Simple Nature - Light and Matter

position relative to the sun at exactly midnight on January 1, 1983and January 1, 1995. The 1983 date is taken to be t = 0.t (s) x y z0 1.784 × 10 12 m 3.951 × 10 12 m 0.237 × 10 12 m3.7869120000 × 10 8 s 2.420 × 10 12 m 8.827 × 10 12 m 0.488 × 10 12 mCompare the time elapsed on the spacecraft to the time in a frameof reference tied to the sun.⊲ We can convert these data into natural units, with the distanceunit being the second (i.e., a light-second, the distance light travelsin one second) and the time unit being seconds. Convertingand carrying out this subtraction, we have:∆t (s) ∆x ∆y ∆z3.7869120000 × 10 8 s 0.2121 × 10 4 s 1.626 × 10 4 s 0.084 × 10 4 sComparing the exponents of the temporal and spatial numbers,we can see that the spacecraft was moving at a velocity on theorder of 10 −4 of the speed of light, so relativistic effects should besmall but not completely negligible.Since the interval is timelike, we can take its square root andinterpret √ it as the time elapsed on the spacecraft. The result isI = 3.786911996 × 10 8 s. This is 0.4 s less than the timeelapsed in the sun’s frame of reference.aa / Light-rectangles, example 8.1. The gray light-rectangle representsthe set of all events such asP that could be visited after A andbefore B.2. The rectangle becomes asquare in the frame in which Aand B occur at the same locationin space.3. The area of the dashed squareis τ 2 , so the area of the graysquare is τ 2 /2.Invariance of the interval example 8In this example we prove that the interval is the same regardlessof what frame of reference we compute it in. This is called“Lorentz invariance.” The proof is limited to the timelike case.Given events A and B, construct the light-rectangle as defined infigure aa/1. On p. 389 we proved that the Lorentz transformationdoesn’t change the area of a shape in the x-t plane. Thereforethe area of this rectangle is unchanged if we switch to the frameof reference aa/2, in which A and B occurred at the same locationand were separated by a time interval τ. This area equals halfthe interval I between A and B. But a straightforward calculationshows that the rectangle in aa/1 also has an area equal to halfthe interval calculated in that frame. Since the area in any frame404 Chapter 7 Relativity