Class 3: Time dilation, Lorentz contraction, relativistic velocity transformations
Transformation of time intervals Suppose two events are recorded in a laboratory (frame S), e.g. by a cosmic ray detector. Event A is recorded at location xa and time ta, and event B is recorded at location xb and time tb. We want to know the time interval between these events in an inertial frame, S ′, moving with relative to the laboratory at speed u.
Using the Lorentz transform,
ux x′=−γ() xut,,, yyzzt ′′′ = = =− γ t , (3.1) c2 we see, from the last equation, that the time interval in S ′ is
u ttba′−= ′ γ() tt ba −− γ () xx ba − , (3.2) c2 i.e., the time interval in S ′ depends not only on the time interval in the laboratory but also in the separation.
If the two events are at the same location in S, the time interval (tb− t a ) measured by a clock located at the events is called the proper time interval . We also see that, because γ >1 for all frames moving relative to S, the proper time interval is the shortest time interval that can be measured between the two events.
The events in the laboratory frame are not simultaneous. Is there a frame, S ″ in which the separated events are simultaneous? Since tb′′− t a′′ must be zero, we see that velocity of S ″ relative to S must be such that
u c( tb− t a ) β = = , (3.3) c xb− x a i.e., the speed of S ″ relative to S is the fraction of the speed of light equal to the time interval between the events divided by the light travel time between the events.
Time dilation Consider a clock at rest in frame S. This clock measures proper time for events that occur at the location of the clock. These events could be ‘ticks’ of the clock. As shown above, when this clock is viewed from a moving frame, the interval between events or clock ticks, will be larger by a factor γ . This effect is called time dilation .
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Lorentz contraction The length of a rod in a frame in which the rod is at rest is called its proper length . When measured in a frame in which the rod is moving, the rod is found to have a length shorter than its proper length. This effect is called Lorentz contraction or Lorentz-Fitzgerald contraction .
Suppose the rod of proper length L0 is at rest in frame S ′, and has speed u in frame S. The length of the rod in S is the distance between the positions of the ends of the rod measured at the same time. Let the
ends of the rod in S be at x1 and x2 at time t. From the Lorentz transformation these ends have positions in S′,
x1′ =γ () x 1 − ut , (3.4) x2′ =γ () x 2 − ut .
We see that the length of the moving rod as measure in frame S is
x2′− x 1 ′ L 0 L=−= x2 x 1 = . (3.5) γ γ
Since γ >1, we see that the moving rod is shorter than the rod at rest.
Relativistic beaming Consider a source of radiation, isotropic in its rest frame S′, which is moving with speed u towards an observer at rest in frame S. What does the angular distribution of the radiation look like when viewed from frame S?
y′ y
x′ x
S′ u S
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Consider a ray at angle θ′ to the x′-axis in frame S ′. In a time ∆t′, the light moves outward a distance c∆ t ′, so that
∆x′ cosθ′ = . c∆ t ′
Similarly in frame S, we have
∆x cosθ = . c∆ t
Applying the inverse Lorentz transformation
x=+γβ( x′′ ct), ct =+ γβ( ct ′′ x ) , (3.6) we find
∆x′ + β γ()∆x′ + β c ∆ t ′ cos θ′ + β cosθ = =c∆ t ′ = . ′ ′∆x′ ′ γβ()c∆+∆ t x 1+ β 1 + βθ cos c∆ t ′
We see that all the light emitted into the forward hemisphere in S ′ (for which 0≤θ′ ≤ 90 ) is emitted into a cone in S with 0≤θ ≤ cos−1 β . At relativistic speeds, the cone opening angle is very small, e.g. if β = 0.99, then θ < 8 . By comparing the solid angle in S to that in S ′, the amplification factor is 11( −β) ≈ 2. γ 2 This relativistic beaming effect is part of the reason why gamma-ray bursts are very bright.
Relativistic velocity transformations Consider a particle moving in frame S with velocity v. The components of the particle’s velocity are
dx dy dz vx=, v y = , v z = . (3.7) dt dt dt
Suppose we want to know the particle’s velocity v′ in frame S ′ moving relative to frame S. For simplicity, we again make the x-axes of the two frames parallel to the relative velocity. The components of v′ are
dx′ dy ′ dz ′ vx′=, v ′y = , v z′ = . (3.8) dt′ dt ′ dt ′
Using the Lorentz transformation, we see that
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dx − u γ ()dx− udt dxudt−dt vux − v′x = = == . (3.9) u u udx uv x γ dt− dx dt− dx 1 − 1 − c2 c2 cdt 2 c 2
Also,
dy dydt v y v′y = = = , (3.10) u udx uv x γdt− dx γ1 − γ 1 − c2 cdt 2 c 2
with a similar expression for vz′.
As an example of the application of the relativistic velocity transformation, consider two colliding protons. In the laboratory frame the velocities of the protons are 0.90c and -0.90c. What is the speed of one of the protons in a frame in which the other proton is at rest?
Consider a frame S ′ moving with velocity u = 0.90 c relative to the laboratory frame. In S ′ one of the protons is at rest, and the other proton, for which vx = − 0.90 c , has velocity
−0.90c − 0.90 c 1.80 vx′ = =− c =− 0.99 c . (3.11) 0.90c×() − 0.90 c 1.81 1− c2
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