try to
understand, whatever we are trying to say
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as far as, this hypotheses of Ether is concerned
see in this particular figure I have shown:
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one particular object which is here, and let
us imagine that I am talking of my Ether medium
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everything is given in this particular Ether
medium, this object I am calling as an object
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O, this v is this speed as measured by person
in the Ether medium, this c is of course,
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is the speed of light also given in the Ether
medium and because I am talking in terms of
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Ether medium therefore, the magnitude of this
velocity must be equal to one upon under root
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epsilon of naught remember, I repeat that
this particular picture is assumed to be in
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Ether medium.
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This velocity v is also in the Ether medium,
this c is also in Ether medium, and because
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this is in Ether medium. Therefore, the magnitude
of the velocity must be given by the fundamental
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constants. Now suppose there is an observer
sitting on this particular object O, what
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is the speed? that it will measure for this
particular light, as we have just observed
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that, this must be given by this standard
velocity addition formula which means that
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if an object is, if the observer is sitting
on this particular object O he or she would
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measure this speed of light, to be different
and that would be given by c minus v, this
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is a standard velocity addition velocity formula;
which I will write as v m the velocity as
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measured, velocity of light as measured in
object frame O, this am I have written just
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to imply that this is the velocity. I would
measure by sitting on this particular object,
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if my hypotheses of Ether is correct.
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Now, I want to make a claim under this particular
hypothesis, is that even if v is same of course,
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the magnitude of c is also same. If I change
the direction of speed of light, then both
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the magnitude and direction of this speed
measured in object frame, will turn out to
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be different. This is what I am trying to
show by very simple vector additional formula;
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let us look at this particular figure see,
this is the same v which I have drawn in the
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figure last transparency, and this is c which
is the speed of light, this is also c which
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is the speed of light, this is also in Ether
frame, this c is also Ether frame, all that
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has become different is that in this particular
first figure, this c is been shown to be directed
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in this particular direction; while in this
second figure c is been shown to be directed
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in this particular direction.
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The magnitude of this particular c magnitude
of this particular c is same, this v both
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magnitude and in direction are same. Again
I mention that, this particular object is
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means is moving with a velocity v, as seen
in the Ether frame and this c, that I am talking
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is also been measured in such frame; my question
is what will be the velocity of this speed?
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as would be measured by an object by an observer
sitting on co-frame or the object frame as
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we have just now said, this formula that we
have to use is v m is equal to c minus v,
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which is this particular formula, which I
have written it here also, this particular
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formula which will give us, what is the speed
of light as measured in these two cases; In
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the first case as, you can see that this will
be v every number, when I am saying v m is
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equal to c minus v; it means that v m plus
v is equal to c, and if you look at this particular
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figure, standard velocity additional formula
v m plus v is equal to c. If I walk in this
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particular direction, then walk in this particular
direction my mid displacement is here.
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This is the way, we remember normally remember
our velocity addition formula therefore, I
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expect here v m plus v is equal to c similarly,
here v m plus v is equal to c. So this v m
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is given by c minus v, as I can see that this
v m is pointed out at this particular direction.
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This v m is pointed out in this particular
direction and as also very clear, the magnitude
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of this v m is much smaller in comparison
to the magnitude of this v m. So this is what
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I am trying to explain saying that, if the
velocity of light is measured in different
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direction, I would find that not only the
direction; which is of direction of measurement
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but, it is magnitude will also become different,
so one conclusion, which I want to draw that,
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if my Ether hypotheses is correct, then if
I am in any other inertial frame, which is
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also in which I am taking the measurement.
I will find that in that frame these magnitude
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of this speed of light, would be direction
dependant depending upon, which direction
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I measure this speed of light, I will find
magnitude to be different. This is what I
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am trying to say, even if I am talking of
the same v, even if y object or inertial frame,
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is the same inertial frame.
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So, this is what I have reiterated in this
particular transparency; that is Ether hypotheses
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is correct, then in O frame one should measure
different values of speed of light, when the
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direction of light makes different angle,
with the direction of motion of the object
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in Ether, and the measures speed should be
consistent with the velocity addition formula,
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now let us imagine, that is object about which
I was just now talking is actually earth and
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for simplicity let us, not go to the too much
of complications as of now, assume that this
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v, this particular earth is moving with a
constant velocity, in this Ether medium which
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I am calling as v.
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We know, that this is not strictly been correct,
because the motion of earth around sun is
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fairly complicated, it also has a spin movement
but, let us ignore all those things. Let us
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imagine, that earth is really in inertial
frame, let us imagine that at this moment
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that it has only velocity in orbital velocity
v, which I assume at moment to be constant,
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because then only I can assume that the earth
is n inertial few. So let us imagine that
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this object is earth about, which I have just
now talked, now this earth let us try to shine
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light into different directions, as we have
just now said, I will find that the magnitude
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of this light meant to this speed of this
light, which are travelling in two directions
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would be different, let us make some specific
cases and try to understand, this particular
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thing you keep better, this is what I said
on earth. If we send two light signals; one
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along the velocity direction and another perpendicular
to it in earth’s frame, then we should travel
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with different speeds when measured on earth.
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So, what I am trying to say is that, let us
choose two directions; one along the direction
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of the motion of earth in the Ether medium,
so this is my earth which is moving in this
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particular direction with velocity assumed
to be constant. Then I sent one light signal
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along this particular direction, which is
the direction of v. I sent another signal
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which is perpendicular to this direction.
So one light signal goes like this, one goes
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like this; Remember these signals are being
emitted, with respect to an observer sitting
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on earth. So this piece of these two will
not be seen but, we what why have mentioned
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earlier would be v m and as, we have discussed
earlier the magnitude of this v m would be
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different, because they are making different
angles related to the velocity v, which is
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the velocity of earth in the eternity.
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Let us, try to make a calculation and try
to find out, what are what are the velocities
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in the perpendicular in the parallel and the
perpendicular direction? and next figure shows,
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the same thing this is assumed to be earth,
which is moving with the velocity v in this
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Ether medium, there is one light signal, which
is being sent perpendicular to the direction
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v, which I am writing as v perpendicular.
There is a light signal, which is being sent
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parallel to v, which I am writing as v parallel.
Remember the velocity of light c as per my
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both hypothesis is c only in Ether way, so
let us suppose this is my c, remember what
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I am measuring is, this let us look at perpendicular
direction; first what I am measuring is this
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particular thing but, this is a result of
two velocities components; one is velocity
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of light c, which I in the Ether medium and
one v, which is because of the fat, that this
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particular earth is moving with a velocity
v in Ether medium.
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Once, I go back to my old equation, which
I had written earlier, which is v m plus v
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is equal to c therefore, this v m in this
particular case would be, my v perpendicular
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so, I must get v perpendicular plus v is equal
to c, so what I am measuring is, this component
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v perpendicular which is the outcome of a
vector subtraction within c and v and let
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us come to the parallel of this. if I come
to this parallel case situation is exactly
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similar except that, this v parallel and v
both are in the same direction but, this v
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parallel must have been obtained, as a result
of subtraction of c m v by the same formula
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v m is equal to c minus v, which I have just
now written here now in this particular case
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my v m, now becomes v parallel and this v
parallel must be given by c minus v or c plus
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v must be given by two parallels. If we lift
here v parallel plus v must be equal to c,
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so this is the velocity component, that I
am measuring, which I have shown in this particular
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figure, by this v parallel vector.
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This v parallel plus this v vector must be
to c, which of course, is given by the fundamental
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constants, one upon under root epsilon of
mu, let us try to find out, what is v perpendicular?
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and what is v parallel? the one most vector
is easy to find it out all. I have done is
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reproduce these two figures, from my previous
transparencies or we realise, that v perpendicular
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must be equal to c minus v same expression,
which I have written earlier. If I look v
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perpendicular, this is v c and if I look at
the magnitude of this, we will realise that
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this makes an right angle triangle, this is
the hypot loose, so c square of magnitude
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must be divided by v perpendicular square
plus v square therefore, v perpendicular square
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must be equal to c square minus v square,
this is the standard theorem therefore, this
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is what I have written is the magnitude of
v perpendicular must be given by under root
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c square minus v square, where I have used
the notation that when I do not write any
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vector sign on the top of it, I mean only
the magnitude effect.
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So, magnitude of c is being represented by
just seeing without a vector sign similarly,
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the magnitude of v is being represented by
v without a vector sign, so this is just the
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magnitude c is also, just the magnitude in
magnitude of v perpendicular is given by under
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root c square minus v square by looking at
this particular triangle, which is a right
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angle triangle. Now let us see that, let v
parallel v parallel is too simpler, because
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v parallel and v both are in the action, so
we realise that v parallel is equal to c minus
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v by same formula. We have discussed earlier,
so v parallel means which would is just simple
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magnitude c minus, so this is what I have
written that v parallel magnitude is c minus
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as we realise and as have been mentioning,
that this v perpendicular and v parallel would
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be of course different in magnitude.
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Now, let us try to do one more thing, we have
just now seen, what would be the velocities
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of just what would be have evaluated, what
would be the velocities, if a light signal
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is sent on earth perpendicular to the v direction,
and parallel to the v direction, let us just
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reverse this direction. Let us say imagine,
that light signal is perpendicular to v but,
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not in the original direction but, to the
opposite direction similarly, for the v parallel
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we will assume, that is not really parallel
but, anti parallel to the velocity that exchange,
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let us evaluate these velocities situation
is exactly similar as before other than that
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I have revert reverted this particular vector,
so v this v perpendicular, which was earlier
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pointing towards up is now.
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Pointing towards bottom, this v parallel which
has pointing towards right hand side, is not
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pointing towards the left hand side, that
is why I have written here v perpendicular
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is the negative sign. Here so this, is v minus
perpendicular, this I have written is v minus
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back ground. Now my idea is to calculate,
this v perpendicular minus and v parallel
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minus, I will use exactly the same terminology
or the same method the one, which I have used
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to calculate v perpendicular and v parallel
all that has happened with the directions
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of the vector, such is again like before these
two figures have been copied from the earlier
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transparencies. Now you realise that in this
case hardly anything has changed, because
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this c is still the hypothesis of this right
angle triangle, this v min minus perpendicular
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is still v minus c changing, the sign has
not changed. This magnitude it is still given
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by magnitude of c square, or the magnitude
of this square is still given by the magnitude
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of c square minus magnitudes of v square by
the standard theorem.
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Therefore, we write this v minus perpendicular
also, as under root c square minus v square,
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which has not changed from my previous value.
However there has been lot of change in v
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minus perpend parallel, because as you can
see from the figure, itself earlier minus
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v by by v parallel was very small but, no
my v parallel has become very very large,
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because to this I must add v to get, my c.
If I go this much then I must revert back
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to get this c and because this magnitude of
c is same, and I have to actually subtract
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this v from the magnitude wise, then only
I will be able to get this c therefore, it
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is very clear that using this particular equation,
by v minus parallel. Now we will be c plus
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v, so what we find out is that this particular
velocity minus parallel direction has increased
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it has become c plus v, I repeat in v minus
perpendicular and v plus perpendicular, there
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was no change as per the velocity direction
is concerned, it means if I am sending a beam
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upwards like this and or in the reverse direction
towards backwards it makes no difference it
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will still give you the same speed of light.
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However, if the light travels in this particular
direction, this will travel with that different
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speed, then when light is travelling in opposite
directions, it makes a difference when light
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speed of light is being measured in the direction
of this, now let us discuss this famous Michelson
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Morley experiment, Michelson Morley was one
of the great experiments, which was defined
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to test the Ether hypothesis and this is a
very very remarkable experiment, because as
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we we shall be seeing it gave a negative result,
and this negative result itself was a pock
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making and changed the way e look at physics,
today this is a very interesting experiment
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using essentially, the same ideas that, we
have just now discussed. Let us look at the
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experimental set up in this transparency I
am show in rough sketch of this particular
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Michelson Morley experiment by s I mean e
have a source of light here.
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Let us assume that, this is a locomotive source
of light, you know how to create a locomotive
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source of light even at that time of Michelson?
it was known how to create a mono locomotive
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source of light? we have a mono locomotive
source of light as light is coming being emitted
23:31.540 --> 23:38.430
by this for the moment, let us assume that
this particular direction is the same direction,
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as the direction v that I have been talking,
which is the direction of velocity earth in
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Ether medium for simplicity. We can discuss
the other cases later now. Here we have what
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I have written m is a half silvered mirror
by half silvered mirror we mean, that it does
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not reflect all the light, it reflects some
of the light approximately 50 percent of the
24:06.560 --> 24:11.860
light and 50 percent of the light allows to
be transmitted.
24:11.860 --> 24:16.940
And if I put this particular mirror at an
angle of approximately 45 degrees from the
24:16.940 --> 24:23.940
direction of the velocity of light, then we
will find that approximately half of it is
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light will travel in this direction, which
was the original direction of the light while
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half of the light or half the intensity of
the light will travel in a direction perpendicular
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to the ocean direction of light, because the
angle of incidence has to be angle of reflection
24:39.760 --> 24:45.230
therefore, and I have taken this angle as
45 degrees or I have designed this particular
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angle as 45 degrees, therefore, this light
beam would, now travel at an angle of 90 degrees
24:51.640 --> 24:58.640
from my original direction. So this beam is
divided into two beams; one travelling upwards,
25:00.330 --> 25:06.070
one travelling towards right, and as I have
said I have assumed that this right direction
25:06.070 --> 25:12.690
is actually the direction of v this direction
therefore, is direction perpendicular to v.
25:12.690 --> 25:19.690
Now, here we have two mirrors; one is m 1,
another is m 2; these two mirrors are the
25:20.770 --> 25:25.650
normal mirrors, they are not half singled
mirrors so, they will reflect most of the
25:25.650 --> 25:30.510
light which is being vigilant on them and
let us assume, that these two mirrors are
25:30.510 --> 25:36.520
being put in such a way, that this light is
normal to them this incident is normally to
25:36.520 --> 25:42.500
them. If the light is incident normally to
them the light will be reflected back and
25:42.500 --> 25:49.500
therefore, this light will go up and will
come back similarly, this particular light
25:49.540 --> 25:54.900
upwards, the right hand side also go and will
get reflected from this particular mirror
25:54.900 --> 26:01.900
m 1 which is also a fully silvered mirror
and this mirror will reflect, the light the
26:02.030 --> 26:07.650
light will go retrace it is path backwards
and will keep upcoming backwards and again
26:07.650 --> 26:12.940
these two sources of light would meet, this
half silvered light because this, mirror is
26:12.940 --> 26:17.290
this half silvered mirror, because this mirror
is half silvered, again some of the lights
26:17.290 --> 26:21.290
of this particular beam will be reflected
in this particular direction, and some may
26:21.290 --> 26:25.830
be transmitted let us ignore that particular
beam, similar to the light which is coming
26:25.830 --> 26:28.920
back from this particular direction part of
the lights will be transmitted parts will
26:28.920 --> 26:33.590
be reflected, let us ignore the reflected
beam in this particular case let us, look
26:33.590 --> 26:40.590
only at this particular beam, which is now
resulting as the super imposition of two beams;
26:41.110 --> 26:47.870
one which is reflected from m 1 and again
reflected from n another one, which is reflected
26:47.870 --> 26:54.870
from m 2 and then refracted from n and now
this particular beam is being observed by
26:54.980 --> 26:59.120
a telescope.
This is what is standard Michelson Morley
26:59.120 --> 27:05.490
experiment, you know about the Michelson there
is Michelson is also devised an interferometer
27:05.490 --> 27:09.800
which is very very known as micro centerferometer
using essentially some or similar concepts
27:09.800 --> 27:16.390
in that sense, the ideas are where we is of
course, here we are testing something which
27:16.390 --> 27:22.740
is very different. Now let me try to calculate,
what will happen to these two beams of light;
27:22.740 --> 27:28.020
one which has gone upwards and got reflected
from m 2 and the one which has gone to the
27:28.020 --> 27:35.020
right and got reflected from m 1. We realise
that when the light as travelling upwards,
27:36.970 --> 27:42.970
if you travel the same speed, as we need for
travelling backwards and of course, both these
27:42.970 --> 27:49.690
speeds will be different from speed of light
in Ether as far as this particular light is
27:49.690 --> 27:55.380
concerned, because this is travelling along
the direction of beam. We realise that this
27:55.380 --> 27:59.960
velocity is going to be different from, this
particular velocity here is just travelling
27:59.960 --> 28:06.960
backwards, because this is in a direction
opposite to v what I will do is to calculate
28:07.960 --> 28:12.890
the time, that this particular light will
take this beam will take to travel from this
28:12.890 --> 28:19.740
distance to this distance and come back, and
the light which travels from this, here this
28:19.740 --> 28:23.840
distance and come back and assume that these
two lengths are same.
28:23.840 --> 28:29.220
Just for easiness let us assume, that that
two lengths are same, so the lengths are same
28:29.220 --> 28:34.270
only thing which is different are the speeds.
So we calculate how much time the light would
28:34.270 --> 28:41.270
take one it is reflected from original beam
to go upwards, and come back here and going
28:41.660 --> 28:46.650
this way and coming back here, because this
these are the two things which are different,
28:46.650 --> 28:50.850
otherwise everything is same just we have
come and joined this particular point, the
28:50.850 --> 28:56.130
two wheels are travelling together. Let us
do that particular calculation and find out
28:56.130 --> 28:59.260
the time distance.
28:59.260 --> 29:04.040
This picture shows what are the velocity directions?
What are the velocity magnitudes? this is
29:04.040 --> 29:08.950
just now, what I have mentioned that, when
it travels upwards it travels with a speed
29:08.950 --> 29:14.660
of under root c square minus v square, when
it travels backwards it travels exactly the
29:14.660 --> 29:20.940
same speed under root c square minus v square
on the other hand, this particular beam when
29:20.940 --> 29:25.620
it travels in this particular direction, it
travels with a single c minus v when it comes
29:25.620 --> 29:28.670
back it travels to the speed of c distance.
29:28.670 --> 29:35.450
But, one of the standard expressions the time
taken by a particular beam will be length
29:35.450 --> 29:41.150
l divided by this speed. Let us first calculate
the time taken for the light travelling in
29:41.150 --> 29:48.150
the parallel hour, the ark which I have shown
to be along the direction of v, which is the
29:48.410 --> 29:55.410
relating speed between the earth and the Ether
medium, as we have said the first beam is
29:56.780 --> 30:01.960
travelling to the right, it travels with the
speed of c minus v, so the time that this
30:01.960 --> 30:08.700
would have taken, if we assume that the length
is l then if you will take a time of l divided
30:08.700 --> 30:15.700
by c minus v, to go from this particular original
mirror let me, write here this was my mirror
30:19.960 --> 30:26.960
m, this was my other mirror fully mirrored
silvered mirror. I have first calculating
30:28.040 --> 30:35.040
the time, that it takes the light beam takes
to travel this way this length I have assumed
30:35.570 --> 30:42.570
it to be l, this is pth as I have said for
c minus v, so the time taken for this will
30:43.560 --> 30:50.560
be l divided by c minus v, then when the light
beam travels this way this speed becomes different
30:54.250 --> 30:57.820
the length is same this speed is c plus v.
30:57.820 --> 31:04.820
So, this time will be l divided by c plus
v, so the time taken for the light would travel
31:04.870 --> 31:11.110
from this point to this point is l divided
by c minus v, the time taken for the light
31:11.110 --> 31:17.410
you call from this point back of this point
is l divided by c plus v, so the total time
31:17.410 --> 31:23.770
taken by the light, to go from this point
to here and come back will be the addition
31:23.770 --> 31:30.120
of these two times, which is what I have mentioned
here l divided by c minus v plus l divided
31:30.120 --> 31:37.120
by c plus v and let us, simplify little bit
we just multiply both the sides by this side
31:38.870 --> 31:45.220
by c plus v numerator and denominator, this
side of also c minus v both numerator and
31:45.220 --> 31:50.780
denominator, then our denominator becomes
c square minus v square, the standard algebraic
31:50.780 --> 31:55.730
simplification they will be c plus v, here
there is c minus v, here look of this side
31:55.730 --> 32:00.650
I have multiplied and divided by c minus 1
v this side I have multiplied and divided
32:00.650 --> 32:05.740
by c plus v numerators and denominator. So
there is c plus v remaining here and there
32:05.740 --> 32:10.990
is a c minus v remaining here, as we can see
very easily this v and this v would cancel
32:10.990 --> 32:17.990
here numerator ought to be remaining is only
2 c divided by c square minus v square.
32:18.380 --> 32:25.380
This multiplied by l will be the time taken
for the light travel, along the parallel r
32:26.500 --> 32:32.310
this r i simplified, it further this si what
I have mentioned l is equal to 2 c divided
32:32.310 --> 32:38.840
by c square minus v square I have taken c
square common from this denominator, so this
32:38.840 --> 32:45.840
c square I have written out here this 2 c
I have taken out here so numerator here becomes
32:47.080 --> 32:53.190
two l multiplied by c but, this 2 c has been
taken out of this particular bracket, so what
32:53.190 --> 32:58.830
is remaining is only one here c square has
been taken out of this particular bracket,
32:58.830 --> 33:04.130
so here what remains is one and this case
divided by c square, so what is remaining
33:04.130 --> 33:11.130
inside this bracket is 1 divided by 1 minus
v square by c square, because there is a c
33:13.600 --> 33:19.140
here and there is a c square, so I have cancelled
one of the c and this expression given by
33:19.140 --> 33:26.140
2 l divided by c multiplied by 1 divided by
1 minus v square by c square.
33:26.720 --> 33:33.400
So, this is the time light would take to travel
along the parallel direction in going from
33:33.400 --> 33:40.030
half silvered mirror to the silvered mirror,
and coming back I simplify this equation a
33:40.030 --> 33:46.930
little further by assuming, that v is smaller
than c much smaller than c, which is I think
33:46.930 --> 33:52.000
correct this speed. We whatever we are talking
is much smaller than the speed of light, so
33:52.000 --> 33:56.380
under that particular approximately on this
particular expression can be further simplified,
33:56.380 --> 34:00.570
and because I have to compare this particular
result with something else so, that is why
34:00.570 --> 34:06.760
I am looking for this particular simplification.
So, what I have done this expression which
34:06.760 --> 34:13.760
I had written earlier 2 l divided by c 1 divided
by 1 minus v square by c square. I have written
34:17.720 --> 34:22.779
this particular expression as 2 l divided
by c, which is still remaining there and one
34:22.779 --> 34:27.990
minus v square by c square to the power of
minus 1.
34:27.990 --> 34:33.440
This is a very standard mathematical way of
writing it, why I write this as 2 l divided
34:33.440 --> 34:39.779
by c 1 minus v square by c square to the power
minus 1, the advantage of writing this, in
34:39.779 --> 34:45.249
this particular form is, that I can use a
binomial expression and expand this into a
34:45.249 --> 34:50.440
series and because this series will contain
higher orders of v square by c square, I can
34:50.440 --> 34:55.990
neglect higher orders and retain only the
first term and by binomial expression this
34:55.990 --> 35:02.990
minus 1 they’ll get multiplied by here therefore,
what will be remaining is 2 l by c 1 plus
35:06.799 --> 35:13.799
v square by c square, so this is an approximation
which we must realise, because this is only
35:14.180 --> 35:20.499
correct and the limit that v is not much smaller
in comparison to c. This is what I have written
35:20.499 --> 35:25.180
in this particular transparency; that this
particular time turns out to be 2 l divided
35:25.180 --> 35:32.180
by c multiplied by 1 plus v square by c square.
Now let us try to calculate the time in the
35:33.799 --> 35:40.799
perpendicular r, now we realised that this
speed and the perpendicular r in both the
35:41.569 --> 35:46.009
directions; whether upwards or downwards is
same, which is given under root c square minus
35:46.009 --> 35:52.490
v square therefore, time can also be obtained,
in a very simple fashion which is just 2 l
35:52.490 --> 35:59.339
divided by under root c square minus v square,
which is I have written the time in perpendicular
35:59.339 --> 35:59.779
direction.
35:59.779 --> 36:04.539
This two factor is, because the beam goes
up and comes back and of course, with the
36:04.539 --> 36:10.769
same speed therefore, this is two l under
root divided by under root c square minus
36:10.769 --> 36:16.700
v square, again I am trying to simplify, this
same time by using the same approximation,
36:16.700 --> 36:21.980
that v is much smaller than c therefore, i
have decided to write this under root in this
36:21.980 --> 36:27.369
particular form, I have taken 2 l by c out
and written this particular thing in bracket
36:27.369 --> 36:33.230
as 1 divided by under root 1 minus v square
by c square, which has taken this c square
36:33.230 --> 36:38.819
out of then, the root once I take c square
out of the root. Then this will become under
36:38.819 --> 36:43.599
root c square which is just c. So what I am
getting here is, just 2 l divided by c multiplied
36:43.599 --> 36:49.059
by c is very important. Ok remember here there
is under root, that is what is the difference
36:49.059 --> 36:52.999
for the earlier expression, now we know that
this particular under root can be written
36:52.999 --> 36:59.619
as to the power of minus 1 by 2. So this is
what I have written here, this is 2 l divided
36:59.619 --> 37:06.619
by c multiplied by 1 minus v square by c square
everything to the power of minus 1 by 2, because
37:07.009 --> 37:11.390
under root is nothing but, to the power of
1 by 2, because this is in the denominator.
37:11.390 --> 37:17.210
So there is a negative sign and this becomes
to the power minus 1 by 2. Now I will do the
37:17.210 --> 37:23.240
same trick, I expand this into binomial series
therefore, this particular factor end of course,
37:23.240 --> 37:29.349
neglect higher order terms just retaining
it, term involving v square then what is happening
37:29.349 --> 37:34.940
I must multiply this by minus 1 by 2 when
I multiply by minus 1 by 2 this minus becomes
37:34.940 --> 37:41.230
plus and this 1 by 2 becomes, now v square
by 2 c square which you remember in the earlier
37:41.230 --> 37:42.900
expression there was no 2 c.
37:42.900 --> 37:49.900
There was only v square by c square, here
we are having v square by v c square. We find
37:50.930 --> 37:56.230
that time perpendicular time taken by the
b perpendicular direction is different from
37:56.230 --> 38:03.230
the time taken in the parallel direction.
this is what I was trying to do, let us calculate
38:05.420 --> 38:08.289
the time difference the time difference, which
I have written here is e parallel minus t
38:08.289 --> 38:12.480
perpendicular of course, you will realise
that time taken in a parallel direction is
38:12.480 --> 38:17.769
more than in perpendicular direction, because
here we had only v square by c square while
38:17.769 --> 38:24.769
here we had v square by 2 c square because,
I have divide by 2, this expression has become
38:25.099 --> 38:29.519
smaller. Now I am taking the difference of
the 2. So I have just taken 2 l by c, which
38:29.519 --> 38:36.450
is a common factor out, and this t parallel
we had 1 plus v square by c square, In this
38:36.450 --> 38:43.190
t perpendicular we had 1 v square minus 2
c v square by 2 c square, because there is
38:43.190 --> 38:48.269
a negative sign everything becomes negative,
so this becomes minus 1 minus v square by
38:48.269 --> 38:49.140
2 c square.
38:49.140 --> 38:57.819
So, this sign 1 plus n minus 1 cancels out
,and you get 2 v square sorry, v square by
38:57.819 --> 39:02.289
2 c square, because there is a v square by
c square and minus v square by 2 square c
39:02.289 --> 39:09.689
square, so this will give me v square divided
by 2 c square, this two factor would cancel
39:09.690 --> 39:12.750
with this 2 and this is what I will be getting
39:13.260 --> 39:30.560
just write it, and explain here 2 l by c
minus 1 minus v square by 2 c square, this
39:30.569 --> 39:37.569
1 cancels out so, what I have will be having
is 2 l divided by c into v square by c square
39:39.940 --> 39:46.940
minus v square by 2 c square, this will give
me 2 l divided by c into v square by 2 c square,
39:52.420 --> 39:59.420
this 2 will cancel with this 2 and this is,
what I will get l divided by c multiplied
40:00.119 --> 40:06.839
by v square by c square. This is what I have
written the time difference between these
40:06.839 --> 40:11.440
2 l by 2 multiplied by v square by c square.
40:11.440 --> 40:18.440
Now, let us come to nex actual experimentation,
see in actual experimentation these angles
40:22.359 --> 40:28.029
that we are talking are not really especially
the mirrors m 1 and m 2, they are never perfectly
40:28.029 --> 40:34.349
normal to the beam, they are made very very
small marginal angle difference, so what happens
40:34.349 --> 40:40.440
the beam when it goes and comes back the different
parts of the beam are slightly deflected more
40:40.440 --> 40:44.430
or less and therefore, you find there is a
continuous path difference between, these
40:44.430 --> 40:50.859
two or this beam m 1 is reflected from m 1
and the beam which is reflected from m 2,
40:50.859 --> 40:55.349
because they take different time route travel
therefore, when they come and super impose
40:55.349 --> 41:01.859
on each other and are being viewed by the
mirror, you will find just because of these
41:01.859 --> 41:04.470
small angle you will find a fringes.
41:04.470 --> 41:09.880
Of the fringes, you will be observed even,
if there is no time difference obviously,
41:09.880 --> 41:14.170
the fringes now because of this particular
time, difference there will be a shift in
41:14.170 --> 41:20.519
the fringes, because now the time difference
will cause a small amount of shift in the
41:20.519 --> 41:27.049
ozone fringe pattern had there not been any
time difference, so what you see is a bright
41:27.049 --> 41:32.999
fringe and a dark fringe which means a bright
portion of light, where there is a light is
41:32.999 --> 41:37.450
bright and there is a darkness, this is what
is called standard fringes, so these fringes
41:37.450 --> 41:44.180
would be seen but, because of this additional
apathetic difference, the fringe pattern would
41:44.180 --> 41:50.559
shift, this is what I have written here, if
the mirrors m 1 and m 2 are slightly inclined
41:50.559 --> 41:56.640
they would create a gradual phase shift, and
one would observe a fringe pattern in the
41:56.640 --> 42:02.499
telescope t, the time difference in the arrival
of the waves in two arms would cause an additional
42:02.499 --> 42:05.049
phase difference, so there is already some
phase difference.
42:05.049 --> 42:12.049
And the time difference will be on the tonsick
but, basic problem is that if I want to really
42:13.519 --> 42:18.829
check by if my hypothesis is correct, then
I must be able to check and show that, there
42:18.829 --> 42:23.609
is a really this much of time difference between
these two beams, if I am able to check it,
42:23.609 --> 42:27.829
if I am able to verify it then I have proved
my hypothesis, if I have not been able to
42:27.829 --> 42:34.700
do it, I have not been able to prove my hypothesis,
so question is very simple but, there is a
42:34.700 --> 42:40.630
problem the problem is that, we have said
whenever even if there was no change in the
42:40.630 --> 42:45.140
the time taken by the light, if the two beams
would have taken the same light even if they
42:45.140 --> 42:49.910
even that there would have been a fringe,
now because of this time difference there
42:49.910 --> 42:54.869
has been an additional shift but, how do I
measure the shift, because it is not possible
42:54.869 --> 43:01.710
for me, to create a situation when these two
beams take the same time; because this beam
43:01.710 --> 43:08.710
this particular whole pattern has been devised
with taking the motion of earth into consideration.
43:09.420 --> 43:14.950
So, it is not really possible for me, to what
I call a point is 0 of this particular fringe
43:14.950 --> 43:19.880
saying that, this was this is where my original
fringe should have been and not, because of
43:19.880 --> 43:25.819
this it has shifted by this measure amount,
because it is not possible for me to create
43:25.819 --> 43:29.819
a situation or it is not possible for me to
find a situation when there is no time difference
43:29.819 --> 43:34.950
between the two beams so, time difference
has been created, because of the motion of
43:34.950 --> 43:41.239
earth in the Ether medium. So how do I find
0, so how do I find there has been a real
43:41.239 --> 43:47.140
shift, so this particular problem was solved
very nicely with a brilliant idea by Michelson
43:47.140 --> 43:54.140
Morley, what they said has been shown in this
particular transpires, they said let us try
43:55.999 --> 43:59.880
to rotate, the whole apparatus what will happen
if I wrote.
43:59.880 --> 44:06.269
Let us assume that we are rotating over this
particular axis normal to this particular
44:06.269 --> 44:13.210
picture, so this particular telescope has
not come here, sorry this particular source
44:13.210 --> 44:19.269
has come here. This particular telescope has
come here, this mirror has been rotated by
44:19.269 --> 44:25.809
90 degree, this particular mirror, which is
I was calling is m 1 or m 2 have now changed
44:25.809 --> 44:32.809
their position, there is one mirror here there
is one mirror here now remember, this was
44:33.880 --> 44:40.880
my original r in, which I was my light was
travelling now, this has gone here after rotation
44:43.009 --> 44:49.420
of 90 degree, so this r which was originally
parallel with the velocity direction has now
44:49.420 --> 44:55.210
become perpendicular to the light direction,
this r which was perpendicular because it
44:55.210 --> 44:58.900
has not been rotated by 90 degrees, this particular
mirror has come here.
44:58.900 --> 45:05.400
So, this particular arm which was earlier
perpendicular has, now become parallel so
45:05.400 --> 45:09.190
because of this particular rotation, an arm
which has originally perpendicular has not
45:09.190 --> 45:14.039
become parallel and the arm, which has parallel
has not become perpendicular. So whatever
45:14.039 --> 45:20.589
is happened the arm in, which the light was
taking earlier a larger time, now after 90
45:20.589 --> 45:26.690
degrees of rotation in that particular arm
time taken, is will be smaller similarly,
45:26.690 --> 45:31.769
in an arm the light time taken was smaller,
now will become larger now that is, what I
45:31.769 --> 45:36.059
am trying to say is correct, then by this
rotation I would be able to see that fringe
45:36.059 --> 45:43.059
is now moving and if I am about to see it
is motion, then I have verified my hypothesis
45:44.480 --> 45:48.960
this is what they did is trying to create
a rotation in this particular Michelson Morley
45:48.960 --> 45:53.329
experiment, to see this particular fringe
shift because they were not able to decide,
45:53.329 --> 45:58.950
they were not able to find out, how to check
the finite equations?
45:58.950 --> 46:04.849
So, what is aid in a rotated position time
difference on the same arm will be same but,
46:04.849 --> 46:08.549
now with a negative sign, because the arm
initiative was positive, now it has become
46:08.549 --> 46:14.339
negative it is taking longer time, so the
time difference magnitude wise will be same
46:14.339 --> 46:20.489
l by c multiplied by v square by c square
but, this time difference will be negative
46:20.489 --> 46:26.640
and if I rotate by 90 degree, then the total
rotation will be just the difference of these
46:26.640 --> 46:33.640
two, which because of negative sign as surely
implies an addition therefore, the total phase
46:33.759 --> 46:40.470
difference the total I am sorry, total time
taken would now be 2 l v square 2 l 2 l multiplied
46:40.470 --> 46:45.430
by v square and divided by c multiplied by
c square, this will be the time difference
46:45.430 --> 46:52.430
between the two arms, this would cause as
an emission is shift in the fringe pattern.
46:52.809 --> 46:58.670
So as I rotate if I keep on rotating, I will
see my fringe is getting shifted and if I
46:58.670 --> 47:05.670
am able to form this shift in the fringe pattern,
as surely I have proven my hypothesis.
47:05.849 --> 47:12.849
Now in order to find out, how much is the
fringe shift I must divide this particular
47:13.190 --> 47:20.190
time difference in comparison to must divide,
by the time period what I say per particular
47:20.509 --> 47:27.380
oscillation and this particular time period,
is inverse of the frequency, so you are aware
47:27.380 --> 47:34.380
this frequency, this speed is given by frequency
multiplied by lambda, which is the wavelength,
47:40.359 --> 47:47.359
this frequency I can write as 1 divided by
t multiplied by lambda therefore, this t would
47:50.430 --> 47:57.430
be given by lambda by c, this is the time
period of one of the oscillation in a harmonic
47:58.239 --> 48:03.950
oscillate, that is what we normally call,
so this time difference must be compared with
48:03.950 --> 48:10.170
the time upon oscillation must be divided
by this, so this is what I have done delta
48:10.170 --> 48:17.170
p minus delta t must be divided by t and this
t would be given by c lambda, if I calculate
48:19.619 --> 48:25.489
this particular thing, this is get as 2 l
divided by lambda multiplied by v square by
48:25.489 --> 48:31.410
c square, so this will be the amount of the
fringe shift that, we would be observing as
48:31.410 --> 48:34.569
a result of this particular.
48:34.569 --> 48:40.769
Of course, this particular experiment has
been performed number of times but, in one
48:40.769 --> 48:46.829
of the original experiments these were the
standard values which were taken this l length
48:46.829 --> 48:52.239
sagginess 11 metres not very large the lambda
was taken approximately as 5.5 difference
48:52.239 --> 48:59.239
of minus 7 and the speed of earth in Ether
medium medium was taken as 10 to the power
49:00.900 --> 49:07.420
minus 4 c, so I give these numbers here I
will find that the fringe shift turns out
49:07.420 --> 49:12.720
to be equal to 0.4, I have just substituted
the numbers in the earlier expression, which
49:12.720 --> 49:16.799
I have written earlier to find that there
will be approximately, the 1/2 of the fringe
49:16.799 --> 49:23.799
shift Michelson Morley had done an error analysis,
and we will very show that, it is a different
49:24.210 --> 49:31.059
shift was really there was existing they will
be able to observe this particular fringe
49:31.059 --> 49:38.059
shift.
However, we never observed in fact this particular
49:38.789 --> 49:44.680
experiment was repeated number of times in
different seasons, because you are never sure,
49:44.680 --> 49:51.130
what is the direction of Ether? Whether earth
is how eth how the earth is floating Ether
49:51.130 --> 49:58.099
different seasons different positions has
never found to be true never found out a differential,
49:58.099 --> 50:05.099
as I mentioned that this experiment has been
repeated by current years, noti would not
50:05.130 --> 50:09.359
say very recent years but, in this particular
century this particular experiment has been
50:09.359 --> 50:16.359
reputed by better precisions, and everytime
we found that, the result is negative. There
50:16.890 --> 50:23.009
was no fringe shift, that was observed appears
there is something really wrong with the Ether
50:23.009 --> 50:28.670
hypothesis, this is what I mentioned that
experiments are performed in various seasons
50:28.670 --> 50:31.140
but, never gave a positive results.
50:31.140 --> 50:37.640
Then there are lot of attempts to explain,
why we are getting negative results of Michelson
50:37.640 --> 50:44.640
Morley? and none of these explanations they
are really found to be very strong would never
50:48.589 --> 50:52.759
believe, that Michelson Morley experiment
could be understood to the failure of Michelson
50:52.759 --> 50:59.670
Morley experiment could be understood by these
explanations, I would like to mention that,
50:59.670 --> 51:04.309
other than Michelson Morley experiment, which
is basically an experiment based on light
51:04.309 --> 51:10.700
there had been other attempts also one of
the standard attempts to look, because if
51:10.700 --> 51:16.749
you remember in our last lecture, we had mentioned
that the basic starting problem was v cross
51:16.749 --> 51:21.249
b, when we are talking about the Lorentz force
and we said that, what is this v that we are
51:21.249 --> 51:26.900
talking at that time we said that probably
this v that would mean is also, that speed
51:26.900 --> 51:32.089
of that particular object that particular
charge in Ether medium ok.
51:32.089 --> 51:38.769
So, if earth is actually eth moving in Ether
medium then, if I put a particular charge
51:38.769 --> 51:45.680
then it would have a v and therefore, in a
given velocity it should experience a force
51:45.680 --> 51:51.579
of v cross v, so there were some attempts
which were made in order to assume a force
51:51.579 --> 51:58.579
on a given charge, all these attempts relate
to a fringing, that there is no Ether, there
52:01.789 --> 52:05.680
is nothing like Ether medium.
52:05.680 --> 52:11.140
So, this is what is the summary of whatever
we have discussed today. Experimental attempts
52:11.140 --> 52:16.979
to look for motion of earth in Ether, we are
not giving successful results, it did not
52:16.979 --> 52:21.529
appear that we could assign an absolute velocity
to an object.
52:21.529 --> 52:22.319
Thank you. .