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wk5.pdfGallilean Transforms MerminTwins and KinematicsRelativity1905 - Albert Einstein:• Brownian motionfi atoms.• Photoelectric effect.fi Quantum Theory• “On the Electrodynamics of Moving Bodies”fi The Special Theory of RelativityThe Luminiferous EtherHypothesis: EM waves (light) travelthrough some medium - The EtherSpeed of light: c = 3 x 108 m/s w.r.t fixed ether.The earth moves at v = 3 x 104 m/sw.r.t fixed ether.fi Speed of light w.r.t earth should depend on direction.The Michelson-MorleyExperimentAn interferometer*LightSourceMirror AMirror BTelescopeBeam ABeam B1/2 Silvered MirrorThe interference fringes should shift.But no effect was observed!What was wrong?The Lorentz-Fitzgerald ContractionSuppose that the ether squashes anyobject moving through it?To counteract the change in light speed,we need: d’ = d √1 - v2/c2Galilean Transformations.KyxzK’y’x’z’t’ = tz’ = zy’ = yx’ = x - vtFrom Head on Collision to Collision at Rest by changing Frames Start from known (“Obvious”): equal-mass head-on elastic collision Relate to elastic collision with one at rest View train frame (5f/s right): transforms into known situation Ut = Us – V (+ u,v to right): Ut = Us - 5 Then transform back to station (5f/s left): Us = Ut + V: U’s = U’t + 5 Result: cue ball stops, target ball rolls on Mermin 2005Equal-mass totally inelastic collision Relate to inelastic collision with one at rest Ut = Us – 5 U’s = U’t + 5 Result: combined mass moves at half speed of incident(Very) Asymmetric Elastic Collision Choose Train Frame to put big mass at rest: Ut = Us – (-10) = Us + 10 U’s = U’t – 10 Result: light ball (nearly) twice speed of heavy ball; heavy ball (nearly) unaffectedFinally, use to solve for Asymmetric head-on Elastic Collision Again Choose Train Frame to put big mass at rest: Ut = Us + 5 U’s = U’t – 5 Result: light ball (nearly) three times speed of heavy ball; heavy ball (nearly) unaffected Example: drop tennis ball on top of basketball rebound matches this situation Lessons from changing frames: An exercise in Gallilean transform for velocities Analysis from the simplest point of view Well-chosen transformation can give non-trivial resultsIn frame K, two charges at rest. Force isgiven by Coulomb’s law.KyxzQ1Q2K’y’x’z’In moving frame K’, two charges are moving.Since moving charges are currents,Force is Coulomb + Magnetism.Principle of relativity:“The laws of nature are the same in allinertial reference frames”Something is wrong!• Maxwell’s Equations?• The Principle of Relativity?• Gallilean Transformations?Einstein decidedfi Galilean Transformations are theproblem.Einstein’s two postulates:1. The principle of relativity is correct.The laws of physics are the same in allinertial reference frames.2. The speed of light in vacuum is thesame in all inertial reference frames(c = 3 x 108 m/s regardless of motion ofthe source or observer).The second postulate seems to violateeveryday common sense!Rocket Light pulse Observerv=0.5 c v=cEinstein says: observer measures the lightas traveling at speed c, not 1.5c.Gedanken ExperimentsA light clock:It ticks every Dt = 2 w/c seconds.One can synchronize ordinary clocks withit.wmirrormirrorphotocellTime DilationOG: Observer on GroundwOT: Observer on TruckOT’s clock as seen from the ground: c = 3 x 108 m/s (ct/2)2 - (vt/2)2 = w2wvwvt/2ct/2Time for one round trip of light, as seenfrom the ground:t = (2 w/c) √ 1 - v2/c2For v = 0.6c, t = (2 w/c) x 1.25All of OT’s processes slow down comparedto OG as seen by OG.Similarly,All of OG’s processes slow down comparedto OT as seen by OT.Length ContractionOG: Observer on GroundwOT: Observer on TruckDevice on truck makes mark on trackeach time clock ticks.As seen from ground:Distance between marks= (time between ticks) x v= [(2 w/c) √1 - v2/c2 ] vwvAs seen from truck:Distance between marks= (time between ticks) x v= (2 w/c) v(To the person on the truck the timebetween ticks is (2 w/c).)(Distance measured on truck) = √ 1 - v2/c2 x (distance measured on ground)As seen from a moving frame, restdistances contract.(L-F contraction)SimultaneityEvents occur at a well defined position anda time (x,y,z,t).But events that are simultaneous (same t)in one inertial frame are not necessarilysimultaneous in another frame.The light from the two flashes reach OGat the same time. He sees them assimultaneous.d dABOT passes OG just as the lights flash.vBut light from B reaches OT first. Sinceboth light beams started the samedistance from her, and both travel atspeed c, she concludes that B must haveflashed before A.vccLorentz Transformations• Flashbulb at origin just as both axes are coincident.• Wavefronts in both systems must be spherical:x2 + y2 + z2 = c2t2 andx’2 + y’2 + z’2 = c2t’2• Inconsistent with a Galilean transformation• Also cannot assume t=t’.yxzy’x’z’vAssuming:• Principle of relativity• linear transformation (x,y,z,t) -> (x’,y’,z’,t’)Lorentz Transformations (section 2.4)x’ = g ( x - v t )y’ = yz’ = zt’ = g ( t - v x / c2 )With g = 1 / √1 - v2/c2 .(Often also define b = v / c . )Time Dilation (again)Proper time: time T0 measured betweentwo events at the same position in aninertial frame.OG’s clock: T0 = t2 - t1, (x2-x1=0)OT’s clock: T’ = t’2 - t’1t’2 - t’1 = g (t2 - t1 - v/c2 (x2-x1) ) T’ = g T0 > T0Clocks, as seen by observers moving at arelative velocity, run slow.vOGOTOT’s friendLength Contraction (again)Proper length: distance L0 between pointsthat are at rest in an inertial frame.OT on truck measures its length to be L0 = x’2 - x’1. This is its proper length.OG on ground measures its length to beL = x2 - x1, using a meter stick at rest(t2 = t1).ThenL0 = x’2 - x’1 = g (x2 - x1 - v (t2 - t1)) = g LOG measures L = L0/ g < L0.Truck appears contracted to OG.vx’1x’2x1x2An applicationMuon decays with the formula:N = N0 e-t/tN0 = number of muons at time t=0.N = number of muons at time t secondslater.t = 2.19 x 10-6 seconds is mean lifetime ofmuon.Suppose 1000 muons start at top ofmountain d=2000 m high and travel atspeed v=0.98c towards the ground. Whatis the expected number that reach earth?Time to reach earth:t = d/v = 2000m/(0.98 x 3 x 108 m/s) = 6.8 x 10-6 sExpect N = 1000 e-6.8/2.19 = 45 muons.But experimentally we see 540


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MSU PHY 215 - wk5

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