1417Chapter 371. From the time dilation equationt=t0(wheret0is the proper time interval,112/,and=v/c), we obtainFHGIKJ102tt.The proper time interval is measured by a clock at rest relative to the muon. Specifically,t0= 2.2000s. We are also told that Earth observers (measuring the decays of movingmuons) findt= 16.000s. Therefore,22.2000s10.99050.16.000s2. (a) We findfrom112/:2211110.14037076.1.0100000(b) Similarly,2110.0000000.99498744.(c) In this case,21100.000000.99995000.(d) The result is211000.00000.99999950.3. (a) The round-trip (discounting the time needed to “turn around”) should be one yearaccording to the clock you are carrying (this is your proper time intervalt0) and 1000years according to the clocks on Earth, which measuret. We solve Eq. 37-7 for:2201y110.99999950.1000ytt
1418CHAPTER 37(b) The equations do not show a dependence on acceleration (or on the direction of thevelocity vector), which suggests that a circular journey (with its constant magnitudecentripetal acceleration) would give the same result (if the speed is the same) as the onedescribed in the problem. A more careful argument can be given to support this, but itshould be admitted that this is a fairly subtle question that has occasionally precipitateddebates among professional physicists.4. Due to the time-dilation effect, the time between initial and final ages for the daughteris longer than the four years experienced by her father:tfdaughter–tidaughter=(4.000 y)whereis the Lorentz factor (Eq. 37-8).LettingTdenote the age of the father, then theconditions of the problem requireTi=tidaughter+25.00 y ,Tf=tfdaughter– 25.00 y.SinceTfTi= 4.000 y, then these three equations combine to give a single conditionfrom whichcan be determined (and consequentlyv):54 = 4= 13.5= 0.9973.5. In the laboratory, it travels a distanced= 0.00105 m =vt, wherev= 0.992candtis thetime measured on the laboratory clocks. We can use Eq. 37-7 to relatetto the properlifetime of the particlet0:22002110.9920.9921/tvdtttccv cwhich yieldst0= 4.4610–13s = 0.446 ps.6. From the value oftin the graph when= 0, we infer thantoin Eq. 37-9 is 8.0 s.Thus, that equation (which describes the curve in Fig. 37-22) becomes0228.0 s1( / )1ttv c .If we set= 0.96 in this expression, we obtain approximately 29 s fort.7. We solve the time dilation equation for the time elapsed (as measured by Earthobservers):tt0210 9990( .)
1419wheret0= 120 y. This yieldst= 2684 y32.6810y.8. The contracted length of the tube would beLL0125.00m1(0.999987)20.0255m.