If the speed of light were smaller than it is, would relativistic phenomena be more or less conspicuous than they are now?

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1 Physis 07 Problem. If the speed of light were smaller than it is, would relatiisti phenomena be more or less onspiuous than they are now? All of the phenomena of speial relatiity depend upon the fator G(): G The square box at the end of this equation indiates it is just text; i.e., not "lie." If were smaller, then G() would differ from unity at muh lower eloities thus making relatiisti effets more onspiuous than they are now. et's plot G() as a funtion of / to see the effet. This olumn is for =3x0 8. Set the range of : Plot: 0, G This olumn is for =3x0 7. Set the range of : Plot: GG G 0.5 GG Notie how G() falls rather slowly Notie the rapid dropoff in GG() When G() is "near", relatiisti effets are not obious. When G() deiates signifiantly from, relatiisti effets are more obious. Physis 07 Problem. It is possible for the eletron beam in a teleision piture tube to moe aross the sreen at a speed faster than the speed of light. Why does this not ontradit speial relatiity? The image of a moing objet reated by the suessie impats of the eletrons on the sreen is not an objet. The information arried by the eletrons moes from the eletron gun to the sreen at a speed below that of light. Adjaent dots on the sreen may light up as if the dot had moed from one position to the next at a speed greater than the speed of light, but in fat, no dot moed. Eah lit dot is produed by a different eletron.

2 Physis 07 Problem.3 Is it a good idea for an athlete trying to set a world reord 00-m dash tie to hae his time taken by an obserer on a moing spaeraft? If the obserer in the spaeraft times the run by wathing a lok on earth, nothing is gained beause the lok and athlete are in the same referene frame (the athlete's speed is so small ompared to that we an ignore his motion relatie to the lok). If the obserer in the spaeraft times the run by wathing a lok in the spaeraft, it appears as if the lok on the earth ran slow, so that, in fat, MORE time elapsed during the run. The spaeraft obserer would atually measure a longer time. This answer ignores the ompliation of length ontration, whih we will get into later in this hapter. Attempting to aount for length ontration would introdue the problem of simultaneity (see the pole-barn paradox in the lass notes). The time measured by the spaeraft obserer would not be useable by an earth obserer. Physis 07 Problem.4 An obserer on a spaeraft moing at 0.7 relatie to the earth finds that a ar takes 40 minutes to make a trip. How long does the trip take to the drier of the ar? Solution: we apply the time dilation equation. An obserer on the spaeraft measures a dilated time t = 40 min (why is it the dilated time?). We need to alulate the proper time, as measured by the drier of the ar. t t0 Assign alues and sole: t 0 t 0.7 t 40 t 0 t t 0 = Something to think about: what time would an obserer who remained stationary on earth measure?

3 Physis 07 Problem.5 Two obseres, A on earth and B in a spaeraft whose speed is x0 8 m/s, both set their wathes to the same time when the ship is abreast of the earth. (a) How muh time must elapse by A's rekoning before the wathes differ by s? (b) To A, B's wath seems to run slow. To bb, does A's wath seem to run fast, slow, or keep the same time as his own wath? Solution (a): by A's rekoning, B's wath runs slow. Suppose A has two idential wathes. Wath is used as the timer and wath is used to proide a time interal. If both wathes are in A's frame of referene, wath will measure the proper time t 0 for the time interal. If A gies wath to B, and B moes relatie to A, A an use wath to measure the time t it takes for wath to tik off the time interal. Sine the moing wath tiks more slowly (aording to A, who is doing the measuring), A's wath must reord a longer time for wath to indiate the same time interal that it indiated when it was in A's referene frame.thus, when the wathes differ by s, A's wath, whih is being used to measure the time t, has tiked more seond. Thus, t=t 0 +; t is always greater than t 0. The two equations we need to sole are the time dilation equation and the relationship t=t 0 +. t t 0 and t t 0 + Sole the aboe two equations for t (et Γ be the square root): t t Γ Result: t t Γ t Γ t Γ t Assign alues to parameters: Or I ould say = and =/3. Calulated t: Γ t Γ t = 3.97 seonds Solution (b): Aording to B, A is in motion relatie to B. Moing wathes (relatie to the obserer) always run slow. Therefore, B laims A's wath runs slow. 3

4 Calulate t: Physis 07 As in problem.5, t Assign alues to parameters: t Γ Γ Γ t = seonds Problem.6 An airplane is flying at 300 m/s (67 mi/h). How muh time must elapse before a lok in the airplane and one on the ground differ by s? Solution: this is just problem.5 with an airplane replaing a roket ship. Just plug in the new numbers. You an use =.998x0 8 if you want and get a slightly different answer. t Physis 07 How fast must a spaeraft trael relatie to the earth for eah day on the spaeraft to orrespond to d on the earth? Solution: an obserer on the spaeraft measures a proper time t 0 = d, and a dilated time t= d for the same eent as it takes plae on the "moing" earth. We sole the time dilation equation for. t 0 Define the alues and plug them in: Problem.7 t 0 t t 0 t t 0 t = t 0 t Or =0.866, beause I let =. 4

5 Physis 07 Problem.8 The Apollo spaeraft that landed on the moon in 969 traeled there at a speed relatie to the earth of.08x0 4 m/s. To an obserer on the earth, how muh longer than his own day was a day on the spaeraft? This is another time dilation problem. We are gien a relatie eloity. We are gien a time interal t 0 of one day on the spaeraft moing relatie to an obserer. We want to find the dilated time t measured by the obserer. et's define our ariables first, so I don't hae to make the time dilation equation into text and then later on re-enter it as an equation t 0 The time here is in days, so my answer t will be in days. We don't need to worry about units here beause eloities hae same units. t 0 t This is the equation for t. Below I will type "t=" to see the answer. t = There are not enough digits to show any effet. Below I will type "t=" again and then type "f" to allow me to show more digits. t = Clik on the number and selet Math, Numerial Format to see that I piked a preision of to display t. The problem asks how muh longer t is than t, so I'd better alulate t-t 0. t t 0 = days or ( t t 0 ) = seonds. Physis 07 Problem.9 A ertain partile has a lifetime of x0-7 s when measured at rest. How far does it go before deaying if its speed is 0.99 when it is reated? Solution: our first job is to figure out that the problem is really asking us to alulate how far a human obserer would obsere this partile to trael. The obserer sees the partile moing at a speed of 0.99, and sees the partiles "lok" dilated aording to equation.3, where t 0 is the time the obserer sees the partile in motion. We need to alulate d=t, where =0.99 and t is gien by eq..3. I'm going to inlude units in this solution. You ould append the file "units.md" to for use with this problem. Instead, I will do the units here for you to see. Define units: m kg M s T Define parameters: m t s 0.99 s Pertinent equations: t 0 t d t d = 0.5 m 5

6 Physis 07 Problem.7 An astronaut whose height on the earth is exatly 6 ft is lying parallel to the axis of a spaeraft moing at 0.9 relatie to the earth. What is his height as measured by an obserer in the same spaeraft? By an obserer on earth? An obserer in the same spaeraft, at rest relatie to the astronaut, measures the proper length, 6 ft, of the astronaut. If the astronaut were lying perpendiular to the eloity etor of the spaeraft, the obserer on earth would also measure his proper length. But with the astronaut parallel to the diretion of relatie motion, the obserer on earth measures a ontrated length. Define parameters, then do alulation =.65 feet Physis 07 Problem.8 An astronaut is standing in a spaeraft parallel to its diretion of motion. An obserer on the earth finds that the spaeraft speed is 0.6 and the astronaut is.3 m tall. What is the astronaut's height as measured in the spaeraft? Solution: this is just problem.7 but soling for 0 instead of. Define parameters: =.65 meters Physis 07 Problem.9 How muh time does a meter stik moing at 0. relatie to an obserer take to pass the obserer? The meter stik is parallel to its diretion of motion. The relatie speed is : m/s 0. 0 We need to sole for the ontrated length, and then determine how long it takes to trael this length at a speed of 0.*. 0 = The equation d=t still works in relatiity, so we an sole it for t: t t = seonds 6

7 Physis 07 Problem.0 A meter stik moing with respet to an obserer appears only 500 mm long to her. What is its relatie speed? How long does it take to pass her? The meter stik is parallel to its diretion of motion m/s First we need to sole 0 for relatie speed. 0 = m/s 0 0 Time to pass obserer: (I all it t 0 beause the eent is timed in the obserer's referene frame. t 0 t 0 = s Physis 07 Problem. A spaeraft antenna is at an angle of 0 degrees relatie to the axis of the spaeraft. If the spaeraft moes away from the earth at a speed of 0.7, what is the angle as seen from the earth? This problem is not assigned or "testable" and the solution is inluded only for "interest." This problem is triky beause when the spaeraft moes, the projetion of the antenna along the spaeraft ontrats, whih means that the apparent length of the antenna also ontrats. You hae to be areful to take both ontrations into aount. It also helps to hae a drawing. et 0 be the length of the antenna when the spaeraft is at rest, and x 0 be the projetion of the antenna parallel to the diretion of spaeraft trael when the spaeraft is at rest. Then when the spaeraft is at rest, the angle of the antenna is found from x 0 0 os θ 0 When the spaeraft is in motion, both 0 and x 0 appear ontrated to an obserer on earth, and the angle an be found from x os( θ) exept that you an't apply the length ontration formula to 0 in order to alulate, beause the antenna has a omponent of length perpendiular to the diretion of motion whih has not been ontrated. et's let y be the projetion of the antenna length perpendiular to the motion. When the spaeraft is in motion, the antenna has a length, a projetion x= os(θ) along the diretion of motion, and a projetion y= sin(θ) perpendiular to the diretion of motion. The angle θ is gien by tan( θ) y x sin θ os θ That last equation looks "irular" (we already know tan=sin/os) exept that we an use our original angle to alulate the numerator, and we an apply the length ontration equation to the denominator, beause it represents the omponent along the diretion of relatie motion. 7

8 tan( θ) 0 sin θ 0 tan θ ( 0 os( θ 0 )) 0 sin( 0) ( 0 os( 0) ) tan( θ) tan( 0) θ atan tan( 0) To sole, plug in alues, and remember to onert angles to radians when you are using Mathad. 0.7 tan θ 0 θ atan θ = 0.4 radians θ θ 360 θ = degrees π θ 0 0 π 360 Physis 07 Problem.7 Dynamite liberates about 5.4x0 6 J/kg when it explodes. What fration of its total energy ontent is this? The total energy ontent of this kg of dynamite is m. The fration is simply f We need to use (SI units). Here's how the units work out: if you hoose the mass to be kg, the units on top are joules, and the m=kg times also gies joules in the denominator, so the result is a pure number, or fration. Then f f = 6 0 Physis 07 Problem.8 A ertain quantity of ie at 0 degrees C melts into water at 0 degrees C and in so doing gains kg of mass. What was its initial mass? It takes 80 alories to melt a gram of ie, and a alorie is equialent to 4.9 joules, so in SI units, it takes =335. joules to melt a gram of ie, or 3.35x0 5 joules to melt a kilogram of ie. et M be the initial mass of the ie, and let be the latent heat of fusion, as aboe. The energy added to melt the ie is M. The mass equialent of this energy is gien in the statement of the problem as m= kg. 8

9 M M( m) M = kg Using the density of ie, you an alulate that this mass would require a blok of ie 6.5 km long, 6.5 km wide, and km high. (Or about 0 miles by 0 miles by 0.6 miles high, if I remember my onersion fators orretly). Yes, ie really does gain mass when it melts. Or rather, the energy that "goes into" the ie is manifested as mass. Similarly, water would "lose mass" when it freezes. No, that's not the main reason why ie floats on water. Physis 07 Problem.9 At what speed does the kineti energy of a partile equal its rest energy? rest_energy KE γ When KE=rest_energy, γ γ γ m m I anelled out the 's. m m Canel the m's and re-arrange: diide by and square both sides Physis 07 Problem.30 How many joules of energy per kilogram of rest mass are needed to bring a spaeraft from rest to a speed of 0.9? Again, we use γ + K At a speed of 0.9, Soling for K gies m m where /= m 0.9 K K = joules per kilogram. 9

10 Physis 07 Problem.3 An eletron has a kineti energy of 0. MeV. Find its speed aording to lassial and relatiisti mehanis. et's begin by onerting MeV to mks units. ev This onerts ev to Joules MeV ev 0 6 This onerts ev to MeV Classial alulation. We use our familiar equations of lassial mehanis. K l m l m kg K l 0. MeV l K l l = m/s m Relatiisti alulation. K rel 0. MeV If you start with γm =m +K and sole for /, you get an "unoffiial" but extremely useful equation that is sure easier to use than soling "by hand" eery time. Make sure this equation is on your 3x5 ard. rel rel = m/s K rel + et's put the two answers side-by-side for omparison: l = rel = Classially-alulated speeds are always too large, although the error is not signifiant for low speeds and low energies. Physis 07 Problem.33 A partile has a kineti energy 0 times its rest energy. Find the speed of the partile in terms of. Be lazy and use our "unoffiial" equation (it's unoffiial beause it is deried, not fundamental). ( K, m) + K , m = Expressed "in terms of :" 0, m = or =

11 Physis 07 Problem.34 The speed of a proton is inreased from 0. to 0.4. (a) By what fator does its kineti energy inrease? (b) The speed of the proton is again doubled, this time to 0.8. By what fator does its kineti energy inrease now? Here's another handy "unoffiial" ariant of our KE equation: K( m, ) K m, We ould do this algebraially, without using the proton mass, but here it is for those who like numbers: m proton (, ) K m proton, 0.4 K m proton 0. = 4.47 (, ) K m proton, 0.8 K m proton 0.4 = 7.39 Doubling the speed does not inrease KE by the lassially-expeted fator of 4, and the disrepany is larger as the speeds get greater. Mathad made this solution ery easy. If you're doing this "by hand," it will take seeral lines of algebra and omputations. Physis 07 How muh work (in MeV) must be done to inrease the speed of an eletron from.x0 8 m/s to.4x0 8 m/s? Solution: a body in motion has a total energy The work done on the eletron is just the differene in the energies E at the two different eloities: To get the work in units of MeV, I ould alulate the numbers out and onert joules to ev to MeV. Or I ould be leer and use the eletron mass in energy units. W m e E W E E γ ( ) Problem.35 E Remember E f -E i =[W other ] i-->f from Phys. 3? W 0.94 MeV Setting MeV= aboe let me stik in my expression for me so that it looked like a unit. (. 0 8 ) MeV m e 0.5 MeV

12 Physis 07 Problem.4 In its own frame of referene, a proton takes 5 min to ross the Milky Way galaxy, whih is about 0 5 light-years aross. (a) What is the approximate energy of the proton in ev? (b) About how long would the proton take to ross the galaxy as measured by an obserer in the galaxy's frame of referene? Part (a). The only way for the proton to "think" it rosses the galaxy in 5 minutes is for the proton to see the galaxy's length ontrated: 0 This equation lets us sole for, and we ould plug into E=γm. Or we ould be leer and note that 0 γ E m γ so that E 0 E m,, 0 0 =.58 E , 5 60, E=.58 joules To get the energy in ev, use the onersion fator ev=.6x0-9 joules: You ould be still more leer and express the proton mass in energy units (938 MeV/ ) and sae multiplying by = Part (b). The proton's energy is about 0 9 ev, or 0 3 MeV. That's about 0 0 times its rest energy of roughly 000 MeV. This is one fast proton. We might as well use proton =. I'm guessing our answer might be off in the 0th deimal plae or so. Double-hek by alulating the proton's speed, if you don't beliee me. If the proton is moing with the speed of light, and the galaxy is 0 5 light years aross, the obserer will say it takes 0 5 years (not 5 minutes) for the proton to ross the galaxy. The obserer say "proton, your time was way too short." The proton will say "obserer, your distane aross the galaxy was way too long. Physis 07 What is the energy of a photon whose momentum is the same as that of a proton whose kineti energy is 0 MeV? E p + Problem.4 E photon p E proton p m proton + p E proton E photon E proton m proton m proton both proton and photon hae same momentum p this is p for photon, whih is equal to p for photon The E in here is total energy; you are gien proton KE.

13 K proton E proton m proton E proton K proton + m proton E photon K proton + m proton m proton here are the speifi alues for this problem: K proton E photon K proton + m proton E photon 0 MeV 37 MeV m proton 938 MeV m proton Find the momentum (in MeV/) of an eletron whose speed is 0.6. p γ Physis 07 Problem.43 If I use mass in "energy units" of MeV/ and in terms of, the answer will ome out in units of MeV/, beause γ is unitless. Or you an work the problem in SI units and onert to "energy units." m eletron 0.5 MeV p eletron p eletron = eletron eletron 0.6 m eletron eletron I hae defined aboe numerially as 3x0 8. When I ombine symboli and numerial work, like I do here, Mathad atually diides by a to alulate the mass of the eletron. When I multiply by =0.6, Mathad "puts bak" one of those "two" 's I diided out. To get the answer in "energy units" I hae to put the other fator of bak in "by hand." You won't enounter this onfusion on an exam. This answer has the fator of / numerially embedded in the result. To see the answer expressed in units of MeV/, I hae to multiply through by. If you do this using pen and paper (see the student solutions manual, aailable on resere at the library), you will get the textbook answer "automatially." p eletron MeV See leture notes let03.ppt for better "penil-and-paper" ersion. 3

14 Physis 07 Problem.44 Find the total energy and kineti energy (in GeV) and the momentum (in GeV/) of a proton whose speed is 0.9. The mass of the proton is GeV/. et's do the momentum first, beause it is "like" the preious alulation. GeV This definition just lets me inlude "GeV" in a problem as if I were writing the units by hand. m proton GeV proton 0.9 p proton proton m proton proton p proton.937 GeV E proton p proton m proton + E proton p proton m proton + E proton =.5 GeV Physis 07 Problem.45 Find the momentum of an eletron whose kineti energy equals its rest energy of 5 kev. First I'll do the algebra. I ould hae also used E=γm, but I already had the Eproton equation aailable for ut and paste from problem 4. K proton E proton m proton K proton.4 GeV E p + p E K E ( m ) E K + p K + p ( m ) K + Now I'll plug in the numbers. kev m eletron 5 kev K eletron 5 kev K eletron m eletron + p eletron p eletron 885 kev m eletron 4

15 Physis 07 Problem.47 Find the speed and momentum (in GeV/) of a proton whose total energy is 3.5 GeV. First think: this proton's total energy is about 4 times its rest energy of GeV, so the speed had better be lose to. et's do the algebra first, working aross the line and then down to sae paper. E γ E γ E 0 γ E E 0 E E 0 E 0 E E 0 E E 0 E E 0 E E GeV E 3.5 GeV To find the momentum, we an use this equation from the preious problem (no sense re-deriing here): p E p E E 0 p E E 0 p E E 0 GeV ( 3.5 GeV) ( GeV) p p p 3.37 GeV Physis 07 Problem.48 Find the total energy of a neutron (m=0.940 GeV/ ) whose momentum is. GeV/. E p + E. GeV GeV E (. Ge) + ( GeV) E (.) + ( 0.940) GeV E (.) + ( 0.940) GeV E.5 GeV 5

16 Physis 07 Problem.49 A partile has a kineti energy of 6 MeV and a momentum of 335 MeV/. Find its mass (in MeV/ ) and speed (as a fration of ). You hae two knowns and two unknowns. None of our equations ontain one of the unknowns expressed only in terms of the knowns. ooks like we are going to hae to sole a system of equations. Here are two equations that look like andidates to me: E K + E p + Put E from the first equation into the seond equation. p K + + Makes me a bit nerous; any time you square an expression, you run the risk of introduing an extraneous root. et's see where this takes us anyway. K + K + p + K + K p K p K m p K K m 335 MeV 6 MeV 6 MeV m ( ) MeV 6 MeV m ( ) MeV 6 m 874 MeV Now that I hae the mass (a bit less than that of a proton or neutron) I an use our handy unoffiial equation introdued in problem 3. There are other ways to get --see the student solution manual. + K Those of you who obserant will notie I quit making my equations "lie." This eliminates the glith mentioned in problem 43. I'm just using Mathad as a symboli text proessor now. + 6 MeV 874 MeV See the student solution manual for a really handy algebra trik that eliminates the potential extraneous root problem. 6

17 Part (a): Physis 07 Problem.50 (a) Find the mass (in GeV/ ) of a partile whose total energy is 4 GeV and whose momentum is.45 GeV/. (b) Find the total energy of this partile in a referene frame in whih its momentum is GeV/. E p + E p m E 4 p m E 4 p m 4 GeV 4.45 GeV m 4 GeV 4.45 GeV 4 m 4.45 GeV m 3.73 GeV Part (b): mass is relatiistially inariant, so we an use the mass from part (a) along with the new momentum. E p + E p + E GeV GeV E GeV GeV GeV E E 4.3 GeV 7