Length
Contraction and Time Dilation
Length
Contraction
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 L_{0}, 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
= L_{0 }(1 – u^{2}/c^{2})^{1/2}.
To
understand this, consider a spaceship. Suppose, initially spaceship is at
rest and for an observer outside the spaceship, the length of the
spaceship is L_{0 }= 50 meters. For the traveller inside the
spaceship, length of the spaceship as seen from inside is say, L_{01}
= 49.8 meters which is less than L_{0 }due to some thickness of
front and rear part of the spaceship. Now, suppose spaceship is moving
with 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 spaceship is hundreds of miles away from him
after some time. Suppose, through his divine vision, he can also see
inside the spaceship, even if 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 velocity u will
appear to be L = L_{0}(1
– u^{2}/c^{2})^{1/2}. Thus, if 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 = L_{0 }(1 – u^{2}/c^{2})^{1/2
}
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 traveller does
not feel any difference in the length of the spaceship.
Now
suppose, the spaceship is moving with the velocity nearly equal to that of
light i.e. v = .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.
Now
suppose, the spaceship is moving with the velocity u = .9999c 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 – .9999^{2})^{1/2
}(u = .9999c implies u/c = .9999)
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.
Time
Dilation
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, 0.000
second. Suppose the traveller in the spaceship has the same type of
digital stopwatch, which is also reset.
Now
suppose, spaceship is moving with the velocity u and the
traveller, who is inside the spaceship, presses the start button of his
stopwatch at some time. The observer outside the
spaceship and at rest, who is observing the traveller 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 the same time, he stops his stopwatch
and then sees the time elapsed in his stopwatch, then he will find that
more than 14 seconds have in his stopwatch.
If
spaceship were moving with the velocity u = c/2 and if 14 seconds (Δt_{traveller
in spaceship} = 14) 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) 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 = .9999c
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 saw that the
traveller inside the spaceship is having a dinner, then all the motions
e.g.
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 (whereas to the traveller inside the spaceship,
everything moves at a normal rate) that is not only do the length shorten
but also the clocks slow 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
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 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 (e.g. 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 man (who was observer and is now being observed by
the traveller inside the spaceship) 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 appear to be contracted to the observer
outside the spaceship and at rest, similarly the distance between the
spaceship and the observer (who is now being observerd by the traveller)
will appear to be contracted to the traveller in the spaceship as
explained in the next example.
Length
Contraction & Time Dilation
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 sec away, suppose 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 second / .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
}
As
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 second 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 second time interval 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 while 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 second × (1
– .8^{2})^{1/2 }= 100 light second
× .6 = 60 light second.
Thus,
for the observer at rest, this journey was of 100 light second, 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 second, which the traveller completed in 125 × (1 – u^{2}/c^{2})^{1/2}
seconds while moving with the velocity u = 0.8c.
As
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 = 0.8c.
^{
} |