WEBVTT
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In our last lecture, we had applied the postulates
of special theory of relativity and eventually
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arrived at Lorentz transformation. We do not
say that we derived Lorentz transformation.
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We just say that, we arrived at it. We presented
a series of arguments on how Lorentz transformation
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could be arrived at. A real test that Lorentz
transformation is believed on is on experiment.
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If we find an experimental support, then only
we believe a Lorentz transformation. After
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that, we took an example, which was an old
example of a light being emitted from the
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origin.
We had earlier found out that, under Galilean
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transformation, the speed of the light does
not remain constant, if we change the frame
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of reference, which was obvious because the
classical mechanics was not designed for that.
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Then we apply Lorentz transformation on the
same example and said that if we apply Lorentz
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transformation, then the velocity of light
turns out to be same in the other frame also,
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which is what Lorentz transformation is based
on or the second postulate of special theory
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of relativity is based on. So, that is the
way we proceeded in our last lecture.
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Just to re-capitulate capitulate, let us write
or let us discuss the Lorentz transformation
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again. These are the equations governing Lorentz
transformation. These are the equations. We
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introduced two symbols; one is beta, which
is V divided by C. Just to remind, V is the
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relative velocity between the frames and C
of course is the speed of the light.
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Then, we defined another parameter called
gamma, which was 1 upon under root 1 minus
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beta square. If we use these two parameters,
the Lorentz transformation equations become
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like this. For X co-ordinate, it is x prime
is equal to gamma multiplied by X minus V
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t. Y prime remains equal to Y and Z prime
remains equal to Z and t prime becomes gamma
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multiplied by t minus V X divided by C square.
We had earlier said that in the non relativistic
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limit, meaning when V is much smaller than
C, this beta is negligible in comparison to
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1 and gamma is essentially close to 1. Then
X prime just becomes X minus V t, which is
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the standard Galilean transformation. These
two equations are common in a Galilean transformation
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Here, if V is very small in comparison to
C, this particular term is neglected or can
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be neglected. This gamma will tend to 1, then
t prime will be tending equal to t, which
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is again the standard classical Galilean transformation.
So, these equations effectively become important,
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only when V is close to C, where this factor
beta square cannot be neglected in comparison
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to 1.
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Today, we will try to discuss some of the
example, some other examples of Lorentz transformation.
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Specifically, two very very important and
popular results arrived out of Lorentz transformation.
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One of them is called length contraction.
The length contraction, as the name suggests,
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means that the length of an object may not
be constant, if an observer in a different
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frame observes the length of the same object.
In fact, as you will be seeing, there is a
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contraction or length becoming smaller is
associated with this phenomenon, which is
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called length contraction. But before we really
derive it, let us understand a few terms,
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because important to derive or to work with
length contraction formula correctly. It is
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important that we are familiar with these
things.
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The first term is the proper length. Proper
length as the name suggests or as we have
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defined of a particular rod is length measured
in a frame in which it is at rest. So, let
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us imagine that we have to measure the length
of this particular rod. This is a rod and
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the length of it you have to measure. Now,
this particular rod is stationary in my own
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frame of reference. So, I do not see any motion
of this particular rod because as I said,
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this is stationary in my frame of reference.
So, whatever is the length that I would measure
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for this particular rod would be called proper
length. If this particular rod was moving
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in my frame of reference, then the length
that I will measure of this particular rod
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would not be call proper length.
So, proper length has a specific meaning that
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it must be measured in a frame of reference,
in which this particular rod is at rest. That
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frame of reference, in which it is rest, is
in which it is at rest, that particular frame
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of reference is called proper frame of reference.
That is what we have written in this particular
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transparency. Proper length of a rod is length
measured in a frame, in which it is at rest
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and proper frame is a frame in which this
rod is at rest. Now, let us assume that this
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particular rod is along the x direction. Remember
this particular x direction is the direction
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of the relative velocity between the frames.
So, let us assume that this particular rod
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is along inclined or it is put along the x
direction or it is put exactly on the x axis
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to be more precise.
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So, this is what I have written. Let us assume
that the rod is placed along the x axis. Then
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we will show just now, that its proper length
is the largest and if we go to any other frame,
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in which there is a relative motion along
the X direction, then its length will turn
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out to be shorter. That is why it is called
length contraction.
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Now, let us suppose that, we have to measure
the length of this particular rod. How do
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I measure? The question is very simple. I
measure the co-ordinates of the two ends of
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the rod. If I measure the co-ordinates of
the two ends of the rod, actually by subtracting
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these co-ordinates, I can always find out
what is the length of this particular rod.
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So, this rod length is to be measured, I can
measure what is the co-ordinate here, I can
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measure what is the co-ordinate here. Then
take the difference of these two and this
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will give me the length of the rod. But the
question which I am posing is that, is time
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important. Let me explain what I mean to say.
Suppose, I measure this particular length,
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this particular end or co-ordinate of this
particular end of the rod, let us say at t
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is equal to 0. Now, this end, this particular
co-ordinate, can I measure at any other time
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is time? Is time important for its measurement?
See, the answer to the question is that, no,
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it is not important, as long as this particular
rod is at rest.
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So, I can measure this co-ordinate now, go
and have a cup of tea, come back and then
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measure the other co-ordinate. Still the difference
of these two co-ordinates will give me the
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length of the rod, because this rod is fixed
in my frame of reference. It is not moving.
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So, this rod is not going to change its co-ordinate
as a function of time. So, it is immaterial
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when I decide to measure the co-ordinate of
this particular end and when I decide to measure
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the co-ordinate of this particular end. But
suppose this particular rod was moving in
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my frame of reference, then the same statement
is not correct. Because if I make measurement
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of this particular end now and make the measurement
of the co-ordinate of this end little later,
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during this particular time this rod has been
moving. So, this rod is moving. I make a measurement
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now and by the time I go the other end make
the measurement, by that time the rod as been
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displaced from its position. So, the co-ordinate
has changed
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Therefore, if you still decide to take the
difference of the two co-ordinates, that will
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not give me the correct length of the rod.
This is what I have said. Measure the co-ordinates
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of the two ends of the rod. Is time important
for the measurement? Yes, only if the rod
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is moving. Then it is important that the two
ends of the rod must be measured at exactly
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at the same time.
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This situation, I have pictures here in the
next transparency. I have depicted a moving
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rod. This red is the position of the rod at
a given time. Let us call time t is equal
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to 0. Now, let us suppose the time t is equal
to 0. I made the measurement of co-ordinate
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of this particular end and find it out that
is equal to X 1. Now, I make some delay. I
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do not immediately make the measurement of
this particular end. But after some time,
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I decide to make the measurement of the other
end of the rod. Then by this particular time,
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this rod is moved ahead. So I, at a later
time, what I would be measuring is probably
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these co-ordinates which is X 2. You can clearly
see that X 2 minus X 1 is not equal to the
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length of the rod. It is a different question.
From these two measurements also, I can find
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out the length of the rod, provided I take
care of this displacement, that the rod has
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taken or the displacement that the rod as
undergone during the difference of the time,
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during which I have to measure. That is a
different question. That also I will give
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an example little later. But at this moment,
what I want to do is, I want to measure the
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length of the rod by just measuring the co-ordinate
differences. For that, if the rod is moving
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in my frame of reference, it is important
that the difference the two co-ordinates must
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be measured in at the same time. Only under
that condition, the difference of the co-ordinates
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will give me the correct length of the rod.
So, this is what I have written. If the co-ordinates
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are measured at two different times, their
difference does not give the correct length.
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That is what I have said.
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There may be other methods of measuring the
length of a moving rod. Some would be discussed
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later. As I have mentioned, if I can calculate
how much is the rod, how much the rod has
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moved during the time difference that I have
taken or the time I have taken in making a
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measurement of one end to the other end. Of
course, I can find out the length of the rod.
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That I will do little later. But if the co-ordinate
difference of a moving rod has to give me
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length, then for a moving rod, the two ends
must be measured at the same time. That is
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what I have written. But the co-ordinate difference
of a moving rod would give correct length,
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only if the two ends are measured at the same
time.
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This is an important point to note, because
this is where I find a very large number of
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students make a mistake. They do not bother
about time. They just decide and they just
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say just x 2 minus x 1 is equal to length,
which may not to be correct. Because if the
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object is moving in the frame of reference,
then x 2 minus x 1 need not give the correct
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length, until these two measurements were
carried out exactly at the same time.
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Now, let us come back to the problem. Let
us assume that this particular rod is at rest
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in S prime frame of reference. We have already
defined what are our S frame and S prime frame
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of reference. Lorentz transformation is applicable
when we impose sudden conditions on S and
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S prime frame of references, which we have
discussed earlier, that the S’ should be
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parallel and the relative motion should be
along x direction. So, that is anyway have
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to be valid. But what I am assuming in this
particular case that it is the S prime frame
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of reference, in which this rod is stationary
It means, it is S prime frame which is the
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proper frame. So, this rod A B, which I am
calling here, A B is at rest in S prime frame
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of reference. Obviously, it means that in
the S frame of reference, this particular
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rod is moving. I can always imagine that I
have put a rod, let us say in a train and
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the train is moving with respect to an observer
on ground. So, the person sitting on the train
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feels that the rod is not moving, but a person
sitting on the ground, and feels that the
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particular rod is moving. Remember, this x
direction is the direction of the relative
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motion. This is in this direction that S prime
is moving relative to S. Now, we have earlier
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agreed that relativity things are simple,
if we define events. Generally, it is always
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a good idea to define events. So, let us suppose
we define events relating to the measurement
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of the length of this particular rod and these
events, I am defining here, calling event
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number 1 and event number 2.
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So, this is my event number 1. Let me just
write to show the rod here. This was my S
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prime frame of reference and this rod was
at rest here. This was my end A, this was
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my end B and this rod is stationary in S prime
frame of reference, being also observed by
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another observer, which is an S frame. In
this S frame, this rod is actually moving
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to the right, as the frame of reference, S
frame is also moving to the right with respect
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to S.
So, my event number 1 is that, I make the
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measurement of the co-ordinate of this particular
end of the rod. Let us assume that this particular
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measurement is being done by an observer in
S frame of reference. So, he makes a measurement
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of this particular, the co-ordinate of this
particular end of the rod. This is my event
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number 1. Then he goes here. Let me now say
goes here, but he also makes a measurement
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of the end B of the rod. This I call as event
2. Now, as this particular rod A B is not
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stationary in S frame of reference, so in
peaceful he cannot go from A to B and make
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the measurement. The measurement of the co-ordinate
here and co-ordinate here have to be done
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at the same time, if the co-ordinate difference
of these two events has to give me a correct
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length of the rod.
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So, this is what I have written here. I have
defined my events. My event number 1 is, observer
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in S measures the co-ordinate of A at the
end of the rod. Let us suppose this rod be
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x 1 and t 1. Normally, we have a tendency
of now including time also as a co-ordinate
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because as you will be seeing going ahead
in the course, we realize that for many purposes,
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time x acts as the another co-ordinate. So,
because this is event number 1, so I have
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put a subscript here 1. So, this event number
1 occurs at a co-ordinate of x 1, at a position
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co-ordinate of x 1 and at a time co-ordinate
of t 1. Similarly, the observer in S measures
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the co-ordinate of the B end of the rod. Let
us suppose, he does it and finds out the co-ordinate
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to be x 2 and the measurement is done at time
t 2. So, the co-ordinate of this particular
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event is x 2 t 2, while the co-ordinate of
the first event is x 1 t 1.
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As we have earlier realized and discussed,
this x 2 minus x 1 will give me correct length
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of the rod, only if t 2 is equal to t 1, because
it is in x frame that the rod is actually
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moving.
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This is what I have written here. x 2 minus
x 1 is equal to l in S, only if t 2 is equal
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to t 1, which I have said. Let us assume that
is equal to t rather writing same symbol.
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So, as I said, let us assume that the t 2
is equal to t 1 is equal to t. Then these
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two co-ordinates, I used Lorentz transformation
to find out the information in S prime frame
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of reference.
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Let me write the Lorentz transformation once
more here. Just to remind you, it was x prime
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is equal to gamma x minus v t. y prime is
equal to y. z prime is equal to z. t prime
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is equal to gamma t minus v x divided by c
square. This is what was my Lorentz transformation.
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Now, let us apply the Lorentz transformation
on the two events. If I apply, let us say
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first event number 2, it means the value of
x was x 2 and time t was t 2. I will get the
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co-ordinate of that particular event in S
prime frame of reference. So, I will get x
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2 prime, the co-ordinate of the second event
in S prime frame of reference.
18:58.909 --> 19:05.889
Similarly, if I substitute here x 2, substitute
here t 2, I will get t 2 prime, which is the
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time at which the second event occurs according
to an observer S prime frame of reference.
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This is precisely what I have written here.
If you go back to the transparency, it is
19:19.139 --> 19:26.139
x 2 prime is equal to gamma x 2 minus v t,
because t 2, I have taken to be equal to t.
19:26.289 --> 19:32.470
This is what I have written earlier. So, this
I have just put it equal to t. Then t 2 prime
19:32.470 --> 19:39.470
will be given by gamma t, which is just t
2 is equal to t minus v x 2 divided by c square.
19:42.149 --> 19:48.950
Exactly the same thing. I put the co-ordinate
position and time here. For the first event,
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put position and time for the first event,
I get these two equations. You can very clearly
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see that, as far as S prime observer is concerned,
t 2 prime is not equal to t 1 prime. See this
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t is same as t, but this x 2 and x 1 there
has to be difference. If the length has to
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have some, if the rod has to have some length,
then x 2 cannot obviously be equal to x 1.
20:17.090 --> 20:21.649
Therefore, this t 2 prime is not equal to
t 1 prime.
20:21.649 --> 20:27.779
So, what it means that, according to the observer
in x prime frame of reference, these two events
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occurred at different time. It should not
be surprising to us because remember earlier,
20:33.460 --> 20:38.759
we have discussed the simultaneity relative.
I am I shall be discussing and I shall be
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harping a number of times saying that, simultaneity’s
relative. It is in S frame that the two events
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occurred on the same time, and hence, they
where simultaneous. According to S prime frame
20:52.500 --> 20:58.230
of reference, these two events did not occur
at the same time. Therefore, they are not
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simultaneous.
So, according to him, the two events occurred
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at a different time. Would it mean that X
2 prime will not agree that X 2 prime minus
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X 1 prime is equal to length? No, that is
not correct because in x prime frame of reference,
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this rod is stationary. So, it is not important
whether the two ends should be measured at
21:21.799 --> 21:27.990
the same time. What is important is that,
we just make measurement at any given time.
21:27.990 --> 21:33.419
So, even if the measurement of one end of
the co-ordinate in S prime frame of reference
21:33.419 --> 21:38.970
was done in a different time from the other
end, the rod being at rest at this particular
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frame of reference, the time, the position
difference will still give me the correct
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length. So, what I want to argue out that
x 2 prime minus x 1 prime is equal to the
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correct length of the rod in x prime frame
of reference irrespective of the fact that
21:56.309 --> 22:02.889
t 2 prime is not same as the t 1 prime.
While in S frame, we could not have done a
22:02.889 --> 22:09.149
similar argument. I have to force the time
to be same in S frame of reference. Only in
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that case x 2 minus x 1 would have given me
the correct length. Now, if I want to find
22:15.659 --> 22:21.000
out a relationship between the length measured
in the S and S prime, all have to do is to
22:21.000 --> 22:28.000
take the differences. So, I know that x 2
minus x 1 x 2 minus x 1 was the correct length
22:30.000 --> 22:37.000
in S frame of reference because these two
times are same. x 2 prime minus x 1 prime
22:38.309 --> 22:45.039
will anyway be correct length in S prime frame
of reference, even though the times are different.
22:45.039 --> 22:52.039
So, I will say that x 2 prime minus x 1 prime
is L prime. The length of the rod in S prime
22:52.460 --> 22:59.399
frame of reference and call x 2 minus x 1
as the length of the rod in S frame of reference.
22:59.399 --> 23:05.009
So, if I subtract these two, we will just
realize that this particular term v t here
23:05.009 --> 23:11.370
would cancel and time being identical, so
you will just get x 2 prime minus x 1 prime
23:11.370 --> 23:15.649
is equal to gamma multiplied by x 2 minus
x 1.
23:15.649 --> 23:22.649
This is what I have written in the next transparency.
We note that t 2 prime is not equal to t 1
23:23.309 --> 23:28.899
prime. Still x 2 prime minus x 1 prime is
equal to L prime. This is because the rod
23:28.899 --> 23:35.899
is at rest in S prime. Measurement time has
to be adjusted to be same in S and not necessary
23:37.590 --> 23:39.590
in S prime.
23:39.590 --> 23:46.279
I have taken the differences. I get x 2 prime
minus x 1 prime is equal to gamma x 2 minus
23:46.279 --> 23:52.039
x 1. As we discuss, this is the length in
S prime frame of reference, this is the length
23:52.039 --> 23:57.049
in S frame of reference, I call this is L
prime and I call this is L. So, this is L
23:57.049 --> 24:02.789
prime is equal to, this L equal to L prime
divided by gamma, which is what I have written
24:02.789 --> 24:09.789
in this particular equation. As we as seen
earlier that, gamma is to be always greater
24:10.070 --> 24:15.279
than 1, because you remember, gamma was equal
to 1 upon under root 1 minus v square by c
24:15.279 --> 24:19.509
square.
24:19.509 --> 24:26.509
So, you are always subtracting some quantity
from 1. So therefore, this particular factor
24:28.480 --> 24:34.309
is always going to be less than 1 and 1 divided
by something less than 1, always make a quantity
24:34.309 --> 24:41.149
greater than 1. So, gamma would always be
greater than 1. Therefore, this L would always
24:41.149 --> 24:43.490
be smaller than L prime.
24:43.490 --> 24:50.490
So, we say, length is contracted. It is called
length contractions. Let us take an example.
24:52.889 --> 24:59.889
Slightly different example, in which, I do
not make the rod along the x direction. But
24:59.999 --> 25:06.360
let it have a slightly different orientation.
So, in this particular example, I have assumed
25:06.360 --> 25:13.059
that this particular rod is line actually
the x prime y prime plane, I still assume
25:13.059 --> 25:17.690
that it is x prime frame of reference, which
is the proper frame of reference for this
25:17.690 --> 25:18.490
particular rod.
25:18.490 --> 25:24.279
So, this rod is actually stationary in an
S prime frame of reference. But only difference
25:24.279 --> 25:29.940
I say that, this rod is no longer along this
particular direction, but making some angle.
25:29.940 --> 25:36.940
You know somewhere here like this. So, my
rod is like that A B and this is S prime frame
25:37.570 --> 25:44.570
of reference. So, I assume that the end A
of this particular rod is at the origin here.
25:44.580 --> 25:50.649
I assume that length of the rod is 1 meter.
So, effectively in S prime frame of reference,
25:50.649 --> 25:55.730
I know the x co-ordinate of A which is 0 0,
x co-ordinate of B which is I can find out
25:55.730 --> 26:02.730
easily, if I know the length and if I know
the angle. What I have to do now is to find
26:02.990 --> 26:08.149
out the length of this particular rod in S
frame. That is what is the question.
26:08.149 --> 26:14.669
So, let may read the statement of the question,
statement of the example. There is the 1 meter
26:14.669 --> 26:20.909
rod A B, the one which I have said. The length
has been given to be 1 meter rod and is kept
26:20.909 --> 26:27.230
stationary in S prime frame of reference,
which is the proper frame of reference, with
26:27.230 --> 26:34.139
end A at origin, which I just now showed in
the picture, in x prime y prime plane making
26:34.139 --> 26:41.139
an angle of 60 degrees. So, I know theta value
which is 60 degrees with x prime frame, with
26:42.090 --> 26:48.259
prime axis. What would be its length in S,
if the relative speed between S and S prime
26:48.259 --> 26:54.779
is 0.6 c? So, I have given some numbers. So,
you can calculate. The answer, a numeric answer.
26:54.779 --> 27:00.710
Also, find the angle that the rod makes with
x axis in S. So, according to S observer,
27:00.710 --> 27:05.450
at what angle this particular rod is inclined
with respect to x axis.
27:05.450 --> 27:11.720
So, this is essentially the question. The
same picture, which I had drawn on this particular
27:11.720 --> 27:17.200
piece of paper, and we have S prime frame
of reference. This particular rod is at rest
27:17.200 --> 27:24.200
with A being at the origin. This end B being
here. This angle being given 60 degrees and
27:26.990 --> 27:33.990
this length is being to be given 1 meter.
Now, like before, I will define the events
27:36.779 --> 27:42.639
and will define the events to show that these
two events occurred at the same time in S
27:42.639 --> 27:49.009
frame of reference. Because it is the S frame
that this particular rod is moving. Also,
27:49.009 --> 27:53.289
because here y co-ordinate is also involved;
see in earlier example, we did not bother
27:53.289 --> 27:58.619
about the y co-ordinate, because y was 0 in
S frame. It had to be 0 in S prime frame of
27:58.619 --> 28:02.119
reference. But here, because this particular
other end of the rod will have a different
28:02.119 --> 28:06.860
value of y than the first end of the rod.
Begins will have a different y than the a
28:06.860 --> 28:13.860
end. Therefore, is better talk about the y
co-ordinate also. So, we will define our events.
28:14.990 --> 28:20.710
So, I said like before, we have to define
events relating to measurement of the ends
28:20.710 --> 28:27.710
of the rod and these events have to occur
at the same time in S. Let the co-ordinates
28:29.080 --> 28:36.080
of these events be again, like before, event
is x 1 y 1 t. So, this co-ordinate occurs,
28:36.450 --> 28:42.720
this particular, sorry, event occurs at co-ordinates
x 1 y 1 and t, according to an observer in
28:42.720 --> 28:49.720
S frame. Similarly, event number 2, I have
put subscript 2, happens at x 2 y 2 and at
28:51.659 --> 28:58.230
time t. These two times I have to put same
to ensure that actually the measurement of
28:58.230 --> 29:03.679
these co-ordinates will give me the correct
length of this particular rod. So, in principle,
29:03.679 --> 29:07.820
as we will be discussing, if I take the difference
of this and take the difference of this and
29:07.820 --> 29:13.490
square and add, take the under root, I will
get the length of the rod in S frame. Again,
29:13.490 --> 29:20.490
we have to carry out the Lorentz transformation
to find the corresponding co-ordinates of
29:20.690 --> 29:26.009
these two events in S prime frame of reference.
The only difference in this particular case
29:26.009 --> 29:30.809
is that I already know the co-ordinates, so
I can write this equation. It makes not such
29:30.809 --> 29:35.600
a great difference, but just ensure that we
understand the situation well.
29:35.600 --> 29:42.600
Now, we applied for the second event the Lorentz
transformation, which is x 2 prime is equal
29:44.879 --> 29:50.019
to gamma x 2 minus v t. Just like in the earlier
example of length contraction.
29:50.019 --> 29:57.019
Only thing I say that, if this particular
length was 1 meter, this angle is theta, then
30:00.379 --> 30:06.649
the x co-ordinate of this particular rod will
be 1 cos theta, because 1 is the length of
30:06.649 --> 30:13.649
the rod. The y co-ordinates of this particular
rod will be equal to 1 sin theta. So, this
30:15.529 --> 30:22.480
is the x co-ordinate, this is the y co-ordinate,
this will be 1 cos theta, this will be 1 sin
30:22.480 --> 30:28.220
theta, because theta has been given equal
to 60 degrees. So, 1 into cos 60, which is
30:28.220 --> 30:35.220
half and this will be half meter and 1 sin
theta will be equal to root 3 by 2 meters.
30:39.269 --> 30:46.269
So, this particular co-ordinate, the y co-ordinate
would be equal to root 3 by 2 meters and while
30:47.519 --> 30:52.799
this x co-ordinate will be equal to half meter.
So, this is what I have written in this particular
30:52.799 --> 30:59.529
transparency, that x 2 prime is equal to gamma
x 2 minus v t, which I know is equal to half
30:59.529 --> 31:06.529
meter and y 2 prime is equal to y 2, which
I know is under root 3 by 2 meters.
31:08.309 --> 31:15.309
Similarly, I can take the Lorentz transformation
of the x co-ordinate, which gives me x 1 prime
31:15.840 --> 31:22.840
is equal to gamma x 1 minus v t. This co-ordinate
is equal to 0. y 1 prime is equal to y 1 is
31:22.989 --> 31:29.179
equal to 0, because I know in S prime frame
of reference and A is at origin. So, x 1 prime
31:29.179 --> 31:36.179
and x 2 prime, I am sorry, x 1 prime and y
1 prime both must be equal to 0.
31:36.419 --> 31:42.820
Like before, we take these two events to be
different in S prime. The time will be different
31:42.820 --> 31:47.399
in S prime. But have not specifically written
in this particular case, the equations involving
31:47.399 --> 31:53.419
the time difference. Refer the time transformation
because as far as he problem is concerned
31:53.419 --> 31:56.820
it is asking is only for the length. It is
not asking for time. So therefore, I am not
31:56.820 --> 32:02.239
written the equations relating to time transformation.
I have just written the x transformation because
32:02.239 --> 32:07.230
for the problem, that is not important. Only
thing I have to ensure, that the time of the
32:07.230 --> 32:12.269
two events is same in S. That, I have to ensure
by putting the 2 times to be same. That is
32:12.269 --> 32:13.710
all.
32:13.710 --> 32:20.139
Like before, I take the difference of these
two, just like what we did in the earlier
32:20.139 --> 32:26.379
case. I will find gamma x 2 minus x 1 to be
equal to 0.5 because remember one was 0.5
32:26.379 --> 32:33.379
and another was 0. So, this difference is
just 0.5. Similarly, y 2 minus y 1 was equal
32:34.330 --> 32:40.629
to root 3 by 2 because one was root 3 by 2
and another was 0. So, this, this. Of course,
32:40.629 --> 32:47.629
for velocity equal to relative velocity equal
to 0.6 c, I get gamma is equal to 1 divided
32:48.580 --> 32:54.789
by under root 1 minus v square by c square,
which is 0.6 square. This comes 0.361 minus
32:54.789 --> 33:01.789
0.36 under root is 0.8. So, this is equal
to 1.25. So, gamma is equal to 1.25. This
33:02.460 --> 33:07.730
0.6 gives you a very clean number for gamma.
So, in many of the problems to make the mathematical
33:07.730 --> 33:12.889
simple, making the algebra simple, we choose
the gamma value, which is something 0.6 0.8.
33:12.889 --> 33:15.850
These gives clean values of gammas.
33:15.850 --> 33:22.850
So, I substitute these values of gamma here
and find the length. As I have said earlier,
33:23.840 --> 33:29.879
the length is given by x 2 minus x 1 plus
y 2 minus y 1 square. Look back at this particular
33:29.879 --> 33:35.299
rod. You find out the co-ordinate of this
particular rod. Even if they are not 0, so
33:35.299 --> 33:39.379
what I have to take the difference of the
x co-ordinate and the difference of the y
33:39.379 --> 33:45.169
co-ordinate, this length plus this length,
this length square plus this length square
33:45.169 --> 33:49.399
will give me this length square. If I take
the under root, I will get the length of the
33:49.399 --> 33:56.059
rod. So, this is what I have written here.
The length of the rod in S is thus given by
33:56.059 --> 33:59.869
x 2 minus x 1 whole square plus y 2 minus
y 1 whole square.
33:59.869 --> 34:06.340
We have just now seen from the previous transparency
here, that x 2 minus x 1 is equal to 0.5 divided
34:06.340 --> 34:13.340
by gamma and gamma is equal to 1.25. So, I
have written here is 0.5 divided by 1.25 square
34:14.530 --> 34:19.510
plus, here there was no gamma, it is just
root 3 by 2 because y prime is equal to always
34:19.510 --> 34:25.869
y. If I calculate this number, we get under
root 0.91, which is approximately equal to
34:25.869 --> 34:32.869
0.95 meter. Hence, the length of the rod in
S frame of reference would turn out to be
34:33.309 --> 34:40.309
approximately equal to 0.95 meter and not
1 meter. I have to calculate the angle and
34:41.940 --> 34:46.139
angle again, if I have to calculate this angle,
I have to calculate this distance, and this
34:46.139 --> 34:51.710
distance, divide by 2 and this will give me
tan theta. So, this is y 2 minus y 1, this
34:51.710 --> 34:56.690
is x 2 minus x 1 and divide the two and that
will be equal to tan theta.
34:56.690 --> 35:01.770
So, this is what I have written here in the
transparency, that tan theta is equal to y
35:01.770 --> 35:08.770
2 minus y 1 divided by x 2 minus x 1. y 2
minus y 1 was root 3 by 2. x 2 minus x 1 was
35:11.030 --> 35:17.030
0.5 divided by 1.25. So, 1.25 comes into the
numerator. If you calculate this number, approximately
35:17.030 --> 35:24.030
turns out to be 2.165, which gives theta approximately
equal to 65.2 degrees.
35:25.099 --> 35:32.099
So, according to an observer in S, this rod
is inclined not in 60 degrees, but 65.2 degrees.
35:36.150 --> 35:40.780
I hope you would have appreciated one particular
aspect from this particular derivation, that
35:40.780 --> 35:47.319
the gamma factor has arrived only in the x
component and not along the y component. Therefore,
35:47.319 --> 35:52.380
actually it is only the x component which
get contracted and not the y component. So,
35:52.380 --> 35:57.230
this is the point which I have mentioned here
specifically, that we note that only the x
35:57.230 --> 36:01.140
component of the rod length is contracted
and not the y component.
36:01.140 --> 36:06.299
Similarly, not even the z component because
y prime tends out to be y and z prime is tends
36:06.299 --> 36:11.140
out to be equal to z. Therefore, there is
no gamma factor involved. Gamma factor involved
36:11.140 --> 36:16.690
is only in the x component and that is the
factor, which is the relative velocity direction
36:16.690 --> 36:19.809
along which the contraction what really occur.
36:19.809 --> 36:26.809
Now, let us taken another example. There is
an observer in S and he finds that a lightening
36:30.430 --> 36:37.430
strikes at a distance of 20 kilometers from
origin at t is equal to 0. So, for example,
36:37.920 --> 36:43.630
I am sitting here in this particular place
and 20 kilometers away from me, lightening
36:43.630 --> 36:50.630
strikes. It could be any other event, but
I am sort of taking as a lightening. So, at
36:51.089 --> 36:56.559
20 kilometers away, some lightening strikes
and that was time t is equal to 0 as far as
36:56.559 --> 37:01.069
I am concerned.
So, I am sitting at my origin and I find out
37:01.069 --> 37:07.920
that at time t is equal to 0, lightening had
struck at a distance of 20 kilometers away.
37:07.920 --> 37:14.160
Remember, at this time, t is equal to 0. According
to me, the origin of S prime was also passing
37:14.160 --> 37:19.970
by my side. So, the observer in S would feel
that the origin of the S prime passing by
37:19.970 --> 37:26.339
his side and this event occurring at t is
equal to 20 kilometers happens exactly at
37:26.339 --> 37:32.099
the same time, which is t is equal to 0.
Now, the question is, at what distance from
37:32.099 --> 37:39.099
origin did this event of lightening occur
as far as observer in S prime frame of references
37:40.609 --> 37:47.609
is concerned? This S prime observer is moving
relative to me with a speed of 0.6 c. So,
37:47.750 --> 37:54.660
let us read the statement again. According
to an observer in S, a lightening strike at
37:54.660 --> 38:00.970
a distance of 20 kilometers from origin at
t is equal to 0. At what distance from origin
38:00.970 --> 38:07.970
did this event occur in S prime, which moves
with a speed of 0.6 c in frame S? Let me first
38:10.240 --> 38:16.579
to give a wrong answer for this particular
example. My experience tell that a very large
38:16.579 --> 38:21.890
number of students have a tendency of giving
this wrong answer. Then we will see what is
38:21.890 --> 38:27.579
the correct answer and why the wrong answer
is not really correct.
38:27.579 --> 38:34.579
So, standard wrong approach to this particular
problem is essentially applying blindly the
38:34.869 --> 38:41.869
formula of length contraction. So, I may start
thinking that t is equal to 0 and t prime
38:42.510 --> 38:49.510
is equal to 0, the origins of S and S prime
were coincident. Let us imagine a stationary
38:50.270 --> 38:56.760
rod in S frame. That in S frame there is a
stationary rod, which is 20 kilometers long;
38:56.760 --> 39:02.020
very long, which is extending from the origin
to the place of lightening. I can always imagine
39:02.020 --> 39:08.579
that there is a huge long rod, which is going
right from origin to 20 kilometers away and
39:08.579 --> 39:15.579
this is stationary in S frame. Now, I expect
that the length of this particular rod would
39:17.569 --> 39:24.569
be contracted in S prime because this S, this
length of the rod is proper in S frame.
39:26.220 --> 39:32.799
So, probably I can apply the length contraction
formula. Let me first just read before I go
39:32.799 --> 39:37.829
to the next transparency. At t is equal to
0 and t prime is equal to 0, the origins of
39:37.829 --> 39:44.640
S and S prime were coincident. Imagine a stationary
rod of 20 kilometers extending from origin
39:44.640 --> 39:51.640
to the place of lightening in S. The length
of this rod would be contracted in S prime.
39:51.740 --> 39:58.740
So, I just now said that gamma is equal to
1.25 for v is equal to 0.6 c. So, gamma is
40:01.210 --> 40:08.020
1.25. So, this 20 should be divided by 1.25
because this is the contracted length and
40:08.020 --> 40:14.089
that answer is 16 kilometers. Therefore, this
particular lightening must have struck in
40:14.089 --> 40:21.089
S prime frame at a distance of 16 kilometers
from origin. Let me read. So, the co-ordinates
40:21.280 --> 40:26.140
of lightening in S prime would be given by
the contracted length 20 divided by 1.25,
40:26.140 --> 40:33.140
which is 16 kilometers. I am again repeating,
this is the wrong answer I have written. Only
40:36.829 --> 40:40.170
wrong answer. Let us try to find out the first
correct answer. Correct answer. Whenever there
40:40.170 --> 40:44.430
is the confusion, apply the Lorentz transformation
directly. You will not making any mistake.
40:44.430 --> 40:48.720
Lorentz transformation. I know the co-ordinate
of this particular event, which is at S is
40:48.720 --> 40:55.520
equal to 20 kilometers, and t is equal to
0. I can always find out what is S prime and
40:55.520 --> 40:59.319
what is t prime by applying Lorentz transformation.
40:59.319 --> 41:06.319
So, this is the correct answer. The co-ordinate
of event in S is x is equal to 20 kilometers
41:07.880 --> 41:14.440
and t is equal to 0. Hence, using Lorentz
transformation, the x co-ordinate in S prime
41:14.440 --> 41:21.440
will be given by, x prime is equal to gamma
x minus v t. x is 20 kilometers and t is 0.
41:24.230 --> 41:31.230
So, this particular quantity becomes 0. 1.25
multiplied by 20, 25 kilometers. In fact,
41:31.770 --> 41:35.170
this length appears not to be contracted,
but extended.
41:35.170 --> 41:39.960
So, according to the correct answer or according
to the answer, which have you obtained from
41:39.960 --> 41:44.770
Lorentz transformation, this particular event
occurred at a distance of 25 kilometers from
41:44.770 --> 41:51.770
the origin. Not at 16 kilometers. Not at 20
kilometers, but at a distance of 25 kilometers.
41:52.309 --> 41:59.309
The length has not contracted. Does not it
look funny? Let us try to find out what is
42:01.020 --> 42:05.770
the reason and why we are getting different
answer. Let us try to understand and let us
42:05.770 --> 42:12.270
try to, we will think that why this earlier
result was wrong and why this particular result
42:12.270 --> 42:19.270
is correct. Why this discrepancy? Look at
time.
42:19.910 --> 42:26.480
Let us first do the time transformation. t
prime is equal to gamma t, which is 0 minus
42:26.480 --> 42:31.059
v which is 0.6 c, relative velocity between
the frame. x co-ordinate of the event, which
42:31.059 --> 42:36.809
is 20 kilometers, 20 into 10 to the power
3 meters divided by c square. If I take c
42:36.809 --> 42:42.089
is equal to 3 into 10 to the power 8 meter
per second, I find t prime is equal to minus
42:42.089 --> 42:47.470
5 into 10 to the power minus 5 seconds. What
is this negative? There is a minus sign here.
42:47.470 --> 42:54.470
So, this thing becomes minus. What does that
minus sign means? It means, this event occurred
42:55.470 --> 42:58.680
before 0 time.
Remember, according to S prime of observer
42:58.680 --> 43:05.650
also, at t prime is equal to 0, its origin
was coincided with the origin of S. This event
43:05.650 --> 43:12.020
t prime, this particular event of lightening
struck according to S prime, at t prime is
43:12.020 --> 43:18.869
equal to minus 5 into 10 power minus 5 seconds.
It means it appeared before. It means, the
43:18.869 --> 43:24.130
origin of S and S prime were not coincided
metal that particular time, when the event
43:24.130 --> 43:28.760
occurred. The two origins become coincided
at a later time at 5 into 10 to the power
43:28.760 --> 43:34.260
of minus 5 seconds after this particular event
occurred in S prime frame of reference. This
43:34.260 --> 43:39.520
is somewhat shocking, but this is what will
be the perception of S prime frame of reference.
43:39.520 --> 43:43.890
Observer sitting on S frame of reference,
that this particular event occurred before
43:43.890 --> 43:50.260
the origins of the two frames were coincided.
This is what I have written.
43:50.260 --> 43:56.569
So, according to S prime, lightening has struck
5 into 10 power minus 5 seconds before the
43:56.569 --> 44:02.319
origins of the two frames coincided.
44:02.319 --> 44:07.609
In S prime, the events of the two origins
coinciding; if you call this as an another
44:07.609 --> 44:13.319
event, then the two origins being coincident,
this event of the two origins coincident and
44:13.319 --> 44:20.319
the lightening striking are not simultaneous,
though in S they are. According to S observer,
44:20.390 --> 44:25.510
lightening striking and the origin of S prime
passing, both occur at time t is equal to
44:25.510 --> 44:32.510
0; same time. Therefore, the events were simultaneous.
According t prime, first the event of lightening
44:32.670 --> 44:39.200
struck before the second event. That is, the
coincident of the origin occurs later. They
44:39.200 --> 44:45.170
were not simultaneous. Simultaneity’s relative,
this is what we are discussing by giving many
44:45.170 --> 44:49.390
examples, that this is something which is
very different from the classical ideas to
44:49.390 --> 44:56.390
understand that simultaneity is relative in
special theory of relativity.
44:58.589 --> 45:03.349
According to S prime, the origin of S is approaching
him with a speed of 0.6 c.
45:03.349 --> 45:10.349
If you are looking with respect to an observer
an S prime frame of reference, this particular
45:11.940 --> 45:18.940
origin S was actually moving towards him.
Only after 5 into 10 to the power minus seconds,
45:21.069 --> 45:27.510
will the origin S reach his origin. So, this
origin is always coming towards him and it
45:27.510 --> 45:32.109
will take 5 into 10 to the power minus 5 seconds
before this origin will really come and meet
45:32.109 --> 45:38.970
him while lightening has struck. Earlier a
lightening was struck here, when the origins
45:38.970 --> 45:43.960
where not coincident. This is what I have
written. According to S prime, the origin
45:43.960 --> 45:50.180
of S is approaching him with a speed of 0.6
c and only 5 into 10 to the power minus 5
45:50.180 --> 45:57.180
seconds after the lightening striking, will
the origin of S reach his origin.
45:58.559 --> 46:03.069
I wanted to make one particular point, which
sometimes, again I find that students are
46:03.069 --> 46:10.069
somewhat confused. See, it is not important
when the observer in S prime gets the information
46:12.030 --> 46:15.640
of lightening because the lightening, the
information will take certain time to reach.
46:15.640 --> 46:22.640
What is important is the time of the event.
Suppose, one of our friends is departing from
46:24.780 --> 46:31.780
the railway station at a particular given
time and I come to know that his train departed
46:32.339 --> 46:37.619
exactly at 10 o clock. This information may
not reach me exactly at 10 o clock. The person
46:37.619 --> 46:44.619
who has gone to see him off may come here,
may take 1 hour to reach my place and may
46:45.160 --> 46:50.280
give me information that the train had departed
actually 10 o clock. It does not mean that
46:50.280 --> 46:55.010
the event occurred at the time I got the information.
Even, it had occurred earlier.
46:55.010 --> 46:59.420
So, what is important is the time of the event.
I know that the train is departed at 10 o
46:59.420 --> 47:04.839
clock, even though I am have got the information
late. So similarly, an observer in S prime,
47:04.839 --> 47:10.500
may have got this information later because
it always take a finite amount of time for
47:10.500 --> 47:15.329
the information to reach him. But all this
time, if we try to calculate and take into
47:15.329 --> 47:19.799
account, he can find out at what time the
event actually takes place.
47:19.799 --> 47:24.039
So, for example, from the railway station,
if somebody sends me a light signal saying
47:24.039 --> 47:28.329
that train is departed, I know how much time
the light would have taken to reach me here
47:28.329 --> 47:32.660
and I can calculate at what time this particular
event would have occurred. So, when I am saying
47:32.660 --> 47:38.680
the time of event, it is important to realize
that this is the time the event actually occurred.
47:38.680 --> 47:44.230
So, this is what I have written. Note. When
the observer in S prime gets information of
47:44.230 --> 47:49.819
lightening is not important. What is important
is the time when the lightening took place
47:49.819 --> 47:51.910
as per his calculation.
47:51.910 --> 47:58.910
Now, as we have said here, that after 5 into
10 to the power minus 5 seconds, the origin
48:02.289 --> 48:07.510
of the two frames will reach and at that time,
according to S prime will be time t prime
48:07.510 --> 48:14.480
will be equal to 0. I can find out what is
the distance here. How do I find the distance?
48:14.480 --> 48:21.059
I know that after 5 into 10 to the power minus
5 seconds, this origin will reach him; will
48:21.059 --> 48:25.680
reach here and I know what is the speed, which
is 0.6 c. So, I can calculate what will be
48:25.680 --> 48:32.010
this distance at the time when lightening
struck. So, I can find out what is the distance
48:32.010 --> 48:39.010
of the origin of S from the origin of the
S prime at the instant lightening struck according
48:39.940 --> 48:43.549
S prime of observer.
Remember, all these things I am talking now
48:43.549 --> 48:47.450
with respect to S prime observer. We should
never confuse the frame of reference. When
48:47.450 --> 48:51.359
I am talking of one frame of reference, let
us assume that we are sitting in that frame
48:51.359 --> 48:56.760
of reference and try to get all the information
in that frame only. So, I calculate this particular
48:56.760 --> 49:02.529
time, 5 into 10 to the power minus 5 seconds
multiplied by the speed with which the origin
49:02.529 --> 49:05.789
is approaching me. This distance is 9 kilometers.
49:05.789 --> 49:12.789
So, let us now recall our ideas, and readjust
our things. Assume that lightening struck
49:12.970 --> 49:19.970
at point P and a stationary rod OP of 20 kilometer
length is at rest in S. As we said earlier,
49:21.960 --> 49:28.089
this length is proper. This length would indeed
appear to be contracted in S prime and would
49:28.089 --> 49:32.299
indeed be given by 16 kilometers. About that
particular aspect there is no problem. The
49:32.299 --> 49:36.490
problem is only that, this particular length
contraction, whether my co-ordinate gives
49:36.490 --> 49:40.650
me the correct length or not. This is what
I have written here.
49:40.650 --> 49:46.220
But according to S prime, the lightening occurred
before his origin coincided with S. Hence,
49:46.220 --> 49:51.049
the co-ordinate of the event does not measure
the length of the rod in his frame. While,
49:51.049 --> 49:55.119
what I had calculated by length contraction
is the length, but the question that has been
49:55.119 --> 49:59.619
asked is, what is the co-ordinate of that
particular event and that co-ordinate is not
49:59.619 --> 50:01.299
equal to the length of the rod.
50:01.299 --> 50:08.299
So, I have given the situation in the picture.
When the lightening strikes, assume that lightening
50:08.760 --> 50:14.460
strikes, and at that instant, the picture,
according to S, the picture that S observer
50:14.460 --> 50:21.359
will draw, he will draw that the origin of
O prime and O were constituted at that time
50:21.359 --> 50:28.359
and at a distance of 20 kilometers, this lightening
struck. This is the way S is going to picturise
50:28.430 --> 50:34.480
the event. But according to S prime, the picture
will be somewhat different. According to S
50:34.480 --> 50:40.960
prime, he will feel that his 9 kilometers
will be away from the origin. This event actually
50:40.960 --> 50:47.960
occurred at 25 kilometers. But then this rod
OP, according to him will be 25 minus 9 kilometers,
50:49.660 --> 50:55.270
which is 16 kilometers, which is indeed the
contracted length that we had observed. So,
50:55.270 --> 50:59.760
that is what is the difference of perception.
According to this observer S prime, he was
50:59.760 --> 51:03.079
9 kilometers away from the origin when the
event occurred.
51:03.079 --> 51:08.410
So, the co-ordinate of this particular event
is actually 25 kilometers, what we have obtained
51:08.410 --> 51:14.250
from Lorentz transformation. P 9 kilometers
away from here and 16 kilometers is actually
51:14.250 --> 51:20.529
length of the rod OP, which is indeed the
contracted length.
51:20.529 --> 51:26.039
Before I end this, let us discuss one more
important aspect of this particular Lorentz
51:26.039 --> 51:31.779
transformation, which is also very popular,
which is called time dilation. For discussing
51:31.779 --> 51:37.260
time dilation, we have to define a proper
time interval, just like we have defined a
51:37.260 --> 51:43.130
proper length. Proper time interval we can
define between any two events. There could
51:43.130 --> 51:47.279
be any two events; event 1, event 2. Now,
we have defined many types of events. We will
51:47.279 --> 51:52.529
also be defining in the course of lecture
many other events. If these two events occur
51:52.529 --> 51:58.319
at the same position in a frame of reference,
then that time interval, which is measured
51:58.319 --> 52:01.520
in that frame of reference, is called a proper
time interval.
52:01.520 --> 52:06.150
So, let us assume that two events appear to
be occurring at exactly the same position.
52:06.150 --> 52:11.230
So, I am sitting here. Something happens here.
Then again I sit at the same point. Something
52:11.230 --> 52:15.480
happens here exactly at the same point. So,
if I take the time difference between these
52:15.480 --> 52:20.339
two events, that is what is called proper
time interval and I measure the time interval
52:20.339 --> 52:27.160
between these two events. What I am trying
to say is that, if anybody else who is moving
52:27.160 --> 52:32.029
relative to me, measures the time interval
between these two events, he will find that
52:32.029 --> 52:36.779
this time interval is larger. That is why
it is called time dilation. Like, the eye
52:36.779 --> 52:40.990
specialist dilates the eye. The ball becomes
bigger.
52:40.990 --> 52:45.789
So similarly, this particular time interval
becomes dilated. It becomes larger. That is
52:45.789 --> 52:51.849
why it is called time dilation. In any other
frame, the time interval between these two
52:51.849 --> 52:56.900
events would appear to be larger than the
proper time interval. That is why this is
52:56.900 --> 52:58.309
called time dilation.
52:58.309 --> 53:03.460
Now, let us assume that it is S frame. We
already discussed what is S frame and S prime
53:03.460 --> 53:10.180
frame. That is, S frame in which the time
interval is proper. It means these two events
53:10.180 --> 53:15.720
must have occurred at the same position in
S frame. If they have occurred at the same
53:15.720 --> 53:20.250
position, it means that their co-ordinates,
x co-ordinates would be same. So, it means
53:20.250 --> 53:27.250
these two events, event 1 and event 2 occurred
at the same value of x. Let us suppose the
53:27.470 --> 53:34.170
time is measured by an observed in S is t
1 and t 2 for these two events. So, t 1 minus
53:34.170 --> 53:38.480
t 2 or t 2 minus t 1, whatever you want to
call and the difference of time interval is
53:38.480 --> 53:43.500
the proper time interval as measured in S.
Because in S, the two events occurred at the
53:43.500 --> 53:46.369
same value of x or same position.
53:46.369 --> 53:53.369
Now, I apply Lorentz transformation. If I
apply Lorentz transformation, exactly like
53:53.559 --> 53:59.789
before, I get t 2 prime is equal to gamma
t 2 minus v x upon c square. This x, I have
53:59.789 --> 54:05.859
taken to be same. So, this x, I have not written
x 2. For the first event, I have written t
54:05.859 --> 54:12.619
1 prime is equal to gamma t 1 minus v x by
c square and this x being same. I take the
54:12.619 --> 54:18.720
difference of these two. Again, these two
will cancel out. I will get t 2 prime minus
54:18.720 --> 54:23.630
t 1 prime is equal to gamma t 2 minus t 1.
54:23.630 --> 54:30.049
This is what I have written here. The time
difference between these two events in S prime
54:30.049 --> 54:35.930
is therefore, given as t 2 prime minus t 1
prime, which is equal to gamma t 2 minus t
54:35.930 --> 54:41.569
1. This t 2 minus t 1 was the proper time
interval, because this was measured in a frame
54:41.569 --> 54:46.529
of reference, in which the two events occurred
at the same position. So, this is proper time
54:46.529 --> 54:52.000
interval. Gamma being greater than 1 and t
2 prime minus t 1 prime will be larger than
54:52.000 --> 54:58.650
the t 2 minus t 1 and therefore, time interval
will be dilated. Many times, this proper time
54:58.650 --> 55:03.420
interval between the two events is written
as tau. This is one of the symbol, which is
55:03.420 --> 55:07.089
very well commonly used. There are one or
two small comments, which I would like to
55:07.089 --> 55:08.269
make.
55:08.269 --> 55:14.109
First, that you will note that to apply time
dilation, only the x co-ordinates of the two
55:14.109 --> 55:19.500
events had been used. Strictly speaking, if
the y co-ordinate of the two events were different,
55:19.500 --> 55:23.890
one event occurred here and another event
occurred at some other height. One event occurred
55:23.890 --> 55:28.170
here and another event occurring at some other
height, but at the same position. Still time
55:28.170 --> 55:32.589
dilation formula is applicable. To strictly
speak that the time interval is not proper,
55:32.589 --> 55:36.789
we will define exactly a correct definition
of proper time interval a little later. But
55:36.789 --> 55:40.789
as far as the application of time dilation
formula is concerned, only the x co-ordinates
55:40.789 --> 55:45.420
of the two events need be same.
So, that is what I have written. We note that
55:45.420 --> 55:50.260
to apply time dilation, only the x co-ordinates
of the two events have to be same. Though
55:50.260 --> 55:55.150
the general definition of proper time interval,
all the co-ordinates have to be same. As I
55:55.150 --> 55:58.470
said, we shall discuss this aspect in detail
later.
55:58.470 --> 56:03.170
The second point, which I want to mention
is that, if I apply the Lorentz transformation
56:03.170 --> 56:07.819
for X co-ordinate, according to S prime observer,
these two events would indeed appear to be
56:07.819 --> 56:12.740
occurring in different positions. Like, in
the length contraction, according to an observer
56:12.740 --> 56:15.799
in S prime, the time difference of the two
events was not same. It was different. Here,
56:15.799 --> 56:22.779
the X will be different. But this is not shocking
because even classically also that what is
56:22.779 --> 56:27.609
what we see. If an event for example occurs
here, it will classically we do not have to
56:27.609 --> 56:28.390
go to relativity.
56:28.390 --> 56:33.130
If an event occurs, let us say a mole here,
somewhere here and one frame was here. Another
56:33.130 --> 56:37.779
frame, let us say here, also supposed to be
here. They observed that these two events
56:37.779 --> 56:43.059
happen exactly at the same value of x. But
the second event when it occurs, the second
56:43.059 --> 56:48.720
observer has moved to the right. Therefore,
the event, the x co-ordinates has change.
56:48.720 --> 56:53.609
Therefore, the x co-ordinates being different
is not at all that shocking. This is also
56:53.609 --> 56:58.710
expected. Classically, it is a time difference
being different in different frame, that appears
56:58.710 --> 57:00.460
to be much more shocking.
57:00.460 --> 57:05.579
So, that is where I will end my lecture, just
by giving a small summery. We discussed two
57:05.579 --> 57:10.260
important consequences of Lorentz transformation;
namely, the length contraction and time dilation.
57:10.260 --> 57:16.420
Then we gave some examples of length contraction
and warned where one can make an error in
57:16.420 --> 57:18.609
direct use of the formula.
Thank you.