Chapter 20 - Relativity - Davidson Physics

Announcements Review The Twin Paradox Length Contraction Final QuestionsChapter 20RelativityRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsReading AssignmentRead section 20.1 - 20.5Homework Assignment 11Homework for Chapter 20 (due at the beginning of class on Tuesday, December 7)Q8, Q12, Q16, Q18, Q20, Q24, Q26, Q29, E6, E10, E12, SP4Relativity

Announcements Review The Twin Paradox Length Contraction Final Questions“Person of the Century”Einstein’s questionWhat would I see if I rode a beam of light?Special relativityFirst introduced in Einstein’s 1905 paper “On theElectrodynamics of Moving Bodies”Based on two deceivingly simple postulatesPostulate I (principle of relativity)The laws of physics are the same for observers in all inertialreference frames (No frame is preferred)Postulate II (constancy of the speed of light)The speed of light in vacuum has the same value c in alldirections and in all inertial reference framesPhotograph of the German-American physicistAlbert Einstein (1879-1955).Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsMeasuring an eventAn event is something that happens, to which an observer can assign three space coordinates and one timecoordinateExamples of possible events include:an explosionthe collision of two particlesthe passage of a pulse of light through a specified pointA given event may be recorded by any number of observers, each in a different inertial reference frameThe relativity of simultaneityIf two observers are in relative motion, they will not, in general, agree as to whether two events aresimultaneousIf one observer finds them to be simultaneous, the other generally will not, and converselySimultaneity is not an absolute concept but a relative one, depending on the motion of the observerRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsTime dilation∆t =∆t 0√1 − v 2 /c 2To distinguish between these two measurements, we use the following terminology:When two events occur at the same location in an inertial reference frame, the time intervalbetween them, measured in that frame, is called the proper time ∆t 0Measurements of the same time interval from any other inertial reference frame are always greaterIn other words, moving clocks are observed to run more slowly than clocks at rest in the observer’s ownframe of referenceThe increase in the time interval between two events from the proper time interval is called time dilationThe Lorentz factorWe define the Lorentz factor γ > 1 asγ =1√1 − v 2 /c 2The time dilation equation then takes the convenient form ∆t = γ∆t 0Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsObservationTime dilation is significant only if the relative speeds are close to that of light∆t =∆t 0√1 − v 2 /c 2Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsObservationTime dilation is significant only if the relative speeds are close to that of light∆t =∆t 0√1 − v 2 /c 2v = 10 m/s (runner)Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsObservationTime dilation is significant only if the relative speeds are close to that of light∆t =∆t 0√1 − v 2 /c 2v = 10 m/s (runner)v/c = 3.3 × 10 −8v 2 /c 2 = 1.1 × 10 −15Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsObservationTime dilation is significant only if the relative speeds are close to that of light∆t =∆t 0√1 − v 2 /c 2v = 10 m/s (runner)v/c = 3.3 × 10 −8v 2 /c 2 = 1.1 × 10 −15v = 10,000 m/s (space shuttle)Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsObservationTime dilation is significant only if the relative speeds are close to that of light∆t =∆t 0√1 − v 2 /c 2v = 10 m/s (runner)v/c = 3.3 × 10 −8v 2 /c 2 = 1.1 × 10 −15v = 10,000 m/s (space shuttle)v/c = 3.3 × 10 −5v 2 /c 2 = 1.1 × 10 −9Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsObservationTime dilation is significant only if the relative speeds are close to that of light∆t =∆t 0√1 − v 2 /c 2v = 10 m/s (runner)v/c = 3.3 × 10 −8v 2 /c 2 = 1.1 × 10 −15v = 10,000 m/s (space shuttle)v/c = 3.3 × 10 −5v 2 /c 2 = 1.1 × 10 −9v = 100,000,000 m/sRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsObservationTime dilation is significant only if the relative speeds are close to that of light∆t =∆t 0√1 − v 2 /c 2v = 10 m/s (runner)v/c = 3.3 × 10 −8v 2 /c 2 = 1.1 × 10 −15v = 10,000 m/s (space shuttle)v/c = 3.3 × 10 −5v 2 /c 2 = 1.1 × 10 −9v = 100,000,000 m/sv/c = 0.3v 2 /c 2 = 0.1Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsQuestionThe Twin ParadoxAn astronaut is chosen to embark on an epic mission while his identical twin stays at home. The astronaut’sjourney involves him traveling in an experimental fully-automated space ship at an astounding speed of 0.9999c toa distant star and then returning home. After blasting off, the astronaut measures the total time of the round tripto be one year. How much time has elapsed on Earth (and for the astronaut’s identical twin)?Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsQuestionThe Twin ParadoxAn astronaut is chosen to embark on an epic mission while his identical twin stays at home. The astronaut’sjourney involves him traveling in an experimental fully-automated space ship at an astounding speed of 0.9999c toa distant star and then returning home. After blasting off, the astronaut measures the total time of the round tripto be one year. How much time has elapsed on Earth (and for the astronaut’s identical twin)?AnswerAbout 70 yearsRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsA paradox?Clocks on a space ship run slower than clocks on EarthFrom the point of view (frame of reference) of the astronaut in the space ship,Earth is moving away from him, so clocks on Earth should run slower than clocks on the ship!It would seem that by symmetry, both the astronaut and his twin will observe the other as aging slower.Who is right?Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsA paradox?Clocks on a space ship run slower than clocks on EarthFrom the point of view (frame of reference) of the astronaut in the space ship,Earth is moving away from him, so clocks on Earth should run slower than clocks on the ship!It would seem that by symmetry, both the astronaut and his twin will observe the other as aging slower.Who is right?QuestionsAre the two cases truly symmetric?Is there anything that distinguishes the astronaut from the twin?Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsA paradox?Clocks on a space ship run slower than clocks on EarthFrom the point of view (frame of reference) of the astronaut in the space ship,Earth is moving away from him, so clocks on Earth should run slower than clocks on the ship!It would seem that by symmetry, both the astronaut and his twin will observe the other as aging slower.Who is right?QuestionsAre the two cases truly symmetric?Is there anything that distinguishes the astronaut from the twin?AnswerThe two cases are not symmetricThe astronaut is distinguished from the twin because the astronaut must accelerateRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsA paradox?Clocks on a space ship run slower than clocks on EarthFrom the point of view (frame of reference) of the astronaut in the space ship,Earth is moving away from him, so clocks on Earth should run slower than clocks on the ship!It would seem that by symmetry, both the astronaut and his twin will observe the other as aging slower.Who is right?QuestionsAre the two cases truly symmetric?Is there anything that distinguishes the astronaut from the twin?AnswerThe two cases are not symmetricThe astronaut is distinguished from the twin because the astronaut must accelerateSpecial relativity does not claim that all observers are equivalent (only that observers at rest ininertial reference frames are equivalent)Because the astronaut must accelerate (speed up, slow down, turn around, etc.), he is not alwaysat rest in an inertial frame of referenceRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsA paradox?Clocks on a space ship run slower than clocks on EarthFrom the point of view (frame of reference) of the astronaut in the space ship,Earth is moving away from him, so clocks on Earth should run slower than clocks on the ship!It would seem that by symmetry, both the astronaut and his twin will observe the other as aging slower.Who is right?QuestionsAre the two cases truly symmetric?Is there anything that distinguishes the astronaut from the twin?AnswerThe two cases are not symmetricThe astronaut is distinguished from the twin because the astronaut must accelerateSpecial relativity does not claim that all observers are equivalent (only that observers at rest ininertial reference frames are equivalent)Because the astronaut must accelerate (speed up, slow down, turn around, etc.), he is not alwaysat rest in an inertial frame of referenceTherefore, there is no paradoxRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsYet another gedanken experimentYou and a friend both decide to measure the length of an objectThe object is at rest in your frame of reference; your friend is seated on a train moving with a constantspeed vRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsYet another gedanken experimentYou and a friend both decide to measure the length of an objectThe object is at rest in your frame of reference; your friend is seated on a train moving with a constantspeed vWhat happens?In your frame of reference:The object is not movingYou measure the length of the object to be L 0Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsYet another gedanken experimentYou and a friend both decide to measure the length of an objectThe object is at rest in your frame of reference; your friend is seated on a train moving with a constantspeed vWhat happens?In your frame of reference:The object is not movingYou measure the length of the object to be L 0 = v∆t, where v is the speed of your friendThis time interval ∆t is not a proper time interval because the two events that define it (yourfriend passes the back of the object and your friend passes the front of the object) occur at twodifferent places so you must use two synchronized clocks to measure the time interval ∆tRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsYet another gedanken experimentYou and a friend both decide to measure the length of an objectThe object is at rest in your frame of reference; your friend is seated on a train moving with a constantspeed vWhat happens?In your frame of reference:The object is not movingYou measure the length of the object to be L 0 = v∆t, where v is the speed of your friendThis time interval ∆t is not a proper time interval because the two events that define it (yourfriend passes the back of the object and your friend passes the front of the object) occur at twodifferent places so you must use two synchronized clocks to measure the time interval ∆tIn your friends’s inertial frame of reference:The object is moving with a speed vTo your friend, the two events that you measured occur at the same place so the time interval ∆t 0she measures is a proper timeTo her, the length L of the object is L = v∆t 0Since you and your friend are moving relative to one another, the time that she measures isdifferent than the time that you measure (time dilation)Length contractionL = v∆t 0 = v∆tγ = L √0γ = L 0 1 − v 2 /c 2Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsLength contractionL = L √0γ = L 0 1 − v 2 /c 2Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsLength contractionL = L √0γ = L 0 1 − v 2 /c 2What are these things?L 0 (the proper length): the length interval measured in the frame of reference in which the object is at restL: the length measured in the frame of reference in which the object is movingRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsLength contractionL = L √0γ = L 0 1 − v 2 /c 2What are these things?L 0 (the proper length): the length interval measured in the frame of reference in which the object is at restL: the length measured in the frame of reference in which the object is movingObservationThe proper length L 0 is always greater than the contracted length LThe length of an object is largest for the observer at rest with respect to the object and smaller forobservers in motion with respect to itRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsLength contractionL = L √0γ = L 0 1 − v 2 /c 2What are these things?L 0 (the proper length): the length interval measured in the frame of reference in which the object is at restL: the length measured in the frame of reference in which the object is movingObservationThe proper length L 0 is always greater than the contracted length LThe length of an object is largest for the observer at rest with respect to the object and smaller forobservers in motion with respect to itWho is right?Both are!Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsMuons and time dilationMuons are created in the Earth’s atmosphere at an altitude of 10-15 kilometers and approach the Earthwith a speed near that of light (about v = 0.998c).Muons are unstable, and quickly decay into other particles (average lifetime of a muon at rest is about∆t 0 = 2.20 × 10 −6 seconds)In this time, a muon would travel a distance d = v∆t 0 = 659 metersSince this distance is small we would expect very few of them to reach the Earth’s surface (but anappreciable number of muons do reach the surface!)We haven’t taken into account time dilationMuons decay by their own internal clockTherefore, in a time interval ∆t 0 = 2.20 × 10 −6 seconds, an Earth clock reads a time interval∆t =t√1 − v 2 /c 2 = 2.20 × 10−6 s√1 − (0.998) 2 = 3.48 × 10 −5 sTherefore, on average, the muons travel a distance(d = v∆t = (0.998c) 3.48 × 10 −5 )s = 1.04 × 10 4 m = 10.4 kmRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsMuons: continuedHowever, a hypothetical observer on the muon would not invoke time dilation, because in their frame ofreference they are not moving!Relativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsMuons: continuedHowever, a hypothetical observer on the muon would not invoke time dilation, because in their frame ofreference they are not moving!For the imaginary observer on the muon, the time it takes before the muon decays is still∆t 0 = 2.20 × 10 −6 secondsHowever, the observed distance (length) that the muon needs to travel to reach the Earth’s surface isshorter (length contraction)√L = L 0√1 − v 2 /c 2 = (10 km) 1 − (0.998) 2 = 0.633 km = 633 metersTo travel that distance would take a timet = L v =633 m0.998 ( 3 × 10 8 m/s ) = 2.11 × 10−6 sThis time is roughly the same as the muon’s lifetime, so both observers agree that many muons will reachthe Earth’s surfaceRelativity

Announcements Review The Twin Paradox Length Contraction Final QuestionsReading AssignmentRead section 20.1 - 20.5Homework Assignment 11Homework for Chapter 20 (due at the beginning of class on Tuesday, December 7)Q8, Q12, Q16, Q18, Q20, Q24, Q26, Q29, E6, E10, E12, SP4Relativity