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Synchronizing a SymphonyDoes it make a difference?Say, oboe at the center of the orchestra keeps thetempo.How long does it take to reach the players at the edge ofthe pit which is 10 meters away?10 m 300 m/s 0.033 s or 33 msPlayers at the edgeare 33 ms late.Since the human ear is capable of discerning thedifference in sounds that are 0.66 ms apart, this is aproblem. The whole orchestra will be out of sync.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Synchronizing a SymphonyWhy does the conductor fix this problem?The conductor uses light, not sound, to synchronize8the orchestra. Light travels at 3 x 10 m/s whichmeans that it takes only 0.000067 ms for the playersat the back of the pit to see the hand gesture.Hence, they are only 0.000067 ms out of sync withthe players from the front of the orchestra.The human ear is only able to discern a difference of0.66 ms in sounds. So, the orchestra sounds in sync.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Thunder and LighteningThunder and lightening during in a storm are generatedat the same time. Assume that the distance between youand the source is d. Calculate the difference in time youobserve between the lightening and thunder.dtl cdandts vs1ddd d( )ts tl vscvsc11 )ts tl d(300 300, 000, 000d[s]ts tl 300Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Next Time:Michelson - Morley: The downfall of the ether!Simultaneity: Will two observers always agreethat two events are simultaneous?Physics 3305 - Modern PhysicsProfessor Jodi Cooley

The End(for today)Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Welcome backto PHY 3305Today’s Lecture:General RelativityAlbert Einstein1879-1955Physics 3305 - Modern PhysicsProfessor Jodi Cooley

AnNouncements-There will be no lecture video (and hence no quiz) on Tuesday,September 13th. We will be doing an in class activity that willcount towards your homework grade. We will meet in FOSC 026that day.-No reading assignment for Tuesday, September 13th.-Homework assignment 3 is due Tuesday, September 13th at thebeginning of class.-Regrades for assignment 2 are due Tuesday, September 13th atthe beginning of class.-Dr. Cooley will be out of town September 8-12th. Mr. Thomas willcover office hours that day.-Extra Credit Opportunity 1 will be due Thursday, September 29th.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Review Question 1An electron moves through the lab at 99% the speed oflight. The lab reference frame is S and the electron’sreference frame is S’. In which reference frame is theelectron’s rest mass larger?A) In frame S, the lab frame.B) In frame S’, the electron’s frame.C) It is the same in both frames.rest mass stationary energy2m E/cPhysics 3305 - Modern PhysicsProfessor Jodi Cooley

Review Question 2Are the (a) kinetic energy and (b) the total energy ofa 1 GeV electron more than, less than or equal tothose of a 1 GeV proton?The total energy of a 1 GeV electron equals thetotal energy of a 1 GeV proton. However, the restenergy of an electron is less than the rest energyof a proton. Thus, the kinetic energy of a 1 GeVelectron is greater than that of the kinetic energyof a 1 GeV proton.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Lecture Video:the Principle of EquivalenceEinstein postulated:A homogenous gravitational field is completelyequivalent to a uniformly accelerated referenceframe.-orThe form of each physical law is the same in alllocally inertial frames.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Which of the following people will observe the sameexperimental results as someone who is standing on a solidlab floor on the moon?a) An observer in a spaceship moving at constantvelocityb) An observer in an elevator on Earth movingupwards at constant velocityc) An observer in an elevator on Earth with a brokencable (i.e. in free-fall)d) An observer in an elevator on Earth movingdownwards at constant velocitye) An observer in a spaceship moving with constantaccelerationPhysics 3305 - Modern PhysicsProfessor Jodi Cooley

Deflection of Light in a Gravitational FieldObserving a beam of light in an accelerated referenceframe.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Deflection of Light in aGravitational Field-In an accelerating frame a light beams appearsto be “falling down” in a parabolic arc-According the the principle of equivalence:there is no physical difference between anaccelerating frame and a gravitational field.-Therefore a light beam will “fall down” in agravitational fieldPhysics 3305 - Modern PhysicsProfessor Jodi Cooley

Video Lecture:Gravitational LensingNote: This picture is anextreme exaggeration of theeffect known as “GravitationalLensing”If you envision space-time as afabric -- photons can not leavethe fabric.Demonstration!Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Some of the most convincing evidence for dark matter comes from a phenomenon known as gravitationallensing. This was first predicted by Einstein in his theory of relativity. The theory predicts that large masses inouter space, such as clusters of galaxies, bend light that travels near them. So, as the light from a distant starpasses by a large mass, its path is distorted by gravity. Gravitational lensing was first observed experimentallyin 1919 when physicist Arthur Eddington observed light from a distant star being bent by the Sun.Some of the best evidence that we have for the existenceof dark matter comes from gravitational lensing.30Where starappears to be1. Given that the mass of the Sun is 1.99 x 10 kg and8its radius is 6.96 x 10 m, calculate the angle ofdeflection for light from a distant star that passes veryclose to the Sun’s surface.4GM2. A ray of light thatpasseswithin a distance of 16 million light years from the centre2of a cluster ofgalaxies is bent by an angledcof 2.0 x 10 radians. Use-5θgravitational lensing to calculate the mass of thecluster.dθMActualposition of star3. Order the following three scenarios according to theangle of deviation (from highest to lowest) for lightthat just passes by the edges of the clusters.Calculate the deflection of light from a distant thatapproaches very close to the surface of the sun. The4GMθ radiusdc of the sun is 6.96 x 10 m.The angle θ (measured in radians) by which light from adistant star or galaxy is bent by a mass M is given by thefollowing formula2-1122the closest thewhere G 6.67 x 10 Nm /kg , d is112 light2comes to the centre of the object, and c is the speed oflight.814a) A cluster of galaxies with a mass of 10 times the7mass of the Sun and a radius of 10 light years.14b) A cluster of galaxies with a mass of 5 x 10 times6the mass of the Sun and a radius of 3 x 10 lightyears.8cluster of galaxies with a mass of 2 x 10146 timesc) Athe mass of the Sun and a radius of 4 x 10 lightyears.4(6.67 10N · m /kg )(1.99 1030 kg)6 8.48 10rad82(6.96 10 m)(3 10 m)Convert to degrees:Physics 3305 - Modern Physics 4.86 10 2.92 104degrees2arcminProfessor Jodi Cooley

LAST TIME:Gravitational Redshift tlowerPhysics 3305 - Modern PhysicsgH thigher (1 2 )cProfessor Jodi Cooley

Relativity and GPS tlowergH thigher (1 2 )cGeneral relativity tells us that a clock runs slower when itis deeper in a gravitational field (closer Earth). So,clocks on a satellite run faster than clocks at groundlevel. In Earth’s non-uniform field .GME1 GME )] tE tsat [1 2 (crErsatApply the Physicsr distance from center of EarthA GPS satellite orbits at an altitude of 2.0 x 107 m and aspeed of 3.0 x 103 m/s. The radius of Earth is 6.4 x 106 m.a) By what fraction must the time be adjusted to accountfor both regular speed dependent and gravitational timedilation?Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Speed dependent time dilationis given by t γ t0 tE !Use Binomial Expansion:1 GME1 GME )] tE tsat [1 2 (crErsat tsat tE !v21 c2 tsat 10 12 (1 1.7 10) tsat32(3.9 10 m/s)(3 108 m/s)2 tE (1 8.5 10 11) tsatGravitational time dilation of the satellite is given by(6.67 1 tE tsat [1 (82(3 10 m/s)14 m310 s2 )5.98 6.4 106 m tE (1 5.26 102410 kg 10 (6.67 14 m310 s2 )5.98 2.64 107 m1024 kg)]) tsatThe gravitational effect makes the satellite’s time larger. Its clock runs faster.The speed-dependent effect makes the satellite’s time smaller. Its clock run slower.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

The fractional change is then: tE tsat 1110 (8.5 10 ) (5.26 10 ) tsat tE tsat 10 4.4 10 tsatIf this effect was not accounted for, how soon would thetime be in error be 10 ns?10 10 9 s 4.4 10tsattsat 23 sPhysics 3305 - Modern Physics10Clocks aboard GPS satellites accountfor the gravitational effect.Velocities and elevations of thereceiver units are calculated in thereceiving units.Professor Jodi Cooley

Einstein’s Postulates:1. The laws of physics areinvariant to observers indifferent inertial referenceframes.2.The speed of light is the sameto all observers, regardless oftheir motion relative to thelight source.As a consequence of Einstein’s Postulates,no information travels faster than the speed of light.This claim is supported by the Lorentz Equations!Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Events simultaneous in one reference frame occur indifferent orders in another reference frame.Is it possible that the effect can precede the cause?NO!The Lorentz transformations forbid it! Events canonly be reversed if they are not causally related.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Event A occurs at space time coordinates (300 m, 2µs).A) Event B occurs at spacetime coordinates (1200 m,6 µs). Could A possible be the cause of B?Yes1200v 6 10 6Physics 3305 - Modern Physics300 m2 106s 2.3 106 m/sProfessor Jodi Cooley

Intuitive example, think of how a shadow moves.Project a shadow of your finger on a wall & wiggle it.The shadow travels faster than your finger. If youwiggle your finger at c, the shadow’s speed will appeargreater than c.Is this a violation of cause and effect?Physics 3305 - Modern PhysicsProfessor Jodi Cooley

ShadowsShadows on a wall.The misconception is that the edge of a shadow"moves" along a wall, when in actuality the change in ashadow's motion is part of a new projection, which willpropagate at the speed of light from the object ofinterference.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Television In the‘olden’ daysIn the ‘olden’ days televisions contained a picture tube, abeam of electrons sent from the back to the front (screen)by an electron gun. When the electrons strike the screen, itcauses a phosphor to glow briefly. To produce an imageacross the entire screen, the glowing beam is electricallydeflected up and down and right and left.The beam may sweep from left to right at a speed greaterthan c. Why is this not a violation of the claim that noinformation may travel faster than the speed of light?Physics 3305 - Modern PhysicsProfessor Jodi Cooley

LightSpotsThe light spots are caused by electrons hitting thephosphors, not the electrons themselves. No electrontravels from one side of the screen to the other.Nothing that can have information about the left-handside of the screen moves to the right-hand side of thescreen. The events (electrons hitting phosphors) are allplanned ahead of time and have no effect upon onanother. Thus, no information travels faster than c.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

Review ProblemA plank, fixed to a sled at rest in frame S, is of length L0and makes an angle Θ0 with the x-axis. Later, the sledzooms through frame S at a constant speed v parallel tothe x-axis. Derive an expression for the length of theplank and the angle it makes with the x-axis according toan observer who remains at rest in frame S. Simplify sothat your final expression for L is in terms of cosΘ0 andyour final expression for Θ is in terms of tanΘ0.Physics 3305 - Modern PhysicsProfessor Jodi Cooley

LxLo cosx θ o L1o cos vθ2 o c 21 v c1 v 2 1 c2 v cy γ tan θo. As the speed increases, the angle θ seen by an observer in S increases. This makes sense in view oftan θo. As the speed increases, the angle θ seen by an observer in S increases. This makes sense in view ofthe fact that the faster the relative motion, the shorter the x component of the plank will be. The y componentact that thedoesnCtfaster change,the relativeshorter the x component of the plank will be. The y componentso themotion,angle is thelarger.Original:Later:nCt change, so the angle is larger.x2.39If it can pass through according to one observer it must be able to pass through according to the other! Therecannot be a collision in one frame and not in the other! ItCs just a matter of when. One argument is that itscan pass throughaccordingto onetoobserverit mustbe ableto nsperpendicularthe directionof relativemotiondonCtthroughchange[itsheight isin eitherot be a collisionin onenot inistheItCs seesjust aa plankmatterwhen. Oneitsat restAnna,so it cannothit. frameAnotherandargumentthatother!while Bobandofdoorwayat the argumentsame angleisθ, thatwith respecttotothethedirectionplank, seesofarelativeplank at motiona smallerdonCtangle,θ0. She alsoheightsees a wallher,contractedensions perpendicularchange[itsis themovingsame towardin eitherframe,′Bobandseesθ seenby Bob.The topwill passthe directionmotionthereforeat anda largerangle thanθ, Anna,at restcannot hit. alongAnotherargumentofisrelativethat whilea plankdoorwayat thethesameangle00y passes through at time zero on both clocks. According to Bob, the topthroughfirst!LetCssay thatbottomrespect to theplank,seesa plankat athesmallerangle, θ0. She also sees a wall moving toward her, contractedwill also (simultaneously) pass through at time zero: t 0. It canCt according to Anna! The top passing throughg the directionof relative motion and therefore at a larger angle than the θ seen by Bob. The top will passvvvugh first! clocks.AccordingtoBob,thetop′′t′ 2 L0(an event) occurs at t 0 and at x′ L0cosθ0. Find t′. t γ ν 2 x t 0 γ v 2 L0 cos θ 0 t ′cc top passing through calso (simultaneously) pass through at time zero: t 0. It canCt accordingto Anna! Thevvv0 earlier. This adds up! We vknow esthroughcosθ002′′00t′ 2 L0event) occurs at t 0 and′at x′ L0cosθ0. Find t′. t γ ν 2 x t c0 γ v 2 L0 cos θ 0 t ′cccx angle than Bob. If it passes through00Anna sees a smallerat onceaccordingtoBob,thetopmustpassthroughfirst2µ go through all at oncevaccordingtoAnna.Ifitdoesnotaccordingto Anna,itaddscan indeedbeknowlargerthatthan ghcosθearlier.Thisup!We000]hole .c2What is the observed length along the y-axis?L L sin θWhat is the observed length along the x-axis?L cos θL γ !v1 (L cos θ )ca sees a smaller angle than Bob. If it passes through at once according to Bob, the top must pass through firstPhysics 3305 - Modern PhysicsProfessor Jodi Cooleyrding to Anna. If it does not go through all at once according to Anna, it can indeed be larger than the

Use geometry to find total length L.!′2 LL L′2yx21v2222L (L0 sin θ0 (1 2 )L0 cos θ0 ) 2c21v222L L0 (1 cos θ0 cos θ0 2 cos θ0 ) 2cL L0!v21 2 cos2 θ0cUse geometry to find Θ.tan θ Physics 3305 - Modern Physics′LyL′xγµ L0 sin θ0 L0 cos θ0θ tan 1 (γµ tan θ0 )Professor Jodi Cooley

THE END(FOR TODAY)xkcd.comPhysics 3305 - Modern PhysicsProfessor Jodi Cooley

You have ALL been learning and doing classical physics for about 20 years NONE of you have had any first hand experience with modern physics Will be introduced to some counter-intuitive concepts Do not expect to be able to do the problems on the very fir