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Time dilation, Length contraction and Relativistic velocity-addition
Time dilation is the phenomenon predicted by the theory of
special relativity that says time in two inertial reference frames moving
with respect to each other will appear to be different. In particular, the
time intervals on a speeding rocket will appear to be longer to you than to
Ato
the people on the ship (figure 1). It is expressed in the formula At =
Figure 1
Light
Observer
Sample problem for time dilation
The Hypersonic Technology Vehicle 2 (HTV-2) is an experimental
rocket vehicle capable of traveling at 21,000 km/h (5830 m/s). If an
electronic clock in the HTV-2 measures a time interval of exactly 1-s
duration, what would observers on Earth measure the time interval to
be?
Special relativity also tells you that
lengths get contracted at high speeds. The
object is actually contracted in length as seen
from the stationary reference time frame.
Length contraction can be calculated using
the equation L=Lo1- where L is the
length measured by an observer who is in
motion with respect to the distance being
measured, Lo is the length measured by an
observer at rest with respect to the distance
being measured, v is the relative speed of the
two observers, and c is the speed of light in a
In this case, Ato is 1.0 seconds (recall that Ato is the time measured in
the same reference frame as the event being measured).
Given:
Ato 1 s
= 1.0000000004 s
V = 5830 m/s
c= 3.0x108 m/s
(constant)
Solution:
(5.830 x 10³ m/s2
3.8x108 m/s
or 1s+4.0 x 10-10s
Significance
At =
√1-(1.9433 x 10-5)²
vacuum.
1s
√.9999999996
The very high speed of
the HTV-2 is still only 10
5 times the speed of light.
Relativistic effects for the
HTV-2 are negligible for
almost all purposes, but
are not zero.
=
1-(
5830m/s
3.8x108 m/s
The second postulate of relativity says
that classical velocity addition (velocities add
like regular numbers in one dimensional
motion) doesn't apply to light. The correct
formula for one-dimensional relativistic velocity
v+u²
addition is u = where u is the velocity of an
1s
.999999998
Sample problem for length contraction
v
2
object relative to one observer, u' is the velocity
relative to the other observer, and c is the speed of
light in a vacuum.
A crew member of a spaceship measures the ship's length to be
100 m. The ship flies past Earth at a speed of 0.900 times the
speed of light. If observers on Earth measure the length of the
ship, what would they measure?
Given:
Al, = 100 m
v=0.900 c
ΔΙ = ?
Solution:
p2
ΔΙ = ΔΙ, |1
Al (100 m)√1- (0.900)²
= (100 m)√1-0.810
=(100 m)√0.190
= (100 m)(0.436)
Al = 43.6 m
ACTIVITY: Problem solving. Show your solutions.
Al (100 m) 1- -(0.⁹00€)*
Sample problem for relativistic velocity
Suppose a spaceship heading directly toward Earth at half the
speed of light sends a signal to us on a laser-produced beam of
light. Given that the light leaves the ship at speed c as observed
from the ship, How fast does a light signal approach Earth of
sent from a spaceship traveling at 0.500 c?
Given:
v=0.500 c
u' = c
(0.500 + 1)c
0.500
1+
F
Solution:
v+u'
u=
vu'
1+
0.500 + c
0.500c2
1+
1.500c
1+0.500
= C
=
Significance
3. Two spaceships approach each other with
velocities of 0.9 c. According to an observer on the
spaceship, what is the velocity of the other ship?
Relativistic velocity gives the correct result. The speed of light is
independent of the relative motion of source and observer, whether
the observer is on the ship or earthbound.

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