Chapter 3 Relativistic Kinematics
Chapter 3 3.1 Relativistic Transformation 3.2 Relativistic Transformation II 3.3 Four Vector, Space Time 3.4 Kinematics: Basics 3.5 Interactive Exercises
3.1 Relativistic Transformations
Relativistic Kinematics
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Relativistic Transformations Lorentz transformations were postulated initially by Lorentz to explain the null results of the MichelsonMorley experiment. These transformations allow an event recorded in reference frame S at (x, y, z, t) to be related to the measurements in reference frame S' at (x', y', z', t'). Note that we do not assume that t = t’. We simply state the Lorentz transformations :
t = (t' + υx'/c2)/(1 - υ2/c2)1/2 x' = (x - υt)/(1 - υ2/c2)1/2 t' = (t – υx/c2)/(1 - υ2/c2)1/2 y = y' y z = z' S
y' (x, y, z, t) (x', y', z', t') x
S' x'
z' v z In examination of the Lorentz transformations the concepts of space and time are mixed. In specifying an event, an observer must record both the position of the event and the reading on his clock located at the position of the event. For example, when event 1 occurs, observers in reference frame S record its position x1 and the time t1; observers in S’ record this event as (x'1, t'1). Similarly the coordinates for event 2 are (x2, t2) in reference frame S and (x'2, t'2) in S’. Dayalbagh Eduacational Institute
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The interval between the two events has two parts: (1) a time interval for S,
Δt = t2 - t1, and a time interval for S’,
Δt' = t'2 - t'1; and (2) a space interval for S,
Δx = x2 - x1, and a space interval for S’,
Δx' = x'2 - x'1. There may also be space-time intervals Δy and Δz. These space-time intervals are also related by the Lorentz transformations: (1) (2) (3) (4)
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Time Dilation The time interval Δt' between two events on a single clock in reference frame S’. Since the clock was on hand for both events in S’, the two events happen in the same place; that is Δx' = 0. If we set Δx' = 0 in Eq. (2), we find
Again the proper time interval Δt’ is measured on a single clock, and it is smaller than the improper time interval Δt. Note that Eq. (4) is not applicable because we have no information about the spatial separation Δx between the two events in the reference frame S. When observers in reference frame S’ wish to measure a rod that is at rest with respect to S, they must observe the two ends of the rod at the same time, so that Δt' = 0. If we set Δt' = 0 in Eq. (1) and solve for Δx'. we find
The proper length Δx, the rest length rod, is greater than the length measured by S'. “Moving rods” appear to be shortened.
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The Relativity Of Simultaneity The relativity of simultaneity is also a special case of Lorentz transformations. Suppose observers in reference frame S read clocks M' and D' simultaneously at t = 0. For these observers, the time interval between the reading of the clocks is Δt = 0. The two clocks M' and D' are separated by a distance x’ - 0 = Δx' according to observers in frame S'. While observers in frame S find clock D' at t = 0, observers in S' find clock D' at an earlier time.
Thus observers in S’ find a time interval Δt' between the event clock M passing M' equal to -υ Δx'/c2. This result is derivable from Eq. (2). With Δt = 0, Δt' = -υ Δx'/c2.
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The Velocity Transformation Finally, we turn to the velocity transformation. In Fig. 1 trains S and S' move relative to each other with velocity Ď….
Fig 1: Observers in reference frames S' and S measure the velocity of particle P.
Observers in reference frame S measure a particle moving in the x-direction with velocity
observers in S’ measure the velocity of the particle to be
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Using the Lorentz transformations, we find the relation between u and u'. Dividing Eq. (1)by (2) yields
or
In terms of u and u', this gives (4) Solving Eq. (4) for u' yields
As a specific example, let the velocity of a particle as measured by observers on train S' equal u' = 0.5c to the right. If train S' travels with a velocity of 0.8c to the right relative to S, the speed of the particle measured by observers in S is
If we use the classical “velocity addition rule,” u = u' + υ, we find u = 1.3c, a value of u greater than c. Using the relativistic transformation for velocities gives a velocity less than c. Dayalbagh Eduacational Institute
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For the special case in which an observer in reference frame S' measures the velocity of light, that is, u' = c, observers in reference frame S measure
Not surprisingly, the relativistic velocity transformation yields the result of Einstein’s postulate: the speed of light is the same for all observers. Example 1 Two spaceships leave the earth. Ship A moves to the right with velocity uA = ⅗c and ship B to the left with velocity uB = ⅘c as measured by observers on earth (Fig. 2).
What is the velocity of ship B with respect to ship A?
Fig 2
The four results of the special theory of relativityTime dilation, length contraction, the relativity of simultaneity, and the velocity transformation are quite foreign to our everyday experience. Dayalbagh Eduacational Institute
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This is understandable because, in the world in which we live, relative speeds of reference frames are only a small fraction of the speed of light. In our everyday life, υ/c << 1 and (1 - υ2/c2)½ is approximately equal to one. When υ/c << 1, the Lorentz transformations reduce to the classical transformation, lengths and time intervals become invariants, and the classical rule for addition of velocities is applicable. The theory of relativity predicts that the speed of light is a limiting speed. If υ could be greater than c, the Lorentz transformations become meaningless since (1 - υ2/c2)½ is then an imaginary quantity.
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Experimental Verification The special theory has made some startling predictions concerning length contraction, time dilation, and velocity addition. In this section we will discuss only a few of the many experiments that have been done to confirm the special theory of relativity. When high-energy particles called cosmic rays enter the Earth's atmosphere from outer space, they interact with particles in the upper atmosphere, creating additional particles in a cosmic shower. Many of the particles in the shower are Ď&#x20AC;-mesons (pions), which decay into other unstable particle called muons. The properties of muons are told in chapter 2. Because muons are unstable, they decay according to the radioactive decay law
where N0 and N are the number of muons at times t = 0 and t = t, respectively, and t1/2 is the half of the muons will decay to other particles. The half-life of muons (1.52 x 10-6 s) is long enough that many of them survive the trip through the atmosphere to the Earth's surface. Let us perform an experiment by placing a muon detector on top of a mountain 2000 m high and counting the number of muons traveling at a speed near ν = 0.98 c (see fig 1a). Dayalbagh Educational Institute
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A
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B
At 2000 m, we detect 1000 muons in At sea level, we detect only 542 muons period t0 traveling at speed near 0.98c. in the same period t0 traveling at speed near 0.98c. Fig. 1: The number of muons detected with speeds near 0.98 c is much different on top of a mountain (a)than at sea level (b) because of the muon's decay. The experimental result agrees with our time dilation equation.
Suppose we count 103 muons during a given time period t0. We then move our muon detector to sea level (see Figure 1b), and we determine experimentally that approximately 540 muons survive the trip without decaying. We ignore any other interactions that may remove muons. Classically, muons traveling at a speed of 0.98c cover the 2000-m path in 6.8 x 10-6 s, and according to the radioactive decay law, only 45 muons should survive the trip. There is obviously something wrong with the classical calculation, because a factor of 12 more muons survive than the classical calculation predicts. We need to consider this problem relativistically. Because the muons are moving at a speed of 0.98c with respect to us on Earth, the effects of time dilation will be dramatic. Dayalbagh Educational Institute
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In the muon rest frame, the period for the muons to travel 2000 m (on a clock fixed with respect to the mountain) is calculated from Equation t' = γt to be (6.8/5) x 10-6 s, because γ = 5 for ν = 0.98c. For the time t = 1.36 x 10-6s, the radioactive decay law predicts that 538 muons will survive the trip, in agreement with the observations. It is useful to examine the muon decay problem from the perspective of an observer traveling with muon. This observer will not measure the distance from the top of the 2000-m mountain to sea level to be 2000 m. Rather, this observer would say that the distance is contracted and would be only 2000 m/ 5 = 400 m. The time to travel the 400-m distance would m/0.98c = 1.36 x 10-6 s according to a clock with a muon. Using the radioactive decay observer traveling with the muons would still 538 muons to survive.
be 400 at rest law, an predict
Therefore, we obtain the identical result whether we consider time dilation or space contraction, and both are in agreement with experiment, thus confirming the special theory.
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Atomic Clock Measurement An extremely accurate measurement of time can be made using a well-determined transition in the 133Cs atom that has a frequency of 9,192,631,770 Hz. In 1972 two American Physicists, J. C. Hafele and Richard E. Keating, used four cesium beam atomic clocks to test the time dilation effect. They flew with one clock eastward on regularly scheduled commercial jet airplanes around the world and another westward around the world. The two other clocks stayed fixed in the Earth's system at U.S. Naval Observatory in Washington, D. C. (fig. 2).
Fig 2 Earth's Rotation
Westward
Eastward
U.S. Naval Observatory (Washington, D.C.)
The trip eastward took 65.4 hours with 41.2 flight hours, whereas the westward trip, taken a week later, took 80.3 hours with 48.6 flight hours. The comparison with the special theory of relativity is complicated by the rotation of the Earth and by a gravitational effect arising from the general theory of relativity. The actual predictions and observations for the time differences are Dayalbagh Educational Institute
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Travel
Predicted
Observed
Eastward
-40 ± 23 ns
-59 ± 10 ns
Westward
275 ± 21 ns
273 ± 7 ns
A negative time indicates the time on the moving clock is less than the reference clock. The moving clocks lost time (ran slower) during the eastward trip, but gained time (ran faster) during the westward trip. This occurs because of the rotation of the Earth, indicating that the flying clocks ticked faster or slower than the reference clocks on Earth. The special theory of relativity is verified within the experimental uncertainties.
Example 1: Assuming a jet airplane travels 300m/s and the circumference of the earth is about 4 x 10 7 m, calculate the time dilation effect expected for a round-the-world trip exclusive of the Earth's rotation and gravitational correction.
Solution: A clock fixed on Earth will measure a flight time T0 of
which is about 37 hours. Because a clock in the airplane will run slowly, an observer on Earth will say time measured on the airplane is where
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The time difference is Because β is such a small quantity, we can use a power series expansion of the square root, keeping only the lowest term in β2.
Velocity Addition An interesting test of the velocity addition relation was made by T. Alväger and colleagues at the CERN nuclear and particle physics research facility. They used a beam of almost 20-GeV (109-eV) protons to impact on a target to produce neutral pions (π0) of more than 6 GeV. The π0 (β ≈ 0.99975) have a very short half-life and soon decay into two γ rays. In the rest frame of the π0 the two γ rays go off in opposite directions. The experimenters measured the velocity of the γ rays going in the forward direction in the laboratory (actually 6º, but we will assume 0º for purposes of calculation, because there is little difference). The Galilean addition of velocities would require the velocity of the γ rays to be u= 0.99975c + c = 1.99975c, because the velocity of γ rays is already c. Dayalbagh Educational Institute
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However, the relativistic velocity addition, with v = 0.99975c being the velocity of the Ď&#x20AC;0 rest frame with respect to the laboratory and u' = c being the velocity of the Îł rays in the laboratory tobe, according to Equation
The experimental measurement was accomplished by measuring the time taken for the Îł rays to travel between two detectors placed about 30 m apart and was in excellent agreement with the relativistic prediction, but not the Galilean one. We again have conclusive evidence of the need for the special theory of relativity. Although we have mentioned only three rather interesting experiments here, there have been literally thousands of cases examined by physicists performing experiments with nuclear and particle accelerators that conclusively verify the correctness of the concepts discussed here. Measurements done in recent years have confirmed that time dilation effects of special relativity to 40 parts in one million by using lasers to repeat experiments similar in principle to that of MichelsonMorley. Dayalbagh Educational Institute
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Quantum electrodynamics (QED) includes special relativity in its framework, and QED has been tested to 12 decimal places. Physicists continue to search for violations, including a challenge to prove that the speed of light on Earth is the same east-to-west as it is west-to-east to withing 50 m/s.
Special relativity & lifetime Lifetime only makes sense in special frame → rest frame of particle Example μ+→ e+ νe νμ ; τ = 2.2 x 10-6 s High energy cosmic ray Earth's atmosphere Many particles from collision with nucleus in atmosphere
μ ? 50 km
μ Decay: e + νs not “seen” Detected ν oscillations Earth's surface
Assume μ traveling at speed of light mean distance to decay = c · τμ = 3 x 108 x 2 x 10-6 ≈ 600 m.
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[M]·[s] [s]
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Time to reach Earth's surface at 50km
How many μ are undecayed after this time? Fraction surviving
Why do we see any μ at sea level? Obviously this calculation is incorrect.
τlab ≠ tμ
Defines τ in rest frame
τlab = γ tμ
Say τ = t2rest - t1rest
That's better!
So: τlab = γ t2lab - γ t1lab
Now let's hypothesize that the cosmic ray μ have an energy of 100 GeV
m2c2
Which any kid knows! Mass of μ is mμ = 106 MeV/c2
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Fraction of μ left at sea level
Fraction un decayed at sea level ~ 90%
Example 2 How do volumes transform? If a container has a volume V' in its own rest frame S', what is its volume as measured by an observer in S, w.r.t which it is moving at speed ν ?
Solution 2 The dimension along the direction of motion undergoes length contraction , but the other two dimensions do not. Thus the volumes transform according to V = V'/γ
Example 3 How do the densities transform? If a container holds ρ' molecules per unit volume in its own rest frame, S', how many molecules per unit volume does it carry in S?
Solution 3
The number of molecules N is invariant. Density is ρ = N/V, So, it transform by ρ = N/V = γ N/V' = γ ρ'
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Example 4 Cosmic ray muons are produced high in the atmosphere (at 8000m) and travel down towards the Earth at very nearly the speed of light (0.998c), 4a) Given lifetime of the muon (2.2X 10-6sec), how far would it go before disintegrating, according to perrelativistic Solution 4a The distance = νt = (0.998 X 3 X 108 m/s) (2.2 X 10-6 s) = 659 m No, they will not make it to ground level 4b) Now answer the same question using relativistic physics Solution 4b Due to time dilation, the muons last longer, so they travel further. i.e. γ =
= 15.8
d = distance travelled = ν ( γt) = γ X 659 = 10,400 m In this case the muons make it to ground level
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4c) Pions are also produced in the upper atmosphere : the sequence is proton hits proton (atmosphere) → p + p + pions then ∏- → μ- + νμ , ∏+ → νμ + μ+ but the lifetime of the pion is 2.6 X 10 -8 sec. Assuming the pions have the same speed (0.998c), will they reach the ground level?
Solution 4c The pions only travel 10,400 X
= 123 m
Example 5 Half the muons in a mono energetic beam decay in the first 600m. How fast are they going?
Solution 5 The decay traversed = 600 m, τ = 2.2 X 10-6 sec the half life t1/2 = (ln 2) τ d = ν ( γt1/2) = γ
= γ
or = gives Dayalbagh Educational Institute
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Example 6 A car travelling 150 km/hr covers a certain distance in 12.00 sec according to the driver's watch. What does an observer at rest on Earth measure for the time interval?
Solution 6 The car's speed relative to earth is 150km/hr. = 1.5 X 105 m/s. The driver is at rest in the reference frame of the car, so we set â&#x2C6;&#x2020;t0 = 12.00 sec in the time dilation formula.
The time measured by an observer on Earth would show no difference from that measured by the driver Using binomial expansion, for x<<1 From time-dilation formula, we have a factor thus
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So, the difference between â&#x2C6;&#x2020;t and â&#x2C6;&#x2020;t0 is predicted to be 9.66 X 10-15 sec, an extremely small amount.
Example 7 A passenger on a high speed spaceship travelling between Earth and Jupiter at a steady speed of 0.75c reads a magazine which takes 10.00min according to her watch. a) How long does this take as measured by Earth-based clocks? b) How much farther is the spaceship from Earth at the end of reading the article than it was at the beginning?
Solution 7 a) The given 10.0-min time interval is the proper timestarting and finishing the magazine happen at the same place on the spaceship. Earth clocks measure
= 15.1 min
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b) In the Earth frame, the rocket travels a distance D= v Δt = (0.75c) (15.1 min) = (0.75) (3.0 X 108 m/s) (15.1 min X 60 s/min) = 2.04 X 10
11
m.
In the space ship's frame , the earth is moving away from the spaceship at 0.75c, but the time is only 10.0 min, so the distance is measured to be D0 = νΔt0 = (2.25 X 108 m/s) (600s) = 1.35 X 1011 m
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