WEBVTT
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Hello, in our last lecture we discussed the
concept of photon. We said that special theory
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of relativity gives us a possibility that
a particular particle may have 0 rest mass
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and may still carry energy and momentum. In
classical mechanics, we can never imagine
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a particle with 0 rest mass, but special theory
of relativity gives us that possibility. And
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we discussed saying that the light can also
be imagined as consisting of particles which
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have 0 rest mass, and they are called photon.
Of course, we have a condition that if the
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particle has a 0 rest mass, it must travel
with speed of light, and we know that light
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travels with speed of light. So, this is consistent
in saying that light consist of particles
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which has photons.
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So, this is what we have said we discussed
how light can be thought of consisting of
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particles known as photons. We also discussed
compton effect in our last lecture, which
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was one of the experimental proof. In which
we treated photon like just any particle and
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assumed that this gets collided or get scattered
with an electron, and treated just this as
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like viewed in classical mechanics discussed
any other collision process or scattering
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process, except for the fact that we utilized
lit-eristic expressions.
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Then we found that the frequency or the energy
of the photon after scattering would get changed
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and we could evaluate that depending upon
the angle of scattering what will be the changed
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wavelength or changed frequency or changed
energy, then we described experimental way
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of determining this change and we found that
the two are consistent. In that sense compt
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on effect is a very interesting experiment,
because it provides us a proof shows a proof
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that light can be really thought of consisting
of particles, which have 0 rest mass which
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carry momentum and energy like any other particle.
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Then we worked out an example, where we saw
that energy and momentum of photon can be
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transformed using standard energy momentum
transformation. See as far as relativity is
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concerned there is nothing special about photon
except for the fact that it has a 0 rest mass.
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There can be particles with different masses
different rest masses similarly, photon is
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also a particle with 0 rest mass therefore,
it does not require any other special treatment
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other than the fact that m naught is equal
to 0. So, all the transformation expressions
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that we have derived earlier should also be
valid for the case of photon and we dealt
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with an example in our last lecture, where
we really saw that we can apply exactly the
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same type of energy momentum transformation
that we can apply to any other particle we
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can also apply to the case of protons. Now,
let us go ahead and discuss what we call as
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a Doppler effect?
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If we change a frame of reference we know
that the speed of light will not change that
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is what we have always said, this is one of
the postulates of relativity but, we have
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just now said that is energy and momentum
will get transformed it means they would change
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if the frame of reference changes, and what
energy and momentum is related to in the case
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of photon is by this expression e is equal
to h nu energy is equal to h nu and momentum
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p is equal to h nu by c; it means, change
of energy, change of momentum, implies a change
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of frequency; it means, if i change my frame
of reference then this is equivalent of saying
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that the frequency of the photon has changed
though this speed has not changed remember.
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So, its frequency has changed, and this change
of frequency once we go from one frame to
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another frame is called Doppler effect. So,
that is what we have written if the energy
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and momentum of the photon changes upon change
of frame this would imply a change in frequency
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in the other frame, this is called Doppler
effect.
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Doppler effect is very well-known in sound
effect, the first example that we have always
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been giving of Doppler effect is a sound,
which many times we see in our daily life
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that a train is coming towards us we find
as if the frequency of these is viseling,
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the frequency of the visel from the train
is appears to be of high pitch higher frequency
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was the train we find that this frequency
is of lower frequency, this a well-known Doppler
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effect and very well explained understood
by the wave theory in the classical mechanics
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case. So, this is similar to that but, they
are certain different ways of treating it,
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because of the fact that I m going to describe
next because of the relativistic effect the
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light has to be treated somewhat differently
from the sound Doppler effect.
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Let us just recapitulate the sound Doppler
effect then we come to the actual expression
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of the light Doppler effect. In sound Doppler
effect we get two different expressions let
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us take simple case, one case in which observer
is stationary and the source is moving, another
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case in which source is stationary and observer
is moving. These two cases in the case of
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sound are treated differently, because of
the main reason that sound requires a medium
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to travel. So, whenever I am talking of the
speed of let us say source, which is emitting
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sound or an observer, which is hearing that
sound the motion or the speed of velocity
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of these source and the observer are always
defined with respect to the medium, and these
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two expressions these two cases when observer
is moving and the source is at rest and the
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source is moving and observer is at rest Corresponds
to different cases will explain this in a
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very simple fashion.
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Suppose, the observer this I m sorry the source
is at rest and this is the entire air medium
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in which is emitting sound waves then all
these sound waves will move out this spherical
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wave front from this particular source. Now,
these wave fronts these are sort of circles
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and I have not drawn very well. So, these
are sort of circles.
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Now these spherical wave fronts move in the
air medium because sound requires this medium
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to travel; sound is essentially mechanical
wave. In which this medium which requires
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medium this vibrations are carry forward in
this particular air. If you do not have this
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air or any other medium we will not have will
not hear the sound. So, this particular medium,
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which I am assuming air these waves are all
generated these wave fronts are created in
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this particular medium. If a person is walking
away or walking towards this or walking away
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from this then the rate at which its sort
of listens at which it comes across these
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wave fronts is going to be different from
the original cases. Now, if you have a different
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situation when this particular emitter of
sound or the source of sound itself is travelling
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but, the person is stationary we will find
that these particular wave fronts they themselves
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are created in different fashion.
Because initially when the source was here
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it created away from something like this and
when it moves towards this particular person
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next time the wave fronts will be centered
according to this person. Here, remember with
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respect to the medium, this person has travel
from here to here. So, now the wave front
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with the circular with particular point in
the center. Sorry, my figure is not very good
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but, essentially it means the wave the disturbance
is created inside the medium that itself is
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different. So, the two situations are different
and the expression of change of frequency
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that we get corresponding to these two cases
that also turn out to be different.
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So, this is what I have written the expression
of changed frequency as for Doppler effect
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sound is concerned depends on whether the
source or the detector which is free.
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The difference is caused because sound needs
a medium to travel the speed or source and
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or detector are defined with respect to the
medium.
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The expression of changed sound frequency
when source is fixed and observer is going
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away with speed v is given by this, means
if we have source which is fixed and this
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observer, which is
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Going away from here like this or observer
is going like this. If nu naught is the frequency
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which is being emitted by this particular
source write nu naught is the frequency which
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is emitted by this particular source the frequency
which will be heard by these two persons person’s,
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which are moving opposite direction their
distance is increasing will be given by this
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expression nu d s is the changed frequency
which I am calling as Doppler shifted.
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Frequency is equal to nu naught which is the
original frequency, which is being emitted
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by the source multiplied by 1 minus v upon
c, where v is the speed of the observer the
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speed with which its going away with the number
speed is related to the medium. So, that is
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what I have said, the expression of changed
sound frequency
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When source is fixed and observer is going
away with a speed v is given by this is the
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new frequency this the frequency which will
be heard by the person. On the other hand
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if the source is self is moving which i have
said is fundamentally different from the earlier
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case then I will get slight different expression,
and this expression will be given by nu d
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s, which is Doppler shifted frequency is equal
to nu naught, which is the original frequency
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divided by multiplied by 1 divided by 1 plus
v by c, where v is the speed with which the
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source is going away from the observer. So,
this is I have written the expression of changed
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sound frequency when the observer is fixed
and the source is going away with speed v
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is given by this. Now, if you are you can
probably realize that the case of light has
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to be different, because as we have said of
course,, whenever olden days we have been
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talking about ether medium that light requires
an ether medium to travel but, special theory
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of relativity discarded that concept, we do
not require a medium light to travel.
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Therefore, we cannot create a situation when
the light source is coming there is no medium
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in which these wave fronts are created therefore,
whether the observer is moving towards source
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or source is moving towards observer these
two things in relativity have to be treated
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exactly identically cannot be any difference
between these two, because these two speeds
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all of we can talk is relative speeds between
the two there is no medium in which the light
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is travelling unlike the sound wave. Therefore,
I must get the same expression in both these
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cases when we are talking of Doppler shift
of Light.
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This is what I have written for light the
two situations are fundamentally similar as
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it does not require a medium to travel. Hence
we expect a single expression to represent
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both the situations. The two situations cannot
be different I can hope you can understand
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the difference between a light and a sound
requiring a medium to travel therefore, I
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can define the velocity of an observer and
a source relative to that medium, while light
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do not does not require a medium to travel.
So, if I talk of velocity and I can only talk
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of relative velocity between observer and
the source, I cannot talk with respect to
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anything else because the there is no ether.
So, let us first consider a case of what we
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call as the longitudinal Doppler effect? This
is the picture, which gives what is called
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a longitudinal Doppler effect?
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Let us first spend little bit of time understand
this we get. So, this figure is essentially.
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Simple, we have just to fix our idea we can
imagine that this source this particular person
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as a frame is ground frame of reference and
assuming than the ground is an inertial frame
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of reference just to fix our idea its always
simple to imagine things, and there is a train
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or a garage which is moving towards right
with a speed or with velocity v. This particular
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light or this particular sorry this particular
garage or this particular train emits light
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in backward direction.
See remember the figure the way we have gone
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in case the light has to reach this particular
observer s which is sitting on the ground
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it has to be emitted only this way if it is
emitted this way this will not be seen by
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the observer as here. So, I am imagining this
particular case that the motion is collinear
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the observer s sitting on the ground these
another observer s prime, which is standing
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on the train which through some older somewhere
shines a light in this particular direction,.
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So, that this particular light is seen by
this particular observer. Now, when I say
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source frequency essentially it means frequency
which is seen in the frame of reference of
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the source. Here, everything is contained
in this particular garage train garage, which
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I am calling as s prime frame of reference.
So, I can say that this frequency nu prime
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is the frequency of light as being measured
by observer s prime so, its frequency which
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is measured s prime frame of reference or
another words this h nu prime is the energy
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of the photon as measured in s prime frame
of reference.
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Now, my question is what will be the frequency
which will be seen of the light? By this particular
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observer as and this particular changed frequency
will be what we will call as Doppler shifted
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frequency. So, all I have done in the same
problem of Doppler effect I have changed the
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language into the relative language or the
language which we have been using of different
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frames. Now, if I have to find out what is
the frequency of this particular light in
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this particular frame has essentially I have
all to find out what will be the energy of
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this particular photon in s frame of reference?
All I need is a energy momentum transformation.
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Once i do the transformation of energy I will
know what is the energy of the photon as measured
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in s frame of reference remember we have said,
photon is like any other particle. So, when
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we can apply momentum conservation then I
go from s prime to s to s prime for any other
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particle I can equally apply also for the
photon. So, my problem is very straight forward
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the Doppler effect in light can be treated
extremely simple; simply all I have to is
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to apply an energy transformation.
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So, that is what I am doing in the next transparency.
So, I have applied energy transformation,
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and if you remember the energy transformation
I am applying a inverse transition this energy
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e in s frame will be given by gamma multiplied
by e prime, where energy e prime is the energy
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s prime frame of reference plus v, which is
a relative velocity between the frames and
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p x prime is the momentum of the particle
in s prime frame of reference. Now, we have
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just now said, that the energy of the photon
as seen in s prime frame of reference h nu
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prime therefore, I can write e prime is equal
to h nu prime. Only thing I would like you
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to be attention to the fact that this particular
photon has to be emitted in minus x direction.
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So, it reaches the observer s just look at
the earlier picture.
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This photon has to be this is the direction
of the relative motions so, this is my plus
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x direction this photon is being emitted in
minus x direction, So, s to reach s therefore,
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this momentum must have a negative sign must
be minus x direction.
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Therefore, I am writing p x prime as minus
h nu prime by c this what I have done I am
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writing p x prime is equal to minus h nu prime
by c, because if this momentum was this particular
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photon was travelling in a plus x direction
then it will not be seen by that because we
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can create various types of situation the
final expression will always turn out to the
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same this situation which I have created is
the one in which this garage is moving forward
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and the photon is moving emitted backwards.
So, that it can be seen by an observer s.
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So, I write p x prime is equal to minus h
nu prime by c and the energy as seen by an
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observer s will be termed as h nu.
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So, e is equal to h nu. In this particular
expression I substitute e is equal to h nu
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and e prime is equal to h nu prime. So, this
was my original expression e is equal to gamma
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e prime plus v p x prime for this e I will
write h nu, which is the changed frequency
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this remains as gamma.
This e prime I will write as h nu prime this
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v is plus x direction but, this p x prime
is minus x direction. So, I will put a negative
19:57.899 --> 20:04.899
sign, p prime p x prime will be given by h
nu prime by c multiplied by v. Therefore,
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if I cancel h I will get nu is equal to gamma
nu prime minus nu prime v by c. So, this is
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what i am writing in the next transparency.
See I have just
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Substituted h nu is equal to h nu prime minus
v h nu prime by c and this gamma have expanded
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written as under root 1 minus v square by
c square, cancelling these h we get an expression
20:57.059 --> 21:03.470
nu is equal to nu prime, because I have taken
this out. So, this will become one minus v
21:03.470 --> 21:10.470
by c divided by under root one minus v square
by c square. So, this is what I can expression
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remember nu prime was the frequency s prime
frame of reference, nu is the frequency in
21:17.820 --> 21:24.820
the s frame of reference. Now, this under
root 1 minus v square by c square I can write
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as under root 1 minus v by c under root 1
minus one plus v by c 1 plus v by c.
21:52.500 --> 21:59.500
You just factorize a square minus b square
is equal to a minus a plus b. Now, in the
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numerator we had one minus v upon c. So, this
will cancel with this root of that will cancel
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with a root of 1 minus v by c giving with
this particular expression which I have written
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here, nu prime nu is equal to nu prime under
root one minus v by c upon under root one
22:17.870 --> 22:23.220
plus v by c. Please remember this root of
this has been cancelled with under root one
22:23.220 --> 22:28.830
minus v by c under root one plus v by c still
remains here and on the numerator there is
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an under root sign remains here. So, the new
expression that I have brought is nu is equal
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to nu prime under root one minus v by c plus
divided by one plus v by c. Now, what I would
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like to just mention that in Doppler effect
if we try to use this slightly different type
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of notation. So, let us not get confuse with
that particular thing and see remember in
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relativity language
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We called nu prime as frequency or h nu prime
as energy measured in s prime frame of reference
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and nu or as a frequency or h nu as energy
measured in s frame of reference but, in the
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language of Doppler effect the frequency,
which was emitted by s prime we call that
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as original frequency, which we call as a
nu naught which is frequency of the source
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and the
Frequency which we have seen can any other
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frame we call that frequency as the Doppler
shifted frequency. So, I am just changing
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this particular thing; I am changing nu to
nu d s and I am changing nu prime to nu naught.
23:35.169 --> 23:40.980
So, if I do that this what I call as a Doppler
effect notation, I get same expression just
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changing this particular thing nu d s is equal
to nu naught multiplied by 1 minus v upon
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c under root 1 minus v square by c square
same expression which I wrote earlier, which
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is equal to nu naught under root 1 minus v
by c divided by 1 plus v by c. This is the
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original frequency this is the Doppler shifted
frequency same thing has been re-nomen clatured
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depending upon the way we want to talk. In
Doppler effect, we talk of original frequency
24:08.639 --> 24:15.639
and the shifted frequency while in relativity
we talk of s frequency in s frame that frequency
24:15.840 --> 24:22.840
in s prime frame of reference. Now, this particular
expression can be simplified little bit, if
24:23.179 --> 24:29.720
we assume that the relative velocity between
the frames is comparatively small in comparison
24:29.720 --> 24:34.600
to c as we given some examples: today we find
this particular Doppler effect in light to
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be quite useful in many of the experimental
techniques that we use. So, and in many of
24:41.539 --> 24:48.169
those cases the relative velocity between
the frames is not that large,. So, we just
24:48.169 --> 24:53.500
approximately this particular expression under
the condition that v upon c is very small
24:53.500 --> 25:00.500
in comparison to one. So, that is what I am
trying to show in the next transparency. See
25:01.210 --> 25:08.210
remember my expression was one minus v by
c under root this was in the numerator..
25:13.909 --> 25:20.909
So, this can be written as 1 minus v by c
to the power of half. In denominator we get
25:25.919 --> 25:32.919
an expression of 1 plus v by c this I bring
to numerator this expression or let me this
25:37.669 --> 25:44.669
one divided by this can be written as one
plus v by c to the power minus half, because
25:47.380 --> 25:54.380
this was in denominator this already under
root. So, once I take numerator the power
25:54.440 --> 26:01.299
becomes negative. So, this I can write as
1 plus v by c to the power minus half. So,
26:01.299 --> 26:08.100
when I multiplied these two at multiply this
by this expression I will get 1 minus v by
26:08.100 --> 26:15.100
c to the power half multiplied by plus v by
c into the power minus half this minus sign
26:18.220 --> 26:25.139
has come call them as denominator. Now, because
we are assuming that v by c is very small
26:25.139 --> 26:30.700
in comparatively v by c is very small in comparison
to 1, I can expand this into binomial expression
26:30.700 --> 26:36.000
and neglect higher order terms retain only
the first two terms and will get first to
26:36.000 --> 26:42.669
first approximation that in the limit of lower
relatives velocity between the frames what
26:42.669 --> 26:44.360
will be the change into frequency.
26:44.360 --> 26:49.889
So, this what I have written this particular
transparency. So, we nu d s is written as
26:49.889 --> 26:56.889
nu naught is equal to 1 minus v by c to the
power half multiplied by 1 plus v by c to
26:58.080 --> 27:05.080
the power minus half, which is approximately
by expanding this. So, this becomes 1 minus
27:05.480 --> 27:12.360
let us say n times n times by c n is half,
which is the power. So, this becomes one minus
27:12.360 --> 27:16.259
v by two c.
Similarly, here is a plus and the power is
27:16.259 --> 27:22.039
minus half. So, this becomes one minus v by
two c. So, this minus half gets multiplied
27:22.039 --> 27:28.620
here for the first term 1 plus n x 1 plus
if you have to expand one plus x to the power
27:28.620 --> 27:35.620
n this, because one plus n x to the first
term. So, this I cannot multiply neglects
27:37.509 --> 27:43.440
second order terms. So, one multiplied by
one will give me just 1, which is this 1 when
27:43.440 --> 27:50.440
this minus v by t c gets two c gets multiplied
by this one I will get minus v upon c similarly,
27:51.360 --> 27:56.379
this minus v upon two c, where it gets multiplied
by this one I will get another minus v by
27:56.379 --> 28:03.379
two c when I add these two I will get one
minus v by c but, just can simply work out
28:04.110 --> 28:09.090
this is very simple way just open opening
up multiplying this brackets we of course,
28:09.090 --> 28:13.110
will have term v square upon 4 c square which
I will neglect because this is second order
28:13.110 --> 28:19.269
term in v upon c.
So, I have find that Doppler effect effect
28:19.269 --> 28:25.679
frequency is given by approximately nu naught
multiplied by 1 minus v by c which happens
28:25.679 --> 28:30.009
to be the same expression which we have derived
I will not derive but, for the case of sound
28:30.009 --> 28:36.100
when we use the expression for the case when
your source was fixed and train was moving.
28:36.100 --> 28:43.100
Now, there is another type of Doppler effect
which does not happen in sound, which is called
28:43.759 --> 28:47.870
transfer Doppler which is purely relative.
28:47.870 --> 28:53.750
See remember in longitudinal Doppler effect
we assume the source is here and train is
28:53.750 --> 28:59.049
moving just in the same line or look at the
situation which is given in this particular
28:59.049 --> 29:04.539
field. Let us assume that this particular
person s is standing here and this train is
29:04.539 --> 29:11.529
train tracking in front of him and this train
is going like that, and this particular train
29:11.529 --> 29:18.529
is emitting sound in a direction perpendicular
remember when I say perpendicular let us be
29:18.889 --> 29:23.320
careful when I say perpendicular its such
that as for this particular observer s is
29:23.320 --> 29:24.960
concerned to him.
29:24.960 --> 29:29.750
It appears that this particular its really
sounds its light I am sorry that this light
29:29.750 --> 29:35.909
is travelling towards him. So, essentially
it is a situation something like this that
29:35.909 --> 29:42.909
this particular train is moving like this
let us just represent it by one single point
29:44.139 --> 29:51.139
here and this observer s is here and this
observer s prime is here, according to s when
29:55.100 --> 30:00.720
this particular observer was just perpendicular
to this particular direction of motion light
30:00.720 --> 30:07.190
is emitted towards same as seen by observer
as see remember according to s prime this
30:07.190 --> 30:12.960
particular light will not really move perpendicular
to this motion but, this particular observer
30:12.960 --> 30:18.769
s receives light towards him in a direction
perpendicular to the direction of the motion
30:18.769 --> 30:25.129
of the train remember observer s is observing
the motion of train in this particular direction
30:25.129 --> 30:32.129
and same observer as finds that light is travelling
in a direction perpendicular to the motion
30:33.809 --> 30:36.580
or perpendicular to the velocity direction
and it reaches the same.
30:36.580 --> 30:43.580
So, this somewhat the same situation, which
is easy now we can find out using again energy
30:44.039 --> 30:49.740
momentum transformation that if this is the
frequency for this h nu prime is the energy
30:49.740 --> 30:55.240
of the photon which is emitted by this s prime
observer what will be the frequency of this
30:55.240 --> 31:02.240
particular photon as will be seen by the observer
s this is what we call as a transfers Doppler
31:03.100 --> 31:09.350
effect in sound we do not see such type of
transfers Doppler effect is purely relativistic.
31:09.350 --> 31:16.350
Phenomenon and let us work it out. We use
a direct energy transformation, which means
31:17.690 --> 31:24.120
e prime is equal to gamma e minus v times
p x, because this is a direct transformation
31:24.120 --> 31:30.129
therefore, there is a negative sign there
is no positive sign remember I have used direct
31:30.129 --> 31:34.700
transformation because I know that p x has
to be 0, because remember x is the direction
31:34.700 --> 31:41.700
of relative velocity and according to an observer
in s frame this particular photon is moving
31:41.960 --> 31:47.559
purely in y or minus y direction perpendicular
to x direction therefore, p x is 0 let me
31:47.559 --> 31:52.309
come back to this particular figure which
I have drawn here in this particular paper.
31:52.309 --> 31:58.470
Here there is a s which is this motion is
in this particular direction. So, this is
31:58.470 --> 32:05.429
by x direction now according to this observer
s this photon is travelling in this particular
32:05.429 --> 32:11.409
direction. Therefore, according to observer
s the x component of the momentum of the photon
32:11.409 --> 32:16.240
is 0. Now, in s prime we faced through a photon
towards I am using the word through if it
32:16.240 --> 32:22.190
has to emit a photon in this particular direction
this particular photon because this is moving
32:22.190 --> 32:25.769
relative to s would not be exactly in a direction
perpendicular to it.
32:25.769 --> 32:32.730
So, let us not get confused it s not p x prime
which is 0 but, its p x which is 0. So, e
32:32.730 --> 32:39.730
prime is equal to gamma e minus v p x and
with p x is equal to 0 of course,, using exactly
32:40.820 --> 32:47.799
the same way e prime is equal to h nu prime
and e is equal to h nu. So, I put for e prime
32:47.799 --> 32:54.799
h nu prime and for e is equal to h nu p x
is 0. So, I just get nu prime is equal to
32:55.950 --> 33:02.950
gamma nu again I change these two by relativistic
notation my Doppler effect notation this nu
33:03.470 --> 33:09.259
prime was original frequency nu naught this
nu is Doppler shifted frequency.
33:09.259 --> 33:16.259
Because this is being observed by observer
s I will call this as nu d s this I will call
33:17.230 --> 33:24.230
as nu naught this what I get the expression
nu d s is equal to nu naught under root 1
33:24.990 --> 33:31.990
minus v square by c square. So, this is as
I said, a purely relativistic effect called
33:35.090 --> 33:41.990
Transfers Doppler effect. If you commence
if you look at the expression that we have
33:41.990 --> 33:48.990
obtained for the case of light for example,
for longitudinal Doppler’s effect we can
33:50.039 --> 33:57.039
very easily see that, if the light source
is moving towards the observer its frequency
33:59.120 --> 34:05.049
will go up if it is moving away its frequency
will go down remember the expression that
34:05.049 --> 34:12.049
we had found out was nu d s is equal to nu
naught 1 minus v by c.
34:17.980 --> 34:24.980
We have shown this particular expression approximately
we put it approximately. Now, clearly if v
34:32.300 --> 34:39.119
is positive it means remember we have taken
the motion to be in minus x direction means
34:39.119 --> 34:46.119
if the source is moving away I am sorry if
the source is moving away from the observer.
34:48.780 --> 34:55.159
Then this particular case one minus v by c
will be smaller than one and therefore, nu
34:55.159 --> 35:01.760
d s will be smaller than nu naught therefore,
if the train is going away from the observer
35:01.760 --> 35:06.849
we find the Doppler shifted frequency to be
smaller. If it is moving towards him then
35:06.849 --> 35:13.140
this v has to be changed sign this will become
one plus v by c and in that case nu d s will
35:13.140 --> 35:19.410
be larger. So, this is in that sense similar
what we have we generally see in the case
35:19.410 --> 35:23.200
of sound the expression is also very similar
to the sound. So, there is nothing very surprising
35:23.200 --> 35:28.859
about it. Only thing which I want to tell
you that as far as the transfers Doppler effect
35:28.859 --> 35:34.440
is concerned in this particular case the change
frequency will always turn out to be smaller.
35:34.440 --> 35:41.200
Now, let us do some comments for the Doppler
effect. In the case of longitudinal Doppler
35:41.200 --> 35:48.200
effect what we found that in the limit of
v less than c the expression is similar to
35:49.210 --> 35:56.010
the sound Doppler effect,. So, it is quite
clear like the normal sound Doppler effect
35:56.010 --> 36:01.630
if the source is moving towards the observer
the frequency of the light will appear to
36:01.630 --> 36:07.260
be increased.
Similarly, if the source is moving away from
36:07.260 --> 36:12.599
the observer the frequency of the sound will
be appearing to be less. So, this is what
36:12.599 --> 36:19.599
have written as the expression nu d s is approximately
equal to n naught multiplied by 1 minus v
36:19.720 --> 36:25.970
by c, which is as we have set expression what
has been similar to observer in the case of
36:25.970 --> 36:32.150
sound for of course,, this specific case and
therefore, we expect essentially the result
36:32.150 --> 36:38.700
be identical that when a v is a positive nu
d s is smaller than nu naught when v is negative
36:38.700 --> 36:43.780
it means the source is moving towards the
observer then in this particular case this
36:43.780 --> 36:49.329
nu naught gets multiplied by factor larger
than one and nu d s turns out to be larger
36:49.329 --> 36:56.329
than nu naught. So, light we see exactly the
similar effect but, in the case of transfers
36:56.390 --> 37:02.270
Doppler effect will always see a reduced frequency
the observed frequency is always lower than
37:02.270 --> 37:03.170
the source frequency.
37:03.170 --> 37:09.150
This is what I have written here in l d e
means longitudinal Doppler effect the observed
37:09.150 --> 37:16.150
frequency is more if the source is moving
towards observer and lower when the source
37:17.119 --> 37:20.200
is moving away. In transfers Doppler effect
the observed frequency is lower in the source
37:20.200 --> 37:27.200
frequency as we have said it has no classical
analogous. I would like to just mention here
37:28.040 --> 37:35.040
is that the transfers Doppler effect can also
be derived or can also be thought to be as
37:35.180 --> 37:38.890
a cause of time valuation.
37:38.890 --> 37:45.890
Let us see how, if this is a carrier this
draw a carrier or just draw a point source
37:50.010 --> 37:57.010
to be more specific and this point source
is moving in particular direction. Now, this
37:57.309 --> 38:01.900
particular source is emitting light we can
always think or we know that light is electro-magnetic
38:01.900 --> 38:06.960
wave. So, when we say in a wave there is a
maximum there is a minimum no there is a maximum
38:06.960 --> 38:13.069
amp we know displacement we call is a in the
mechanical wave here in terms of electro-magnetic
38:13.069 --> 38:15.099
wave this will be a point, where we have magnetic
field, which is highest or electric field
38:15.099 --> 38:20.800
which is highest or whatever you want to talk
about it when there is a maximum and there
38:20.800 --> 38:23.569
is a minimum.
So, displacement is maximum and minimum that
38:23.569 --> 38:30.569
is what the wave is light. Now, difference
between two consequent maximum one after the
38:31.849 --> 38:38.849
other. So, this is a time difference between
these two is not we call it as time period
38:38.930 --> 38:41.300
in a classical wave. Now, if the wave is being
emitted its being emitted here so, it is goes
38:41.300 --> 38:46.619
like this spherical wave fronts goes like
this there is another way we emitted,. So,
38:46.619 --> 38:53.619
again its spherical wave front goes away from
this. Now, both these the time difference
38:53.859 --> 39:00.349
between emission of the first maximum and
the second maximum that time difference both
39:00.349 --> 39:05.510
are being measured by this frame of reference
when I am measuring the frequency of time
39:05.510 --> 39:10.630
period in s prime frame of reference, and
because these are being measured in s frame
39:10.630 --> 39:16.869
itself because that is the source which is
emitting there the light therefore, when its
39:16.869 --> 39:22.380
emitting first maximum the second maximum
the time difference between these two is proper
39:22.380 --> 39:28.099
time difference, because both these time the
the time difference between these two is being
39:28.099 --> 39:34.859
measured only in s prime frame of reference.
So, therefore, the t prime or the time period
39:34.859 --> 39:39.970
in s prime frame of reference is a proper
frame of reference therefore, if is a proper
39:39.970 --> 39:45.720
time interval sorry therefore, an observer
of aground will find the time interval between
39:45.720 --> 39:52.020
these two events emission of first maximum
second maximum will be dilated, because in
39:52.020 --> 39:57.569
that frame of reference when the first emission
was there it started from here the second
39:57.569 --> 40:02.420
emission according to s prime occurred when
this particular source moved away. So, first
40:02.420 --> 40:06.010
maximum I was emitted from one particular
point this second maximum was emitted from
40:06.010 --> 40:11.300
a different point,. So, these two events emission
of first maximum emission of second event
40:11.300 --> 40:16.890
maximum according to s did not occur at the
same point by that time the source has been
40:16.890 --> 40:22.280
away but, in the first frame of reference
both of them occurred at the same place therefore,
40:22.280 --> 40:29.280
time interval between these two emissions
of maximum is proper in s prime frame of reference.
40:29.650 --> 40:35.500
Therefore, in s frame of reference this will
appear to be dilated, this what I have written
40:35.500 --> 40:38.440
consider transfers Doppler effect,
40:38.440 --> 40:45.440
The two consequent consecutive maximum minimum
in the field of electro-magnetic wave are
40:45.530 --> 40:50.569
produced at same place in s prime. If the
time interval between them, which we call
40:50.569 --> 40:56.420
as time period its proper its prime frame
of reference it is to be s prime frame of
40:56.420 --> 41:03.420
reference and therefore, in s frame it may
be dilated. So, this is what is a picture
41:04.329 --> 41:06.799
I had said its coming from here.
41:06.799 --> 41:10.940
So, when first maximum was emitted and the
second maximum was emitted by this particular
41:10.940 --> 41:15.569
person according to s prime he was at the
same position but, according to s when first
41:15.569 --> 41:21.780
was emitted was here and second was emitted
are they the man has moved away therefore,
41:21.780 --> 41:26.740
these two events did not occur in the same
place therefore, in s frame the time interval
41:26.740 --> 41:33.740
will be dilated. So, using time dilation we
can write that time interval or time period
41:35.329 --> 41:42.329
in between these two emissions in s frame
will be equal to gamma time t prime and as
41:42.410 --> 41:47.349
we know by wave theory this time period is
inversely proportional to one divided by frequency.
41:47.349 --> 41:49.930
So, this t I can write as one upon frequency,
41:49.930 --> 41:55.720
This t prime I can divide it by as 1 minus
1 divided by nu prime, because this is a frequency
41:55.720 --> 41:59.790
in s prime frame of reference its t prime
it is is also an s prime frame of reference,
41:59.790 --> 42:05.680
which gives me exactly same as nu prime is
equal to gamma nu exactly the same expression,
42:05.680 --> 42:12.230
which we have earlier found out by using energy
transformations. So, what I am trying to say
42:12.230 --> 42:19.230
is that transfers Doppler effect can be thought
as arising due to time dilation phenomena
42:21.000 --> 42:25.710
and because there is classically no time dilation
therefore, classically we do not have any
42:25.710 --> 42:26.200
transfers.
42:26.200 --> 42:32.329
Doppler effect, let us say one or two examples
and try to at the end of the lecture we try
42:32.329 --> 42:39.329
to give some idea how the Doppler effect can
be used let us first take very simple example,
42:39.690 --> 42:46.690
the question is what is the required speed
relative to an observer? If a source of gamma
42:46.980 --> 42:53.980
ray of energy 14.4 k e v we say k e v means
k volts one thousand volts; it means, 14.4
42:55.200 --> 42:59.910
multiplied by thousand electron volts if its
energy is to be increased by ten to power
42:59.910 --> 43:05.390
minus six electron volts we just like to we
would appreciate the numbers.
43:05.390 --> 43:11.420
This is 14.4 k e v 14.4 multiplied by thousand
electron volt and this is 10 to the power
43:11.420 --> 43:16.410
minus 6 electron volt with these small change
you know line or you know many, because this
43:16.410 --> 43:21.680
order of thousand electron volt and this is
10 to the power minus 6 electron volt. So,
43:21.680 --> 43:28.680
I want to change
sending or emitting electro-magnetic waves
gamma rays 14.4 k e v, and I want to change
43:36.069 --> 43:43.069
simple very very slightly of the order lines
of order minus line you know of that order
43:43.980 --> 43:50.980
I want to shift its energy by 10 to the power
minus 6 electron volt, then I want to apply
43:53.470 --> 43:58.799
I give it to Doppler shift, what is the speed
data I would require? So, that if we can see
43:58.799 --> 44:05.780
we increased by 10 to the power minus 6 very
small very tiny amount,. So, as we have seen
44:05.780 --> 44:11.950
the frequency has to be increased; it means
this source must move towards the observer.
44:11.950 --> 44:18.049
Therefore this particular source must come
towards you person who is observing this frequency.
44:18.049 --> 44:21.059
So, this you can see as am sitting here i
want to observe it increased by ten to the
44:21.059 --> 44:25.369
power minus six electron volt therefore, the
source which is emitting this particular gamma
44:25.369 --> 44:31.450
rays must come towards me, and as we were
seen that this speed is definitely not very
44:31.450 --> 44:37.359
small in comparison to c,. So, I will use
the expression that we just now derived for
44:37.359 --> 44:41.260
the case of v much smaller than c.
44:41.260 --> 44:45.260
So, this what I have written because the energy
has to be increased the source should move
44:45.260 --> 44:52.260
towards it should not be absorber should be
observer then I use that particular expression
44:53.380 --> 45:00.119
h nu I have multiplied by h nu d s must be
equal to h u naught into one plus v by c.
45:00.119 --> 45:03.880
Here I have used plus expression, because
I already increased the frequency therefore,
45:03.880 --> 45:10.740
v has to be taken as minus v. So, sign has
to be changed alright put the expression this
45:10.740 --> 45:16.720
h nu naught I take on the left hand side this
becomes nu prime d s minus h nu naught h u
45:16.720 --> 45:23.410
equal to h nu naught multiplied by v by c.
This change I want 10 to the power minus 6
45:23.410 --> 45:29.790
electron volt. So, I will put 10 to the power
minus 6 here this I know its 14.4 k e v v
45:29.790 --> 45:35.589
by c I can calculate this speed, which is
turning out to be 2.08 centimeter per second.
45:35.589 --> 45:39.540
These are very reasonable speed you know very
easily calculating manipulated in the laboratory
45:39.540 --> 45:46.540
can be observed in the laboratory therefore,
Doppler effect provides you a facility by
45:46.589 --> 45:52.359
which you can shift the energy of sources
by very very small amount and therefore, as
45:52.359 --> 45:57.410
we have seen just now it can be used for some
very great studies which are generally not
45:57.410 --> 46:03.609
possible by any other mis to resolve that
type of small what used to call in spectroscopy
46:03.609 --> 46:08.740
hyperfine fields structure. Here the energy
levels are different very very by small amount
46:08.740 --> 46:15.740
that can be done by you know by Doppler shifting
the source the the gamma rays source in source
46:16.119 --> 46:21.700
case the gamma rays and 14.4 k e v by the
way is a very standard source, which people
46:21.700 --> 46:28.700
use very often when cobalt 57 decays to iron
57 emits a radiation of 14.4 kilo electron
46:32.880 --> 46:36.859
volt.
Now, these are certain implications and that
46:36.859 --> 46:43.280
implications I would like to discuss today
what is emission and absorption? Let us as
46:43.280 --> 46:47.549
go back to atomic physics and let us go to
very very simple thing we do know want to
46:47.549 --> 46:54.549
talk it is not a force on atomic molecular
physics we do not want to talk too much in
46:57.450 --> 47:00.099
details about the spectroscopy. Let us just
assume that we are doing something like a
47:00.099 --> 47:01.510
hydrogen atom or something like that, which
we know very well when there is a transition
47:01.510 --> 47:06.490
takes place its emits a photon of course,,
this concept of transition is track is much
47:06.490 --> 47:11.609
later but, they were let us let us try to
see the implication of this particular concept
47:11.609 --> 47:17.450
of the photon and also the Doppler effect
as far as these are concerned. Suppose, there
47:17.450 --> 47:24.450
is a transition then as a result of transition
the photon is emitted we know that this the
47:26.010 --> 47:27.790
way photon is emitted in atomic theory.
47:27.790 --> 47:33.040
Now, because we know that photon has a momentum
if we strictly speaking whenever we are writing
47:33.040 --> 47:39.359
an expression for finding out the energy of
the photon as a result of certain transition,
47:39.359 --> 47:46.359
this momentum must also be conserved. Which
essentially means that if we have let us say
47:48.940 --> 47:55.940
an atom which undergoes a transition and emits
a photon and let us assume that this particular
47:57.559 --> 48:03.579
atom was at rest in a given frame of reference
this this photon has a momentum, which is
48:03.579 --> 48:10.579
given by h nu by c, because initial momentum
was 0 therefore, final momentum also has to
48:10.589 --> 48:17.569
be 0; it means, this atom must required back
in this atom requires back then this will
48:17.569 --> 48:22.460
also gains some energy some kinetic energy
and if it gains some kinetic energy that kinetic
48:22.460 --> 48:26.690
energy also has to be supplied by the transition
energy, because once the transition has taken
48:26.690 --> 48:33.690
place that energy has to be partly utilized
for creating the recall energy and partly
48:33.799 --> 48:38.490
for giving energy to the photon; it means,
the energy of the photon will turn out to
48:38.490 --> 48:42.390
be slightly smaller than the transition energy
available to it.
48:42.390 --> 48:46.799
This what I have written a consequence of
photon having a momentum is that emission
48:46.799 --> 48:51.930
and absorption also obey momentum of conservation
this is one as to be realized.
48:51.930 --> 48:58.930
Let us consider a case of emission of a photon
and let us assume that all require speed of
49:00.780 --> 49:05.010
non-relativistic let us not consider a very
high energy transitions at the moment therefore,
49:05.010 --> 49:09.760
we can use partly the classical expressions
of energy momentum kinetic energy and momentum
49:09.760 --> 49:14.109
relationship. So, let us suppose there is
transition which takes place from n street
49:14.109 --> 49:19.349
to m street energy corresponding to this particular
level was e n corresponding to this level
49:19.349 --> 49:26.349
was e m,. So, this is a total energy available
to me as a result of transition. This energy
49:26.630 --> 49:33.630
we now used for two in two ways it will give
energy to the photon and also it will give
49:36.329 --> 49:42.720
recoil energy to the atom we of course,, I
must write e n minus e m is equal to h nu
49:42.720 --> 49:49.250
plus k r where k r is the kinetic energy of
recoil, and I must get a second equation which
49:49.250 --> 49:55.130
is the momentum conservation, which is h nu
by c is equal to momentum of recoil.
49:55.130 --> 49:59.630
Of course, this momentum and kinetic energy
in the classical case are related by very
49:59.630 --> 50:05.940
simple expression k is equal to v square by
two m and let us assume that m n is the mass
50:05.940 --> 50:11.970
of atom, which I am calling mass of nucleus
or mass of atom whatever you want to call
50:11.970 --> 50:16.859
it it depends on what generally nucleus most
heavy that is why I have written m n. So,
50:16.859 --> 50:23.030
p r can be written as under root two m n into
k r. Now, I can substitute this k r by writing
50:23.030 --> 50:24.450
this p r k by two m.
50:24.450 --> 50:30.789
N and therefore, we can find out an expression
for the frequency of the emitted photon, which
50:30.789 --> 50:35.450
I have written. So, this is the expression,
which actually is a quadratic nation. So,
50:35.450 --> 50:40.539
there is h nu is here h nu square and in case
of you have to solve this particular quadratic
50:40.539 --> 50:45.920
equation to get correct expression of h nu
but, we must realize that this h nu will be
50:45.920 --> 50:52.920
slightly smaller than e n minus e m of course,,
I must also see that generally if we take
50:55.020 --> 50:59.470
for some hydrogen atom case the energy that
we are talking of the order the ground state
50:59.470 --> 51:05.089
energy of hydrogen atom is minus 13.6 electron
volt atom of ten electron volt, Well m c square
51:05.089 --> 51:12.089
of case of photon, which is nucleus one of
the lightest nucleus is of the order 940 m
51:12.390 --> 51:13.089
e v.
51:13.089 --> 51:19.410
So, 10 power 9 electron volt. So, this expression
is extremely negligible therefore, most of
51:19.410 --> 51:22.930
the time we neglect recoil specially, when
we are talking of some hydrogen atom but,
51:22.930 --> 51:28.680
we are talking of some nuclear transitions
in that particular the recoil can be significant
51:28.680 --> 51:33.789
and therefore, we may we must account for
this particular thing similarly, if we are
51:33.789 --> 51:40.789
talking of transition of photon if a photon
has to absorb it has to not only supply energy
51:41.000 --> 51:45.650
for the atom to go to higher energy state
but, this atom this photon when is absorbed
51:45.650 --> 51:48.960
the atom will start moving front because momentum
has to be cancelled, because initially there
51:48.960 --> 51:55.960
is a non-0 momentum once photon is absorbed
the same momentum must carry y an atom therefore,
51:57.569 --> 52:01.630
the expression will become h nu is equal to
e n minus e m, Plus k r exactly.
52:01.630 --> 52:07.650
In a same fashion h nu upon c will be equal
to p r and this particular equation will get
52:07.650 --> 52:12.960
e n minus e m is equal to h nu divided by
1 minus h nu upon two m n c by c square.
52:12.960 --> 52:18.780
So, c here there was a plus expression here
there was a smaller expression the e n minus
52:18.780 --> 52:25.780
e m will be slightly larger than e n minus
e m. Now, my question is that what we call
52:26.140 --> 52:32.700
as a resonant absorption is it possible that
we come across a situation when a photon is
52:32.700 --> 52:38.210
emitted as a transition a particular transition
and a photon is used for causing similar inverse
52:38.210 --> 52:43.619
transition in some other atom; it means, a
photon has been created when photon has gone
52:43.619 --> 52:48.930
from n x state to m x state now this photon
has come out now is being absorbed exactly
52:48.930 --> 52:54.339
the similar atom and you want this particular
atom to go from m state to n state. It possible
52:54.339 --> 52:59.309
we have just now said that in the first case
the energy of the photon will be slightly
52:59.309 --> 53:04.440
smaller than the transition energy in the
later case we require the energy to slightly
53:04.440 --> 53:09.869
larger than e n minus e m. See first quantum
mechanics also gives you that there is all
53:09.869 --> 53:14.549
the photons have a particular bandwidth you
cannot have a perfect monochromatic photon.
53:14.549 --> 53:19.329
So, it is. So, happens that if the line width
turns out to be larger than the difference
53:19.329 --> 53:25.010
between the between these recoil energies
of that order. Now, we can sort of resonant
53:25.010 --> 53:30.039
absorption can still take place but, in the
case of many nuclear transitions it is not
53:30.039 --> 53:35.130
possible to do that. So, for example, if this
is your emission line, which has this is e
53:35.130 --> 53:40.770
naught is e n minus e m and this is absorption
line they have finite land widths and this
53:40.770 --> 53:45.309
is, because of the recoil this is because
the differences because of the recoil and
53:45.309 --> 53:51.690
then this recoil energy this is very large
comparison to this land width its hardly a
53:51.690 --> 53:58.690
resonant absorption taking place, on that
is way comes to Doppler effect can the recoil
53:59.309 --> 53:59.450
energy.
53:59.450 --> 54:03.069
Whatever the recoil energy whatever is the
recoil energy can be compensated by providing
54:03.069 --> 54:05.520
a slight amount of Doppler shifting can I
increase the energy or can I decrease the
54:05.520 --> 54:12.520
energy of the photon by giving the sources
certain amount of energy because of the Doppler
54:14.780 --> 54:16.630
shift can be shifted.
54:16.630 --> 54:21.160
In 1950 moon was a scientist who carried out
some experiments and where he could demonstrate
54:21.160 --> 54:25.289
the resonant absorption by imparting the speeds
to the gamma ray of about seven into 10 to
54:25.289 --> 54:30.730
the power 4 centimeters per second remember.
We are talking of gamma ray, which are energies
54:30.730 --> 54:36.630
not that 13.6 electron we have not talk earlier
or found. So, the in this particular case
54:36.630 --> 54:41.770
it required energies out of the 7.10 to the
power 4 centimeters per second and the he
54:41.770 --> 54:46.819
could observe a resonant of absorption in
situation in which it was not possible.
54:46.819 --> 54:52.359
By this particular case resonant absorption
is interesting. See what happened in much
54:52.359 --> 54:57.930
later time most observed that he can really
certain situation he can create I mean that
54:57.930 --> 55:04.059
certain ways in which he can create situation
in which the total recoil is taken by the
55:04.059 --> 55:08.809
solid and total recoil can essentially be
neglect can be can be made essentially recoilless
55:08.809 --> 55:13.430
and when he observed this particular thing
he tries to call this particular thing we
55:13.430 --> 55:17.549
can see could see recoilless emission then
what is a why we are talking recoilless. So,
55:17.549 --> 55:24.549
much? The idea is that in that particular
case in the case of recoilless emission all
55:25.710 --> 55:31.240
these lines, which are which have high per
field structure in which you know find that
55:31.240 --> 55:35.900
the lined the energy differences are extremely
small they can always studied by slowing applying
55:35.900 --> 55:40.910
a Doppler shift giving energy of the order
millimeter per second or centimeter per second
55:40.910 --> 55:47.500
and therefore, can we can scan all those energy
level diagrams even though the energy sources
55:47.500 --> 55:53.150
have high energy and gap between these energy
level is extremely fine but, is still I can
55:53.150 --> 55:58.730
measure by shifting the energy slowly varying
the energy in photon by very small amount
55:58.730 --> 56:04.140
by giving it a speed and therefore, these
photons can get Doppler shifted.
56:04.140 --> 56:10.500
So, Mossbauer was awarded nineteen sixty one
noble prize in physics and to opened up possibilities,
56:10.500 --> 56:17.339
where one can see small changes in the energy
levels caused by external effect using Doppler
56:17.339 --> 56:23.289
effect. Lastly see speaking I will give quickly
an example which is very interesting is called
56:23.289 --> 56:26.650
laser cooling of a gas.
56:26.650 --> 56:32.569
Normally lasers or always thought for heating
something normally use laser to or cutting
56:32.569 --> 56:38.319
something or building something it is never
thought that a laser can also used to cool
56:38.319 --> 56:45.319
something but, it is interesting that laser
can be used to cool gases let us see how?
56:45.520 --> 56:52.000
See we realize that when we talk of a gas
the temperature of the gas is related to the
56:52.000 --> 56:58.450
root means square values of the speeds of
the molecules. If the root means square speeds
56:58.450 --> 57:03.630
is high the temperature is high if they are
smaller then the temperature is smaller. So,
57:03.630 --> 57:10.630
if by sum means I can reduce the r m speed
root minutes per square speed of the gas molecules
57:11.650 --> 57:18.260
the gas molecule. Now, let us see how Doppler
effect helps in particular case? Let us we
57:18.260 --> 57:21.859
see the particular situation take a very very
specific case.
57:21.859 --> 57:28.859
Let us imagine that this particular thing
is a source of laser its emitting from in
57:29.910 --> 57:36.910
this particular direction and there is a gas
molecule, which is moving towards it. If it
57:37.740 --> 57:44.740
moves towards it and if I adjust the such
a fashion that for a given transition the
57:45.309 --> 57:50.380
energy whatever is the energy required the
energy of the photon is slightly smaller than
57:50.380 --> 57:56.369
that so, energy the photons which are emitted
have a slightly smaller energy than the transition
57:56.369 --> 58:03.369
energy. Then what will happen than this particular
atom in its frame this particular photon is
58:03.730 --> 58:10.730
moving towards him this particular atom will
fine than the energy of this photon is slightly
58:11.630 --> 58:18.630
larger and if it becomes equal to the transition
energy this photon will get absorbed by this
58:19.069 --> 58:24.650
particular atom, and as a result of this absorption
the photon of this momentum of this photon
58:24.650 --> 58:30.289
will be transferred to this atom and this
photon is moving in this direction this atom
58:30.289 --> 58:37.289
will slow down its momentum will go down.
Now, if an atom was moving not in this particular
58:37.809 --> 58:39.200
direction.
In that particular direction in that particular
58:39.200 --> 58:46.200
case it will not get absorbed, because according
to this particular person in this particular
58:47.559 --> 58:52.779
frame the energy of the photon will not equal
to the transition energy which happens in
58:52.779 --> 58:59.220
this particular case therefore, it will selectively
absorb photons only which are coming towards
58:59.220 --> 59:02.809
it, and therefore, all the atoms which are
moving in this particular direction there
59:02.809 --> 59:09.809
is speeds are going to be slow down. Now,
I want speed of the atoms is go slow down
59:09.920 --> 59:15.710
not one direction but, in all the directions.
So, what I do I create once force of laser
59:15.710 --> 59:22.710
photons here and this generated by laser another
source here, third source here, fourth source
59:23.420 --> 59:29.900
here, fifth on the top sixth at the bottom
all moving towards gas. So, if a photon is
59:29.900 --> 59:35.770
moving in this particular ay gas atom is moving
this particular direction it is a speed will
59:35.770 --> 59:39.710
slow down by the absorption for this photon
if it is moving in this particular direction,
59:39.710 --> 59:42.690
this speed of this will reduce by absorption
on this photon.
59:42.690 --> 59:48.359
If it is moving in this particular direction
is speed of this particular thing will get
59:48.359 --> 59:53.339
reduce, because an absorption go down from
this particular direction and overall the
59:53.339 --> 59:58.579
speed of this can be reduce. So, let us just
quickly read whatever i have written temperature
59:58.579 --> 1:00:03.710
is related to the r m s speed of the molecules;
the speed can be decreased if a photon is
1:00:03.710 --> 1:00:09.950
absorbed moving in a direction opposite to
that of the gas atom.
1:00:09.950 --> 1:00:14.470
Let us take one dimension motion and take
the transition from e 1 to e 2 expressive
1:00:14.470 --> 1:00:21.470
transition choose a photon source with h nu
less than e 2 minus e 1 only those atoms moving
1:00:22.190 --> 1:00:28.359
towards the source can absorb the photon causing
the reduction in speed, this is what is called
1:00:28.359 --> 1:00:35.359
the system is called six laser is called optical
molasses create a situation in all six directions.
1:00:35.400 --> 1:00:40.069
I was also say that when the photon is absorbed
that also will emitted back this you call
1:00:40.069 --> 1:00:45.440
as a spontaneous emission but, this is spontaneous
is under mean all the direction therefore,
1:00:45.440 --> 1:00:50.309
it increase certain sort of temperature but,
we can show that overall the goes will go
1:00:50.309 --> 1:00:56.609
down Chu Cohen-Tannoudji and Philip got noble
prize in nineteen ninety seven for the laser
1:00:56.609 --> 1:00:58.289
cooling and related experiments.
1:00:58.289 --> 1:01:03.720
Now, I will summarize of whatever discussion
today’s lecture, we discussed longitudinal
1:01:03.720 --> 1:01:08.640
and transverse Doppler effect as applied to
the light and we gave some examples and its
1:01:08.640 --> 1:01:10.580
applications.
Thank you.