Welcome back to Physics 211

Welcome back to Physics 211 Today’s agenda: review from last lecture: events, distance in spacetime, principle of relativity time dilation, length contraction

Reminder • HW12 due Tue/Wed this week • MPHW 7 due Friday 11 pm

Final • Friday 12 Dec 2:45 pm - 4:45pm here • Comprehensive. (NOT Newtonian gravity). One sheet of formulae allowed • Practice questions: waves and relativity + solns online soon • Review session – Tuesday 9 Dec 10:30-11:30 Stolkin

Review … • (Newtonian) principle of relativity violated by light -- moves with a fixed speed independent of velocity of frame of reference! • If want to keep intact this principle and laws of electromagetism need radical revision of concepts of space and time

Review (2) • Focus attention on events – points (x,t) in spacetime corresponding to physical happenings • Special relativity deals with how events are perceived from different inertial frames • Consider separation between two events. Dx and Dt will not be the same when measured from different frames

Review (3) • But possible to find a distance in spacetime between the 2 events which does take same value in all frames Ds2=c2Dt2-Dx2 • Dscalled spacetime distance • Notice: minus sign crucial – distinguishes time from space

Summary -- Special Relativity • Notions of absolute time and distance discarded for absolute spacetime • Necessary for different (inertial) observers to agree on laws of physics • principle of special relativity

Interpretation of spacetime distance • Imagine a frame of reference that moves with a particle  Dx=0 • SoDs2=c2Dt2-Dx2= c2Dt2 • Thus spacetime distance Ds=cDt – just proportional to time elapsed on clock in such a frame • Sometimes call spacetime distance the proper time

so … • For an observer who moves between 2 points (events) in spacetime spacetime distance is time recorded on her clock (multiplied by c)!

An professor times his journey to work. His wrist watch tells him he has been travelling for 1000 seconds when he arrives. What is the distance in spacetime travelled by the professor ? • 0 • 3x108x1000 m • 1000 secs • depends on speed at which he travels

An professor times his journey to work. His wrist watch tells him he has been travelling for 1000 seconds when he arrives. What is the distance in spacetime travelled by him as measured by a stationary grad student ? • 0 • 3x108x1000 m • 1000 secs • depends on speed at which he travels

An professor times his journey to work. His wrist watch tells him he has been travelling for 1000 seconds when he arrives. Is the time elapsed as measured on a clock at home greater or less than 1000 s ? • same • greater • less • depends on speed at which he travels

Time dilation speed of moving observer ? Q ct cDtO=DsPQ lab frame ? P x x,t measured in lab frame

Time dilation continued Lab frame Ds2=c2DtL2-DxL2 =DtL2(c2-v2) spacetime distance same for lab and moving observer  c2DtO2 =DtL2(c2-v2)  DtO=DtL(1-v2/c2)1/2 • Time experienced by moving observer is less than that recorded in lab

Notice • For v/c=1/100 Dtmov/Dt=0.99995 • For v/c=1/2 Dtmov/Dt=0.87 • For v/c=0.99 Dtmov/Dt=0.14 • v must be less than c! • Time runs infinitely slowly as v approaches c!

Length contraction • Now imagine observer at rest with lab moving at speed –v • Distance gone by lab in time Dt as seen by observeris Dx=vDt • But since Dt is smaller than that recorded in lab so must be Dx! • i.e Dx = Dxstationary (1-v2/c2)1/2 – lengths of moving objects are shorter

Summary • Consider 2 events in space and time • In different inertial frames observers will not agree on either the space separation nor the time interval between those events • But they can all compute the distance in spacetime – the spacetime interval and get same number c2Dt2-Dx2

Comments • Take c to infinity – spacetime distance becomes ordinary time (times c) • all time separations and distance measurements for different frames agree • Newtonian rules once again apply!

Imagine a trip to a nearby star (and back) • Say 4.3 light years away (that means it takes light 4.3 years to reach us • d=4.3x(number secs in year)x3x108 m • d=4.3x31536000x3x108=4x1016 m!! • can travel at 95% speed light • outward trip: 2 events • Leave Earth • Arrive Star

Earth frame • Time between 2 events in Earth frame is just 4.3/0.95=4.53 years • Distance between 2 events in Earth frame is 4.3 light years • Can compute spacetime distance! Ds=(4.53*4.53-4.3*4.3)1/2=1.4 light years

Starship frame • Space separation of events =0 ! • What is time separation ? = proper time =spacetime distance / c =1.4 years! • Thus while 4.53 years elapse for trip on Earth only 1.4 years for astronauts! • time dilation again

Another way of thinking … • From point of view of astronauts the remote star and Earth are moving past them at –vS. • In the time DtS the star arrives at their position. • Thus the distance to the star appears contracted to astronauts – distance to star in starship frame = 4.3x(1.4/4.53)

Return Trip • Suppose ship returns to Earth • Astronauts will have aged 2x1.4=2.8 years • While people on Earth will be 2x4.5=9.0 years older!

Twin Paradox • Paradox: surely relativity implies that I could have considered the astronauts to be at rest and the Earth to have moved • But then I would have predicted that it would be the people on Earth that would have aged only 2.8 yrs !! • Contradiction …

Resolution • Actually, situation not symmetric between astronauts and people on Earth • To come back the starship must have accelerated • Not true for people on Earth. • This period of acceleration spoils symmetry between two and resolves paradox.

Picture 2 paths in spacetime not equivalent – one is bent ! max ageing on straightest path ct starship Earth x

Welcome back to Physics 211