Length Contraction and Time Dilation |
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or L = L_{0 }(1 – 1/4)^{1/2
}(u = c/2 implies u/c = 1/2) or L_{0 }(3/4)^{1/2 } or .866l_{0}
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 L_{01} = 49.8 meters that
is traveler 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 = L_{0 }(1 – u^{2}/c^{2})^{1/2 } or L = L_{0 }(1 – .99^{2})^{1/2
}(u = .99c implies u/c = .99) or L = L_{0 }(1 – .9801)^{1/2 } or L = L_{0 }(.0199)^{1/2 } or L =.141L_{0}
or .141 × 50 = 7.05 meters
For the traveller inside
the spaceship, the length of the spaceship as seen from inside
will still be L_{01} = 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 = L_{0 }(1 – u^{2}/c^{2})^{1/2 } or L = L_{0 }(1 – .9999^{2})^{1/2
}(u = .99c implies u/c = .99) ^{
} or L =.0141L_{0}
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 L_{01}
= 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 = L_{0 }(1 – u^{2}/c^{2})^{1/2 } or L = L_{0 }(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 L_{01}
= 49.8 meters. Note that, as per 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.
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 – u^{2}/c^{2})^{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 in the spaceship has the same type
of digital stopwatch, which is also reset. Now suppose, spaceship is moving with
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 and suppose he finds 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 (Δt_{traveller 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 Δt_{observer at rest} = Δt_{traveller in spaceship} / (1 – u^{2}/c^{2})^{1/2 } Or
Δt_{observer at rest}
= 14 seconds / (1 – 1/4)^{1/2}
(u = c/2 implies u/c = 1/2 or u^{2}/c^{2 }
= 1/4) Or
Δt_{observer 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 (Δt_{traveller 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 Δt_{observer at rest} = Δt_{traveller in spaceship} / (1 – u^{2}/c^{2})^{1/2 } Or
Δt_{observer at rest}
= 14 seconds / (1 – .99^{2})^{1/2}
(u = .99c implies u/c = .99) Or
Δt_{observer 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 = .99c Δt_{observer at rest}
= 14 seconds / (1 – .9999^{2})^{1/2}
(u = .9999c implies u/c = .9999) Or
Δt_{observer 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 a
dinner, 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) appear to be slower
than normal, while to the traveller inside the spaceship,
everything moves at a normal rate, that is not only do the
length shortens 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 – u^{2}/c^{2})^{1/2}
seconds, which is more than 1 second. Suppose, the traveller has the divine vision
and he is now seeing the observer 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 observer is having a 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) appear to be faster than normal, while to the man (who
was observer and now being observed by the traveller inside the
spaceship), everything moves at a normal rate. Thus, for the
traveller, velocity of time outside the spaceship appear to be
increased by the factor 1 / (1 – u^{2}/c^{2})^{1/2}.
If 14 seconds passed inside spaceship, which is moving
with the speed u = c/2,
14/ (1 – u^{2}/c^{2})^{1/2 }or
16.156 seconds will have passed outside the spaceship.
Just as the length of the spaceship appears to be contracted to
the observer outside the spaceship and at rest, distance between
the spaceship and the observer (who is now being observerd by
the traveller) will appear to be contracted as explained in the
next example.
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 = 3 × 10^{8}
meter/sec × 1sec = 3 × 10^{8} meter).
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 / .8c = (100 × 3 × 10^{8} meter) / (.8
× 3 × 10^{8} meter/second)
=125 seconds.
Thus, the stopwatch of the observer will show 00:02:05:00 i.e. 2
min, 5 sec. Now from Δt_{observer at rest} = Δt_{traveller in spaceship} / (1 – u^{2}/c^{2})^{1/2 } Δt_{traveller in spaceship} = Δt_{observer at rest} × / (1 – u^{2}/c^{2})^{1/2 } The spaceship is moving with the
velocity u = .8c. Thus, (1 – u^{2}/c^{2})^{1/2
}= (1 – .8^{2})^{1/2 }= .6 Thus, Δt_{traveller in spaceship}
= Δt_{observer at rest} × .6 = 125 × .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 = .8c, the
spaceship will travel .8c × 75 sec = 60 light sec (or 60 × 3 ×
10^{8} meter). Thus, the traveller inside the spaceship will say, 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, journey was of 100 light seconds (which he 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 = L_{0
}(1 – u^{2}/c^{2})^{1/2 }= 100
light seconds × (1 –
.8^{2})^{1/2 }= 100 light seconds × .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 – u^{2}/c^{2})^{1/2 }light seconds, which the traveller completed in 125 × (1 – u^{2}/c^{2})^{1/2} seconds moving with the velocity u = 0.8c.
u = .8c implies (1 – u^{2}/c^{2})^{1/2} =.6. Thus, for the traveller, this journey was of 100 × .6 = 60 light seconds, which the traveller completed in 125 × .6 = 75 seconds while moving with the velocity u = .8c. ^{ } |