TIME DILATION In the Lorentz Transformations the interval between 2 events is INVARIANT . Δx 2 +Δy 2 +Δz 2 − c 2 Δt 2 =Δx 2 +Δy 2 +Δz 2 − c 2 Δt 2 Now for 2 events in S’ at the same place (e.g.clock ticks), we have x’ = 0. This is the clock rest frame and the time interval between 2 such events is the proper time denoted by . So: Divide by then: Therefore the time interval is longer than that in the rest frame - TIME DILATION c 2 Δτ 2 = c 2 Δt 2 − Δx 2 − Δy 2 − Δz 2 Δτ Δt 2 =1 − v 2 /c 2 dx dt S = v Δτ =Δt 1 − v 2 /c 2 Δτ< Δt Δt 2 Δτ
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TIME DILATION
In the Lorentz Transformations the interval between 2 events is INVARIANT.
Δx2+ Δy2
+ Δz2− c2
Δt2 = Δx′2+ Δy′2
+ Δz′2 − c2Δt′2
Now for 2 events in S’ at the same place (e.g.clock ticks), we have x’ = 0. This is
the clock rest frame and the time interval between 2 such events is the proper time
denoted by . So:
Divide by then:
Therefore the time interval is longer than that in the rest frame - TIME DILATION
c2Δτ2
= c2Δt2 − Δx2
− Δy2− Δz2
(Δτ
Δt
)2
= 1 − v2/c2
(dx
dt
∣∣∣∣S′
= v
)
Δτ = Δt√
1 − v2/c2
Δτ < Δt
Δt2
Δτ
Around the World with Atomic Clocks
Predicted relativistic time gain for a flying
clock after a non-stop equatorial
circumnavigation of the earth at various
altitudes. The area within the hatched lines
is below detection thresholds with a
portable Cs clock.
Hafele & Keating, Science 177, 167, 1972
Gravitational “red
shift”
Earth’s rotation Time dilation
Where is clock at rest on Earth;
v is the ground speed of the aircraft
Results (ns) East West
144 ± 14 179 ± 18 Gravity
Prediction -184 ± 18 96 ± 10 Rotation/ Kinematic
-40 ± 23 275 ± 21 Net
Measurement -59 ± 10 273 ± 7
τ − τ0 =
[gh
c2−
(2RΩv + v2
)/2c2
]τ0
τ0
Muon Storage Ring Experiment – CERN 1977
From PS accelerator3.1 GeV/c - beam
Magnet
Electric quadrupole focusing unit
Muons circulate inside vacuumpipe, momentum = 3.098 GeV/c = 29.33, = 0.994
π−→ μ−
+ ν̄μ
γ β
Muon Storage Ring Experiment – CERN 1977
Muon Decay Spectrum in Flight
(Bailey et al, Nature 268, 301, 1977)
Decay: μ−→ e− + ν̄e + νμ τ0 = 2.2 μs
LORENTZ CONTRACTION
Should be called Lorentz-Fitzgerald contraction
Consider a rigid rod of length l0 at rest in frame S’, moving with velocity v w.r.t. to S.
x1 x2
S
O x
x’1 x’
2
S’
O’ x’
l0
v
In S , use light signals as shown to measure x1 and x2 at the SAME TIME
In S’ use a metre rule to measure l0.
From the Lorentz transformations:
x′
2= γ(x2 − vt)
Δx = l0/γ
x′
1= γ(x1 − vt)
Δx′= γΔx
Δx′= x′
2− x′
1= l0
What Would You See?
At rest v = 0.5 c
v = 0.95 c v = 0.99 c
Computer generated graphics show the visual appearance of a three-dimensional lattice of rods and balls moving towards you at various speeds. The lattice only becomes distorted as v approaches c.
A view of a box at rest and a view
of the box photographed by a
single observer shows a distorted
shape.
(a) The appearance of a train at rest as seen
in a photograph. (b) The appearance of that
train as it moves at 0.9c past a camera.
What Would You See?
Galaxy Example
Event 1
(x1, t1)
Event 2
(x2, t2)
S
O x
y
Galaxy
Proton, v
Diameter d (in S) = 105 light years
Eproton = 1019 eV
Mp = 938 MeV/c2
= E/Mpc2 = 1.066 x 1010
S ´
O’ x’
v
y’
Galaxy
Proton at rest in S ´
What is t´?
t´ = t/ = 296 s
d ´ = d/ = 8.87 x 1010 m
The Train in the Tunnel Problem
A train of proper length 2L = 500 m approaches a
tunnel of proper length L = 250 m. The train’s
speed u is such that = 2. An observer at rest
with respect to the tunnel measures the train’s
length to be contracted by a factor of 2 to 250 m
and expects the whole train to fit in the tunnel. An
observer on the train knows that the length of the
train is 500 m, and that the tunnel is contracted by
a factor of 2 to 125 m. Thus the observer on the
train argues that the train will not fit into the
tunnel. Who is right?
Frame S (tunnel)
(a)
(c)
E1, t = 0
E2, t = L/u
x
x
0
0 L
-L
-L
L
Frame S’ (train)
(b)
(d)
E1, t’ = 0
E2, t’ = L/u
x’
x’
0
0-2L
-2L L/2
-L/2
The train’s speed u is such that =
2.
This problem demonstrates once more the importance of the concept of in relativity theory.
The Moving Rod Problem
A rod of proper length L (the length in its rest frame) points along the x axis but moves in a direction making an angle of 45° to this axis. A platform, also parallel to the x axis, lies in the rod’s way, but a slit of proper length 1.1L has been cut out of it, so that the rod can easily fit through if it travels at non-relativistic speed. What happens if its speed is 0.9c?Analyze the problem from both reference frames.
xy
v
platform
rod
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