Ch. 38: Special Relativity End of Chapter Problem Solutions

Transcription

1 Ch 3: Speial Relativity End of Chapter Problem Solutions 1 Chasing Light In order to arry out the onversions in this exerise, we use the standard method of multiplying by unity You do not hange the value of any quantity when you multiply it by one For example: 1 year 1 = seonds Using this method, the solutions to parts (a) through (d) are as follows: (a) Continental drift v 3 entimeters year 1 year 1 meter = meters seond 316 seonds entimeters All units anel (In aneling units, we make no distintion between singular and plural, suh as seond and seonds) v = (b) Highway speed 3 v 0 kilometers hour 1 hour meters 3 = 300 meters seond 36 seonds 1 kilometer One again, all units anel: v = 96 () Supersoni plane v kilometers hour 1 hour meters 3 = 300 meters seond 36 seonds 1 kilometer (d) Earth in orbit v = 6 7 v = 300 meters seond 1 kilometer 3 30 kilometers seond meters v = 4 (e) Most everyday phenomena involve speeds that are suh small frations of the speed of light that speial ativity is not required to desribe them or their onsequenes Fast Computation A single omputation yle requires a round trip transmission between memory and proessor Take the size of the omputer to be L Then the total distane traveled during 1 1 yles in one seond is L Suppose that these signals travel with the maximum possible speed, namely the speed of light Then we have: S-1

2 Ch 3: Speial Relativity End of Chapter Problem Solutions 1 L = = 1 seond 300 meters seond Solve for L: L = = 4 15 meters 015 millimeters 3 Examples of the Priniples of Relativity EACH of the idential experiments should give the same result in the uniformly moving train as in the losed freight ontainer Note that the mass of a proton (d) is an invariant, independent of speed, as embodied in Eq 3-1 This will be true for independent experiments arried out by both observers Also note that in eah referene frame Newton s Seond Law of motion (f) is valid only at partile speeds muh smaller than the speed of light, and departures from this law will be the same at the same higher speeds measured in eah frame 4 Riding to Alpha Centauri Your spaeship is unpowered, so it onstitutes an inertial referene frame (see Setion 3-5) Therefore you will disover that all experimental observations (and all the resulting laws of physis) are the same as determined in your unpowered spaeship as they are on the surfae of Earth The one exeption is that on Earth you experiene the vertially downward aeleration of gravity This aeleration is absent in your unpowered spaeship Assume that all stars that we see with the naked eye are atively at rest with respet to Earth Then looking out the windows of your spaeship, you see hanges in the olor of stars, beause of the Doppler Shift desribed in Setion 3-14 The olors of stars diretly ahead of you are shifted toward the blue, while olors of stars diretly behind you are shifted toward the red Preview: Exerise 35, titled The Headlight Effet, shows that stars approahing you will also look muh brighter than they did when you were at rest, while those behind are muh dimmer A less obvious result of the same headlight effet predits that stars in your field of view shift forward in angle toward the diretion of your motion As a result, stars appear bunhed up ahead of you and spread out toward your rear 5 Deduing a Speed Two nanoseonds is the differene between the time of the 60-meter travel of light and the 60-meter travel of the proton pulse at the as-yet-unknown speed v Sine t = s/v, we have: 60 meters 60 meters 9 seond = v Solve for v and set = 300 meters/seond After some manipulation, the speed v of the proton pulse is found to be or v , = 606 = v = 97 meters seond 6 Eruption from the Sun The solution is similar to that of Exerise 5 Let T be the time between arrival of the light signal and the partile pulse Then: S-

3 Ch 3: Speial Relativity End of Chapter Problem Solutions light-minutes light-minutes T = / The speed of light is, by definition, one light-minute of distane per minute of time In other words it has the value unity in the equation above Therefore: T = 64 minutes minutes = 56 minutes So you have 56 minutes after the initial light pulse to seek shelter from the later pulse of partiles 7 Synhronizing a Clok We need to know the distane L of your position from that of the synhronizing lok at the origin of the oordinate system This distane L is equal to the square root of the sum of the squares of the three oordinates: 1/ L ( ) ( ) ( ) = kilometers = 60 kilometers So you need to set your lok to a time T equal to the time it takes light to travel 60 kilometers The proess involves a hange of units by our usual method of multiplying by unity: 60 kilometers 60 kilometers 3 meters T = = = 300 meters seond 1 kilometer 4 seond This is the time to whih you quikly set your lok when you reeive the synhronizing 4 pulse The proedure may be simplified by setting the lok to seonds ahead of time and starting it when the synhronizing pulse arrives Earth s Surfae Inertial? (a) Moving at 099, a proton will over meters in a time t given by: meters meters t = = = meters seond 336 seond (b) In this time the proton ontinues to move horizontally while it falls vertially a distane y given by: meters seond 336 seond meter ( )( ) y = gt = = () This distane is less than the lassially-omputed diameter of the proton itself and far smaller than any possible adjustment of the aelerator slits (for example) Conlusion: The aeleration of gravity has negligible onsequenes for this experiment In effet, Earth s surfae is an inertial frame for the purpose of analyzing results of this experiment 9 Light Clok for a Faster Roket The wristwath time (proper time) τ between events A and B as reorded on the roket lok is the time it takes for the light pulse to travel up 7 meters and bak down 7 meters, namely: 7 meters 14 meters τ = = = 300 meters seond 47 seond In the laboratory frame, in ontrast, the light from event A slants upward along a hypotenuse of length L given by: S-3

4 Ch 3: Speial Relativity End of Chapter Problem Solutions 1/ L ( ) ( ) = meters = 5 meters Then the light is refleted from the traveling mirror and slants downward an equal distane L to event B The total laboratory time t for this trip between events A and B is then: L 50 meters t = = = 300 meters seond 17 seond Hene the ratio of laboratory time to wristwath time between events A and B is given by Where and When? Use Equation 3-: t 17 seond = = 36 τ 47 seond ( τ ) = ( t) ( x) (a) In the laboratory frame the two firerakers explode in the same plae Therefore the laboratory time between these explosions is the wristwath time τ = 1 years We are also told that in the roket frame the time between the explosions is t = 13 years From these numbers we an use Eq 3- to ompute the separation x between the explosions in the roket frame: or 1/ ( x) = ( t) ( τ ) ( ) ( ) ( )[ ] 1/ x = t τ = 300 meters seond years 3 5 meters-years = seond This expression has mixed units, so we need to arry out a onversion: meters-years 316 seonds x = = seond 1 year This is the distane between the explosions in the roket frame, meters (b) In the frame of the roket observer the laboratory observer moves a distane x in a time of 13 years From the first (unonverted) equation for x above, we have 3 5 meters-years x seond v = = = 1 meters seond t 13 years or a little more than one-third the speed of light 11 Traveling to Vega (a) Joelyn travels at 099 during her 6 light-year trip to Vega, speed and distane measured with respet to the Earth-Vega frame The speed of light = 1 light-year/year, by the definition of light-year Therefore the time for her trip as measured in the Earth frame is given by: S-4

5 Ch 3: Speial Relativity End of Chapter Problem Solutions distane 6 light-years time = 63 years speed = 099 light-year year = (b) Of ourse, inhabitants of Earth annot verify that Joelyn has arrived at Vega until the light or radio signal reporting her arrival travels bak to Earth at speed, whih will take an additional 6 years Thus the total time from Joelyn s departure until word of her arrival returns to Earth is = 53 years () Joelyn ages an amount equal to the proper time τ between departure from Earth and arrival at Vega, as given in Eq 3-3: ( ) τ = = = 1 v t t years=371 years So she arrives at Vega muh younger than observers on Earth would predit from nonativisti mehanis 1 Travel to the Dog Star (a) Mya travels to the Dog Star at the speed 075 = 075 light-years/year So as reorded by Earth observers, Mya takes the time for the outward trip given by: 7 light-years toutward = = 116 years 075 light-years year This is the Earth-time at whih Mya arrives at Sirius (b) Mya now orbits the Dog Star slowly The word slowly tells us that her wristwath time will be essentially the same as the time reorded in the Earth-Vega rest frame t orbit = 7 years So the elapsed Earth-time at whih Mya starts home is: tstart home = toutward + torbit = 16 years () Then the return trip at the same speed as the outward trip will take the same Earthtime as the outward trip So the total Earth-time from departure to return is t arrive home = 16 years years = 30 years (d) For Mya the outward trip takes the proper (wristwath) time Using Equation 3-3: ( ) τ = = = outward 1 v t years years=77 years (e) For Mya, the wristwath time lapse during whih she orbits slowly is 7 years, the same as the Earth-linked time Therefore she starts home at a time on her wath given by: τ start home = 77 years + 7 years = 147 years (f) For Mya, the homeward trip takes a time equal to that of the outward trip, 77 years Therefore the elapsed wristwath time from departure from and return to Earth is t arrive home = 147 years + 77 years = 4 years This ontrasts with the elapsed time of 30 years on Earth loks 13 Fast-Moving Muons (a) Start with Equation 3-3: S-5

6 Ch 3: Speial Relativity End of Chapter Problem Solutions τ = 1 v t Square both sides and solve for v / Substitute values τ = 16 miroseonds and t = 16 miroseonds from the statement of the problem v τ = 1 = = 099, t v or 0995 = (b) Moving at this speed, in one half-life a muon pulse will over a distane given by: d = v t = = meters () If there were no ativisti time strething, the distane moved would be ten times less, or approximately 40 meters (d) In ten half-lives a pulse of muons will move a distane equal to 4 kilometers (e) In ten half-lives the number of muons in the initial pulse will be redued by a fator of = 4 Then an initial pulse of muons will be redued to a final population of 9 4 at Earth s surfae 14 Lifetime of a Fast Partile The partile lifetime in the rest frame of the detetor is given by: 3 distane 5 meter t = = = speed meters seond seond Use Equation 3-3 to obtain the orresponding wristwath time for the partile itself = = = τ v t 1 (099) 353 seond 446 seond 15 Living a Thousand Years in One Year The aging of the traveler will be the same on both the outward trip and the return trip The orresponding (but different) time lapse on Earth will also be the same on outward trip as on return trip (a) You want to be one year older while Earth ages 00 years Therefore we have, from Eq 3-3: τ = = t 3 1 v, from whih 6 v = 1 = , and v= (b) You age one year, aording to the statement of the problem () It does not matter how you exeute the trip, provided the aeleration you experiene is not too great Very large aeleration an lead to two results: (1) your death, or () the need to use general ativity to analyze the motion This is beause, stritly speaking, speial ativity deals only with inertial (free-float) referene frames S-6

7 Ch 3: Speial Relativity End of Chapter Problem Solutions 16 Birthdays The astronaut s wristwath time lapse between his 1th and 1st birthday parties is 3 years The Earth time lapse between these parties is given as 5 years (a) Start with Equation 3-: ( τ ) = ( t) ( x) Now the speed of light = 1 light-year/year, by the definition of light-year Substitute the numbers given in the statement of the exerise ( ) ( ) ( ) x = t τ = 1 light-year year 5 9 years = 4 light-years (b) From the statement of the problem t = 5 years Therefore, using the result of part (a), the speed of the spaeship in the Earth frame is: x 4 light-years = = 0 t 5 years 17 Relations between Events Sine t and x are both measured in the units of years, Eq 3- simplifies to the form τ = t x, for events separated by a time-like interval Also, Equation 3-1 simplifies to the form σ = for events separated by a spae-like interval x t Between events 1 and, the time separation is 5 years and the spae separation is 3 lightyears Between events 1 and 3, the time separation is 3 years and the spae separation is 5 lightyears Between events and 3, the time separation is years and the spae separation is lightyears From these oordinate separations we an omplete the requested list: (1,) timelike yes (1,3) spaelike no (,3) lightlike yes 1 Proper Distane and Proper Time (a) Spaelike separation Start with Equation 3-1: ( σ ) ( x) ( t) = (3-1) For a given σ, we are free to hoose values of x and t arbitrarily, as long as the differene of squares on the right side of 3-1 is equal to ( σ) Eah different suh hoie represents spae and time separations in a different referene frame moving with respet to other frames So hoose t = 0 Then in the frame with that zero time separation, x = σ, as we were asked to show (The statement of the problem asks for distane, so we hoose the plus sign when taking the square root) (b) Timelike separation Start with Equation 3-: S-7

8 Ch 3: Speial Relativity End of Chapter Problem Solutions ( τ ) ( t) ( x) = (3-) In this ase, for a given τ, we are free to hoose values of x and t arbitrarily, as long as the differene of squares on the right side of 3- is equal to ( τ) Eah different suh hoie represents spae and time separations in a partiular referene frame moving with respet to other frames So this time hoose x = 0 Then in the frame with zero spae separation, t = τ, as we were asked to show (The hoie of plus sign for the square root reflets our belief that the two loks run in the same diretion) () Lightlike separation By definition, when two events are separated by a lightlike interval, the two terms on the right side of 3-1 or 3- are equal, so that we have Eq 3-13: ( σ) ( τ) = = 0, (3-13) so both the proper time and the proper distane between the two events are both equal to zero Now, when the spae parts of Eqs 3-1 or 3- are equal to the orresponding time parts, the resulting equality an be expressed: ( x) = ( t) This equation desribes the motion of a flash of light in either the positive or negative x- diretion, whih is just the path that moves a single light flash between these two events separated by a lightlike interval 19 Symmetri Relativity of Simultaneity In the following rude diagrams, T represents the train, R represents the position of the rider on the train, E represents the position of the earth observer, and * represents the positions of the light flash that results from the lightning strikes at the two ends of the train The following diagrams represent the various positions at different times in the rest frame of the train TIME t 1 : TTTTTTTTTTRTTTTTTTTTT * * E TIME t TTTTTTTTTTRTTTTTTTTTT * * E TIME t 3 : TTTTTTTTTTRTTTTTTTTTT ** E You an see from the diagram at time t that the flash from the left end of the train arrives first aording to the ground observer 0 Boosting the Speed Start with Equation 3-17 for the energy of a moving partile: S-

9 Ch 3: Speial Relativity End of Chapter Problem Solutions E = m ( 1 v ) (a) The inrease in energy in hanging speed from v/ = 00 to v/ = 009 is: The rest energy of the eletron is: so our result is m 1/ m m E = 1/ 1/ 1 ( 009) 1 ( 00) m m = = (041 03) m =00009 m ( 911 kilogram)( 300 meters seond) = 31 = 0 14 joule, E = joule Notie that at these atively low speeds, this result is almost the same as given by Newtonian mehanis, in whih the kineti energy K has the definition: 1 v K = mv = m, so that the differene in energy would be ( 009) ( 00) K = m = m = m (b) The inrease in energy in hanging speed from v/ = 09 to v/ = 099 is m m E = 1/ 1/ 1( 099) 1( 09) m m = = ( ) m =06 m From the value of m alulated in part (a), this is equal to E = joule Notie that this hange in energy is almost 300 times that required to go from v/ = 00 to 009, even though the inrease in speed is the same in both ases At high speed, this alulation annot be approximated using the Newtonian expression for kineti energy, whih would yield S-9

10 Ch 3: Speial Relativity End of Chapter Problem Solutions ( 099) ( 09) ENewton = K = m = = m m NEWTON IS WRONG HERE! Compare this wrong number with the orret result 06 m for the ativisti analysis 1 Light Bulb Radiating Mass 7 A 0-watt bulb emits 0 joules/seond (most of it as heat) In one year, 316 seonds, the 0 watt bulb dissipates total energy E tot given by the equation joules 316 seonds 316 joules E tot = = seond year year The mass equivalent of this energy is obtained by dividing the total energy by From the equation = E/m we see that the units of an be expressed as joules/kilogram Therefore tot 9 E 316 joules year m = = = joules kilogram 351 kilogram year This is about 35 mirograms/year, demonstrating one again the immense amount of energy stored in matter Proton Crosses Galaxy One year is equal to 1 minute = = 60 seonds 1 year seonds 57 5 minutes Therefore the diameter D of our galaxy is equal to 5 57 minutes D = 5 light-years = 57 light-minutes 1 year The problem statement demands that the proton ross the galaxy in τ = 1 minute on its own wristwath This must mean that it travels very lose to the speed of light as observed in the galaxy rest frame Therefore it takes slightly more than t = 57 minutes to ross the galaxy in the rest frame of the galaxy (You an verify that this is so by substituting τ and x = D into Eq 3-1 You will find that the numerial value of t is dominated by D) Equation 3-17 ates the energy of the proton to the ratio of time lapses in different frames These times are those for the proton to move between two events, in this ase events of departure from one edge of the galaxy and arrival at the other edge t 57 minutes E = m m = 57 m τ 1 minute The rest energy of the proton is approximately 9 eletron-volts So the energy of our 19 galaxy-spanning proton is approximately 5 eletron-volts or about joules Exerise 51 reports that some osmi rays have been deteted with twie this energy 3 Converting Mass to Energy Adding up the initial masses, we have S-

11 and the final masses add up to: Ch 3: Speial Relativity End of Chapter Problem Solutions m( p) + m( F) = 075u u = u m( α ) + m( O) = u u = u The derease in the sum of the masses of omponent nulei during the reation 3 is m = 71 u Aording to Equation 1-9 in Setion 17, the ation between atomi mass units and kilograms is 1u kilogram = Therefore the number of kilograms lost during the reation we are analyzing is kilogram m = 71 3u = kilogram u The energy eased is this mass hange times E = m = kilogram 900 joules kilogram = 1300 joules Using the onversion fator between joules and MeV, we have: 1 MeV E = = 160 joule joules 17 MeV 13 4 Aspirin-Powered Automobile A mass of 35 milligrams has an equivalent energy equal to (reall that = E/m has the units joules/kilogram): 1 kilogram E = m = 35 milligrams 900 joules kilogram 16 3 milligrams 16 = 93 joules From the data provided in the exerise, we have, for the driving distane L: 175 kilometers liter L = joules liter joules= kilometers This is about 6 times the distane from Sun to Earth We an hope that in the distant future an advaned ivilization will not need the long-exhausted oil reserves of Earth to power their vehiles! 5 Converting Energy to Mass (a) The kineti energy of eah train is (1/)mv, so the total kineti energy of the two trains before the ollision is equal to: KE = mv = 6 kilograms = tot 6 4 joules 150 kilometers 3 meters 1 hour hour 1 kilometer 3 36 seonds (b) Divide by = E/m with the units joules/kilogram: S-11

12 Ch 3: Speial Relativity End of Chapter Problem Solutions tot KE 4 joules m = = = joules kilogram This is equal to 0116 milligram kilogram 6 Eletrially Aelerated Eletron The kineti energy of the eletron is to be equal to one perent of the rest energy of the eletron: ev = 001 m The rest energy of an eletron is MeV = 511 eletron-volts Therefore times this is 511 eletron-volts By the definition of the eletron-volt, this is the voltage through whih the eletron must be aelerated 7 Powerful Proton We are given that the proton has a kineti energy equal to N times its rest energy From Equations 3-19 and 3-0: m ( 1 v ) 1/ m = Nm Canel the ommon fator m and rearrange to yield (a) Solve this equation for the speed 1 ( 1 v ) 1/ = N + 1 (A) ( + ) 1/ 1/ 1 N N v = 1 = ( N + 1) N + 1 (B) (b) Substitute Equation 3-3 for τ into Eq 3-15 to obtain: x m x mv p = m = = τ t v v ( 1 ) ( 1 ) 1/ 1/ (C) Use Equation A to eliminate the square root in the denominator of the right side of Eq C Then use Equation (B) to eliminate v in the numerator We obtain: ( ) 1/ p = N N + m Relativisti Chemistry (a) The reation emits energy That means that the reatants have more energy than the produts So the resulting water will have less mass than the original hydrogen plus oxygen gases (b) The desribed reation with 4 kilograms of hydrogen eases E = 1 joules of energy The orresponding mass equivalent is: 1 E joules m = = = joules kilogram S kilogram=11 milligrams () We do not want to put several metri tons of reatant and produt on a deliate hemial balane A smaller amount is OK, provided the mass ratio of hydrogen to oxygen is 1 to But here we ompute the frational hange using the larger amounts,

13 Ch 3: Speial Relativity End of Chapter Problem Solutions whih will be the same frational hange for the smaller amounts The sum of reatants in part (b) is + = 9 kilograms The produt water has about -5 kilograms less mass than the reatants The frational differene hange is approximately This is too small by approximately two orders of magnitude to be deteted by the most sensitive hemial balane 9 Finding the Mass (a) Start with Equation 3-1: ( m ) = E ( p) Substitute for the energy E from Equation 3-0 for the kineti energy E = m + K The square of m on both sides anel The result is: Solve for m: ( ) 0= m K + K p p K m = K For slow partile speed, this redues to the first term, whih beomes m, as expeted (b) From the statement of the problem: ( ) ( ) 11 MeV 550 MeV 1331 MeV 75 MeV 56 MeV m = = = 550 MeV 13 But 1 MeV = 160 joules and has the units of E/m or joules/kilogram, so we have MeV 160 joule m = = joules kilogram 1 MeV 1 kilogram The mass of the eletron is kilogram So the mass of our partile is m m e = = 911 kilogram 1 kilogram This is approximately the mass of a muon 30 A Box of Light (a) Every one of us will make a different estimate of the eletriity used per member of the population We hose kilowatts per person, whih inludes street lighting, muniipal transportation, and so forth Then the total use for a ity of million is 16 million kilowatts (b) One hour has 3600 seonds and one watt is one joule/ seond so the total energy used by our ity in one hour is S-13

14 Ch 3: Speial Relativity End of Chapter Problem Solutions 3 watts 1 joule seond E = 16 6 kilowatts 36 3 seonds 1 kilowatt 1 watt = joules The mass equivalent is: 13 E 576 joules m = = = = joules kilogram That is how muh the mass of the box would inrease kilogram 064 gram 31 Creating a Proton-Antiproton Pair (a) Reall that total energy onsists of rest energy m plus kineti energy (see Eq 3-0) Beause the two protons (with equal rest mass) are fired in opposite diretions with equal energy, we know that their momenta are equal and opposite Therefore, the total momentum before the ollision is zero, and so the total momentum after the ollision must also be zero We know also that the total energy before the ollision must be equal to the total energy after the ollision So the lowest possible energy (rest energy plus kineti energy) of the four partiles after the ollision results if all four are at rest Then the total energy both before and after the ollision is just four times the rest mass of one proton (b) After the ollision the total energy is four times the rest energy of one proton Therefore before the ollision, the total energy is four times the rest energy of one proton But before the ollision the total energy is two times the rest energy of eah proton plus two times the kineti energy of eah proton Therefore, the kineti energy of eah inident proton is equal to the rest mass of one proton () The kineti energy of eah inident proton is approximately 1 GeV, whih is equal to the rest energy of the proton, as derived in part (b) This result is reasonable, sine in the zero-total-momentum frame all of the kineti energy goes into reating new rest mass (provided the produts are at rest after the ollision) 3 Really Simultaneous? The seond Lorentz transformation Eq 3-5 reads: t = t ( v x ) ( 1 ( v ) ) This equation ates the time differene between two events in the primed roket oordinates to the time separation and x-separation between the two events in the unprimed laboratory oordinates Now suppose that two events our at the same time in the laboratory frame ( t = 0) and at the same x-oordinate in the laboratory frame ( x = 0) Then the equation tells us that the two events will also our at the same time they will be simultaneous in the roket frame ( t = 0) (a) and (b) The onditions just derived are satisfied for the pair of events in part (a) and all three pairs of events in part (b) So, we onlude that events are simultaneous in both the laboratory and roket frames for the pair of events in part (a) and for all three pair of events in part (b) () The generalization is that if two events are simultaneous in the laboratory frame and our at the same oordinate along the diretion of ative motion (in our ase the x- diretion), then the two events are simultaneous in the roket frame You an use the inverse Lorentz transformation (3-7) to show that the reverse is also true S-14

15 Ch 3: Speial Relativity End of Chapter Problem Solutions 33 Transformation of y-veloity Aording to the statement of the problem, the partile moves in the roket y -diretion with speed v y = y t We use the equation for the inverse Lorentz transformation (Eq 3-7) Divide x by t The resulting expression simplifies beause ( 1 v ) 1/ anels in the numerator and denominator, and beause x = 0 in this problem We obtain: v x v t t t x = = = v This makes sense, beause the two events our along the roket y axis, whih moves with a speed v in the laboratory frame In a similar manner (but this time without the anelled square root): y y 1/ 1/ vy = = ( 1 ( v ) ) = v y( 1 ( v ) ) t t 34 Transformation of Veloity Diretion Apply the inverse Lorentz transformation, Eq 3-7 to an expression for the tangent of the angle in the laboratory frame y y y t tanφ = = 1 ( ) = 1 ( ) + + ( v ) ( v ) 1/ 1/ x x v t x t v v y 1/ v sinφ 1/ ( 1 ( v ) ) ( 1 ( v ) ) x + osφ + = = v v v v 35 The Headlight Effet (a) Use the inverse Lorentz transformation equation 3-7 and set x t = v = φ ( ) x os, ( v ) ( ) ( ) ( ) 1 ( )/ x t x + v t x t + v osφ = = = t + v x + v x t osφ + v = 1+ osφ Optional: The result of Problem 34 an be put into the same form as this result when v =, by setting tanφ = sinφ osφ and using the trigonometri identity: and simplifying the result φ + 1 os sin, (b) Consider a ray of light that moves in the roket frame perpendiular to the diretion of ative motion: φ o = π The osine of this angle is equal to zero Then the result of part (a) beomes the desired equation: S-15 φ os φ o = v () Substitute v = 099 into the equation in part (b) to obtain φ o = 1 degrees 36 Eletron Shrinks Distane In this ase the evauated tube is at rest in the laboratory frame and is ontrated as observed in the roket (moving eletron) frame Modify Equation 3- to represent this situation:

16 Ch 3: Speial Relativity End of Chapter Problem Solutions ( ) x = v x = = =00153 meter = 153 entimeter meters meters 37 Passing Time The length L of the spaeship in the laboratory frame is ated to its passage time by the equation, L = vt That is one equation in the two unknowns L and v A seond equation for the same two unknowns is Eq 3- for the Lorentz ontration: L = v L 1 (a) Substitute for L from the first equation and square both sides, Solve for v and substitute known values: L v = = ( ) vt = 1 v L meters 1/ 1/ [ t + L ] 7 ( ) meters seond v = seond (b) From the first equation above and our new value for v: ( ) 1/ L = 1 v L = 50 meters It is interesting that even at a speed this lose to that of light the Lorentz-ontrated length is still half of the rest length 3 Transformation of Angles (a) The inverse Lorentz transformation Equation 3-7 tells us that y = y Use the Lorentz ontration equation 3-: ( ) 1/ x = 1 ( v ) x These length transformations allow us to alulate the tangent of the angle in the laboratory frame: y y tanφ tan φ = = = 1/ 1/ x v x v ( 1 ( ) ) ( 1 ( ) ) (b) The length of the meter stik rekoned by the laboratory observer is the square root of the sum of the squares of its omponents in that frame 1/ 1/ ( ) ( ) ( 1 ( ) ) L = x + y = v x + y 1/ 1/ = [ x + y x ( v ) ] = 1 ( ( v ) ) os φ meter This makes qualitative sense: The laboratory length is less than one meter beause the x- omponent of the length is Lorentz ontrated Chek the limiting ases: As v goes to zero, L goes to the value unity, whih is orret for a meter stik at rest S-16

17 Ch 3: Speial Relativity End of Chapter Problem Solutions As φ goes to zero, the meter stik lies along the x-diretion, so the formula redues to the usual form for the Lorentz ontration As φ goes to 90 degrees, the meter stik moves sideways, and the laboratory observer measures its length to be the proper length: one meter () Optional In the transformation in this exerise, time does not enter We are NOT desribing the oordinate separations between two events Instead, the roket observer and the laboratory observer eah independently measures the x-omponent of the meter stik using different pairs of events, eah pair hosen to be simultaneous in the frame of that observer 39 Traveling to the Galati Center (a) Yes, speial ativity tells us that the wristwath time τ between two events (two events onneted by a timelike interval) an be made as small as desired by hoosing the appropriate frame One an analyze this either in terms of the time-strething ation between τ and t between events (Eq 3-3) or in terms of the Lorentz ontration of distane (Eq 3-) between Earth and the galati enter (b) The galaxy enter is light-years distant This means it will take light t = years to reah the enter as measured in the galaxy rest frame We want to get there in wristwath time τ = 30 years From Equation 3-16: From whih τ 30 years v 1 t = = years v ( ) 3 6 = = Use approximation (3-1) to take the square root: v ( ) 1/ = = = Limo in the Garage The limo, rest length 30 meters, must be ontrated to 6/3 = meters at speed The required Lorentz ontration fator is therefore (using Eq 3-): x meters v x = = = 30 meters , from whih v = Bakfire Use Equation 3-31: u + v u = 1 + u ( v ) In the statement of the problem we are given everything exept u, the veloity of the bullet with respet to the roket Solve this equation for u and substitute the values given uv u = = = = 0 1uv S-17

18 Ch 3: Speial Relativity End of Chapter Problem Solutions The minus sign tells us that the roket observer measures the bullet to be moving in the negative x diretion 4 Separating Galaxies (a) The observer on Galaxy A will observe our galaxy to be reeding with speed 03 (b) The formal analysis of this part is idential to that of problem 41 Choose the positive x diretion as the diretion of motion of Galaxy A, and its speed with respet to us is v = + 03 We measure Galaxy B to be moving in the negative x diretion with veloity u = 03 Then the final formula in the solution to problem 41 is adapted to give the veloity of Galaxy B with respet to Galaxy A uv u = = = = uv Deaying K o Meson Start with Equation 3-31: u + v u = 1+ uv We are given that v = 09 The maximum and minimum values of u our when the daughter mesons move along the x diretion of ative motion The two answers are ± ± u = = 1± ± 0745 u + = 0990 and u 0 44 Transit Time Assume that with respet to the Earth frame the spaeship moves in the positive x diretion and the mirometeorite moves in the negative x diretion Then we an solve this exerise the same way we solved problem 41 Substituting values from the present exerise into the seond equation of the solution to problem 41, we have veloity of the roket with respet to Earth = v = 0, veloity of mirometeorite with respet to Earth = u = 0 The result is: = uv u = = = = 09 1 uv Now we know that the mirometeorite is moving at a speed of 09 with respet to the spaeship in the spaeship rest frame We know that the ship is 350 meters long in its rest frame, so that the time t it takes for the mirometeorite to pass in the spaeship frame is: 350 meters 350 meters t = = = 45 Listening to the Traveler Use Equation 3-33: meters seond 1 6 seond 1/ 1/ 1v 900 o f = f = fo 09 0 MHz 1 v = = 9 MHz S-1

19 Ch 3: Speial Relativity End of Chapter Problem Solutions 46 Speed Trap From Figure 34-1 in Chapter 34 we selet the wavelengths of 55 nanometers for green light and 650 nanometers for red light, assuming pure spetral olors, λ = 55 nanometers and λ red = 650 nanometers green We need to adapt Equation 3-33 for motion in whih the soure (the red stoplight) omes toward us, the observer Do this by reversing the sign of v that appears in two plaes in Equation 3-33 The result is: f 1+ v = 1 v 1/ f o Here f o is the frequeny of the emitted red light from the stoplight and f is the green light pereived by the approahing motorist Beause of the inverse ation f = / λ between frequeny and wavelength, we an rewrite this equation as: Square both sides Solve for v /: f green λred 1+ v f red λgreen 1 v = = S-19 1/ λ red 650 nanometers 1+ v = = 153 = λ green 55 nanometers 1 v v = = = So the driver approahes the red stoplight at 01 What would the fine for speeding amount to? 47 Redshift Fator z From the statement of the problem: ( z) λ = 1+ λ = 7 λ observed emitted emitted From the inverse ation between wavelength and frequeny in the equation f = / λ, we onlude that f = emitted observed Equation 3-33 deals with the ase we are onsidering, the soure moving away from us: Solve for v /: f 7 = = fobserved 1 1 v femitted 1+ v v = = So the distant supernovas are moving away from us at a speed of 096 aording to the speial ativity interpretation 1/

20 Ch 3: Speial Relativity End of Chapter Problem Solutions 4 Reeding Galaxy The wavelength is inreased, so the frequeny is dereased This means that the galaxy is moving away from us (answer to part b) (a) Beause of the inverse ation f = / λ between frequeny and wavelength, we an rewrite Equation 3-33 as: Square both sides of this equation λ = = f observed emitted 1 v femitted λobserved 1+ v 1/ Solve for v /: λ emitted 513 nanometers 1 v = 0955 λ = = observed 55 nanometers 1+ v v = = This is the speed of reession of the galaxy, aording to speial ativity 49 Exodus from Earth (a) Rasmia s data fixes her age at the moment she was disassembled on Earth Data orretly transmitted does not age, so when resurreted on Ziron Rasmia starts her new life from the same age as she left Earth (You might be tempted to say that the proper time for transmission by light or radio waves is zero, (Equation 3-13), equal to the zero aging of Rasmia s data during transmission You would be right, but data need not age if it is properly error-orreted Therefore Rasmia would not have aged upon arrival at Ziron even if the data were physially transmitted between Earth and Ziron at a speed muh less than the speed of light) (b) In the Earth-Ziron rest frame the transmission of data has taken 0 years, so both Earth and Ziron age 0 years during this transmission () Rasmia s obituary an reah Earth after her departure by a time 0 years (outward transmission of Rasmia data) years (Rasmia s added lifetime in the Earth-Ziron frame) + 0 years (return transmission of her obituary) = 350 years (d) The total time is five generations, or 00 years, on Ziron plus 0 years eah for the outward transmission of data sets and the return transmission of reports, a total of 0 years on Earth 50 Eletron in Orbit Use Newton s seond law of motion F = ma (Eq 3-) with the aeleration of an objet in irular orbit (Eq 5-34) and the fore between two harges (Eq -4), alling e the harge on the eletron and Ze the harge on the nuleus Assume fore and aeleration are radially inward where Eq -7 equals o Ze mv F = k ma, r = = r (A) 1 9 k = = 99 newton-meter oulomb (B) 4πε S-0

21 Ch 3: Speial Relativity End of Chapter Problem Solutions Rewrite the seond and fourth entries in equation (A) and substitute the expression mvr = given in the statement of the problem: kze = mrv = v, or kze v = Substitute values for k from equation (B) and the following values: e = oulomb, 34 h 663 newton-meter-seond = = = π π 6 34 newton-meter-seond The result is 6 ( ) v = 17 meters seond Z (a) For hydrogen, Z = 1 and the speed is less that one one-hundredth of the speed of light () So speial ativity need not be used in the Bohr theory of the hydrogen atom (b) For the inner orbit of uranium, Z = 9 and v = 00 meters seond (d) This is two-thirds of the speed of light, so speial ativity should be used in the Bohr theory of the inner orbit of the uranium atom 51 Super Cosmi Rays This exerise is similar to problem, and the solution is similar but not idential 7 One year is equal to 316 seonds Therefore the diameter D of our galaxy is equal to seonds 5 1 D = light-years = 316 light-seonds 1 year Equation 3-17 ates the energy of the proton to the ratio of time lapses in different frames These times are those for the proton to move between two events, in this ase the events of departure from one edge of the galaxy and arrival at the other edge E = m t τ, or τ = t ( Em) The rest energy m of the proton is approximately 9 eletron-volts, so 0 E eletron-volts 9 m = eletron-volts = 11 The proton moves at a speed very lose to the speed of light, so t is approximately equal to the time it takes for light to ross the galaxy We onlude that S-1

22 Ch 3: Speial Relativity End of Chapter Problem Solutions 1 t 316 seonds τ = = 316 seonds 11 Em ( ) This is approximately half the one minute transit wristwath time for the proton of energy 5 19 derived in problem You an verify the assumption that t is lose to the time it takes for light to ross the galaxy by substituting values for τ and x = D into Equation Synhronization by a Traveling Clok (a) Yes, we an define synhronization any way we want Different methods of synhronization, however, will lead to different experimental results, whih will require different interpretations if we want our laws of Nature to be onsistent (b) Evelyn s method is one possible way to synhronize separated loks Note, however, that the result will depend on the speed of the traveling lok () Moving at the speed v = 5 5 kilometers/hour, Evelyn s traveling lok overs the distane D = one million kilometers distane between Big Ben and Little Ben in a time t given by The speed of light is 6 D kilometers t = = = hours 5 v 5 kilometers hour ( ) ( meters seond 36 seonds hour) ( 3 kilometers meter) = = 9 kilometers hour 5 so at v = 5 kilometers/hour, Evelyn s traveling lok is moving at v= Equation 3-3 tells us that the wristwath time lapse τ on Evelyn s traveling lok during the trip is ( v ) t ( ) 1/ 1/ τ 6 6 = 1 = hours hours 014 hour, where we have used the approximation (3-1) So the differene between the reading on Little Ben and the nearby lattie lok is t τ = hour 36 milliseonds hour =0770 milliseond (d) Redo the alulation of part () for a speed of the traveling lok 00 times as great: 6 D kilometers t = = = v 5 kilometers hour 3 hour=7 seonds, and v= 0463 Then ( v ) t ( ) 1/ 1/ τ = 1 = seonds=09 7 seonds=64 seonds and t τ = 7 seonds 64 seonds = 0 seond S-

23 Ch 3: Speial Relativity End of Chapter Problem Solutions So, for these partiular speeds, when the speed inreases by a fator of one thousand we see that the differene in times that appear on Little Ben and the nearby lattie lok also inreases by approximately a fator of Down with Relativity General: Speial ativity helps us to find the spae and time ations between pairs of events as reorded in referene frames in uniform ative motion The following solutions diret our attention to the pairs of events used in eah ase (a) The two events: two sequential tiks of a single moving roket lok whih are reorded on two different loks in the laboratory frame in whih the lok moves Relativity tells us that the wristwath time between tiks reorded on the moving lok is less than the time between the pair of tiks reorded on the two different laboratory loks Now think of a different pair of events: two sequential tiks of a single laboratory lok, whih are reorded on two different roket loks as the laboratory lok passes Here too ativity tells us that the wristwath time between tiks reorded on the laboratory lok is less than the time between the pair of tiks reorded on the two different roket loks The two results are weird but they are not inonsistent, beause they refer to different pairs of events (b) Think of two ases: Case one: Two firerakers explode at the two ends of a moving (roket) meter stik, explosions that are simultaneous in the laboratory frame The laboratory-measured distane between these simultaneous explosions is the laboratory measure of the length of the moving meter stik Relativity tells us that the distane between explosions will be less as measured by the laboratory observer than the one-meter length of the stik measured in the roket frame in whih it is at rest Case two: Two firerakers explode at the two ends of a meter stik at rest in the laboratory frame In this seond ase the two explosions are timed to be simultaneous in the roket frame Of ourse the meter stik moves as observed in the roket frame Nevertheless the definition of length is the same: the distane between the simultaneous explosions at the ends of the stik as measured in the roket frame Here again ativity tells us that the distane between explosions will be less as measured by the roket observer than the one-meter length of the stik measured in the laboratory frame in whih it is at rest Eah of these two ases is weird but they are not inonsistent, beause they refer to different pairs of events () How do we measure the speed of light? One way is to measure the time between two events: the arrival of a light two different loations, say two posts both at rest in our referene frame Let that measurement be made in the laboratory frame Now set up a roket frame moving in the same diretion as the flash of light How does a roket observer measure the speed of this flash? By measuring the time between a different pair of events, the arrival of the light flash at two different loations, say two posts both at rest in the roket frame Speial ativity tells us that the speed will have the same value as measured in both frames Again, this is weird but not inonsistent (d) What is reality? This is not a sientifi question but a philosophial one Different people will have different opinions about it Our opinion is that, stritly speaking, siene does not deal with reality (sorry!), but rather devises models that predit experimental results and ompares these preditions with observations Speial ativity satisfies this definition of a siene (e) Sara, in this ase you are right about our presentation of speial ativity! Coordinates are for the onveniene of people; oordinates are not fundamental Speial ativity an S-3

24 Ch 3: Speial Relativity End of Chapter Problem Solutions be expressed in invariant form, for example desribing the separation between all pairs of events using wristwath time or proper distane Values of these invariant quantities do not depend on the partiular inertial referene frame in use Suh a desription is more fundamental than the oordinate-dependent desription given in our hapter 54 The Photon as a Zero-Mass Partile (a) The equation E = p follows from Equation 3-1: ( m ) = E ( p), when we set m = 0 (b) The total momentum is zero before the deay Hene the total momentum must be o zero after the deay This tells us that in the rest frame of the π meson the two gamma rays must have momenta of equal magnitude (and therefore equal energy) but opposite diretion () The energy of eah outgoing gamma ray is (135/) MeV = 675 MeV 55 Pair Prodution with Gamma Rays (a) The threshold energy of eah photon is equal to the rest energy of one eletron, or about 0511 MeV (b) The total momentum of the system p sys = 0 Therefore, the mass M sys of the system before the ollision is the rest mass of the nuleus plus twie the energy of eah gamma ray divided by M = m+ E sys p After the ollision the momentum of the system still has zero momentum, p sys = 0 Then the mass M sys of the system is just the sum of the rest masses of the individual partiles (beause eah of the partiles is at rest): M = m+ m sys () Mass without mass? As the mass m of the nuleus goes to zero, we have before the ollision a system of two oppositely-moving photons, a system that has mass equal to the mass of the two eletrons at rest after the ollision For example, two oppositely-moving photons would provide a gravitational attration to a distant orbiter John Arhibald Wheeler and his graduate student Rihard Feynman one devised a theoretial objet, whih they alled a geon, omposed of nothing but photons Eah photon orbits the struture held together by the mutual attration of all the other photons Further study predits that geons are unstable: they spontaneously fly apart after some time Probably this is the reason that geons have not been observed in Nature 56 Resonant Absorption of a Gamma Ray (a) The first equation follows immediately from the onservation of energy, equating the total energy before the ollision to the total energy after the ollision The seond equation equates the momentum of the photon before the ollision with the momentum of the exited nuleus after the ollision The equation for momentum p is obtained by rearranging Equation 3-1 and applying the result to the produt nuleus ( m ) = E ( p) S-4 e

25 Ch 3: Speial Relativity End of Chapter Problem Solutions beomes p = 4 ( E m ) 1/ (b) Use the first onservation equation to eliminate E m* from the seond onservation equation Solve for m* : ( p ) m = m m+ E * () For E p / << m, we have m* = m This makes sense, beause in this ase the inoming photon adds negligible mass-equivalent energy to the initial nuleus 57 Photon Braking Using the notation of the figure, the onservation of energy (inluding rest energy) yields: E = m + E (A) M p The rest mass of a photon is zero (problem 54), so the ation between its energy and its momentum (substituting zero mass into Equation 3-1) is, for positive momentum, E = p And the onservation of momentum for the present reation tells us: pm = Ep (B) We have two equations, (A) and (B) in the three unknowns E p, p M, and E M A third equation is Equation 3-1, applied to the inoming nuleus: ( M ) E ( p ) M M = (C) Substitute for p M from equation (B) into equation (C) and solve the result for the square of E p : ( ) E = E M (D) p M Solve equation (A) for E p, square both sides, and substitute for E p squared from equation (D) After some manipulation and aneling, the result is: E ( + ) M m M = Chek this result by examining limiting ases: Suppose that M is equal to m This means that no photon was emitted and the initial partile of mass M was already at rest Our solution says that in this ase E M beomes m, the rest energy of the final partile, as expeted At the other limit, suppose that M >> m, whih means that a photon of very high energy must have been emitted In this ase E M approahes the limit M /(m), whih says that the inoming mass must have been very large, whih simply restates our limit 5 Limo and Garage Paradox All paradoxes in speial ativity that we know about hinge on the ativity of simultaneity Look at figures (b) and () for problem 40 In the garage frame Event A m S-5

26 Ch 3: Speial Relativity End of Chapter Problem Solutions front door loses (with the bak end of the limo just inside the front door) ours before Event B rear door opens (with the front end of the limo just inside the bak door) Look at figure (a) for problem 40, the rest frame of the limo In your mind keep the limo at rest but piture the garage moving to the left and rapidly approahing the limo The garage is Lorentz-ontrated along its diretion of motion Now onsider Events A and B Clearly the bak end of the garage will reah the front end of the limo (Event B) before the front end of the garage reahes the bak end of the limo (Event A) In brief, the time sequene of Events A and B are reversed with respet to the two referene frames So Carman is orret: In Carman s limo rest frame the limo annot fit into the moving garage If you prefer a more formal proof of possible reversal of time ordering of events, look at the seond Lorentz transformation in Eq 3-5: t = t( v x ) 1/ ( 1 ( v ) ) Suppose that t is positive One an find pairs of events (that is, values for t and x) for whih the numerator on the right-hand side beomes negative For these pairs of events a positive time separation t in the laboratory frame orresponds to a negative time separation t in the roket frame In other words the time order of the two events in the two frames is reversed The limo and garage paradox is an example of this reversal of the order of events For Garageman the bak end of the limo enters the garage before the front end of the limo leaves the garage For Carman the front end of the limo leaves the garage before the bak end of the limo enters the garage 59 Twin Paradox Cyril Allen is a sophistiated riti Think of a very long tape measure strething between Earth and the distant star a thousand light-years from Earth (this star assumed to be at rest with respet to Earth) Posted along this tape measure are a series of loks synhronized with respet to the Earth lok aording to the method desribed in Setion 3-5 The Earth observers all agree that Cyril s lok runs slow as he flies along the tape, so that, for suffiient speed, he is still alive when he reahes the distant star Now think of a very long rod stiking out the front of Cyril s spaeship and at rest with respet to that spaeship In thought, the rod is long enough to reah from Earth to the distant star Along this rod are posted observers with loks synhronized in Cyril s spaeship rest frame After Cyril reahes speed, Lorentz ontration will result in the distant star appearing muh loser, lose enough that Cyril an reah it in his lifetime In Cyril s rest frame, the lok on Earth runs slow with respet to his hain of loks, and so reads less than his lok when he reahes the distant star Conlusion: Both sets of observers agree that Cyril an reah the distant star in one lifetime We have seen that to the Earth observers Cyril s lok runs slow It is also true that for Cyril the lok on Earth runs slow as read by his fellow observers This symmetry does not ontradit the agreement by both sets of observers that Cyril an reah the distant star while still alive 60 The Runner on the Train Paradox A rider on the train observes the runner s wath and the lok mounted on the trak move away from him at the same ative speed and in the same diretion (That these two ative veloities are the same is part of the statement of the problem) So we know from the start that the runner and an earthbound lok are stationary with respet to one S-6