WEBVTT
00:19.970 --> 00:26.970
Hello, so let us recapitulate what we had
discussed in our last lecture. We discussed
00:28.869 --> 00:35.869
two important consequences or I would say
two most popular consequences of Lorentz transformation;
00:36.839 --> 00:43.239
one is length contraction, another is time
dilation. We had defined something called
00:43.239 --> 00:49.440
a proper length that is the length of an object
measured in a reference, frame of reference,
00:49.440 --> 00:54.390
in which the object is at rest; we called
that particular length as a proper length.
00:54.390 --> 01:01.150
We said that if we go to any other frame of
reference, the x-component of this particular
01:01.150 --> 01:06.600
length will appear to be contracted in a different
frame of reference. Of course, as we all know
01:06.600 --> 01:12.330
that the direction of x is always chosen along
their relative velocity between the frames.
01:12.330 --> 01:16.119
The other aspect that we had discussed was
time dilation.
01:16.119 --> 01:22.150
Similarly, we had defined a quantity called
a proper time interval, which is a time interval
01:22.150 --> 01:29.150
between two events in a given frame in which
the two events occur at the same position.
01:29.900 --> 01:35.290
This proper time interval appears to be dilated
or supposed to be enhanced or increased, if
01:35.290 --> 01:40.110
you go to any other frame of reference, in
which these two events do not appear at the
01:40.110 --> 01:46.320
same position. Then eventually we give some
examples relating to the length contortion.
01:46.320 --> 01:53.320
Today, we will try to give some more examples,
initially to discuss various other aspects
01:53.630 --> 01:59.840
of Lorentz transformation. Let us, let us
go to one example which is probably again
01:59.840 --> 02:02.400
a simple example.
02:02.400 --> 02:09.119
Let us assume that there is an observer standing
on a platform, a train platform, the type
02:09.119 --> 02:16.119
of platform that we normally are quite aware
of. Now in that particular platform a train
02:17.530 --> 02:23.049
passes by it, let us assume that platform
is also on a frame of reference which is inertial.
02:23.049 --> 02:28.230
Though we all know that earth is not strictly
speaking a inertial frame of reference but,
02:28.230 --> 02:32.980
we for all practical purposes we can treat
it and let us assume that this particular
02:32.980 --> 02:37.730
platform is on an inertial frame of reference.
And let us also assume that the train which
02:37.730 --> 02:44.620
is passing by the platform is also moving
with constant velocity therefore, it also
02:44.620 --> 02:49.220
represent n inertial frame of reference. So
there is no acceleration involved, so we can
02:49.220 --> 02:53.180
always treat one frame reference is s and
s another is s one s prime.
02:53.180 --> 02:59.420
Now both the observer an observer on platform
another observer sitting in a compartment
02:59.420 --> 03:05.020
in a train both of them want to measure the
length of the platform assuming that this
03:05.020 --> 03:10.569
particular is moving to relativistic speed.
The session means that the speed is very high
03:10.569 --> 03:16.819
close to speed of light. Then let us do have
some discussion about the measurement of the
03:16.819 --> 03:20.470
length of the platform.
So this is what I have written as the subject
03:20.470 --> 03:26.010
of the example. Measurement of a length of
a platform, by an observer stationary on the
03:26.010 --> 03:32.080
platform and another that is another observer
moving in a train with relativistic speed
03:32.080 --> 03:38.510
relative to of course, platform. Assume botht
rains to be inertial.
03:38.510 --> 03:45.510
This figure approximately represents the situation
in this particular case. Let us assume that
03:45.930 --> 03:50.480
is the platform, platform number you know
the beginning of the platform is A; the end
03:50.480 --> 03:56.739
of the platform is B and there is an observers
standing on the platform. You can assume that
03:56.739 --> 03:59.640
is origin you can assume it to be atman called
an.
03:59.640 --> 04:06.620
We can move that thing and this represents
the frame of reference S. This is a train
04:06.620 --> 04:12.040
which is moving velocity v which is we are
assuming to be relativistic speed very high
04:12.040 --> 04:19.040
speed. And observer is standing or sitting
in a proper a particular seat in this particular
04:19.470 --> 04:24.810
compartment and which called this frame of
reference as S prime frame of reference. And
04:24.810 --> 04:31.810
as we said that both observer S and S prime
want to measure the length of the platform
04:32.090 --> 04:38.460
A B. Now let us define events. As we always
said that the relativity is much easier to
04:38.460 --> 04:42.690
solve problems if we define events and write
their coordinates including time.
04:42.690 --> 04:49.620
So that is what we will do. We will define
two events relating to this measurement of
04:49.620 --> 04:54.729
length I mean similar thing which we have
done, also in the case of time dilation and
04:54.729 --> 05:01.729
dilation. So we come to the fact when we define
the events, my first event is defined when
05:03.430 --> 05:10.430
the origin of S prime coincides with A end
of the platform. Essentially means that whatever
05:11.050 --> 05:15.810
is origin of S prime we can take as one of
the corner let us move this particular corner
05:15.810 --> 05:22.810
coincides with point A. Because train is moving
a relative to B at a later time the symbol
05:22.910 --> 05:28.160
if we take for example, this as the origin
of this particular S prime frame of reference
05:28.160 --> 05:34.820
it makes so difference what we take. Then
this point coincides with point B of the platforms,
05:34.820 --> 05:39.470
so that is my event number 2.
So I repeat event number A is that origin
05:39.470 --> 05:46.290
of S prime coincides with A, event number
B at 2 is origin of S prime coincides with
05:46.290 --> 05:51.350
end B of the platform. So these are my two
events. Let us try to write the coordinates
05:51.350 --> 05:55.139
of these events in both S and S prime frame
of reference.
05:55.139 --> 06:02.139
So here I am defining my events E 1 event
one, the origin of S prime coinciding with
06:02.370 --> 06:08.790
end A of platform. E 2 the origin of S prime
coinciding with end B of the platform. We
06:08.790 --> 06:13.460
can choose any origin in S prime. It does
not make a difference, because what matters
06:13.460 --> 06:19.240
is only the difference but, for thinking proposal
we can another one end or somewhere in the
06:19.240 --> 06:22.440
centre or at any other point.
06:22.440 --> 06:29.440
Let us write the events or before that let
me raise some questions. Is the coordinate
06:30.259 --> 06:34.300
difference equal to the length of the platform
in any of the frames? This is a question number
06:34.300 --> 06:41.300
1 that I imposing. So if we take the difference
of X let us assume that AB is a wrong X direction
06:41.400 --> 06:45.060
because that is what we have any way assumed.
If we want to apply Lorentz transformation,
06:45.060 --> 06:49.770
because relative velocity direction is always
chosen along the x direction or other x direction
06:49.770 --> 06:55.520
is always chosen along the relative velocity
direction. So what is a ordinate difference
06:55.520 --> 07:01.630
that is x 2 minus x 1? Is it equal to does
it give equal to other length in any of the
07:01.630 --> 07:07.050
frame of reference S or S prime? Question
number 2, the time difference between these
07:07.050 --> 07:12.259
two events; it means we take the time t 1
when event number 1 occurs, time t 2 when
07:12.259 --> 07:18.530
event number 2 occurs. We take time difference
t 2 minus t 1 is proper in any of the two
07:18.530 --> 07:23.160
frames that I have just now described namely
S and S prime.
07:23.160 --> 07:30.160
Let us have a little bit of discussion. We
realise that two events occur at different
07:33.150 --> 07:39.169
ends of the platform. One occur at end A another
in occurs at end B, this what we have already
07:39.169 --> 07:46.169
said. And neither in S nor in S prime the
occur at the same time. So there is always
07:46.330 --> 07:51.039
you going to be a time difference between
event 1 and event 2 whether it is S which
07:51.039 --> 07:55.240
observes it or whether it is S prime observes
it.
07:55.240 --> 08:01.419
And if you remember we have said it earlier
that in object which is moving and if we measure
08:01.419 --> 08:05.500
the coordinates of the two ends then that
is not going to give you correct length. It
08:05.500 --> 08:09.979
is going to be a coordinate differences going
to give you correct length only in a situation
08:09.979 --> 08:16.080
when the object is at rest, because then only
x 2 minus x 1 is corrected otherwise if there
08:16.080 --> 08:21.160
is a time difference; during a time difference
of measurement the object has moved and the
08:21.160 --> 08:25.569
coordinate difference would not actually give
the correct length. So we realise that at
08:25.569 --> 08:32.240
this particular situation also it is only
the S frame that is the platform frame that
08:32.240 --> 08:38.610
it is only that platform frame that the platform
is not moving. In the train frame the platform
08:38.610 --> 08:42.610
is actually moving and am assuming the length
of the platform.
08:42.610 --> 08:47.290
So if the coordinates of the two events has
to give me the length of the platform then
08:47.290 --> 08:52.050
this platform must be raised at rest, because
the two events are occurring at two different
08:52.050 --> 08:59.050
times. And that is only the frame S, so I
conclude that in frame S the length of the
09:02.139 --> 09:08.459
platform will be given by the coordinate differences
x 2 minus x 1. This is what I have written.
09:08.459 --> 09:12.779
Let us just read, the two events occur at
the two different ends of the platform.
09:12.779 --> 09:19.779
However, times for two event is different
both in S and S prime. We just now discussed.
09:20.059 --> 09:25.139
Hence the coordinate difference will be equal
to the length of the platform only that frame
09:25.139 --> 09:29.499
in which platform is at rest, that is something
which I have been probably over emphasizing.
09:29.499 --> 09:34.529
But, this is something which is to be understood,
because this is where people normally tend
09:34.529 --> 09:38.569
to make a mistake. They just write tried x
2 minus x 1 equal to length. X 2 minus x 1
09:38.569 --> 09:45.569
is length only in case the object is not moving
or the time should be same. So hence the coordinate
09:49.379 --> 09:53.160
difference will be equal to length of the
platform only in that frame in which platform
09:53.160 --> 09:55.889
is at rest and that is the frame S.
09:55.889 --> 10:02.889
Now if you go to S frame, suppose we are standing
on a platform and you watch the train coming
10:05.360 --> 10:12.360
you realise that event number one actually
occurred at point A. Let us write here let
10:13.040 --> 10:20.040
say this is A this is B. So if suppose my
observer was stranding here, what he would
10:24.189 --> 10:30.839
observe that train is coming from this side
and passing by the platform and going to the
10:30.839 --> 10:36.319
other side. And let suppose the front end
is origin of S prime, so when this particular
10:36.319 --> 10:43.319
train reaches this particular object this
particular point event one occurs. When the
10:45.279 --> 10:52.129
same train comes and reaches at point B event
number two occurs.
10:52.129 --> 10:59.129
So position of event number one is A; position
of event number two is B. And we have discussed
11:00.049 --> 11:05.040
the position difference of this would be giving
the length of the platform. But, we realise
11:05.040 --> 11:12.040
that these two events actually occur at different
location in S prime S frame. It means that
11:12.799 --> 11:17.540
the time difference between them is not proper,
because for the proper frame for the time
11:17.540 --> 11:22.720
difference to be proper these two events must
occur at the same position.
11:22.720 --> 11:28.610
But what an observer S prime would be feeling?
If you look from the point of view an observer
11:28.610 --> 11:35.600
S prime, he would see completely different
situation. He would feel or she does not see
11:35.600 --> 11:42.600
his motion. Let us suppose this my origin
of the train; he will feel that this particular
11:43.480 --> 11:50.480
platform A B is actually approaching towards
him, because this person observer sitting
11:53.360 --> 11:59.160
in S prime does not notice his or her own
motion. So according to him the platform is
11:59.160 --> 12:03.920
actually coming and going by this side. This
is a very common thing if you are moving in
12:03.920 --> 12:09.059
a fast train and you see a particular station
at least a train does not stop, it appears
12:09.059 --> 12:13.889
as if one end of the platform is coming towards
you, then it passes by and the second end
12:13.889 --> 12:18.160
of the platform reaches you.
So in fact this person standing here or sitting
12:18.160 --> 12:24.189
here will observe exactly the same thing.
That end A of the platform first reached him,
12:24.189 --> 12:30.790
that end B of the platform be at him. So according
to an observer an S prime the first event
12:30.790 --> 12:36.869
that is A point meeting this particular point,
that occurs at this particular point event
12:36.869 --> 12:43.220
number two that is point B meeting this particular
point, again occurs at the same point. So
12:43.220 --> 12:48.279
as far as the observer sitting on the train
is concerned he is sitting comfortably or
12:48.279 --> 12:53.489
standing comfortably at his place, he notices
end of A of the platform coming towards him
12:53.489 --> 12:57.879
passing by it, then end B of the b of the
platform coming towards him passing by it.
12:57.879 --> 13:04.069
He is standing at his origin or he is sitting
at his origin. For him both the events point
13:04.069 --> 13:09.949
A, coinciding with him, point B coinciding
with him occurs at the same position. Hence
13:09.949 --> 13:14.199
the time interval that will be measured by
this observer in S prime frame of reference
13:14.199 --> 13:18.290
would really be a proper frame of reference
proper time interval.
13:18.290 --> 13:25.290
So that is what I have written in S the two
events occur at two different locations. What
13:25.549 --> 13:31.019
an observer S prime would feel? An observer
S prime would feel as I said that they occur
13:31.019 --> 13:36.359
at the same position wherever the person is
sitting or standing. Hence time difference
13:36.359 --> 13:41.999
is proper in S prime frame of reference. So
therefore, in S frame this time difference
13:41.999 --> 13:42.910
must be dilated.
13:42.910 --> 13:49.910
Now let us go to frame S and try to write
the coordinates. We just now agreed that if
13:52.559 --> 13:58.149
we are looking with respect difference to
the frame of reference S, which is the person
13:58.149 --> 14:05.149
standing on the platform the two events the
difference of two events will give the actual
14:06.790 --> 14:07.389
length.
14:07.389 --> 14:14.389
Because according to this observer this event
occurs let us say at X 1 t 1 this event occurs
14:16.759 --> 14:23.759
at X 2 t 2,now clearly X 2 which is here X
1 which is here, both these events the position
14:31.699 --> 14:37.329
difference X 2 minus X 1 this minus this will
actually give me the length of the platform.
14:37.329 --> 14:44.329
So I must have X 2 minus X 1 is equal to l
which is the length of the platform. That
14:44.449 --> 14:51.449
is what I have written here. That in S the
length is proper and would be given by l is
14:51.579 --> 14:57.119
equal to X 2 minus X 1 where l I have taken
as the length of the platform as measured
14:57.119 --> 15:01.929
in S frame of reference.
However this particular observers had another
15:01.929 --> 15:07.249
way of measuring the length, provided he knows
the relative velocity of the train. If he
15:07.249 --> 15:12.989
knows with what speed the train is passing
by it, then what he could have done, he could
15:12.989 --> 15:19.989
have just measure the time difference between
the two events event one and event two. But,
15:20.919 --> 15:24.009
for that he must to have he must have the
information of speed and he must have the
15:24.009 --> 15:28.829
information of the time differences.
So if he knows the time difference t 2 and
15:28.829 --> 15:35.379
t 1, t 2 minus t 1 rather, then if you multiplies
this speed of the, speed of the train, he
15:35.379 --> 15:40.600
should still be able to get the length of
the platform. Or that is another way of determining
15:40.600 --> 15:45.369
the length of the platform. If I know the
speed of the train and I know how much time
15:45.369 --> 15:50.660
the train took how much time the train took
to go from here to here, that particular time
15:50.660 --> 15:56.379
and if I know the time difference between
these two, then the speed multiplied by the
15:56.379 --> 15:58.959
time would give me the length of the platform.
15:58.959 --> 16:04.439
So this what I have written here that this
length would also be given by v multiplied
16:04.439 --> 16:09.749
by t 2 minus t 1 which is time difference
between these two events. But, as we have
16:09.749 --> 16:14.989
just now said that this time difference must
be dilated, because it is in the S prime frame
16:14.989 --> 16:21.929
of reference, in which the time in interval
is proper. Therefore, this t 2 minus t 1 must
16:21.929 --> 16:28.749
be given by gamma times t 2 minus t 1 prime,
where as we all know gamma depends on the
16:28.749 --> 16:35.579
relative speed of the train and is actually
given by gamma is equal to 1 upon under root
16:35.579 --> 16:42.579
1 minus v square by c square.
So if I know the relative speed of the train,
16:45.319 --> 16:49.809
I also know the gamma. Once I know the gamma
I can substitute in this particular value
16:49.809 --> 16:55.139
gamma and I know that t 2 prime minus t 1
prime that is a time difference measured in
16:55.139 --> 16:59.529
S prime frame of reference which is actually
the proper time interval between these two
16:59.529 --> 17:05.490
events. That multiply by gamma must give me
t 2 minus t 1. So this is equal to gamma times
17:05.490 --> 17:10.630
tow. If you remember we have mentioned that
normally we have been writing the proper time
17:10.630 --> 17:15.439
interval between two events as tow. So this
is gamma times tow.
17:15.439 --> 17:22.439
Now let us go with respect to the frame of
reference of S prime. We agreed that in S
17:25.689 --> 17:32.070
prime frame of reference the time interval
was really proper. But, what about position
17:32.070 --> 17:38.570
difference, what about X 2 minus X 1?
We just now seen that, as far as an observer
17:38.570 --> 17:43.309
S prime is concerned the two events occur
at the same position; it means X 2 minus X
17:43.309 --> 17:47.460
1 must be equal to 0. If that is the reason
we have called that this time interval is
17:47.460 --> 17:53.840
a proper time interval, because both the events
occurred at the same position. Is I say as
17:53.840 --> 18:00.070
I say means that X 2 minus X 1 is 0. X 2 prime
rather X 2 prime minus X 1 prime is 0. So
18:00.070 --> 18:04.259
if we take the coordinate difference in S
prime frame of reference that coordinate difference
18:04.259 --> 18:11.259
is 0. So again I remind you that X 2 prime
minus X 1 prime is not equal to the length.
18:11.289 --> 18:17.740
So because actually in the S prime frame of
reference split from which is moving, therefore,
18:17.740 --> 18:23.700
just taking the coordinate difference and
putting equal to length can we disastrous.
18:23.700 --> 18:28.029
So this is what I have written here. Time
interval is proper in S prime frame of reference
18:28.029 --> 18:34.210
I was actually equal to t2 prime minus t1
prime and difference in X coordinate X 2 prime
18:34.210 --> 18:40.240
minus X 1 prime is equal to 0, which is definitely
not equal to the length prime; it means the
18:40.240 --> 18:46.720
length measured length of the platform measured
in S prime frame of reference. Now suppose
18:46.720 --> 18:50.950
you are sitting in a, you are sitting in a
train and want to find out the length of the
18:50.950 --> 18:57.850
platform. What way for example, you can adopt?
You can have a stop watch in your hand and
18:57.850 --> 19:03.960
then end A comes you can start the stop watch,
when end B arise you can stop the stop watch,
19:03.960 --> 19:09.240
measure the time difference. Once you measure
the time difference between these two events
19:09.240 --> 19:14.230
that would be t 2 prime minus t 1 prime and
as I have agreed this is a proper time interval.
19:14.230 --> 19:19.740
And if I know what is the speed of the platform
relative to me an S prime, which essentially
19:19.740 --> 19:26.630
is a same speed as an observer in S will notice
of the train. If I know that particular V
19:26.630 --> 19:30.730
then I can multiply the time difference between
these two objects.
19:30.730 --> 19:36.210
These two events by velocity and I will get
the length as observed in S prime frame of
19:36.210 --> 19:41.740
reference. So this is what I have written
here the last line that L prime which is the
19:41.740 --> 19:47.250
length measured length of the platform measured
in S prime frame of reference will be given
19:47.250 --> 19:53.870
by V which is relative velocity between the
two frames multiplied by t 2 prime minus t
19:53.870 --> 20:00.679
1 prime and as we had agreed that t 2 prime
minus t 1 prime is tow which is the proper
20:00.679 --> 20:07.289
time interval between these two events. Therefore,
this L prime can be written as V times tow
20:07.289 --> 20:14.090
which is t, where tow is the proper time interval
or time interval measured in this frame of
20:14.090 --> 20:14.909
reference itself.
20:14.909 --> 20:19.720
Let us just try to come sort of compare out
notes with reference to S and S prime frame
20:19.720 --> 20:26.720
of reference. We just now said that L prime
must be given by V times tow we have agreed
20:28.509 --> 20:34.850
that this tow is a proper time interval, therefore,
what an observer in S frame that is the platform
20:34.850 --> 20:41.090
frame would observe would be dilated time
interval. Therefore, this tow must be given
20:41.090 --> 20:47.870
by t 2 minus t 1 divided by gamma. Where t
2 and t 1 are the times measured in the platform
20:47.870 --> 20:50.440
frame of reference not the S prime frame of
reference.
20:50.440 --> 20:57.440
In the s prime frame of reference, where platform
is at rest. So this time interval divided
20:57.500 --> 21:03.659
by gamma must give me the proper time interval,
because t 2 minus t 1 was actually dilated.
21:03.659 --> 21:10.659
Tow is the shortest time interval. And if
we agreed we have said earlier that V multiplied
21:10.659 --> 21:17.659
by t 2 minus t 1 is actually the length measured
in S frame of reference. So I can write V
21:17.940 --> 21:23.610
t 2 minus t 1 as the length as measured in
S frame. This terms out to be equal to L divided
21:23.610 --> 21:30.610
by gamma which tells that length of the platform
as will be measured in a frame of reference
21:30.929 --> 21:37.929
S prime would turn out to be contracted. We
have told that gamma is greater than 1, therefore,
21:38.370 --> 21:43.080
L divided by gamma will give you a smaller
length. So this length will be contracted
21:43.080 --> 21:49.049
length as expected because remember L is the
proper length of the platform. Had it not
21:49.049 --> 21:54.129
been proper length, then you could not be
apply this particular formulae. Then we apply
21:54.129 --> 21:58.039
length contraction formulae one of the length
must be a proper length. Then we applied time
21:58.039 --> 22:02.490
dilation formulae, one of the time interval
must be proper time interval. Then only I
22:02.490 --> 22:06.950
can use that particular formulae, otherwise
I cannot use it.
22:06.950 --> 22:13.950
One thus sees the contracted length in S prime
as expected. Now let us discuss this another
22:15.259 --> 22:22.259
example. This example in some form or the
other has been discussed in many textbooks.
22:23.779 --> 22:30.500
This is actually real life example and before
invent of special theory of relativity this
22:30.500 --> 22:35.639
was considered as a problem, which people
were never able to understand. It is only
22:35.639 --> 22:41.250
special theory of relativity, which provided
a solution to this particular issue. Therefore,
22:41.250 --> 22:46.840
it has a historical importance. There are
large number of particles, some of which are
22:46.840 --> 22:51.309
stable; some of which are unstable. Unstable
means they decay and change into some other
22:51.309 --> 22:52.659
particles.
22:52.659 --> 22:59.659
Now mu meson is one such particle. The lifetime
of this particular particle is 2 micro second
22:59.990 --> 23:06.990
if this particular particle is kept at stationary
is or a at rest in a particular frame of reference.
23:08.720 --> 23:12.700
What essentially be the lifetime is it means
there is a statistical process of decay. It
23:12.700 --> 23:17.610
is like radioactive decay. Then there is a
particular chance, you can never predict.
23:17.610 --> 23:21.840
Where a particular radio a nucleus will go
and read your in decay. It may take some time
23:21.840 --> 23:27.559
it may take a different time but, you can
always define something which is called the
23:27.559 --> 23:32.139
lifetime which represents in some way and
approximately in you can call it in average
23:32.139 --> 23:36.600
type of behaviour.
So essentially, the idea is that if lifetime
23:36.600 --> 23:42.669
is very large then particle is much more stable;
if the lifetime is very small compare in comparison
23:42.669 --> 23:47.990
to that the particle is like into decay comparatively
very soon. So this particular particle has
23:47.990 --> 23:53.559
a very, very small lifetime with just 2 micro
seconds, 2 into 10 upon minus 6 seconds is
23:53.559 --> 24:00.559
extremely small. Now these particles appears
are created at the upper atmosphere of the
24:01.230 --> 24:06.100
i.
Now then they are created they are created
24:06.100 --> 24:12.820
with a very large speed. The speed is essentially
close to speed of light, which in this particular
24:12.820 --> 24:17.190
problem we have taken to be a approximately
0.998c just to give some number approximately.
24:17.190 --> 24:24.190
So it is a very large speed essentially close
to speed of light. Now we can calculate classically
24:24.889 --> 24:31.889
how much time it will take for this particular
particle to reach earth. Then we can approximately
24:32.749 --> 24:38.909
estimate that out of how many of these particular
particles which are created at the upper poseur
24:38.909 --> 24:43.649
how many of them would reach earth. And that
number turns out to be extremely small as
24:43.649 --> 24:50.649
we just now see. But, people have found that
on earth you get a very large number of mu
24:50.850 --> 24:54.029
mesons or.
What you called a mu meson shower from the
24:54.029 --> 25:00.279
upper atmosphere. People could not understand
that if their lifetime was only 2 micro second,
25:00.279 --> 25:04.740
we expected very small number of these mu
mesons to actually have reached earth, before
25:04.740 --> 25:09.950
that they would have decayed into something
else. So how it is possible that such large
25:09.950 --> 25:16.740
number of particles can really come and reach
earth in spite of the fact that their lifetime,
25:16.740 --> 25:22.470
so small. And this particular problem was
actually solved by special theory of relativity
25:22.470 --> 25:28.149
by virtue of time dilation.
So let us now read the problem, the incoming
25:28.149 --> 25:34.799
primary cosmic rays create mu meson in the
upper atmosphere. The life time of mu mesons
25:34.799 --> 25:39.210
at rest is 2 micro second, if it is at rest
if you create this particular mu meson in
25:39.210 --> 25:45.789
laboratory at rest the lifetime will turn
out to be 2 micro second. If the mean speed
25:45.789 --> 25:51.830
of mu meson is 0.998c, let us talk about everything
in terms of n average. What fraction of mu
25:51.830 --> 25:56.279
meson created at the height of 20 kilometres,
let us take that no they have created a distance
25:56.279 --> 26:02.039
of 20 kilometres above the means level how
many of them would really reach the sea level;
26:02.039 --> 26:08.429
so that is essential the problem. Now let
us assume that we focus our attention on one
26:08.429 --> 26:09.870
particular mu meson.
26:09.870 --> 26:13.970
Because we are looking at the average behaviour
and we will take one particular mu meson which
26:13.970 --> 26:20.970
has exactly the same lifetime as 2 mu meson
few more micro second. Now if the particular
26:22.669 --> 26:29.669
is at rest then we expect that after 2 micro
second it will decay. So if I this particular
26:32.860 --> 26:37.240
mu meson was created at the upper atmosphere
of the earth and if there is a person sitting
26:37.240 --> 26:42.269
on mu meson or my laboratory was moving along
with the mu meson assuming of course, its
26:42.269 --> 26:45.519
speed to be constant because we applying special
theory of relativity.
26:45.519 --> 26:50.700
Let us assume that mu meson has the same speed
as travel through the earth, it remains constant
26:50.700 --> 26:57.590
as 0.998c. So if it is always the same then
this particular observer would find that this
26:57.590 --> 27:04.590
particular mu meson would decay into micro
second. In just like the train problem and
27:05.279 --> 27:08.100
the platform problem of course, my platform
is from here 20 kilometres long from here
27:08.100 --> 27:14.580
to the top of the atmosphere.
Now this person who is coming along with the
27:14.580 --> 27:20.429
mu meson would conclude that 2 micro second
after 2 micro seconds this particular particle
27:20.429 --> 27:25.419
will decay and because the mu meson is at
rest in that particular frame of reference,
27:25.419 --> 27:31.389
so the event number one creation of mu meson;
event number two decay of mu meson, they will
27:31.389 --> 27:36.379
occur at the same position and therefore,
this time interval will be proper time interval.
27:36.379 --> 27:43.379
However on the earth event number one occurs
20 kilometres above, which is the upper end
27:44.059 --> 27:47.960
of the atmosphere and the event number two
occurs wherever indicates somewhere much down
27:47.960 --> 27:53.789
much below down.
So according to an observer on earth this
27:53.789 --> 27:59.690
time difference between these two events must
be dilated. And because c is and the speed
27:59.690 --> 28:04.499
of the particle is very close to c therefore,
you would find that the time dilation affect
28:04.499 --> 28:11.139
will be very, very large. And therefore, affectively
an observer or on earth would feel that the
28:11.139 --> 28:16.980
life time of the mu meson has increased and
that is why he could observe a such large
28:16.980 --> 28:18.940
number of fraction of mu mesons the reaching
earth.
28:18.940 --> 28:25.940
Now let us do the numerical calculation. Decay
time interval 2 into 10 to the power minus
28:26.240 --> 28:32.879
6 seconds which is 2 micro second is proper
in moves on frame. In earth frame it would
28:32.879 --> 28:39.879
appear to be dilated. If we take c is equal
to 0.998c and use this expression of gamma
28:40.700 --> 28:45.440
that we have just now written gamma is equal
to 1 upon under route 1 minus v square by
28:45.440 --> 28:52.440
c square. We put this number of v which has
0.998c, you will get gamma is equal to 15.82
28:53.669 --> 28:59.470
approximately 16 up of that order. It is a
very large gamma 15.82.
28:59.470 --> 29:06.210
Therefore, decay time or you can say life
time of mu mason in earth frame of reference
29:06.210 --> 29:11.429
will be just this factor multiplied by 2 into
10 to power minus 6 second which terms out
29:11.429 --> 29:17.679
to be approximately 3 into 10 to power minus
5 second one order of magnitude larger 15
29:17.679 --> 29:23.580
times larger. Means lot of difference or if
you want to talk of the tow will if decay
29:23.580 --> 29:30.580
these essentially follows the same law or
same rule as a nuclear decay obvious which
29:30.889 --> 29:33.710
is an exponential behaviour.
29:33.710 --> 29:38.509
Which is given here at the bottom end of this
particular transparency, that N naught is
29:38.509 --> 29:42.860
the number of particles with which we had
started, then total number of particle which
29:42.860 --> 29:49.860
remain at a time t is given by N divided by
N naught into e raise for minus t by a tow.
29:50.470 --> 29:54.389
Or if we calculate which is the fraction of
the particles which are which will be reaching
29:54.389 --> 30:01.159
earth. Now if we take this particular life
time we can calculate how much time it will
30:01.159 --> 30:05.139
take in S frame of reference which I am calling
as a earth frame of reference. How much time
30:05.139 --> 30:11.259
it will take for this particular particle
to reach earth, that will be 20 kilometres
30:11.259 --> 30:15.029
which is here divided by what is the speed
of the particle.
30:15.029 --> 30:20.559
So normally if this particular mu meson was
suppose to reach earth, according to an observer
30:20.559 --> 30:26.779
on earth that observer must have travelled
sorry this particular moves on must have travelled
30:26.779 --> 30:33.779
for time 6.68 into 10 to power minus 5 seconds
approximately 7 into 10 to power minus 5 second.
30:35.230 --> 30:40.210
So if I have to find out what is the fraction
remember this calculation I am doing in earth
30:40.210 --> 30:45.190
frame of reference, if I have to find out
what is a fraction which must reach earth
30:45.190 --> 30:50.039
that will be N upon N naught will be given
by e raise for minus t by tow therefore, t
30:50.039 --> 30:54.519
I will substitute 6.68 into 10 to power minus
5 second.
30:54.519 --> 30:59.249
And for tow will substitute the value which
we have obtained in my last transparency which
30:59.249 --> 31:06.249
is 3.164 into 10 to power minus 5 second.
So I substitute these numbers and next transparency
31:07.440 --> 31:13.429
I have given this fraction turns out to be
0.12 which means approximately 12 percent
31:13.429 --> 31:17.749
of the mu mesons may be able to reach earth
others would have decayed in between. Because
31:17.749 --> 31:21.889
this is sort of a statistical process in which
some will decay some will decay later some
31:21.889 --> 31:26.529
will decay earlier.
But if we do the if we had done the classical
31:26.529 --> 31:31.470
calculation; it means for this particular
time instead of you for this particular tow
31:31.470 --> 31:38.409
instead of using 1 sorry 3.164 into 10 to
power minus 5 you would have used 2 into ten
31:38.409 --> 31:45.409
to power minus 6, then this number would have
turned out to be 3.12 into 10 to power minus
31:45.570 --> 31:50.700
15, see remember this is an exponential factor.
Exponential factor is a very fastly decaying
31:50.700 --> 31:55.889
function. And you have see that how much the
fraction is changed. Earlier this was a calculation
31:55.889 --> 31:59.899
which people were doing earlier before special
theory of relativity and they would have said
31:59.899 --> 32:03.980
that 1 into 10 to power 15 approximately one
out of 10 to power 15 mu mesons want to reach
32:03.980 --> 32:09.460
earth approximately.
I am just ignoring the factor 3 just to calculate
32:09.460 --> 32:15.519
order of magnitude. That many number want
to reach earth while according to relativity
32:15.519 --> 32:21.330
12 percent 12 out of 10 would reach earth.
The number has become so large and that is
32:21.330 --> 32:25.700
what was experimentally observed which people
could not understand earlier and only after
32:25.700 --> 32:29.869
special theory of relativity came people could
understand it. That was considered as one
32:29.869 --> 32:35.749
of the great success of special theory of
relativity. Just a question, what would an
32:35.749 --> 32:42.389
observer a mu meson frame conclude, how you
identify this event in terms of these two
32:42.389 --> 32:47.600
events in terms of a mu meson frame of reference?
As you can probably guess from that platform
32:47.600 --> 32:50.549
frame of reference, platform example which
we have given just now.
32:50.549 --> 32:55.679
That according to the observer on mu meson
the distance between the earth and the upper
32:55.679 --> 33:01.460
atmosphere will appear to be reduced or contracted.
So that observer will feel that this mu meson
33:01.460 --> 33:05.259
is had not actually travel 20 kilometres is
distance but, has travelled much smaller amount
33:05.259 --> 33:09.710
of distance, that is why it has on decayed.
33:09.710 --> 33:14.960
Now let us go back to our old example which
we have discussed number of times in different
33:14.960 --> 33:21.960
ways. That is the light emission from the
centre of train compartment. In our first
33:22.110 --> 33:27.110
time when we had discussed we had discuss
that instead of light we are throwing two
33:27.110 --> 33:33.309
balls, one person sitting at the centre of
a train compartment throws two ball one in
33:33.309 --> 33:40.269
a direction of the motion of the train towards
the front wall, another towards the back wall.
33:40.269 --> 33:45.860
And we defined two events when the ball reaches
the front wall and the ball reaches the back
33:45.860 --> 33:52.860
wall. We concluded that in the frame of reference
of train which we called as S prime the two
33:54.889 --> 34:01.169
events were stationary. Then we did a classical
Galilean transformation and we found out that
34:01.169 --> 34:08.169
in S frame which is the ground frame of reference
also these two events appear to be simultaneous.
34:08.429 --> 34:12.600
Second time when we discuss this particular
problem was when we assumed that instead of
34:12.600 --> 34:18.520
the ball somebody is shining light, one towards
the front of the front wall, another towards
34:18.520 --> 34:25.500
the back wall. We concluded at that time that
these two events in the frame of reference
34:25.500 --> 34:30.690
of S prime which is the compartment frame
of reference that the light reaching the front
34:30.690 --> 34:35.260
wall light reaching the back wall occurs at
the same time and are simultaneously.
34:35.260 --> 34:42.210
Then at that time we discussed with reference
to the ground frame and said that if second
34:42.210 --> 34:46.369
postulate of special theory of relativity
was correct; it means if the speed of light
34:46.369 --> 34:52.240
is same in S and S prime frame of reference
then the observer in S frame of reference
34:52.240 --> 34:57.460
which is the ground frame of reference would
feel that these two events are not simultaneous.
34:57.460 --> 35:02.000
But, at that time we did not calculate time
difference because we did not have Lorentz
35:02.000 --> 35:04.770
transformation.
Now that we have our Lorentz transformation.
35:04.770 --> 35:10.490
Let us actually try to calculate the time
difference between the these two events, which
35:10.490 --> 35:15.430
we had discussed just now as seen in the ground
frame of reference which I am calling as S
35:15.430 --> 35:18.309
frame of reference.
35:18.309 --> 35:23.819
So let us look at this picture again, this
is the picture or this particular train compartment
35:23.819 --> 35:28.410
this person standing at the canter of the
platform, he shines the light. Let us assume
35:28.410 --> 35:33.430
this two hands are just for making picture
clearer. I have put them with a large hand
35:33.430 --> 35:38.390
long hands but, let us assume that these are
just at the same point and the light is emitted
35:38.390 --> 35:43.200
at this particular direction. This I am calling
as a front wall, light is being emitted towards
35:43.200 --> 35:49.700
back direction opposite direction to the motion
of the train and this I am calling as a back
35:49.700 --> 35:55.099
wall my event number one was light pulse reaching
here; event number two light pulse reaching
35:55.099 --> 36:02.099
here, which I had we had agreed in S prime
frame of reference, these two events are simultaneous.
36:03.650 --> 36:08.010
And now my question is that what would be
the time interval between these two events
36:08.010 --> 36:15.010
as will be observed by an observer S standing
on ground. We had earlier defined events,
36:15.040 --> 36:20.220
let us redefined them. Let us rewrite them,
event number one light reaching the front
36:20.220 --> 36:24.829
wall of the train; event number two light
reaching the back wall of the train. So these
36:24.829 --> 36:30.859
are my events which we had already defined.
Let us again reiterate. Now what I will do
36:30.859 --> 36:35.010
because I want to do some mathematics using
Lorentz transformation. Let us be clear and
36:35.010 --> 36:40.690
write coordinates of this event one and event
two. And because the initial information has
36:40.690 --> 36:45.299
been given to me in S prime frame of reference
which is the frame of reference of the train
36:45.299 --> 36:48.770
in fact everything has been described with
respect to that observer.
36:48.770 --> 36:53.589
So let us first write the coordinates in S
prime frame of reference which is the train
36:53.589 --> 37:00.589
frame of reference. This is what I am doing
in my next transparency, so because this is
37:00.760 --> 37:06.339
S prime frame of reference or my coordinates
are putting prime. Now that we know that lengths
37:06.339 --> 37:11.260
are different in two frames. So this length
which is been measured the length of the compartment
37:11.260 --> 37:17.290
itself, that is in S prime frame of reference
so that length also I am calling as L prime.
37:17.290 --> 37:24.290
So we agree that is this is my train compartment
and observer is just sitting standing on half
37:29.849 --> 37:36.849
the way, then in this particular frame of
reference this length is L prime, so the first
37:38.910 --> 37:44.359
coordinate and this particular person is standing
at the origin, so we assume we get that this
37:44.359 --> 37:50.859
particular event number one will occur at
a distance of L prime by 2 positive side I
37:50.859 --> 37:56.819
am taking this is positive side. This is negative
side and this will occur at minus L prime
37:56.819 --> 38:02.369
by 2.
So this is what I have written here that X
38:02.369 --> 38:09.369
1 prime occurs at plus L prime by 2 ;X 2 prime
occurs at minus L prime by 2. Even primetime
38:12.720 --> 38:17.900
of the event number one is a time taken for
the light to travel from here to here. We
38:17.900 --> 38:22.589
know that this time this light travels of
the speed of light c. So this particular time
38:22.589 --> 38:26.589
will be given by actually the length is L
prime by 2 the distance it has to travel is
38:26.589 --> 38:33.589
L prime by 2. So the time taken will be L
prime by 2c. So time will be L prime by 2c.
38:34.349 --> 38:39.799
Similarly, this light which has been emitted
backwards also moves a distance L prime by
38:39.799 --> 38:45.220
2 and travels with the same speed c.
So therefore, this time interval will also
38:45.220 --> 38:52.190
be L prime by 2c. So according to an observer
in S prime frame of reference this event number
38:52.190 --> 38:59.190
one occurs at a vary of X L prime by 2 and
a time L prime by 2c. Event number two occurs
38:59.319 --> 39:05.660
at a coordinate of minus L prime by two and
a time L prime by 2c. These two times are
39:05.660 --> 39:10.730
same, these two events therefore, are simultaneous
in S prime frame of reference.
39:10.730 --> 39:17.089
This is what I have written here this particular
transparency X 1 prime is equal to plus L
39:17.089 --> 39:23.670
prime by 2, t 1 prime is equal to L prime
divided by 2c, X 2 prime is equal to minus
39:23.670 --> 39:30.670
L prime by 2, t 2 prime is equal to L prime
by 2 c remember we have not use any transformation
39:31.130 --> 39:35.890
to write these equations. I do not need any
transformation. If we have all the information
39:35.890 --> 39:40.700
in the same frame and all these basic information
has been given in S prime frame of reference.
39:40.700 --> 39:44.520
I do not need any transformation.
39:44.520 --> 39:51.099
Let us try to write the coordinates of these
two events in S frame. Let me first write
39:51.099 --> 39:58.099
the event number one, event number one was
light reaching the front wall this occurred
39:59.549 --> 40:02.760
at L prime by 2 and L prime by 2 c.
40:02.760 --> 40:09.760
And if you remember the Lorentz transformation
gives you X prime is equal to gamma X minus
40:10.260 --> 40:15.970
v t. Now the information has been given in
X prime frame of reference and I want to find
40:15.970 --> 40:20.520
in S frame of reference, therefore, I must
use inverse Lorentz transformation, which
40:20.520 --> 40:27.520
is given by X is equal to gamma X plus v t
prime X prime plus v t. So we said the prescription
40:30.799 --> 40:37.799
is change un-primed to prime un-prime to prime
and prime to un-prime change the sign from
40:39.130 --> 40:41.520
minus v to plus v, change the sign of v.
40:41.520 --> 40:46.510
So this is my inverse Lorentz transformation.
So I use the inverse Lorentz transformation
40:46.510 --> 40:53.510
here to find out X 1 is given by gamma L prime
by 2 plus v L prime by 2c. This is my value
40:55.859 --> 41:01.539
of X prime, this is the value of my t prime
L prime by 2c. So what we have written is
41:01.539 --> 41:08.520
X prime plus v t. This what is comes to. So
I take gamma L prime by 2 out of this, so
41:08.520 --> 41:13.869
this becomes gamma L prime by 2 multiplied
by 1 plus v by c or.
41:13.869 --> 41:19.299
Similarly we write inverse transformation
for time. Inverse transformation for time
41:19.299 --> 41:26.299
will be given by t is equal to gamma t prime
plus v X prime by c square. Again using inverse
41:29.170 --> 41:36.170
transformation we go back here to the transparency
t1 is equal to gamma times L prime by 2c,
41:37.450 --> 41:43.400
this was the time plus, plus because this
is a inverse transformation. v relative speed
41:43.400 --> 41:50.400
in the frames, x which is the l prime by 2
divided by c square. I take gamma l prime
41:51.869 --> 41:58.869
by 2c out of this bracket. This becomes 1
plus v by c. Of course, we have assumed that
41:59.700 --> 42:03.210
the origins are coincident with the light
where was emitted.
42:03.210 --> 42:07.770
So at that instant of time the observer which
is sit standing on the wrong was also at its
42:07.770 --> 42:14.770
origin and now just passing by it. Now let
us write the coordinate of event number two
42:14.819 --> 42:19.260
in S frame of reference. You realize that
everything will be same here exactly similar
42:19.260 --> 42:23.799
except that that as L prime by 2 we change
sign. So this appears here and this appears
42:23.799 --> 42:29.250
here this we change sign. So that is what
I am writing in the next transparency.
42:29.250 --> 42:36.250
Event number two in S frame, so X 2 that is
X coordinate of the second event will occur
42:36.530 --> 42:41.789
at gamma minus L prime by 2, that is where
it observe with that was X value plus v L
42:41.789 --> 42:46.450
prime by 2 c. Remember time is same though
event occurred at different location, but,
42:46.450 --> 42:51.510
time was same. So this L prime by 2 c, so
this remains the same thing. I take gamma
42:51.510 --> 42:58.510
L prime by 2, common and you get minus v plus
v by c. We take t 2 which is gamma t prime
43:01.539 --> 43:08.539
which is same as L prime by 2c minus v into
X prime, which is minus L prime by 2, because
43:09.730 --> 43:13.940
this event occurred at the back of the train.
So this is a minus L prime by 2 divided by
43:13.940 --> 43:20.940
c square.. I take gamma L prime by 2 c I out,
I get in bracket 1 minus v by c.
43:21.630 --> 43:26.809
So as you can see that this time this time
is different from the other time. The in earlier
43:26.809 --> 43:33.700
case for the event one there was positive
sign here and this time interval was positive.
43:33.700 --> 43:39.809
This was one plus v by c. Here it is one minus
v by c. So these two intervals are not simultaneous
43:39.809 --> 43:46.210
in S frame of reference, which is the ground
frame of reference as we had discussed earlier
43:46.210 --> 43:51.559
but, now we know what is the time difference.
Actually I can calculate t 2 minus t 1 and
43:51.559 --> 43:57.940
find out how much difference in the time was
observed as according to S observer on the
43:57.940 --> 44:03.690
ground observer between these two events.
44:03.690 --> 44:08.039
So that is what I have written in this particular
transparency, that we see that t 2 is less
44:08.039 --> 44:13.780
than t 1. Because there is a negative sign
here, one minus v by c. So t 2 is less than
44:13.780 --> 44:19.430
t 1. Hence in S event two occurred before
event one. This also we had discussed earlier,
44:19.430 --> 44:24.099
somewhat qualitatively without Lorentz transformation.
We had said that if speed of light should
44:24.099 --> 44:28.220
remain same, then actually happen that because
the light which travels a backward has to
44:28.220 --> 44:32.049
travel a smaller distance.
So if will hit the wall first the light which
44:32.049 --> 44:37.250
is emitted in front has to travel larger distance
but, with a same speed. Therefore, that event
44:37.250 --> 44:42.510
will occur later so that particular time will
be larger. So this is what we had qualitatively
44:42.510 --> 44:48.760
discussed. Now we know their numerical numbers.
So hence in S event two occurred before event
44:48.760 --> 44:54.039
one as first qualitatively discussed earlier
the time difference you can just calculate
44:54.039 --> 44:58.319
because that one factor will cancel out.
So that one only the v by c factor thing has
44:58.319 --> 45:04.559
to be taken. And this will turn out to be
gamma L prime v by c square. So this is the
45:04.559 --> 45:10.690
time interval between the two events as observed
in S frame of reference while in S prime frame
45:10.690 --> 45:17.690
of reference this time interval was 0. Both
these events occur at the same time.
45:18.990 --> 45:23.440
I have another question, is the coordinate
difference that we have calculated X 2 minus
45:23.440 --> 45:30.260
X 1 will be equal to the length of the compartment
in S? I think all should be able to give quick
45:30.260 --> 45:35.960
answer that no it cannot be. The reason is
that these two events though occurred at two
45:35.960 --> 45:42.539
different ends of the compartment, but, that
compartment was moving and unless these two
45:42.539 --> 45:48.670
events occurred in S in the same time X 2
minus X 1 will not give me correct length.
45:48.670 --> 45:53.900
X 2 prime and X 1 prime difference will give
me the length. In S prime frame of reference
45:53.900 --> 45:57.619
but, remember in that would go would go frame
of reference anywhere the compartment was
45:57.619 --> 46:01.230
at rest.
So even the time difference tow, in this in
46:01.230 --> 46:06.410
the present example in S prime frame of reference
these two events occur at the same time but,
46:06.410 --> 46:11.400
in that frame of reference the compartment
was anywhere at rest even if they would not
46:11.400 --> 46:15.460
have occurred at the same time is still X
2 prime minus X 1 prime would have given the
46:15.460 --> 46:20.880
correct length but, not in S frame of reference.
In S frame of reference these two events occurred
46:20.880 --> 46:27.880
at different time therefore, X 2 prime minus
X 1 prime is not the correct length. Now can
46:28.099 --> 46:35.099
we guess whether we are getting an over estimate
or an underestimate of the length? Let us
46:35.140 --> 46:36.770
look into the picture.
46:36.770 --> 46:41.260
Anyway we let us first I have calculated the
difference in the coordinates of the two events.
46:41.260 --> 46:46.539
This delta X will be given by X minus X 2,
because the event one was on front of the
46:46.539 --> 46:52.849
wall. So this turns out to be gamma L prime
divided 2 into 2. If we take the X coordinate
46:52.849 --> 46:59.220
transformation which we have done just now.
The result will turn out to be gamma L prime
46:59.220 --> 47:06.150
so according to the S observer these two events
occur at a spacing of gamma L prime along
47:06.150 --> 47:08.920
the X direction.
47:08.920 --> 47:14.460
The X 1 minus X 2 is an overestimate of length
of the train because t 1 is greater than t
47:14.460 --> 47:21.460
2 and can we correct it. Let me just show
you this particular picture. Let us suppose
47:21.460 --> 47:28.460
this was the position of the train, when event
number two occurred which was in the earlier
47:28.680 --> 47:35.680
event, so this was X 2; event number one occurred
when the light reached here. This event occurred
47:38.680 --> 47:45.270
at a later time, how much was the time difference
was delta t which we have just now calculated.
47:45.270 --> 47:52.180
Now if I take the X 1 event according to ground
observer this occurred here because the train
47:52.180 --> 47:58.349
has been moved and this train has moved by
distance of v delta t during the time interval
47:58.349 --> 48:04.589
in which the two events occurred according
to S frame of reference. According to observer
48:04.589 --> 48:09.579
in S frame this event occurred earlier, this
event occurred later and during this particular
48:09.579 --> 48:15.200
time this train was actually moving. And how
much distance you would have move is the speed
48:15.200 --> 48:19.789
of the train multiplied by the time difference.
This was the additional distance that this
48:19.789 --> 48:26.789
particular train moved and therefore, X 1
minus X 2 actually gave me gamma L prime which
48:27.779 --> 48:33.349
we have just now obtained. and if I want to
determine the correct length what I must do?
48:33.349 --> 48:38.730
From this gamma L prime I must subtract v
delta t, then I will get the correct length
48:38.730 --> 48:41.210
as measured in S frame.
48:41.210 --> 48:48.210
This what I have done in the next transparency.
My L is equal to delta X minus v delta t.
48:50.819 --> 48:57.819
As we have just now seen delta X is v gamma
L prime, v is here delta t is gamma L prime
48:58.619 --> 49:05.619
v upon c square. We just redistribute the
numbers it becomes gamma L prime to bracket
49:08.010 --> 49:13.160
1 minus v square by c square. If you know
gamma is equal to 1 upon under route v square
49:13.160 --> 49:17.470
minus c square and there is no under route
here. So this whole quantity will become L
49:17.470 --> 49:21.950
prime under route one minus v square by c
square which is 1 by gamma.
49:21.950 --> 49:28.799
So I get length as L prime by gamma. So as
I expected this length turns out to be a contracted
49:28.799 --> 49:35.799
length. This is what I had expected that because
in S prime frame of reference the length was
49:36.069 --> 49:40.900
proper hence it is in this particular frame
of reference that the length will turn out
49:40.900 --> 49:41.760
to be contracted.
49:41.760 --> 49:48.760
Now let us go to the conclusion, in this particular
lecture, we have essentially discussed some
49:50.010 --> 49:54.250
examples of Lorentz transformation. In the
next lecture, we will obtain the velocity
49:54.250 --> 49:55.849
transformation.
Thank you.