Consequences of the special
theory of Relativity (II)
Relativity of Length
L0
E: Emitter
R: Receiver
E R
v
v
E R
E R
E R
L+vΔt1
L-vΔt2
[See Applet]
Length Contraction
• For the outside observer:
– It takes Δt1 for the light to do the first part of
the trip and Δt2 for the second part.
cΔt1 = L+vΔt1
cΔt2 = L-vΔt2
Δt1 = L/(c-v)
Δt2 = L/(c+v)
Δt = Δt1 + Δt2
= 2(L/c)(1-v2/c2)-1
Time Dilation effect: Δt = γΔt0 with
We deduce: L = L0√(1-v2/c2)= γ-1L0
L0: rest length or proper length ; objects appear contracted in the direction of motion
Example
What length will observers measure for a spaceship moving relative to them
at 0.99c if its occupants determine the length of the ship to be 50m ?
L = 50γ-1 = 50 x (1-(0.99)2c2/c2) = 7.1 m
A word on time dilation
and length contraction
• The two descriptions should be equivalent
[see Muon decay]
• Both observers see the time dilation and
length contraction affecting the other
observer [see Twin Paradox].
• In both cases, if v<<c  L≈L0 and Δt ≈Δt0
(back to classical mechanics)
Muon (µ) decay
• Muons originates from cosmic-rays
Experiment: 1000 muons traveling at 0.98c
detected at 2000m, ~540 muons “survived” (did not
decay) at see level
Radioactive Decay Law:
-ln2.t/T
N(t)=N0(t)e
with T=1.52µs, the half-life of µ
It takes 2000/(0.98c)=6.8µs for the µ’s
to travel 2000m.
According to classical physics,
N(6.8µs) ~ 45 µ’s only should be left
Time Dilation !
• For the observer on earth, time appears
dilated, t = γt0 , e.g. the µ appears to live
longer than its rest half-life T: Tearth=7.64 µs
• Muons surviving at sea level:
-ln2.(6.80/7.64)
N(sea level) = N(7.64µs) = 1000 e
~ 540
Length Contraction !
• One should be able to get the same results by solving the
problem with length contraction
• You’re a muon ! You are to travel at 0.98c a distance of
L0=2000m measured by an earth observer, how much
distance will you measure according to your meter stick ?
L = γ-1L0  L = √(1-(0.98)2) x 2000 = 398m
• The µ travels the distance in: t = 398/0.98c = 1.35µs
• After this time, N(left) = N(1.35µs) = 1000e-ln2.(1.35/1.52) ~ 540
SAME RESULT !
Note: Don’t forget to use T=1.52 µs, the half-life of the µ at rest, since we
are in the referential of the µ !
Space - Time
• Nothing can travel faster than the speed of light.
Therefore, there are regions of space and time
that cannot be reached (CAUSALITY).
Simultaneity / Non-simultaneity
Lightning at x3
(Left)
x1,x2,x3 stationary
(Right) x1,x2,x3 moving
The “Stanley/Mavis” problem
Stanley:
The two events “lightning strike hits (A)”
and “lightning strike hits (B)” are simultaneous.
Mavis:
The two events “lightning strike hits (A’)”
and “lightning strike hits (B’)” are NOT simultaneous.
ct
ct
v
Stanley
(in his own
ref frame)
c
A
Mavis
(in Stanley’s
ref frame)
c
Stanley
B
c
x
A
c
Mavis
B
x
ct’
Mavis
(in her own
ref frame)
c
A’
c
Mavis
B’
x’
Spacetime interval
• Length contraction / Time dilation
– Length and time are not the same in two inertial frames moving
with respect to one another at a sizeable fraction of c
• Is there an invariant quantity that measures the same in
both reference frame? Yes.
Δs2 = (Δx2+Δy2+Δz2) - (cΔt)2 ; Δs2 (in K) = Δs’2 (in K’)
• Using only 1-D of space (x): two events described by
(x1,t1) and (x2,t2) can be expressed by:
Δs2 = (x1-x2)2 - c2(t1-t2)2 (SPACETIME INTERVAL)
[interval]2 = [separation in space]2 - [separation in time]2
Spacetime interval
• Lightlike interval:
– Δs2 = 0 (Δx2 = c2Δt2): the two events can only be
connected by a light signal
• Spacelike interval:
– Δs2 > 0 (Δx2 > c2Δt2): the two events cannot be
causally connected.
• Timelike interval:
– Δs2 < 0 (Δx2 < c2Δt2): the two events can be causally
connected.
The twin paradox
• Two twins: one travels, one stays at home. The
traveler ages slower than its twin, who stayed on
earth. But, relatively speaking, who is moving ?
• The paradox: both are moving with respect to
the other, therefore they should age the same
way !
• Answer: The one that is really “moving” (e.g.
feeling the effect of time dilation) is the one
accelerating and decelerating. [See textbook]
Exercise
Determine the decrease in length of a space shuttle (L0=50m)
traveling 17,000 miles/h relative to the earth observers.
v = 17000 mi/h = (17000 x 1.61 x 103) / 3600 = 7.6 x 103 m/s
Length contraction: L = γ-1 L0 = L0 √( 1 - (7.6 x 103 )2/(3.0 x 108)2 )
= L0 √ ( 1 - 6.4 x 10-10 )
Contraction = L0 – L = 1.6 x 10-8 m !!!!