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Length Contraction and Time
Dilation |
Length Contraction |
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Material bodies contract when they are
moving and this shortening is only in the direction of the
motion. If the length of the body is L0, when it is
at rest i.e. u = 0, then when it moves with the speed u along
its length, then new length L is given by L
= L0 (1 – u2/c2)1/2.
To understand this, consider a spaceship. Suppose, initially
the
spaceship is at rest and for an observer outside the spaceship,
the length of the spaceship is L0 = 50 meters. For
the traveller inside the spaceship, the length of the spaceship as seen from
inside is say, L01 = 49.8 meters which is less than L0
due to some thickness of the front and the rear parts of the spaceship.
Now,
suppose the spaceship is moving with the velocity u along the direction
of its length. Suppose the observer, outside the spaceship and at
rest, has some divine vision so that he can see the spaceship,
even if it is many light years away from him.
Suppose, through his divine vision, he can also see inside the
spaceship, even when it it is far away from him. For this observer,
who is outside the
spaceship, the length of the spaceship, which is now moving with
the velovity u will appear to be L = L0(1 – u2/c2)1/2.
Thus,
if the spaceship is moving with the velocity equal to half the velocity
of light i.e. u = c/2, then to that observer, the length of the
spaceship will appear to be equal to L, where L = L0
(1 – u2/c2)1/2
or L = L0 (1 – 1/4)1/2
(u = c/2 implies u/c = 1/2)
or L0 (3/4)1/2
or .866l0
or .866 × 50 = 43.3 meters
Thus, to an observer outside the
spaceship and at rest, the length of the moving spaceship, seems
to be less than when the spaceship was at rest. For the
traveller inside the spaceship, the length of the spaceship as
seen from inside will still be L01 = 49.8 meters that
is traveller does not feel any difference in the length of the
spaceship.
Suppose, the spaceship is moving with
the velocity nearly equal to that of light i.e. u = .99c for
example. Then, to the observer outside the ship and at rest, the
length of the spaceship will appear to be equal to L where L = L0
(1 – u2/c2)1/2
or L = L0 (1 – .992)1/2
(u = .99c implies u/c = .99)
or L = L0 (1 – .9801)1/2
or L = L0 (.0199)1/2
or L = 0.141L0
or 0.141 × 50 = 7.05 meters
For the traveller inside
the spaceship, the length of the spaceship as seen from inside
will still be L01 = 49.8 meters.
Suppose, the spaceship is moving with
the velocity u = .9999c. Then, to the observer
outside the spaceship and at rest, the length of the spaceship will
appear to be equal to L, where L = L0 (1 – u2/c2)1/2
or L = L0 (1 – .99992)1/2
(u = .99c implies u/c = .99)
or L =.0141L0
or .0141 × 50 = .705 meters or 70.5 centimeters.
For the traveller inside the spaceship,
the length of the spaceship as seen from inside will still be L01
= 49.8 meters.
If spaceship were travelling with the
velocity of light i.e u = c, then, to the observer outside the
ship and at rest, the length of the spaceship will appear to be
equal to L, where L = L0 (1 – u2/c2)1/2
or L = L0 (1 – 1)1/2
(u = c implies u/c = 1)
or L = 0.
For the traveller inside the spaceship,
the length of the spaceship as seen from inside will still be L01
= 49.8 meters.
Note that, as per the theory of relativity it is impossible for a
body to move with the velocity of light. Thus, the length of the spaceship
can never appear to be zero to an outside observer.
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Time Dilation |
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If Δt time interval is passed in a clock
moving with the velocity u, then in the same type of clock,
which is at rest, Δt / (1 – u2/c2)1/2
time interval will
have passed.
Suppose, an observer outside the
spaceship has a digital stopwatch, which is reset so that it
shows the time as 00:00:00:000 that is 0 hour, 0 minute, 0.00
second. Suppose, the traveller inside the spaceship has the same type
of digital stopwatch, which is also reset.
Now suppose, the spaceship is moving with
the velocity u and at some time, he presses the start button of his
stopwatch. The observer outside the spaceship and at rest, who
is observing the traveller inside the spaceship through his
divine vision, presses the start button of his stopwatch at the
same time, so that both the stopwatch start simultaneously.
Now suppose, some time later, the
observer at rest, sees the time elapsed in the stopwatch of the
traveller through his divine vision and suppose, he finds that the stopwatch of the traveller is
showing 00:00:14:000 i.e. 0 hour, 0 minute 14.000 seconds have
elapsed in his stopwatch and at that
very time,
he stops his own stopwatch and then sees the time elapsed in his
stopwatch, then he will find that more than 14 seconds have
passed in
his stopwatch.
If spaceship were moving with the
velocity u = c/2 and if 14 seconds (Δttraveller in
spaceship =
14) seconds had passed in the stopwatch of the traveller inside
the spaceship (i.e. The stopwatch of the traveller was showing
00:00:14:000), then in the stopwatch of the observer, time
elapsed will be
Δtobserver at rest = Δttraveller
in spaceship /
(1 – u2/c2)1/2
Or
Δtobserver at rest
= 14 seconds / (1 – 1/4)1/2
(u = c/2 implies u/c = 1/2 or u2/c2
= 1/4)
Or
Δtobserver at rest
= 16.156 seconds i.e. stopwatch of the observer will show
00:00:16:156
If spaceship were moving with the
velocity u = .99c and if 14 seconds (Δttraveller in
spaceship =
14) seconds had passed in the stopwatch of the traveller inside
the spaceship, then in the stopwatch of the observer, time
elapsed will be
Δtobserver at rest = Δttraveller
in spaceship /
(1 – u2/c2)1/2
Or
Δtobserver at rest
= 14 seconds / (1 – .992)1/2
(u = .99c implies u/c = .99)
Or
Δtobserver at rest
= 99.232 seconds i.e. stopwatch of the observer will show
00:01:39:232 (1 min,
39.232 sec)
If spaceship were moving with the
velocity u = .9999c
Δtobserver at rest
= 14 seconds / (1 – .99992)1/2
(u = .9999c implies u/c = .9999)
Or
Δtobserver at rest
= 989.968 seconds i.e. stopwatch of the observer will
show 00:16:29:968
(16 min, 29.968 sec)
Thus, time itself will appear to be
slower in a moving spaceship. In other words
moving clock ticks
slower.
If the observer outside the spaceship
and at rest sees that the traveller inside the spaceship is having dinner,
then all the motions (eg, taking food in the spoon, thrusting
the spoon in the mouth, pulling the spoon out of the mouth,
chewing the food, drinking a glass of water) of the traveller appear to be
slower
than the normal,
while to the traveller inside the spaceship,
everything moves at the normal rate, that is not only does the
length shorten but also the clock slows down. That is when the
clock inside the spaceship records 1 second elapsed, as seen by
the traveller inside the spaceship, the same type of clock near
the observer outside the spaceship and at rest shows 1 / (1 – u2/c2)1/2
seconds, which is more than 1 second.
Suppose, in addition to observer at
rest, the traveller also has the divine vision
and the traveller is now watching the person who was observer
and who is outside the spaceship
and at rest. For the traveller, time is moving with the same
normal speed in both the cases i.e. when the spaceship is moving
along at a uniform speed as well as when the spaceship is at
rest.
Now, if the person who was the observer, is having dinner,
then to the eyes of the traveller, all the motions (eg, taking
food in the spoon, thrusting the spoon in the mouth, pulling the
spoon out of the mouth, chewing the food, drinking a glass of
water) of that person appear to be faster than
the normal, while to the person (who
was observer and is now being observed by the traveller inside the
spaceship), everything moves at the normal rate. Thus, for the
traveller, velocity of time outside the spaceship appear to be
increased by the factor 1 / (1 – u2/c2)1/2.
If 14 seconds passed inside spaceship, which is moving
with the speed u = c/2,
14/ (1 – u2/c2)1/2 or
16.156 seconds will have passed outside the spaceship.
To the observer outside the spaceship and at rest,
the length of the spaceship appears to be contracted.
Similarly,
to the traveller inside the spaceship, the distance between
the spaceship and the observer (who is now being observerd by
the traveller) appear to be contracted. This is explained in the
next example.
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Length Contraction and Time Dilation |
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Suppose the
spaceship is 100 light seconds away from the observer and is
moving with velocity u = .8c towards the observer.
(one light second is the distance
travelled by light in one second i.e. 1 light second = 299792452
meter/sec × 1sec = 299792452 meters).
Suppose, when
the spaceship is 100 light seconds away , the traveller inside
the spaceship resets & starts his stopwatch and at the same
time, the observer at rest does the same.
When the spaceship reaches to the observer, time elapsed
in the stopwatch of the observer will be 'distance / velocity' =
100 light seconds / 0.8c
= (100 × 299792452 meters) / (0.8
× 299792452 meters/second)
=125 seconds.
Thus, the stopwatch of the observer will show 00:02:05:00 i.e. 2
min, 5 sec.
Now from Δtobserver at rest =
Δttraveller in spaceship
/ (1 – u2/c2)1/2
Δttraveller in spaceship
= Δtobserver at rest × / (1 – u2/c2)1/2
The spaceship is moving with the
velocity u = 0.8c. Thus, (1 – u2/c2)1/2
= (1 – 0.82)1/2 = 0.6
Thus, Δttraveller in spaceship
= Δtobserver at rest × 0.6 = 125 × 0.6
= 75 seconds that is, for the traveller, he completed the
journey ( which was of 100 light seconds for the observer
outside the spaceship and at rest), when in his stopwatch 75 seconds
had passed that is, for the traveller, in actual, 75 seconds had passed inside the spaceship during his journey and
during these 75 seconds, moving with the velocity u = 0.8c, the
spaceship will travel 0.8c × 75 sec = 60 light sec (or 60 ×
299792452 meters).
Thus, the traveller inside the spaceship will say,
the journey was of 60 light seconds (which he completed in 75 seconds moving with the
velocity u = .8c). But the observer at rest will say, the journey
was of 100 light seconds
(which the traveller completed in 125 seconds moving with the velocity u =
.8c).
This contradiction disappears, if we
consider the length contraction that is as the spaceship was
moving towards the observer, the distance between the spaceship
and the observer, as seen by the traveller will be L = L0
(1 – u2/c2)1/2 = 100
light seconds × (1 –
0.82)1/2 = 100 light seconds ×
0.6 = 60 light seconds.
Thus, for the observer at rest, this journey was of 100 light
seconds, which the traveller completed in 125 seconds moving with
the velocity u = 0.8c.
For the traveller, this journey was of 100 × (1 – u2/c2)1/2
light seconds, which the traveller completed in 125 × (1 –
u2/c2)1/2 seconds moving with
the velocity u = 0.8c.
u = 0.8c implies (1 – u2/c2)1/2
= 0.6. Thus, for the traveller, this journey was of 100 × 0.6 = 60 light seconds,
which the traveller completed in 125 × .6 = 75 seconds while
moving with the velocity u = 0.8c.
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