WEBVTT
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In our last lecture, we had discussed the
various postulates of special theory of relativity,
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there were two postulates, we will sort for
revise them. We also discussed about Galilean
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transformation, we discussed what is a transformation
and how the classical Galilean transformation
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look like, we formalize that transformation.
We discussed some examples, specially with
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the idea of time in mind, we saw that in the
classical mechanics time is always treated
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to be same in all the frames. We also saw
that, if there two events which are appear
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would be simultaneous in one frame, it means
they occur at the same time. Then another
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frame also they will appear to be simultaneous
this is ensured by the fact that time is same
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in both the frame of reference, so this is
what we have discussed in a brief lecture.
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That postulates of special theory of relativity,
Galilean transformation and the concept of
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time, and took some examples to show that
it is equality in different frames, assures
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that simultaneous events in a frame are also
simultaneous in another frame. Just to recapitulate,
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let me just tell again the postulates of special
theory of relativity.
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First is the laws of physics are same in all
inertial frames of reference, no preferred
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inertial frame exist, we have always said
that all inertial frames the laws of physics
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should always remain same. You cannot have
something like ether, you cannot say something
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absolute rest, you cannot talk about absolute
velocities all you have to talk is only relative
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in terms of a given frame. You must have something
really physical to which you must attach your
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frame of reference and only with reference
to that you can talk about speeds velocities.
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Then the second postulate was that the speed
of light C is same in all the inertial frame,
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I must add whenever I am saying speed of light,
it means the speed of light in vacuum. Of
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course, when we talk of medium the light of
speed of light may get reduced, but when we
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are talking of the speed of light C is always
the speed of light in vacuum.
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Again, I recapitulate Galilean transformation,
this is what we call as a direct transformation,
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it means if we know the coordinates of an
event which happen at X, Y and Z. And it happens
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at a time t, then the coordinate of the same
event as appearing to another frame of reference
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X prime is given by this equation, X prime
is equal to X minus v t. Y prime is equal
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to Y, Z prime is equal to Z. And of course,
we have said that implicitly, we have assumed
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that t prime is equal to t.
Similarly, we talk of inverse transformation
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we says that, if we know the coordinates of
an event in X prime frame of reference which
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is X prime, Y prime and Z prime, and it occurs
at a time t prime. Then the coordinate of
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the same event in X frame would appear to
be as X is equal to X prime plus v t prime,
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Y is equal to Y prime, Z is equal to Z prime,
t is equal to t prime.
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Of course, this transformation required a
special set of axis about which we had discussed
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earlier that we assume that Y and Y prime
X is are always parallel, Z and Z prime X
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is are always parallel, the relative motion
is only along the X direction. And the origin
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of X prime moves along the X axis, time is
measured from the time, when the origins of
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the two frames are constant; this is what
we have discussed. Inverse transformation
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can always be obtained from direct transformation
by putting v is equal to minus v.
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We also discussed the velocity transformation,
it means I know the velocity components in
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X frame, U X, U Y, U Z, we can find out what
will be the velocity components in X prime
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frame of reference, which turns out to be
U X prime, U Y prime, U Z prime. And similarly
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we discussed about the inverse velocity transformation,
you also said that this is exactly same as
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the normal relative velocity concept, with
which we are very familiar with classical
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mechanics.
Today, I will try to give some example, and
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try to say that how Galilean transformation
and second postulates of special theory of
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relativity do not go along with each other;
it means it is not consistent with the second
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postulate of special theory of relativity.
We have mention about this by passing by in
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our first lecture, when we are trying to do
this special theory of relativity, we have
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measured that from the normal traditional
relative velocity, and we will never find
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this speed of light to be constant.
But, once we have formalize the Galilean transformation,
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let us try to look with the eyes of Galilean
transformation try to see how this particular
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transformation is inconsistent with second
postulate of special theory of relativity.
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Will also give an example, we are we will
see that if second postulates of special theory
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of relativity is correct, then simultaneity
is also relative, it means in a frame if the
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two events appear to be simultaneous in another
frame they may not, these are the two example
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let us start from them.
Before, we really actually come to the transformation
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which is consistent with the second postulate
of special theory of relativity.
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So, this is what I have written here, we now
show that the Galilean transformation is not
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consistent with second postulate; we also
show that simultaneity of event is also relative
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under the second postulates.
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Let us go back to the old example which we
have discussed in our last lecture, see what
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we have done in that particular lecture that
we assume that a ball is thrown making a particular
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angle in the X, Y plane in the X frame. We
try to find out the coordinate of that particular
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ball at time t is equal to 2 seconds in that
particular frame, then we found out what are
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the coordinates of the same event that is
finding the ball at t is equal to 2 seconds,
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in X in X frame, in X prime frame.
And the eventually found out what are the
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velocity components, so this is other problem
which we have discussed. Now, I have modified
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that particular problem, instead of ball we
thrown let image that we throw a pulse of
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light, most of the things remain same other
than we have changed some numbers to make
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it look really somewhat more realistic.
Otherwise, the problem is more or less identical
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of what we did for that particular ball bought
problem in last lecture. So, here we have
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a pulse of light, let us assume that this
pulse of light is highly localized, so you
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can determine to a great accuracy, what is
the position of the pulse. So, we through
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a pulse of light from the origin at the same
time, when this O prime origin was also constant,
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which happens to be t is equal to 0, which
also gives t prime to be equal to 0, most
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of the things as I said have remain same.
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So, we assume that it is thrown in X, Y plane
and it makes an angle of tan inverse 3 by
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4 with X axis, this angle remains same as
we have done in the last problem. What we
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have said now that find the position of the
pulse in S at t is equal to 2 into 10 power
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minus 6 seconds that is 2 micro second to
make number somewhat more realistic, instead
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of 2 seconds I have made it 2 micro seconds,
because the speed of the light is very large.
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And of course, I assuming that the pulse is
highly localized, so I can really determine
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it is position to reasonable amount of accuracy.
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The question is essentially identical, assume
that there is another observer in S prime
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frame which is moving relative to S with speed
of 0.6 c, I have changed to the relative speed
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also to make it somewhat more realistic from
the point of special theory of relativity.
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Assume that the two frames obey the condition
of the Galilean transformation, which we have
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described just now. Again find the speed of
the light pulse and its coordinates at t is
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equal to 2 into 10 power minus 6 seconds in
S prime frame of reference under Galilean
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transformation.
So, we assume that the Galilean transformation
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is still valid, and try to find out this coordinates
and also this speed in S prime of reference,
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as you can see that this problem is very identical
to the problem, that we have just now discussed
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in our last lecture about the ball.
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The method that we are going to adopt is essentially
identical; we find the X component of the
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speed and the Y component of the speed. We
have discussed last time, that if time theta
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is 3 by 4 then sin theta will be 3 by 5, and
cos theta will be 4 by 5.
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Therefore, we know that U X is equal to U
cos theta and U Y is equal to U sin theta
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standard way of taking resolving a vector
long X and Y direction. So, we will get U
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X to be equal to 0.8 C, U Y is equal to 0.6
C and because the light is through in X Y
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direction U Z is equal to 0. So, this remains
essentially identical of what we have done
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earlier, expect that this speed is now C,
and not a classical speed.
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We can find out the coordinates at t is equal
to 2 into 10 power minus 6 second that is
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2 micro second, again by simple application
of formula that the X coordinate will be U
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X multiplied by t which is 0.8 C is the value
of U X time is 2 into 10 power minus 6 second
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we multiplied by that 2 and we get 480 meters.
Similarly, Y will be equal to U Y multiplied
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by t, evaluated by U Y which turns out to
be 0.6 C and t is 2 into 10 power minus 6
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seconds, so Y component turns out to be 360
meters; and of course, Z components is 0.
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Of course, in this calculation I have taken
the speed of light as 3 into 10 power 8 meter
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per second to make our problem simple.
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So, we see that the coordinate of this particular
light pulse at 2 micro second will be 480
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meters, 360 meters and 0. Question is that,
what will be the coordinate of this particular
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light pulse as be viewed in S prime frame
of reference assuming Galilean transformation,
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we apply standard Galilean transformation,
this conclusion we have just now given.
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X prime will be equal to X minus v t, X we
have now calculated 480, v which is the relative
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velocity between the frames is 0.6 C, time
t is 2 micro seconds we put it here, I will
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get 120 meters. Y prime remains same which
is equal to Y, which is equal to 360 meters.
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So, according to observer in S prime frame
of reference, the coordinate of the same event
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will 120 meters, 360 meters and 0. So, the
two observers will find out that their coordinates
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are different in their frames, one is 480,
360, 0, under that 120, 360, 0 standard thing
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expected in classical mechanics. Now, if I
want to find out the speed like we did in
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the case of ball, as seen in S prime of reference
this displacement I must divide by time, and
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the classical mechanics if we remember we
always calculated time, assume time to be
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same.
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So, we take 120 divide by 2 micro second which
is the same time, as seen in S frame and we
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will find out U X prime to be 0.6 into 10
to power 8 meter per second, U Y prime will
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be 360 divided by 2 into 10 power 8 which
is 1.8 into 10 power 8 meter per second. As
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you can see that this U X prime as changed
in fact, it has become smaller than U X obviously,
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you would have expect that the speed of light,
in this particular frame will be reduced if
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Galilean transformation was correct.
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Calculated the speed of pulse in S prime which
is given by standard formula of finding out
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the length of the vector, under root of U
X prime square plus U Y prime square plus
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U Z prime square, this turns out be approximately
1.9 into 10 to the power 8 meter per second.
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While in S the same speed was 3 into 10 to
the power 8 meter per second, so as we can
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see that if Galilean transformation was correct,
then the speed of light will turn out be different
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in different frame, as was will turn out in
different frames. And this is something which
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was against the second postulates of special
theory of relativity.
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So, this is what I have written, we see the
speed of light is different in S prime violating
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second postulates, if we have to find out
the transformation in which the speed of light
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is maintained as C, even in S prime, S prime
of reference. Then, either this X prime or
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t prime or both X prime and t prime have to
be changed, from what has to be obtained from
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the Galilean transformation. So, we must have
a transformation in which still X prime divided
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by t prime, and corresponding calculating
the Y prime U Y prime eventually leads to
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a velocity which is equal to speed of light.
So, probably we required, we change in X coordinate
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as well as in Y coordinate. Let us look at
the second example which we have done in the
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last time, which was an example in which a
particular observer sitting in a train, of
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course two box one towards the motion of the
train, another against the motion of the train.
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This particular observer concludes that the
two events, event number 1 the ball hitting
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the front wall, and event number 2 ball hitting
the back wall occurred at the same time.
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We also concluded that the ground observer
would also notice that the times are same,
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and according to the ground observer also
these two events will appear to be simultaneous,
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it means appearing at the same time. Now,
again we replace this particular experiment,
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this particular example and changed the ball
to a light, a pulse of light, most of the
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problem is this exactly identical.
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So, I have said that an observer is exactly
half way in a running compartment of length
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L prime, as we eventually be seen that the
length have also become frame dependent, so
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because, this S prime of reference I have
decided to call it L prime. So, the length
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as measured in S prime of reference is L prime
and of course, in the classical mechanics
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Galilean transformation will assume that this
length is same, as seen in by the ground observer
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also.
He shines light instead of throwing balls
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at t prime is equal to 0, which travels both
in the front and the back direction, from
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direction means the direction of the motion
of the train, and the back direction is opposite
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to that. We define events exactly as before,
event number 1 light reaches the front wall,
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event number 2 light reaches the back wall.
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I put this particular figure that this observer
is sitting in the train, which is moving with
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respect to the ground observer with the speed
v, this person has one light source here,
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one light source here which shines light or
at the single line that does not make a difference
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shines light which goes this way which is
opposite to that direction of the motion,
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which I call as back wall and like why going
this direction which I am calling it as a
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front wall. So, light moving to the right,
a light moving to the left, this is towards
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a front wall, this is towards back wall.
And our events are event number 1, this particular
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light pulse which in here, event number 2
this light pulse which in here at the back
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wall. Both this events are being observed
by another observer as on the ground, which
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we assume as inertial. Let us look at the
motion, both from point of the S prime, and
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S when we have a light source rather than
two balls been thrown.
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Let us go to the frame of S prime, if we go
to the S prime of reference, this was my train
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light source was let us assume a single light
source, this go in this particular direction,
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this particular light goes in this particular
direction. So, each light pulse actually travels
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a distance of L by 2 according to the observer
in S prime, the speed of light is C, so the
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time taken for each of the event is L prime
by 2 C, and then we replace L by L prime as
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we have just now said. Now, the distance travelled
by both the pulses this and this are same
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which is L prime by 2, both of them travels
with the same speed C, so the time taken to
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be L prime divided by 2 C.
19:23.299 --> 19:28.769
Now, what I have written here both events
are simultaneous in S prime that is the occur
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at the same time, and the time is given by
t prime is equal to L prime divided by 2 C.
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Of course, it assumes that the instant when
the light pulses were thrown, on when the
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light shine the light at that time, time t
prime was equal to 0. So, the observer in
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S prime frame of reference would feel that
these two events are simultaneous, now let
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us look to with respect to a observer in S
frame.
19:57.210 --> 20:04.210
If the observer sitting here in S prime, S
frame in the ground and if the second postulates
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of special theory of light, a special theory
of relativity is correct, then according to
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this observer also this light pulse will be
travelling with speed C, this light pulse
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will also be travelling with speed C. Remember,
in the classical mechanics, in the classical
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velocity transformation the speeds were different,
but if the second postulates of special theory
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is correct, then this is speed must be C,
this speed also must be C, because the speed
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of light is a frame independent quantity,
both the observer will feel exactly the speed
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to be identical.
So, according to the ground observer this
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pulse which was, which originated from this
particular center was actually travelling
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with the speed C, the one which was going
backwards also travel with the same speed
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with C. But by that time the pulse like, pulse
of light moves from this particular position
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towards this particular wall, this wall has
moved ahead; because the train is actually
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moving. So, it takes a certain amount of time
for the light to reach from this particular
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point to this particular point, and according
to the ground observer during this particular
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time, this train has moved ahead and gone
somewhere here.
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So, by the time this light will go and hit
this particular wall, it has to travel a larger
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distance on the other hand, this pulse when
it was trying to move towards the back wall,
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this back wall was approaching towards this
particular source of light. So, eventually
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this light has to travel is smaller distance,
before hitting this particular wall, so remember
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the same concept was used also by Galilean
observer when we are throwing the balls. The
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only difference in that particular case was
that according to that observer, this ball
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was thrown with larger speed and this ball
was thrown with the smaller speed.
22:01.419 --> 22:05.859
But, now both the light pulses travelling
with the same speed, if the second postulate
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is correct, so what the observer on the ground
conclude, the observer on the ground will
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conclude that this particular light will reach
this particular wall later. Then here, because
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both the light pulses are travelling with
the same speed, but this travels the smaller
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distance to reach the wall, while this travels
the larger distance to reach the wall. Obviously,
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these two events cannot be simultaneous, this
event even number 1 will occur later, then
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the event number 2 that it reach in this particular
wall.
22:40.429 --> 22:46.049
This is what I have written here, the two
events again would have to turn out to be
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simultaneous, if we had used the classical
velocity as in the case of balls.
22:50.409 --> 22:55.619
However, under the second postulates, the
speed of light is still C in both the directions,
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but it has to travel in larger distance to
reach the front wall than the back wall, as
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just now I have explained. Hence, event 2
occurs before event 1, so the two events are
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not simultaneous, so simultaneity is also
relative. Now, let us come back out discussion
23:16.690 --> 23:23.690
on time, see time is often related to simultaneity
of two events, when I say that a particular
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event happens let us say class starting, and
taking another event, always a class starting
23:30.379 --> 23:37.379
at 9 o clock. Essentially we mean that, when
watch shows 9 o clock and when the class starts,
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these two events are simultaneous.
Or will we say that a train starts from a
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given station, at least at 10 o clock it means
when watch shows 10 o clocks, and when the
23:48.289 --> 23:54.019
train starts these two events are simultaneous.
We have just now seen that simultaneity is
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relative, it means there would be another
frame in which these two events may not appear
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to occurring simultaneous or appearing to
be occurred at the same time. It means we
24:07.999 --> 24:13.100
probably we have to question the same basic
equation which we have written, which we have
24:13.100 --> 24:20.100
sort implicitly assumes so far, which is t,
which is equal to t prime, so this is what
24:20.320 --> 24:21.739
I have written here.
24:21.739 --> 24:28.340
Time is often related to the simultaneity
of two events, we have just seen that simultaneity
24:28.340 --> 24:34.779
also depends on frame under the second postulate.
Therefore, probably we have reason to believe
24:34.779 --> 24:41.779
that t prime is not equal to t or t is not
equal to t prime; so probably one has to relook
24:41.899 --> 24:48.899
at the concept of time, and we cannot use
a simple equation like t prime equal to t.
24:49.700 --> 24:55.950
So, now we are again back into dark, we do
not know where to start, all we have said
24:55.950 --> 25:02.229
that probably time cannot be taken to be same
in all the frames; so let us start collecting
25:02.229 --> 25:09.229
all our evidences, and try to reach a new
transformation, this transformation is call
25:09.739 --> 25:12.039
Lorentz transformation.
25:12.039 --> 25:19.039
So, let us just before starting some observation,
we are as I said we are essentially as dark,
25:21.009 --> 25:27.289
so we have to start collecting all the clues
that we can get, I must mention that we are
25:27.289 --> 25:32.970
not deriving Lorentz transformation, this
things are not derivable, so when we say something
25:32.970 --> 25:38.330
to be derive, we can derive it from certain
fundamental laws. So, we are only collecting
25:38.330 --> 25:45.330
arguments we are not deriving it, these observations
whatever we have seen, whatever we feel should
25:47.460 --> 25:51.769
to be put into the set of certain equations,
but whether these the equations are correct
25:51.769 --> 25:57.029
or not depends on whether they can be explained
in the experiment.
25:57.029 --> 25:59.690
Like when we say of the Newton’s law of
motion, Newton’s law of motion we do not
25:59.690 --> 26:05.049
derive in our high schools we have never pull
how to derive the laws of motion, they became
26:05.049 --> 26:10.109
a fundamental law. We assume Newton’s laws
of motion and try to explain various things,
26:10.109 --> 26:15.100
and that is a way we have developed our physics.
Similarly, Lorentz transformation we are just
26:15.100 --> 26:20.889
collecting our evidences, some of these evidences
may not look very strong, some of the observations
26:20.889 --> 26:26.850
may not very strong, but eventually we believe
in them. Because, they are able to explain
26:26.850 --> 26:33.129
a large number of experiments which we cannot
otherwise explained, that is why we believe
26:33.129 --> 26:34.739
in Lorentz transformation.
26:34.739 --> 26:41.549
So, the first point about Lorentz transformation
is that, the transformation should be linear;
26:41.549 --> 26:47.549
I must mention that the thing I am taking
this particular treatment is given in Decimix
26:47.549 --> 26:54.080
book which I will be the reference will be
given in the next. So, what I mean by it linear
26:54.080 --> 26:58.419
transformation, if you see this particular
equation, we have written X prime equal to
26:58.419 --> 27:03.789
one constant which I am calling it as B X
X, because this is relating X to X this is
27:03.789 --> 27:10.789
B X Y relating X to Y, B X Z, B X t and C
1, we have written as X prime is equal to
27:11.619 --> 27:18.619
B X X X plus B X Y Y plus B X Z Z plus B X
t t, were all this B is are some constants
27:21.499 --> 27:28.039
which we to have determine. By linear mean,
we mean that this only one single power of
27:28.039 --> 27:31.679
X which is involved, there is single power
of Y which is involved, single power of Z
27:31.679 --> 27:36.789
which is involved, single power of t involved.
We do not have a turn involving, let us say
27:36.789 --> 27:43.789
X square or X cube or Y square are term like
X Y or term like X C, we expect the transformation
27:44.609 --> 27:51.609
to be linear, I will give you the reason why
do we expect like that. You look at Y prime,
27:51.970 --> 27:56.210
we have written exactly similar type of equation,
expect this concepts are different which in
27:56.210 --> 28:02.340
general would be different, this I have written
as B Y X, this Y represents this Y, this X
28:02.340 --> 28:09.340
represents this X, B Y X X plus B Y Y Y plus
B Y Z Z plus B Y t t plus C 2. Similarly,
28:12.659 --> 28:19.659
Z prime we have written as B Z X X plus B
Z Y Y plus B Z Z Z plus B Z t t plus constant
28:23.369 --> 28:26.700
C 3.
Then time we have written, which earlier was
28:26.700 --> 28:33.700
always equal to t, t prime is equal to B t
X X plus B t Y Y plus B t Z Z plus B t t t
28:35.349 --> 28:42.340
plus C 4, let me first given argument why
we expect this particular transformation to
28:42.340 --> 28:43.440
be linear.
28:43.440 --> 28:50.440
Let us suppose we had a transformation which
involve X square term, if we had a transformation
28:52.409 --> 28:59.409
which involved X square term and let us suppose,
let us just put a rod which is length 1 meter,
29:03.700 --> 29:10.700
which is placed between X is equal to 0, and
X is equal to 1. If this rod is put between
29:10.940 --> 29:17.749
X is equal to 0 and X is equal to 1, then
X 1 will be equal to 1 and X 0 will be equal
29:17.749 --> 29:24.749
to 0. Now, X 1 minus 0 will be equal to 1,
if I take X 1 square this will be 1, X 0 square
29:29.619 --> 29:35.259
will equal to 0; now let us suppose we displace
this rod, and put this particular rod now
29:35.259 --> 29:42.259
between X is equal to 1, and X is equal to
2, same 1 meter rod. Now, let us call this
29:42.529 --> 29:49.529
X 3, X 3 will be equal to 1, X 3 let we put
X 4 will be equal to 2, X 3 will be equal
29:54.450 --> 30:00.570
to 1, because this now at X is equal to 2
meters, and this is at X is equal to X is
30:00.570 --> 30:05.149
equal to 1 meter.
If we take X 4 square this would be 4, if
30:05.149 --> 30:12.149
we take X 3 square this is equal to 1, so
if by transformation term involved X square
30:13.809 --> 30:20.309
will find that X square X 1 square minus X
1 square is 1, while X naught X 4 square minus
30:20.309 --> 30:27.309
X 3 square is 3. So, this particular length
would appear to be 1 meter in X prime frame
30:27.619 --> 30:32.460
of reference, if we had a term like this,
and this particular length will turn out to
30:32.460 --> 30:39.349
be 3 meters. It means if this lens rod was
put between X is equal to 0 and 1, its length
30:39.349 --> 30:44.929
will turn out to be different, from the 1,
from the case from the rod was put between
30:44.929 --> 30:50.840
X is equal to 1 and X is equal to 2.
This looks somewhat wizards, this does not
30:50.840 --> 30:56.499
look somewhat reasonable that if we just placed
our rod, and displaced by 1 meter, the length
30:56.499 --> 31:01.239
of the rod in a different frame will appear
out to be different; after all whatever we
31:01.239 --> 31:05.519
have taken as origin could have been a slightly
different origin. We expect this space to
31:05.519 --> 31:10.529
be homogeneous, I do not expect the length
of a rod to be different in different frames
31:10.529 --> 31:17.529
based on where I put the rod, if the length
I can expect that you know the length of 1
31:18.729 --> 31:24.690
meter may not appear to be 1 1 meter, it may
appear to 1.2 to or 0.8 meter. But if I displace
31:24.690 --> 31:29.320
the rod, the same length 1.2 or 0.8 whatever
we had observed earlier, same length should
31:29.320 --> 31:33.509
be observed in X prime frame of reference,
by displacing the rod I should not be able
31:33.509 --> 31:38.739
to, I should not measure a different length.
Because, the space is homogeneous I do not
31:38.739 --> 31:43.729
expect length to be dependent on the fact
where I put this particular rod. Suppose,
31:43.729 --> 31:48.269
this is particular pen if I displace it here,
it should not appear that in a different frame
31:48.269 --> 31:53.399
of reference its length has changed, whatever
might be the length larger or smaller, but
31:53.399 --> 31:58.200
whatever was the length here, same should
be the length observed here also. So, therefore,
31:58.200 --> 32:05.200
I avoid all higher order terms and we expect
linearity, so with the first clue that we
32:05.239 --> 32:11.219
have the transformation must be linear, it
must involve only single powers, should not
32:11.219 --> 32:18.219
involve power like X square etcetera, etcetera.
So, the simple most transformation in linear
32:18.529 --> 32:22.539
is of this particular type which have just
now mentioned.
32:22.539 --> 32:29.299
This is what I have said linearity is essential
to maintain homogeneity of space, the length
32:29.299 --> 32:34.289
of the rod should not depend on the origin
chosen. And this is the reference of the book
32:34.289 --> 32:39.219
from which we have taken this particular derivation;
this is introduction to special relativity
32:39.219 --> 32:46.219
by Robert Resnick, by Wiley Eastern 1988.
Now, let us collect more evidences, what we
32:46.289 --> 32:52.539
will realize see earlier when we do use Galilean
transformation, we had arbitrarily put S and
32:52.539 --> 32:59.539
S prime frame; and taken a assertions a simply
a realistic view of the axes. Now, you realize
33:01.349 --> 33:06.999
that this simplistic view help us in getting
eliminated a large number of constant and
33:06.999 --> 33:10.229
therefore, making transformation equation
simple.
33:10.229 --> 33:17.229
I remind again, your two frames S and S prime,
this was origin O, this was origin O prime,
33:26.340 --> 33:33.340
this was my X axis, this was Y axis, this
is Z axis, this is Z prime axis, this is Y
33:33.340 --> 33:39.989
prime axis, X and X prime axis are always
constant. And the time was time t is equal
33:39.989 --> 33:46.629
to 0 was measured, when O prime was constant
with O, this are the special set of axes that
33:46.629 --> 33:51.919
we have taken for our Galilean transformation.
Now, let us see that by checking these set
33:51.919 --> 33:58.009
of axes, we are ready to get that we can eliminate
a large number of constants.
33:58.009 --> 34:05.009
So, this what in have written special choice
of axis make many coefficient 0, first making
34:06.840 --> 34:11.639
these particular origin choosing these origins
appropriately, remember we have chosen these
34:11.639 --> 34:18.639
origins that O prime, when it coincides with
O. Then and that particular movement the time
34:20.820 --> 34:26.810
is measured to be 0, so let us assume that
at the time when the origins of the two frames
34:26.810 --> 34:32.279
were constant, and event occurred whatever
might be the event, it occurred exactly at
34:32.279 --> 34:33.359
the origin.
34:33.359 --> 34:38.520
So, it means at that particular when the event
occurred, the coordinate of that particular
34:38.520 --> 34:45.520
event in S frame was X is equal to 0, Y is
equal to 0, Z is equal to 0, because it occurred
34:46.849 --> 34:52.919
at origin, and the time limit t is equal to
0. Because, this special choices that we have
34:52.919 --> 34:59.539
taken, we expect that this event in X prime
of reference would also occurs at the origin,
34:59.539 --> 35:03.539
because it was constant at that particular
time event, and because time was also measured
35:03.539 --> 35:07.890
at that particular instant of time; so time
in the particular frame will also turn out
35:07.890 --> 35:12.569
to be 0.
So, I expect that in S prime frame of reference
35:12.569 --> 35:19.569
also the event will turn out to be at 0, 0,
0 and at time t prime is equal to 0. Let us
35:22.220 --> 35:27.640
put these conditions in this particular transformation
equation which we have just now written, and
35:27.640 --> 35:30.760
let us see that we can see get read of some
constants.
35:30.760 --> 35:37.180
So, I have written let us imagine that an
event occurs at the origin in S at t is equal
35:37.180 --> 35:42.740
to 0, the event would also appear to occur
at the origin of S prime at t prime is equal
35:42.740 --> 35:47.390
to 0; it mean if I substitute X is equal to
0, Y is equal to 0, Z is equal to 0, t is
35:47.390 --> 35:54.390
equal to 0, then I must X prime equal to 0.
So, remember, if you look at this equation
35:54.770 --> 36:01.630
this will give you 0 is equal to 0 plus 0
plus 0 plus 0 plus C 1, which means that the
36:01.630 --> 36:08.630
C 1 must be 0 that cannot be a constant term.
Similarly, we can put these conditions in
36:09.119 --> 36:15.430
the Y Z coordinate system also, Y Z transformation
equation also, and we will conclude that C
36:15.430 --> 36:18.680
2 is also equal to 0, and C 3 is also equal
to 0.
36:18.680 --> 36:24.630
Similarly, for time if the event occurred
at the time at t is equal to I put it 0 here,
36:24.630 --> 36:29.240
occurs at Y is equal to 0 I put 0 here, occurs
at Z is equal to 0 here, it occurred at t
36:29.240 --> 36:36.240
is equal to 0 I put it here; then the watch
of S prime observer also measured time t prime
36:38.289 --> 36:45.289
is equal 0; so if I put this is equal to 0
I will get C 4 also to be 0. So, I conclude
36:46.380 --> 36:51.619
that by choosing these special set of axis,
this special conditions on the two frames
36:51.619 --> 36:58.020
I am able to get read of the constant C 1,
C 2, C 3, and C 4.
36:58.020 --> 37:04.049
So, this is what I written here, and the transformation
equation now reduced to this, so remember,
37:04.049 --> 37:10.109
so many constants out of these C 1 count down
as become 0, C 2 has been removed is become
37:10.109 --> 37:15.569
0, C 3 has been removed it has become 0, C
4 has been removed it has become 0. So, know
37:15.569 --> 37:21.869
we are still left with the 16 constant still
very large number of the constant, and let
37:21.869 --> 37:28.869
us see if we get read of some other constants.
Now, let us start looking again the same axis,
37:29.329 --> 37:36.329
and trying to see how this special tries to
make that axis can help us of some more constants.
37:38.200 --> 37:43.369
So, we try to fix the place appropriately
which we have already done by making this
37:43.369 --> 37:50.369
special choices which is given here.
Let us now imagine that an events occurred
37:51.880 --> 37:58.180
at a arbitrary time in X Y plane, so remember
this is your X Y plane, this is your X direction
37:58.180 --> 38:04.980
this is your Y direction, so this is X Y plane.
Now, we realize that X frame of reference,
38:04.980 --> 38:11.980
this X prime X axis is constant for X, Y prime
axis is parallel to Y, so we expect that this
38:13.020 --> 38:20.010
X prime, Y prime plane is exactly same as
X Y plane, we have the same planes. Because,
38:20.010 --> 38:25.539
this particular lines moves in that particular
plane, the Y prime direction moves in this
38:25.539 --> 38:31.720
plane, so X Y plane is same as X prime Y prime
plane.
38:31.720 --> 38:38.720
It means that, if an event occurs in X Y plane
an observer in X frame of reference will also
38:38.799 --> 38:45.799
observe it to be is X prime and Y prime plane,
what is X Y plane characteristic of, X Y plane
38:50.569 --> 38:57.569
is characteristic by the fact that Z is equal
to 0, X Y plane means Z coordinate 0. Now,
39:01.660 --> 39:08.660
X prime Y prime plane also means Z prime equal
to 0, so what we conclude from this particular
39:10.099 --> 39:17.099
discussion that if an event has occurred with
Z equal to 0 means at the X Y plane, it would
39:17.500 --> 39:22.750
appear to an observer in the X prime also
to occur at the X prime in the X X prime Y
39:22.750 --> 39:27.750
prime plane, it means according to that observer
Z prime will also equal to 0.
39:27.750 --> 39:32.910
So, let us put this equation at the condition
in our equations, and let see what happens,
39:32.910 --> 39:39.910
this is the third equation which I have put,
I have put Z is equal to 0 here and I expect
39:40.960 --> 39:47.960
that if Z is equal to 0 I must also equal
to Z prime equal to 0 or this is possible
39:48.220 --> 39:55.220
only remember this event occur at any arbitrary
value if X, Y and t. But so long it occurs
39:55.440 --> 40:02.440
in with coordinate Z is equal to 0, the coordinate
of the same event must also have Z prime axis
40:02.510 --> 40:08.720
equal to 0. So, irrespective of value of X,
Y and t this equation must always be true
40:08.720 --> 40:14.730
that is only possible, if this coefficients
is 0, this coefficient is 0, this coefficient
40:14.730 --> 40:20.880
is 0, this coefficient is 0, so one short
we get read of these three constants.
40:20.880 --> 40:27.880
This is what I have written, this is possible
only if B Z X, B Z Y, B Z T are all set to
40:28.349 --> 40:34.740
0, then only it is possible that an event
which occurs when Z is equal to 0 will also
40:34.740 --> 40:41.740
appear to an observer in S prime to occur
at the Z prime equal to 0. Similarly, we now
40:42.990 --> 40:47.049
look at the Z plane, so let us come back to
this particular equation remember at Z plane,
40:47.049 --> 40:54.049
the X Z plane, you also realize that X Z plane
is same as X prime Z plane, Z prime plane.
40:56.150 --> 41:03.150
So, if an event occurs in X Z plane implying
Y equal to 0, then the event would also appear
41:05.299 --> 41:10.940
in S prime frame of reference occurred in
X prime Z prime plane, it must occur with
41:10.940 --> 41:17.650
Y prime is equal to 0, exactly the same condition
that we have applied for X Y plane, so we
41:17.650 --> 41:20.089
put exactly the same thing, the same condition.
41:20.089 --> 41:25.019
That corresponding to Y equal to 0, you must
have Y prime is equal to 0 therefore, you
41:25.019 --> 41:32.019
get read off constant this constant, this
constant, this exactly the same argument that
41:32.039 --> 41:39.039
we have used for X Y plane, so we get read
of another three constants. Now, we remember
41:40.710 --> 41:46.480
there is one plane which is different and
that is Y Z plane, this plane Y Z plane, and
41:46.480 --> 41:51.910
Y prime Y prime Z prime plane, these plane
is constant only at t is equal to 0, at later
41:51.910 --> 41:58.599
time this plane goes there parallel, but they
were moving relative to each other. So, it
41:58.599 --> 42:05.250
was different for X Z and X Y planes which
remains always identical at the frames, but
42:05.250 --> 42:10.450
as per the Y Z plane is concerned they were
identical exactly at t is equal to 0, but
42:10.450 --> 42:15.440
at later time they have moved each other.
And how much they have moved, they have moved
42:15.440 --> 42:20.680
off by the distance v times t, were v is the
relative speed between the frame.
42:20.680 --> 42:27.680
So, an observer in the S frame would find
that this particular Y prime Z prime plane
42:27.740 --> 42:34.700
has moved with respect to Y Z plane in a time
t with a speed v, so the distance between
42:34.700 --> 42:41.700
them is v t. So, let me repeat an observer
in S frame would find that the Y prime Z prime
42:42.839 --> 42:49.839
plane has moved with respect to Y Z plane
by a distance of v t during time t, this time
42:50.279 --> 42:57.279
t is being measured in S frame, all the observations
are being made in the S frame; let us see
42:57.289 --> 43:01.900
what does they put, this puts condition on
our constants.
43:01.900 --> 43:08.900
So, this is what we I have written, let us
look at the Y Z plane at the time t is equal
43:09.869 --> 43:15.289
to 0, if an event occurred in this plane it
would also occurred at the Y prime, Z prime
43:15.289 --> 43:20.950
plane observer in the S prime. But, at the
later time the X coordinate of this particular
43:20.950 --> 43:25.740
event will be shifted by v t as observed in
the S frame.
43:25.740 --> 43:32.319
Therefore, the transformation equation must
look as follows, taken as quite a bit of jump
43:32.319 --> 43:39.029
here which is X prime is equal to B X X into
X minus v t, because if I put X is equal to
43:39.029 --> 43:45.089
v t then I must get X prime is equal to 0.
Remember, this equation if it put X prime
43:45.089 --> 43:49.529
is equal to if we put X is equal to v t, then
I must get a prime at this particular point
43:49.529 --> 43:56.039
Y prime Z prime plane and other constants
must turn out to be 0. So, we can find this
43:56.039 --> 44:03.039
constants to be 0, we find this constant to
be 0, and X prime we expect to be after for
44:03.589 --> 44:10.589
B X X multiplied by X minus v t. So, as we
can see that this special choice of the axis
44:11.260 --> 44:16.660
have let to, let us into a large number into
a large number of amount of simplification
44:16.660 --> 44:20.160
getting rid of, so many constants.
44:20.160 --> 44:27.160
So let us clean up all our equations and we
obtain the following equations X prime is
44:28.990 --> 44:35.990
equal to B X X into X minus v t, Y prime is
equal to B Y Y times Y, Z prime is equal to
44:37.000 --> 44:44.000
B Z Z times Z, t prime is equal to B t X multiplied
by X plus B t Y multiplied by Y plus B t Z
44:47.210 --> 44:54.210
multiplied by Z plus B t t multiplied by t;
we still we have 4, 6, 7 constants. Let us
44:55.140 --> 45:01.190
see how we are going to evaluate this constants,
but I hope you are able to appreciate that
45:01.190 --> 45:07.210
the special set of axis that we have chosen
know that they are not really that the special
45:07.210 --> 45:13.539
in general enough. But just by choosing this
set of axis we are able to simplify in large
45:13.539 --> 45:20.359
amount of we are able to simplify our transformation
equations otherwise we do not know how to
45:20.359 --> 45:26.349
go about it.
Now, let us look at the symmetry arguments,
45:26.349 --> 45:31.789
the symmetry arguments are very interesting
arguments which often can work out very well
45:31.789 --> 45:38.789
in physics. See when I am talking about the
X axis, X axis is sort of the well defined
45:40.609 --> 45:45.950
axis, because X axis is the direction of the
relative speed, see suppose you are just looking
45:45.950 --> 45:52.670
at the motion of two frames, and you have
to define the relative velocity direction,
45:52.670 --> 45:58.660
you have to define the X direction how will
proceed, will proceed by taking the direction
45:58.660 --> 46:03.799
of the relative motion is the X direction.
But once we have chosen X direction, the Y
46:03.799 --> 46:07.339
direction whether I took this way, whether
I took this way or whether I took another
46:07.339 --> 46:09.130
angle, it does not make a difference.
46:09.130 --> 46:15.430
For example, if I take this my X axis, I could
have taken this is my Y axis or I could take
46:15.430 --> 46:20.500
this is my Y axis making an angle. So, long
it is perpendicular take this is as Y axis,
46:20.500 --> 46:26.930
Y axis was under my control there is no reason
of specifying or choosing a particular specific
46:26.930 --> 46:33.539
Y value Y axis, X axis is definitely X axis
is the direction along which the relative
46:33.539 --> 46:37.690
velocity occurs. So, this is the particular
direction which is the unique direction in
46:37.690 --> 46:43.410
along, which sort of symmetry is broken, because
that direction is the unique direction defines
46:43.410 --> 46:49.930
by the relative velocity directions.
But, the same thing, I can choose Y axis anyway
46:49.930 --> 46:56.380
I like of course, perpendicular to X then
between X and Y, Z axis has perpendicular
46:56.380 --> 47:01.400
to X axis and Y axis, and must follow the
written rule, whatever is the standard rule
47:01.400 --> 47:05.779
for that transformation we were choosing the
axis. But, on the other hand this Y axis could
47:05.779 --> 47:12.779
have been any were along this particular plane,
so I do not expect my physics laws to change
47:13.190 --> 47:20.000
depending on what I choose this is my Y axis.
So, there has to symmetric in that particular
47:20.000 --> 47:26.859
direction, so let us imagine now, let us imagine
that we have chosen some particular Y axis
47:26.859 --> 47:33.010
correspondingly we have chosen Z axis, so
my minus Y prime axis, minus Z prime axis
47:33.010 --> 47:38.690
all this axes are known.
Now, let us assume an event and X naught,
47:38.690 --> 47:45.690
Y naught, Z naught and X naught, minus Y naught,
minus Z naught, this is your X direction,
47:47.500 --> 47:53.609
this is Y direction, this is a direction as
I say X direction is a unique direction, which
47:53.609 --> 47:58.170
I cannot change, because that depends on the
relative velocity. Y direction I could choose
47:58.170 --> 48:02.490
anything I have chosen one particular thing
which is happening which is around in this
48:02.490 --> 48:06.289
particular direction, let us suppose anything
that two events which are occurring here and
48:06.289 --> 48:13.289
exactly with the negative value Y. So, this
occurs X naught, Y naught, Z naught, this
48:15.039 --> 48:22.039
occurs at X naught, minus Y naught, and Z
naught all, so all that difference between
48:23.069 --> 48:29.000
the coordinates of these two events is the
Y coordinate, otherwise all other two coordinates
48:29.000 --> 48:34.410
are same.
Now, looks back at our time equation which
48:34.410 --> 48:39.799
we have written earlier, which is t prime
is equal to B t X X plus B t Y Y plus B t
48:39.799 --> 48:46.799
Z Z, now if I change Y to minus Y and B t
Y is not 0, then t prime will turn out to
48:50.339 --> 48:57.339
be different for different frames; this Y
change in sign of Y will change the time.
48:58.410 --> 49:03.119
But, of course nothing is special note this
particular Y, what I have called is this particular
49:03.119 --> 49:10.119
Y not positive, I could have chosen my Y axis
just an opposite to that, now let my X axis
49:10.130 --> 49:14.740
let to change, but this Y axis I now taken
as negative of that.
49:14.740 --> 49:20.990
And that event would have now appear to occur
at minus Y naught, what event appear be occurring
49:20.990 --> 49:26.039
to plus Y naught, may have occurred at minus
1, because this just depends on us how I choose
49:26.039 --> 49:32.460
my axis. So, I do not expect that physics
will change, because just I have taken different
49:32.460 --> 49:37.329
set of Y axis, so the time of event the other
person do not know the choose of their axis,
49:37.329 --> 49:43.700
it is time note down to be different, depending
upon what I have chosen is that X Y axis or
49:43.700 --> 49:49.240
what I have chosen my minus Y axis. So, I
do not expect that by reversing the coordinate
49:49.240 --> 49:56.240
Y naught, I would expect the time to change
similarly, if I assume two events to be occurring,
49:57.920 --> 50:04.009
at exactly the same value X naught, Y naught
and only Z naught being different.
50:04.009 --> 50:11.009
And occurring at X naught, Y naught, minus
Z naught I do not expect the time to be different,
50:17.849 --> 50:22.759
because these are essential symmetry arguments,
the coordinate that I have chosen have just
50:22.759 --> 50:28.339
depend upon one particular choice for which
I do not have any prior information how to
50:28.339 --> 50:34.480
choose the particular, that was totally arbitrary.
Therefore, just by taking the choice of the
50:34.480 --> 50:41.480
axis time of the event should not change,
as we viewed in time frame of reference. Remember,
50:41.680 --> 50:46.529
I remain along the X direction is the unique
direction, X direction things can change,
50:46.529 --> 50:52.019
because there is the special reason to choose
the X direction, but not for Y and not for
50:52.019 --> 50:58.380
Z. So, I expect that these two coordinates
will lead to constants B t Y and B t Z must
50:58.380 --> 51:02.880
be equal to 0, in order to maintain symmetry.
51:02.880 --> 51:07.289
So, this is what I have written, imagine the
two events occurring at the same time in S
51:07.289 --> 51:12.200
frame, let the coordinate of the two events
be X naught, Y naught, Z naught, and X naught,
51:12.200 --> 51:18.130
minus Y naught, and Z naught. If t prime depends
on Y, then time of these two events would
51:18.130 --> 51:20.329
appear to be different in different frame.
51:20.329 --> 51:26.480
But, what we call Y axis could have also been
termed as minus Y, the choice of X axis is
51:26.480 --> 51:32.569
unique, as it is determine by the direction
of relative velocity in our choice of axis,
51:32.569 --> 51:39.559
but not Y and Z hence, the transformation
equation must appear as follows.
51:39.559 --> 51:46.559
It means, I must have these two terms also
becoming 0, therefore I get read of two future
51:49.480 --> 51:56.480
constants, so now we are left with only 1,
2, 3, 4, and 5 constants.
51:57.049 --> 52:03.819
This is what is I am left with, as we can
see that simplified reasonable well, purely
52:03.819 --> 52:10.819
by physics arguments, and special set of axis
transformation equation is simple for. I would
52:11.490 --> 52:17.119
just like to remain you that Galilean transformation
is also a special case of this equation, remember
52:17.119 --> 52:24.119
in Galilean transformation B X X was 1, B
Y Y was 1, B Z Z was 1, B t X was 0, B t t
52:26.789 --> 52:33.789
was 1. So, Galilean transformation also a
special case of this remember, till now we
52:34.410 --> 52:39.559
invoke any special or any postulate is the
special theory of relativity, this transformation
52:39.559 --> 52:44.640
equation whatever we have discussed are general
enough, so long we take the set of axis that
52:44.640 --> 52:51.599
we chosen. So, this expected to generally
to way that the classical rule or under special
52:51.599 --> 52:57.579
theory of relativity, we expect the same equations
to become.
52:57.579 --> 53:02.809
So, let go to the summary of whatever we have
discussed today, we discussed using examples
53:02.809 --> 53:09.119
that we have to attack the equation, which
makes time same in two frames. We then discuss
53:09.119 --> 53:14.769
the form of transformation equations without
invoking any postulates of relativity only
53:14.769 --> 53:20.640
making time relative, only we have taken consider
the fact that this time could also be related,
53:20.640 --> 53:25.470
we have not invoke any things of special theory
of relativity. See in our next lecture, we
53:25.470 --> 53:29.700
will invoke now the conditions of special
theory of relativity, and we will determine
53:29.700 --> 53:34.960
all the constants that we have not so for
determine, and we will arrive at what we call
53:34.960 --> 53:36.769
today Lorentz transformation.