Einstein's special relativity the essentials

VCE Physics Unit 3: Detailed study Einstein's special relativity the essentials Key knwledge and skills (frm Study Design) describe the predictin frm Maxwell equatins that the speed f light depends nly n the electrical and magnetic prperties f the medium thrugh which it is passing, and nt n the speed f the surce r the speed f the medium; cmpare the predictin frm Maxwell equatins f an abslute speed f light with the classical principle f relativity, i.e. n abslute zer fr velcity; all velcity measurements are relative t the frame f reference; describe Einstein tw pstulates fr his special thery f relativity and cmpare them t classical physics the laws f physics are the same in all inertial (nn-accelerated) frames f reference the speed f light has a cnstant value fr all bservers regardless f their mtin r the mtin f the surce; interpret the results f the Michelsn rley experiment in terms f Einstein secnd pstulate; apply simple thught experiments t shw that the time interval between tw events differs depending n the mtin f an bserver relative t the events (nn-simultaneity) length cntractin f an bject ccurs in the directin f its mtin when bserved frm a different frame f reference explain the cncepts f prper time (t ) and prper length (L ) as quantities that are measured in the frame f reference in which bjects are at rest; explain the unfamiliar nature f mtin at speeds appraching c by mathematically mdelling time dilatin and length cntractin using the equatins t = 2 2 ½ tã and L = L /ã where ã = (1 v /c ) 2 apply Einstein predictin by shwing that the ttal mass-energy f an bject is given by E tt = E k + E rest = mc where m = m ã and s kinetic energy, E k = (ã 1)m c 2 ; explain that mass can be cnverted int energy and vice versa, E = Ämc 2 explain the impssibility f mtin faster than light in terms f relativistic mass m = m ã at speeds appraching c. What fllws is a summary f the essentials f the curriculum as they relate t the way we have discussed relativity in class and in the text. Nt everything in the text is required fr the exam, but it is there t help us get a gd understanding f relativity and the better ur understanding the better we will find the exam! Sectin 1 Tw ideas Einstein did nt want t give up: 1) Galilean & Newtnian Relativity: There is nthing special abut zer velcity. Velcity can nly be measured relative t sme ther inertial frame f reference. A frce prduces a change f velcity, nt velcity itself (as Aristtle and everyne had thught). That is, a frce prduces acceleratin (a = F/m). Ntice that acceleratin is nt relative, but abslute fr any inertial frame f reference. (An inertial frame f reference is ne with a steady velcity.) That is, all bservers in inertial frames will measure the same acceleratin in any frame. 2) Maxwell, hwever, prduced electrmagnetic equatins which suggested that a) light is an electrmagnetic wave and b) that all 8 electrmagnetic waves travel at a fixed speed f 3 10 m/s, regardless f speed f surce (nt t surprising sund is same) r f bserver (very surprising nt like sund). This was quite cntrary t the principle f relativity! Maxwell and thers thught that this meant there must be an aether which was fixed in space and t which this speed f light must be relative. Michelsn and Mrley tried t measure the Earth s speed relative t the aether. Their experiment attempted t measure the difference in speed as the light went acrss the path f the Earth s mtin cmpared t back and frward alng the path. They fund n difference in speed with directin f the Earth s mtin as they shuld have if light really had a speed fixed in the aether. This result culd nt be explained by the physics f the time. Althugh it was cnsistent with the Maxwell predictin abut the speed f light, it was quite at dds with the principle f relativity which said velcities were always relative. Here we need t mentin Newtn, wh realised that when we wrk ut relative velcities we are assuming that space is straight and time is universal. This means that space is like a big bx with three dimensins (the x, y, z crdinates fr example) and that time wuld be measured exactly the same by all clcks wherever they were. Newtn had n reasn t believe therwise, but had the genius t see that this is an assumptin that we make. These are the assumptins we make when we wrk ut relative speeds (because we add r subtract distances and times). Einstein realised that perhaps this assumptin was nt crrect when it came t large distances and high speeds. Sectin 2 Einstein s crazy idea Einstein felt that bth the principle f relativity and Maxwell s equatins were t elegant and t well funded t be wrng. He therefre explred the cnsequences f keeping them bth, despite the apparent cntradictin. Einstein s pstulates were: - the laws f physics are the same in all inertial frames f reference - the speed f light has a cnstant value fr all bservers

VCE Physics 12 Unit 3 DS Einstein s Special Relativity Essentials Page 2 T get a feel fr the implicatins, we fllw Einstein and d sme Gedanken experiments, ntably in ur Gedanken train with its flashing light in the middle. The imprtant thing t ntice is that the light flashes are travelling at c fr bth inside and utside bservers. This is what is different if we take Einstein s pstulates at face value. Als nte that while we can t actually see the light flashes travelling (as suggested by the diagram) we can wrk ut where they were, when, s ur Gedanken bservers d all that fr us! Clearly, A and B see the light reach the ends f the carriage at the same time and return t them at the same time. But cnsider the situatin as seen by C: As the train mves frward, C sees the backward flash meet the ncming back f the carriage befre the frward flash catches up t the frnt f the carriage. Clearly then, C sees the light flashes hitting the frnt and back walls at different times, while A and B see the light hitting at the same time. S what was simultaneus in ne frame f reference was nt simultaneus in the ther frame. This illustratin indicates that time is relative. In ther wrds, there is n big universal clck ut there smewhere ticking away the real time while all ur ther clcks d strange things. THERE IS NO REAL TIME. This is hard t get ur minds arund because we are s used t there being a crrect time. Certainly we adjust ur clcks fr ur time zne, but that is nly a cnvenience thing s we always expect midday (by the Sun) t ccur at arund 12 nn. That a mving clck might inherently tell a different time t ne at rest seems very dd because we never experience anything like that. Hwever, experiments with atmic clcks in aerplanes and space vehicles have thrughly verified Einstein s predictins abut time. In thse situatins we are nly talking f micrsecnds r millisecnds, but atmic clcks are accurate t millins f times greater than that. The whle GPS system wuld literally be miles ut if it didn t take int accunt Einstein s relativistic time. Nw cnsider the flashes reflected frm the frnt and back walls f the train returning t A and B. All bservers must agree that these flashes return t C at the same time. It wuld be ttally illgical if C saw the frward and rear flashes reach A and B at different times. Here we need t be clear abut what we mean by events. An event is smething that happens at a certain time and a certain place. The flashes reaching the frnt and back walls are tw different events they happened in different places and culd therefre als happen at different times. The flashes returning t A and B again are ne event because they happened at the same place. If A and B see them as ne event then C must als see them as ne event. Events which happen at different places must be different events, events which happen at the same place at different times are clearly als different events. Different bservers may find different distances and different times between different events, but all must agree abut what happened in a single event that is, A and B saw the tw flashes arrive as ne event, nt tw. T get a feel fr what this all means, cnsider the fllwing example. It is a little lng-winded, but if yu fllw it step by step it shuld help yu t understand hw events which are simultaneus in ne frame may nt be in anther. Q.1 Tw flashes f lightning just happen t hit the tw ends f a fast mving train at the same time. What is wrng with this statement? These questins are all answered in the fllwing text, but see if yu can answer them first befre reading n! Well, yes it is very unlikely! But the real prblem is that we have nt specified in what frame the events ccurred simultaneusly. As we nw knw, if they ccurred simultaneusly in the train frame they wuld nt have been simultaneus in the grund frame. Let s see why: Situatin 1) Imagine the flashes were simultaneus in the grund frame. That is, an bserver n the grund but at the centre f the train (nt shwn in picture) as the lightning struck, will have seen the flashes arrive tgether and then deduce, because the distances t the ends were equal, that the flashes must have been simultaneus. The by, wh happened t be at the rear end f the train, will have seen the rear flash first, but will be able t calculate (if he survived the lightning) frm the delay in seeing the ther flash (and the length f the train) that the flash at the frnt end must have ccurred at the same time (by his clck) as the ne at his end. Alternatively, the by and the girl (at the frnt end) culd recrd the times n their clcks and then the tw f them culd get tgether at their leisure t cmpare the times n their tw clcks (which are in the same frame f reference). They will find that the tw flashes ccurred at the same times n their clcks. Q.2 But hw will the bserver inside the train see the flashes? Will they be simultaneus? Or which ne will be first?

VCE Physics 12 Unit 3 DS Einstein s Special Relativity Essentials Page 3 We can answer this questin simply by lking at the situatin frm the grund bservers frame (by and girl). As the train is mving away frm the by, he will see the light frm the rear flash take lnger t reach the train bserver (in middle f carriage) than the light frm the frnt flash will (it desn t have t g as far as the train is mving tward it). The by therefre deduces that the train bserver will see the frnt flash befre she sees the rear ne. The girl, at the frnt f the train will deduce exactly the same thing. S what abut the bserver in the train? She must als agree that the frnt flash was first. Ntice that bth sets f bservers must agree abut what the train bserver saw she either saw ne flash first r saw them simultaneusly n bserver can see a different rder fr the flashes at ne place. They can, hwever, see the time between events (flashes arriving fr example) differently. In the train based bserver s frame the flashes have cme the same distance (half the length f the train) and s (because she saw the frnt ne first) she deduces that indeed the frnt flash must have ccurred befre the rear ne. S they (the lightning strikes) were nt simultaneus fr her. Time is relative! We deduced this by arguing frm the grund frame, but the train bserver must agree that they were nt simultaneus. Observers in different frames may measure different times, but they must agree n the rder in which any ne bserver saw the events which ccurred at the same place (that is, the flashes arriving at the centre f the train). Situatin 2) What if it was the train bserver wh saw the flashes simultaneusly? (This was nt the same pair f lightning flashes in situatin 1.) Q.3 If the train bserver sees the flashes simultaneusly, hw did the by and girl n the grund see the flashes? (Remember that all ur bservers always see the speed f light as the same.) Nw everyne must agree that the train bserver saw the flashes tgether that was ne event. S when did the tw separate events, the tw lightning flashes, ccur in the grund frame? The train bserver says they happened at the same time, but what abut the by and girl? (Try t decide n yur answer befre reading n!) In this diagram (previus page) we are lking at the situatin frm the frame f reference f the train the by and girl are speeding by in the ther directin (t the left). It just s happens (this is the advantage f Gedanken experiments) that at the instant the train bserver sees the flashes, the girl is ppsite her and s als sees the flashes tgether. (Nte that neither f them say the lightning flashes were at the same time because they saw them at the same time they knw that the light takes time t reach them. The questin is hw much time.) The train bserver, as we said, deduces that the lightning flashes hit the train at the same time (because they travelled the same distance). But what abut the girl? She is mving t the left (as seen by the train bserver). In the train frame ur bserver will deduce that the girl will decide that the left hand flash ccurred first (althugh she saw them at the same time). Q.4 Hw did the girl deduce that the left hand flash was first? As the girl is mving past the centre f the train, she sees the flashes tgether. But she will then realise that she must have been clser t the right hand end when the flashes ccurred, s if she saw them tgether the left hand flash must have ccurred first because the light had mre distance t travel t reach the centre f the train (in her frame). Here is anther way f lking at the same situatin: This time we stay in the grund frame. The by and girl bth see the flashes reach the train bserver at the same time (just as did the train bserver this is ne event). They knw that the light is travelling tward the centre f the train at the same speed frm bth directins. In the diagram this is represented by the V shaped lines and the flashes getting clser t the centre. The train is als mving, but nt as fast as the light. We want t knw when the light flashes left the ends f the carriages. These are the tw pints where the flashes cincide with the

VCE Physics 12 Unit 3 DS Einstein s Special Relativity Essentials Page 4 ends f the train. The lightning blts are shwn at thse events. We can see that the left hand flash must have ccurred befre the right hand ne because it had further t g t reach the spt where the centre f the train is when the bservers all see the light reach the girl in the centre f the train. Q.5 In this diagram the (grund based) girl was nt at the centre f the train when the light reached the train bserver. Des that make any difference t ur analysis? The answer is n, because any bserver in the grund frame will see the same thing. Depending n hw clse they are t the centre f the train they may see the flashes (tgether) reach the train bserver at different times, but when they calculate the lk-back time they will all deduce the same interval between the lightning flashes and the light arriving at the centre f the train the left hand interval being lnger than the right hand interval. Sectin 3 Time is nt what it seems In rder t understand time dilatin yu need t wrk thrugh the argument in the text. Unfrtunately, n the exam yu wn t have t explain where the time dilatin frmula cmes frm, yu ll just have t use it. But t use it prperly yu need t understand it! t = t ã where ã = 1/ (1 v²/c²) The mst imprtant thing in using the time dilatin expressin is t understand the distinctin between relative time and prper time. Prper time, t, is the time interval between tw events as measured by smene in the same frame as the events. The relative time, t, is the time as measured by smene mving relative t the frame in which the events ccurred. Sme useful values f gamma and speed (it might be wrth jtting sme f these n yur A4 sheet fr the exam). Here (diagram belw) is anther situatin that can help us srt ut time dilatin and the relatinships between different events. The train is mving alng the track (at relativistic speed!). As it passes the girl (event 1) the clcks in the train and n the grund bth read 7:00. As it passes the by further dwn the line (event 2) the train clck reads 7:15 and the by s clck reads 7:20. This is as we expect because time ges slwer in the mving frame. See if yu can answer these questins befre lking at the answers which fllw. v/c as % gamma ã 10% 1.005 20% 1.02 66% 1.33 88.6% 2.00 99% 7.09 99.5% 10 99.9% 22.4 Q.1 Which f these clcks is reading the prper time? Hw d yu knw? Event 1 Event 2 Q.2 Frm the pint f view f the persn in the train the wrld utside is mving by in the ppsite directin (t the left). S why desn t she see the external clck at event 2 reading less than hers (abut 7:11) because f time dilatin? Or des she? Q.3 What is the value f gamma in this situatin, and s what speed (relative t light) is the train mving at? The prper time is the time as measured in the frame f reference in which the tw events ccur at the same place. This is clearly nt true f the girl and by in the grund frame. It is true f the girl in the train hwever. In her frame the tw events ccurred at the same place by her table in the train but at different times, 15 min apart. S the girl in the train has the prper time. At event 2, the train passing the by, the girl in the train and the by must agree abut what they see. They will bth agree that the train clck says 7:15 while the by s clck reads 7:20. Bth grund bservers must agree abut the time recrded frm their frame they can cmpare clcks and times later. S, n, the train bserver des nt see the clcks saying anything different t the grund bservers. But desn t she see time ging mre slwly fr the peple n the grund? Yes! Relativity is quite cnsistent in that there is n special frame f reference the train is as gd as the grund. S hw cme she sees ne clck reading 7:00 and the ther 7:20? Desn t that mean that 20 min have elapsed? N! The by and girl grund bservers are nt at the same place and s fr the train bserver, events 1 and 2 were different in bth time and place as far as the grund frame was cncerned. The train bserver did nt see the grund girl s clck ging slw, just that the girl s and the by s clcks were reading different times. If she culd see the grund girl s clck it wuld indeed read abut 7:11 as she passes the by. The fact that thse tw clcks wuld be different in her frame is because the time between different events is different when bjects are in mtin. Remember that she sees the light frm the girl s clck cming at her at speed c, whereas the grund girl will see that same light take lnger t reach the train girl (because the train is mving away frm the light) hence when the train clck says 7:15, hers will nly read abut 7:11. But the train girl seeing the grund girl at 7:15 and the grund girl seeing the train girl at 7:12 are tw different events, nt the same ne, and the tw bservers will indeed see different times elapse between

VCE Physics 12 Unit 3 DS Einstein s Special Relativity Essentials Page 5 thse tw events. The time in the prper frame (that is, the frame in which bth events ccurred at the same place ie, the train carriage) was 15 min. The dilated time was 20 min and s ã = t/t 20/15 = 1.33. Frm the table abve we see that the speed f the train must have therefre been 66% f c. What if the gamma value was nt n ur table? We simply have t use the gamma equatin t slve fr v! Hwever, it is handy t re-express this equatin in terms f v. First, let s make a simplificatin by defining the speed relative t light as V (Capital V). S V = v/c. This means that we can express the gamma equatin as: ã = 1/ (1 V²). If we square bth sides and crss multiply this becmes 1 V² = 1/ã² A little mre algebra gives: V² = 1 1/ã² (It is handy t put this n yur A4 sheet) T slve ur previus prblem we put ã² = (4/3)² = 16/9 and s V² = 1 9/16 = 7/16 giving V = 7/4 = 0.661, that is V is 66% f c, as we said. Dn t frget t unsquare the V² t get V when yu use this equatin r maybe express it as V = (1 1/ã²). Sectin 4 Time and Space If time is bent by fast travel, it is nt t surprising that space is als. In rder t understand situatins like the previus train clck and grund clck situatins we really need t understand the length cntractin as well as the time dilatin. Einstein s length cntractin equatin: l = l /ã where ã = 1/ (1 v²/c²) Ntice that this time the prper length is divided by gamma rather than multiplied by it. Times appear lnger in the mving frame, and lengths appear shrter. Mving bjects appear shrunk but nly in the directin f mtin, nt in the ther tw dimensins. Again, the prper length is the length as measured by smene in the frame which is at rest relative t the length being measured. Remember that this applies bth t bjects and distances. As we watch a space ship travelling by at 88.6% f c, where ã = 2, the space ship will lk half as lng as nrmal. But f curse the space ship crew will see us lking half as wide as nrmal als. Furthermre, if we see them travel frm the Earth t the Mn (in a cuple f secnds) a distance f 380,000 km, they will say they nly travelled 190,000 km, BUT that is because the distance between the Earth and Mn t them is nly 190,000 km, whereas we measure it as 380,000 km. Space is relative! The faster yu g the mre shrunk it appears t be. (Light phtns g s fast, space is shrunk t nthing!) Q.4 The electrns whirling arund in the new synchrtrn near Mnash will be mving at clse t 99.99999% f c where ã is abut 2000. The actual length (circumference) f the strage ring is 216 m. Hw lng is the ring in the frame f reference f the electrns? Hw lng d the electrns take t get arund the ring (frm ur frame)? Mving bjects are cntracted in the directin f mtin The apparent length f the ring as seen by the electrns is nly abut 11 cm! The time t get arund is 216 m divided by c there is -9 n needed t wrry abut the 0.000001% difference frm c! This is 720 ns (ns = nan secnds = 10 s). Q.5 But wait a minute, dn t the electrns nly see their path as 11 cm? S hw cme they dn t just take 11 cm c = 37 ps -12 (picsecnds = 10 s)? Think hw yu wuld answer this questin befre reading n. T the electrns the ring is rushing by at 99.99999% f c. And yes it nly takes 37 ps fr it t g by. Hwever, that is what we expect frm relativity. Observers (and even electrns) in different frames see different times between the same events (being at ne pint n the ring and then being there again). Q.6 We can wrk ut the time in the electrn frame a different way. What is it? Frm ur frame we see the electrn clck ging slw by the gamma factr. Divide 720 ns by 2000 and what d we get? Relativity might be strange, but it is cnsistent! Q.7 Imagine a 500 m lng space ship is travelling past an even lnger tape measure we have stretched ut in space. The tape is at rest relative t us. The ship is travelling at 86.6% f c (s ã = 2). We see the frnt f the ship pass the zer mark n the tape and start ur (super fast) stp watch. Hw lng is it befre the back f the ship passes the zer mark? And where is the frnt f the ship at this time? (Again, think ut yur answer nw.) Yu will prbably have fund this questin easy enugh the ship appears t be 250 m lng and travelling at ½ c, s it will take 417 ns. If smene takes a pht f the ship in frnt f the tape it will be clear that as the back passes zer the frnt will be at the 250 m mark. The apparent length has cntracted.

VCE Physics 12 Unit 3 DS Einstein s Special Relativity Essentials Page 6 Q.8 Ok, the ship tk 417 ns t pass the zer pint n the tape. And when the back passed the 0 mark the frnt was at the 250 m mark. Des that mean that the ship is really 250 m lng? What will the space ship captain think abut this? Let s cnsider things frm the captains pint f view. He sees the tape measure flying by at ½ c. And he sees its length cntracted. He knws his ship is 500 m (that was its prper length). S when the back f the ship passes the zer mark, where is the frnt f the ship? At the 1000 m mark! This is because while the ship is 500 m lng, the tape appears cntracted and s if the captain gets tw f his crew t recrd the psitins f the frnt and back at the time the back passes the zer mark, the frnt will be ppsite the 1000 m mark. That s all well and gd, but hang n, didn t we see the frnt ppsite the 250 m when the back passed zer? What s ging n? This highlights the imprtance f understanding what nn-simultaneity really means. The pint is that when we said where is the frnt at the same time when the back was at zer we regarded thse tw events (back at zer and frnt at 250) as simultaneus. But these tw events were nt simultaneus fr the space ship captain! The frnt being at 250 will have ccurred befre the event f the back being at zer. T the captain, it was the frnt being at 1000 m and the back at zer which were simultaneus. Ntice that bth bservers in this example can use the tape measure t measure the prper length f the ship and get the same answer! We actually knw that the ship nly appears t be half its prper length because we knw abut relativity. S we duble the 250 m and get 500 m. The captain knws that the tape measure is ging at ½ c and s knws that it appears shrunk t half its length. S he halves the 1000 m t get 500 m. Sectin 5 E = mc² This simple little equatin can be interpreted (and mis-interpreted) in varius ways. What it really says is: Ttal energy f an bject = relativistic mass c². But we need t realise that m here is gamma times the rest mass: m = ãm S E = mc² can be written as E tt = ãmc² Nw when at rest ã = 1, s the ttal energy f an bject at rest is nt zer in fact it is m c² which fr an rdinary bject is a very big number! Einstein realised that mass is a frm f energy. And that energy has mass. The kinetic energy, the energy due t mtin, is therefre the difference between the rest mass and the ttal energy: E K = E tt m c² = mc² mc² = ãmc² mc² = (ã 1)m c² As the last expressin in this set implies, it can be handy t think f the part f ã abve ne as representing the kinetic energy. Fr example: What is the kinetic energy f an electrn if its relativistic mass is 3m? Answer: 2m c² That is, twice its rest mass energy. (Of curse E = ½mv² des NOT give us the crrect answer unless the speed is well belw c.) K -31 Q.9 An electrn f rest mass 9.1 10 kg is accelerated in an electric field and given 100,000 ev f energy (1 ev, r electrn-vlt -19 is the energy an electrn gets ging thrugh a 1 V battery and is equal t 1.6 10 J). Hw fast is it ging? First wrk this ut just using Newtn s mechanics, then using relativity. Then cmpare yur answer with that belw. Newtn says we use the expressin E K = ½mv² (actually Newtn never said this, but it fllws frm his laws f mtin). The energy -19-14 8 is 100,000 1.6 10 J = 1.6 10 J Putting this, and the electrn mass int the kinetic energy expressin gives v = 1.9 10 m/s. This is well ver half f c. -14 Using relativity, we again knw that the kinetic energy is 1.6 10 J. This enables us t find ã, as E K = (ã 1)m c² S ã 1 = E /m c² = 0.1954 and s ã = 1.1954 K 8 Using ur expressin fr V in terms f ã we find V = 0.548 and s v = 0.548 c = 1.64 10 m/s which is a little ver half f c, and a little less than the value fund using cnventinal mechanics. This reflects the fact that the mass f the electrn at this speed is 1.1954 times its rest mass, and therefre the increase in energy des nt speed it up as much as it wuld if it had n increase in mass. The faster it ges the heavier it gets and s the less acceleratin it has. This is why it can never reach the speed f light. Try repeating this prblem starting with a kinetic energy f 1,000,000 ev. Cnventinal mechanics says the speed shuld be 10 the 8 previus speed, ie. 6 10 m/s, twice the speed f light! See what the relativistic speed is. It is wrth nting that the 1,000,000 ev used here is ne millin electrn vlts r 1 MeV. An electrn accelerated by 1 millin vlts wuld have this energy. Our Van de Graaff will accelerate electrns by abut 300,000 V, that is they will have 300 kev f energy. 9 By cmparisn, the synchrtrn will accelerate them t 3 GeV. (A GeV is a Giga ev r 10 ev) Yu can see why the electrns in the synchrtrn are ging at very clse t c!