4.1 The speed of light and Einstein s postulates

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1 hater 4 Secial Relativity 4. The seed of light and Einstein s ostulates 4.. efore Einstein efore Einstein, the nature of sace and time was not generally thought to need discussion. Sace and time were assumed to be comletely di erent things. They were considered to be absolute, i.e. the same for all observers. Newton s idea of time is summarised in this quotation from his Princiia : Absolute, true and mathematical time, of itself and by its own true nature, flows uniformly on, without regard to anything external. (actually the fact that Newton wrote this down exlicitly rather than taking it for granted is an examle of his genius). Light is a wave motion. There is no ossible doubt about this, because light shows interference e ects in exeriments like Young s slits. Other kinds of waves that we are familiar with roagate in a hysical medium. There is some kind of material, whose oscillations constitute the wave motion. Examle: Sound waves travel through air, which is the hysical material that oscillates. Air is comressible, and sound consists of variations of the density of air. Examle: Waves travel on the surface of water. The hysical roerty that oscillates is the water level. y analogy, it was thought in the 9th century that light must consist of the oscillations of some hysical material, called the luminiferous aether. It was not known what this material was, but it was assumed that it must exist. It is exected theoretically, and observed exerimentally, that the observed seed of sound in air deends on the motion of the observer relative to the air. For examle, suosealoudseakeremitssoundwhichisdetectedbyanobserverwho is stationary with resect to the loudseaker; the distance between them is d. If the air is stationary with resect to the loudseaker and observer, then the time taken for a ulse of sound to travel from the loudseaker is d/v,wherev is the seed of sound in stationary 94

2 air. Now suose the same observation is made in a strong wind, with the air moving at seed u in the direction from the loudseaker to the observer, the distance between loudseaker and observer being fixed, as before, at the value d. Now a ulse of sound will travel more quickly, and will arrive in the shorter time d/(v + u). If the wind blows in the oosite direction at seed u, thentheulsewilltakethelongertimed/(v u). The wavelength of the sound is also changed by the motion of the air. If the loudseaker emits sound at frequency f, and the observer is stationary with resect to the loudseaker, then the observer also hears the frequency f. utwavelength = seed/frequency. So with the wind blowing from the loudseaker to the observer, the wavelength is =(v + u)/f, and with the wind blowing from observer to loudseaker it is =(v u)/f. The number of wavelengths that fit into the distance between loudseaker and observer is in the first case, and in the second case. d/ = fd/(v + u) fd/(v u) Therefore, if the aether exists, theseedoflightshoulddeendontherelativemotion of the observer and the aether. And the wavelength should change corresondingly. Amajortheoreticaldiscoveryinthe9thcenturywasMaxwell s equations, whichgavea mathematical descrition of time-varying electric and magnetic fields. Maxwell s equations redict the existence of electromagnetic waves, and also redict the seed of these waves. The redicted seed is the same as the observed seed of light. This imlied that light consists of electromagnetic waves. ut the derived seed comes straight out of Maxwell s equations, which are di erential vector equations, with no reference to either the coordinate system in which the vectors are defined, or to the motion of an observer relative to these coordinates! 4..2 The Michelson-Morley exeriment Nature resents us with a wonderful way of doing an exeriment with light just like the exeriment with the loudseaker and the observer in a strong wind. The Earth moves in a nearly circular orbit around the sun, at a seed of about 3 4 ms. So if there is a aether in the universe, resumably there is an aether wind rushing ast us at this seed. If a light source and an observer are fixed relative to the Earth, then the time taken for alightulsetotravelfromthelightsourcetotheobserverwillbea ectedbytheaether wind in the same way as in the loudseaker examle. Similarly, the number of wavelengths 95

3 that fit into the distance d between light source and observer will be fd/(c + u) whenthe aether wind is blowing from source to observer, but fd/(c u) if it is blowing the other way, with c the seed of light relative to the aether and u the seed of the aether wind. The Michelson-Morley exeriment is designed to detect the aether wind using the ideas exlained above. It is too di cult to measure the travel time of light ulses with su - cient accuracy, so instead the exeriment uses interference e ects to detect the changes of wavelength that should be caused by the aether wind. Schematic of the Michelson-Morley exeriment. The icture shows a schematic of the exeriment. Monochromatic light from a source strikes a beamslitter consistingofahalf-silveredmirroratanangleof45.thisslitsthelight into two beams which are reflected by two normal mirrors so that the beams recombine at the beam-slitter. The recombined beams are observed in a telescoe. ecause the light is monochromatic the two beams can interfere with each other. To make it easier to observe interference, one of the normal mirrors is slightly tilted so as to roduce interference fringes in the field of view. The aether wind e ects are exected to di er for the two beams roduced by the beam slitter. The aaratus is designed to be rotated. Rotation should change the aether-wind e ects in the two arms, and this should be observed as a shift of the fringe attern. Numerical calculations (see the standard text books, e.g. Klener and Kolenkow, An Introduction to Mechanics, McGraw-Hill) show that the fringe shift should be easily observable. In site of this, no e ect at all is observed, within exerimental error. The exeriment has been reeated many times in the ast years, with greater and greater recision, but no e ect has ever been detected. onclusion: the e ects that would be exected from the aether theory are not observed. This casts serious doubt on the existence of the aether. 96

4 4..3 Einstein s ostulates Einstein ostulated that the aether does not exist. One reason is to ensure that Maxwell s equations are equally valid for all observers, no matter what their relative velocity. This imlies that all such observers observe the same velocity of light. According to Einstein, this is why the Michelson-Morley exeriment roduces a null result. Einstein s ostulates are: The laws of hysics have the same form in all inertial systems. The velocity of light in emty sace is a universal constant, the same for all observers. An inertial system is a coordinate system in which all isolated bodies, not acted on by any forces, move with a uniform velocity. Di erent inertial systems move relative to each other with some uniform velocity. So di erent inertial systems are like di erent observers moving relative to each other with constant velocity The relativity of simultaneity Einstein s ostulates demand a comletely new view of time. They show that two events which are simultaneous according to one observer may not be simultaneous according to another. Examle: Amanwithaflash-lamstandsinthecentreofarailwaycarriage. Thecarriage is art of a train travelling along the track. At a given instant, the man makes the lam emit a air of light-ulses, one of which travels forward to the front of the carriage, while the other travels backward to the back of the carriage. The man with the lam sees both ulses travel with the same seed c, and,sinceheisinthecentreofthecarriage,hesees the two ulses arrive simultaneously at the two ends of the carriage. The same events are observed by another man standing beside the track. According to him, the light ulses do not arrive simultaneously at the two ends of the carriage. This is because the whole carriage is moving forward, so the backward-travelling ulse has less far to go than the forward-travelling ulse - the back of the train comes to meet it. The arrivals of the light ulses at the front and back of the carriage are thus simultaneous for one observer (on the train) but not simultaneous for the other (beside the track). The consequence of Einstein s ostulates illustrated by this examle is called the relativity of simultaneity. It comletely destroys Newton s idea of absolute time flowing uniformly on. 4.2 Inertial systems and the Lorentz transformation In order to understand the imlications of Einstein s ostulates, we must examine the relationshi between the descritions of an event reorted by observers who move relative to each other. To do this, we first define what is meant by an inertial coordinate system 97

5 and an event. Using these definitions we will summarise the Galilean transformation relating the secifications of an event according to the ideas before Einstein. Finally we derive the Lorentz transformation required to satisfy Einstein s ostulates Events and transformations An event is something that haens at a articular oint in sace at a articular time. For examle, the emission of a shar ulse of light by a man with a lam is an event. The arrival of the ulse at the end of a railway carriage is an event. The emission of an alha-article by a radioactive nucleus, and the decay of an unstable elementary article like a muon or a ion are events. In order to secify an event, we need to give its osition in three-dimensional sace, and its time. For resent uroses, it is most convenient to use artesian coordinates, so that an event is secified by the four numbers (x, y, z, t). The same event can be observed by observers who are in motion relative to each other. That is, each observer has their own artesian coordinate system and their own clock. Then for a given event, the secifications of that event by two observers will be di erent: they give di erent numbers (x, y, z, t) and(x,y,z,t )totheevent. Ouraim is to give the relationshi between these two secifications. This relationshi is a linear transformation, likethosewedealtwithearlier. Thustherearematriceswhichgive (x,y,z,t ) in terms of (x, y, z, t), or vice versa. These will be 4 4matrices,similarto the 3 3matriceswhichdescriberotations Inertial systems In Newtonian mechanics, a secial significance is given to isolated bodies, i.e. bodies that exerience no forces because they do not interact with the rest of the world. Newton s first law says that isolated bodies move with a uniform velocity. However, this statement needs to be made more recise, because velocity has to be relative to some observer. It is more comlete and exact to state Newton s first law in the form: There are observers for whom all isolated bodies move with a uniform velocity. Such observers are called inertial observers. Here is the definition of an inertial coordinate system: An inertial coordinate system is a system of coordinates such that all isolated bodies move with uniform velocity in the coordinate system. In other words, an inertial coordinate system is a coordinate system in which an inertial observer is at rest. For brevity, an inertial coordinate system is usually just called an inertial system. 98

6 4.2.3 The Galilean transformation onsider two inertial observers, moving relative to each other with seed v. Oneobserver, called S, hasaartesiancoordinatesysteminwhichtheartesiancomonentsofosition are called (x, y, z). The observer S has a clock fixed at the origin of the coordinate system, which measures time t. Theotherobserver,S,hasaartesiancoordinatesysteminwhich ositions are denoted by (x,y,z ), and a clock fixed at the origin of the coordinate system, that measures time t.thelengthandtimeunitsusedbythetwoobserversareidentical, as are their measuring sticks and clocks. The x and x axes of the two systems are arallel to each other, and the same is true of the y and y axes, and the z and z axes. The origins of the two inertial systems coincide at t =,andtheclocksofthetwosystemsare synchronised so that at t =,t =. ThedirectionofmotionofS relative to S is along the ositive x-axis. There is an event, which, according to S, is at (x, y, z, t), and according to S is at (x,y,z,t ). What is the formula relating (x, y, z, t) to(x,y,z,t )? According to Galileo and Newton, at time t, the origin of the inertial system S is at the osition x = vt as measured by S. So at time t, ositionsmeasuredins are simly o set by this dislacement x = vt along the x-direction. We therefore have: x = x vt y = y z = z t = t. (4.) Equations 4. together constitute the Galilean transformation.. It can be reresented as a4 4matrix: v A which would oerate on a four-dimensional column vector (x, y, z, t) suchthat x v x y z A = y A z A t t The Lorentz transformation According to Einstein s ostulate, the Galilean transformation cannot be correct. Suose a light ulse is emitted from the oint x = y = z = attime t =. 99

7 This event is also secified by x = y = z = attime t =; that is the origins of the two frames S and S coincide at t = t =. Now consider the event in which the light-ulse arrives at the oint x = x on the x-axis of inertial system S. SinceobserverS sees the light-ulse travel with seed c, thetimeofarrivalis So observer S secifies the arrival event as t = x /c. (x,,,x /c). According to the Galilean transformation the same arrival event as secified by S is (x vt,,,x /c). This means that according to S the seed at which the ulse travelled was (x vt)/(x /c) =c v(ct/x ). ut x = ct, sotheseedoftheulseins is c v. ThiscontradictsEinstein sostulate. It is ossible to modify the Galilean transformation in such a way that Einstein s ostulate is true. However, there are strong limitations on the modifications that can be made: The equations exressing (x,y,z,t ) in terms of (x, y, z, t) mustremainlinear. The reason is that the two coordinate systems are inertial. A body moving with uniform velocity in one system must also move with a uniform velocity in the other. This can only be so with linear relations. The symmetry roerties of sace must be resected. For examle, an event that is observed to lie on the x-axis in S must also lie on the x-axis in S. Suose that y = z =andy 6=and/orz 6=. Thenthetransformationwouldbreakthe symmetry exected for rotations around the x-axis. Similarly, events on the x -axis in S must lie on the x-axis in S. As another examle of symmetry, suose we have two events which in S are secified by (x,y,z,t )and(x, y,z,t ), so that they di er by reflection in the x-z lane. Then the two events must have the same value of x in S ;otherwisethetransformationwouldbreakthesymmetryexectedfor reflections in the x z lane. In light of this, consider the equation for x in terms of x, y, z and t. Sinceitislinear,we must have: x = Ax + t + y + z, where A,, and are coe cients that may deend on the relative velocity v. (Thereis no additive constant on the RHS, because we must have x =whenx = y = z = t =.) y symmetry, we must have = = ; this follows from the requirement that events in S related by reflection in a lane containing the x-axis must have the same value of x.

8 For similar reasons, the equation for t in terms of x, y, z and t must be: t = x + Dt, with no terms in y and z. Symmetryalsodictatesthatwemusthavey = y and z = z. The most general form of the transformation allowed by symmetry is thus: or, in matrix form, x y z t A = x = Ax + t (4.2) y = y (4.3) z = z (4.4) t = x + Dt. (4.5) A D A x y z t A. (4.6) Exressions for A,, and D can be obtained in the following four stes:. The observer in S sees the origin of the coordinate system S moving along the x-axis with seed v. ThismeansthatifaneventoccursattheoriginofS (for which x =, by definition), then the x and t secifying this event must satisfy x/t = v. From eq. 4.2, this requires that = Av. (4.7) 2. Similarly, the observer S sees the origin of S moving with velocity v. Thismeans that an event at the origin of S (for which x =, by definition) must have x /t = v. From eq.s 4.2 and 4.5, this means that /D = v. omaringthiswitheq.4.7,we see that: D = A. (4.8) 3. A light ulse emitted at t = t =fromtheoriginalongthex-axis must be observed to travel at seed c in both S and S. onsider the event in which the light-ulse arrives at oint x at time t. InS,thissameeventissecifiedbyx and t.wemust have both x/t = c and x /t = c. Dividing eq. 4.2 by eq. 4.5, we have: x t = c = Ax + t x + Dt = Ac + c + D. Now we use eq. (4.7) and (4.8) to exress and D in terms of A: c = c v +c/a. Solving this for /A, weget: = v A. (4.9) c2

9 4. Finally, consider a light ulse emitted at t = t =fromtheoriginalongthey axis. This ulse must be observed to travel with seed c in both S and S.onsider the event in which the light-ulse arrives at oint y on the y-axis. Since x =, we have x = t, y = y, t = Dt. Now the distance travelled by the ulse in S is (x ) 2 +(y ) 2,sothat: (x ) 2 +(y ) 2 (t ) 2 = c 2 = 2 t 2 + y 2 D 2 t 2 = 2 + c2 D D, 2 using the fact that y/t = c. Now write = Av and D = A to obtain: Solving for A, weobtain: c 2 = v 2 + c2 A 2. A = v2 /c 2. (4.) (Note that we must take the ositive square, since for small v/c we must recover the Galilean transformation, so that A! asv/c!.) From the four eq.s 4.7, 4.8, 4.9 and 4. we have exressions for A,, and D, whichare: A = v2 /c 2, = v v2 /c 2 = v/c 2 v2 /c 2, D = v2 /c 2. Putting these exressions into eq.s 4.2, 4.3, 4.4 and 4.5 we have: x = y = y z = z x vt v2 /c 2 t = t vx/c2 v2 /c 2. These equations are known as the Lorentz transformations. In relativistic hysics, the quantity / v 2 /c 2 occurs very frequently, and it is usually called : v2 /c. (4.) 2 Using, we have: x = (x vt) y = y z = z t = (t vx/c 2 ). (4.2) 2

10 and in terms of matrices x y z t A = v v/c 2 A x y z t A. (4.3) Note that when the velocity v becomes very small comared with the velocity of light c, thelorentztransformationreducestothegalileantransform,asweexect Invariance of the velocity of light The Lorentz transformation formulae were derived above by considering light ulses directed along the x-axis and along the y-axis. ut in fact the Lorentz transformation ensures that the velocity of light travelling in any direction is the same in all inertial frames. Let a light ulse be emitted from the origin at t =. The ulse radiates in all directions. In the frame S, attimet, theulsehasreachedalltheoints(x, y, z) onthesurfaceofa shere of radius ct; i.ealltheointsforwhich x 2 + y 2 + z 2 c 2 t 2 =. In the frame S,thesamelight-ulseshouldhavereachedalltheointsx, y, z given by (x ) 2 +(y ) 2 +(z ) 2 c 2 (t ) 2 =, where (x,y,z,t )arerelatedto(x, y, z, t) bythelorentztransformation. From eq. 4.2, we have: (x ) 2 +(y ) 2 +(z ) 2 c 2 (t ) 2 = 2 (x 2 2xvt + v 2 t 2 )+y 2 + z 2 2 c 2 (t 2 2vxt/c 2 + v 2 x 2 /c 4 ) = 2 (x 2 c 2 t 2 + v2 c 2 (c2 t 2 x 2 )) + y 2 + z 2 = 2 (x 2 c 2 t 2 )( v 2 /c 2 )+y 2 + z 2 = x 2 + y 2 + z 2 c 2 t 2. This demonstrates that the set of events secified by (x, y, z, t) ins such that x 2 + y 2 + z 2 c 2 t 2 =,issecifiedby(x,y,z,t )ins such that (x ) 2 +(y ) 2 +(z ) 2 c 2 (t ) 2 =. This is the same thing as saying that a light-ulse sreads out radially with seed c in both inertial frames, and hence in all inertial frames. What we have in fact shown is that the quantity x 2 + y 2 + z 2 c 2 t 2 is invariant under Lorentz transformations. Note that aart from the c and minus sign, this quantity actually looks a bit like the length of a four-dimensional coordinate vector (x, y, z, t). 3

11 4.2.6 Four-vectors The Lorentz transformations mix u sace and time comonents in a manner analogous to the way in which rotations mix u sace comonents. We have already written them as matrix transformations on a column vector. Four-vectors are objects which transform via the Lorentz matrices when boosted between inertial frames. y analogy with normal vectors, four-vectors define a direction in sace-time and have an invariant length. Scalar roducts of four-vectors are invariant under Lorentz transformations. The only di erence (aart from the fact that they are a four-dimensional not three-dimensional) is the minus sign which aears in front of the fourth term. This is taken into account (or exlained by, if you refer) by the fact that the metric (see section 3.9) of sace-time is not just the identity matrix, but is given by: G = A so that the scalar roduct of two four-vectors is: V W = V T GW = (V V 2 V 3 V 4 ) = V W 2 + V 2 W 2 + V 3 W 3 V 4 W 4 A The sace-time coordinates of a oint are one imortant four-vector. In fact, to kee Lorentz transformations dimensionless, it is usual to write the coordinate four-vector as: x y z A ct W W 2 W 3 W 4 A and the Lorentz transformations as x y z A = ct v/c v/c A x y z ct A. (4.4) We will use this form of the transformation most often from now on. There are other four-vectors in hysics - most imortantly the energy-momentum four-vector, which we will meet later. 4

12 4.3 onsequences of the Lorentz transformation 4.3. Length contraction efore Einstein, it was assumed without question that the length of an object is a fixed quantity, irresective of how the object moves. According to the Lorentz transformation required by Einstein s ostulates, this is not correct. To see this, we have to define carefully what we mean by the length of a moving object. Suose we have a stick lying arallel to the x-axis, and suose first that the stick is at rest. Then we note the x-coordinate of the left-hand end of the stick, call it x A,and the x-coordinate of the right-hand end, call it x. The length of the stick L is then the di erence of x and x A : L = x x A. Since the stick is at rest, the coordinates x A and x do not have to be measured at the same moment in time, because they do not change with time. Now suose the stick is moving to the right arallel to the x-axis, and hence arallel to its own length, with velocity v. At some instant in time, we measure x and x A,anddefine the length to be the di erence of the two: L = x x A. Since the stick is moving, it is clearly essential that the measurements of x A and x are made at the same instant of time. If we did not insist on this, the measurement of the length of a moving object would not be well defined. Now we ask: for a given stick, does the length L defined by the above rocedure deend on the velocity of the stick? And how is the length L related to the length L measured when the stick is at rest? To answer this, let S be the inertial system in which the stick is at rest, with S moving along the ositive x-axis with seed v relative to inertial system S. onsider the two events in which the x-coordinates of the two ends of the stick are measured simultaneously in inertial system S. The four-vector coordinates of these two events in S are x A x A and A. ct ct According to our definition of the length of a moving object, the times t of the two measurements are identical. These two measurement events, as secified in inertial system S, are at x A x A and A. ct A ct 5

13 Since the measurement events are simultaneous in S, theyarenotsimultaneousins. However, since the stick is at rest in S,themeasurementsofx A and x needed to determine the length do not have to be simultaneous in S. These four vectors are related via the Lorentz boost between the two frames (eq. 4.4): x A v/c x A A = A A ct A v/c ct and x ct A = v/c v/c A x ct A So, looking at the first element: x A = (x A vt) x = (x vt). Subtracting the two equations, we have: L = x x A = (x x A )= L. From the definition of,eq.4.,this gives: L = L v2 /c 2. (4.5) The length L of a moving object is therefore shorter than the length L of this object when it is at rest by the factor v 2 /c 2.Thise ectiscalled relativisticlengthcontraction. Note: Relativistic length contraction has nothing to do with forces acting on the object. It is not shortened by some force that is comressing it! The contraction occurs simly because if it did not occur the velocity of light in frames S and S would not be the same. The length L of an object measured when it is at rest is called the rest-length Orientation of a moving stick Length contraction occurs only along the direction of motion. Astickofrest-lengthL oriented along the y-axis, still has exactly the same length L if it moves in the x-direction. This means that a rectangular object moving along the x-axis with its edges arallel to the x- andy-axes, contracts only along the x-direction. As an alication of these ideas, consider a stick of rest-length L at rest in inertial system S.Thestickisorientedatanangle to the x-axis, as measured in S. What is the length of the stick and at what angle is it inclined to the x-axis, when observed from inertial system S? 6

14 To answer this, let the ositions of the two ends of the stick in system S be at the ositions (x A =, y A =)and(x = L cos, y = L sin ). Let the ositions of the two ends of the stick be measured simultaneously at the instant t =ins. ThenfromtheLorentz transformation, eq.(4.4), we have: and ct A L cos L sin ct A = A = v/c v/c v/c v/c A x A y A A From eq.(4.6) we get x A =,y A =,andfromeq.(4.7)weget x = L cos = x y = L sin = y. From this the length of the stick measured in S is: L =(x 2 + y 2 ) 2 = alel 2 cos 2 = L ale v 2 c 2 cos2 A (4.6) x y v 2 + L 2 c 2 sin 2 The angle at which the stick is inclined to the x-axis in S is given by: = tan (y /x ) = tan ( tan ). So the moving rod is both contracted and rotated Time dilation 2. A (4.7) Now consider the rate at which a moving clock runs. A clock is at the origin in inertial system S,which,asusual,moveswithvelocityv along the ositive x-axis relative to inertial system S. The moving clock records the lase of a certain time t,andwewant to know the corresonding time lase recorded in S. To ut it more concretely, suose the moving clock ticks every second. What is the time interval between the ticks according to an observer at rest in S? To answer this question, it is helful to have the Lorentz equations exressing x, y, z and t in terms of x, y, z and t, i.e. the inverse equations. These inverse equations can be obtained several ways: for examle, by finding the inverse matrix to the Lorentz matrix, by treating the Lorentz equations as simultaneous equations and solving them, or by noting 7 2

15 that S moves with velocity v relative to S,sowecangettheinverseequationssimly by swaing x and x, y and y, z and z,andtand t,whilechangingthesignofv. We therefore have: x v/c x y z A = y A z A. (4.8) ct v/c ct Now consider two ticks of the clock which occur at the instants t = t A and t as recorded in frame S. Since the clock is at the origin in S,thetwoticksconstitutetwoevents secified in S by A and A ct A ct The corresonding events as secified in S are at times t A and t given, according to eq.s 4.8, by: ct A = ct A,ct = ct. So the two ticks of the clock searated by the time di erence t = t t A in S are searated by time di erence t in S given by: t = t t A = t. Since =/ v 2 /c 2 is greater than unity, this means than t> t.so,forexamle, if the clock ticks once a second, then when it is moving with velocity v astationaryobserver measures a time greater than second between ticks. One exerimental examle of time dilation comes from the decay of muons roduced by cosmic rays in the uer atmoshere. The muon is an unstable analogue of the electron, having a lifetime of 2.2 µs whenmeasuredinaninertialsysteminwhichisisatrest. Travelling at the seed of light c =3 8 ms we might exect them to travel a maximum distance of only 6 m before decaying. However, in fact they succeed in travelling several km from the uer atmoshere to the Earth s surface. This discreancy is exlained by time dilation. See e.g. Halliday & Resnick ( Fundamentals of Physics ) for details The relativity of simultaneity We already saw that events that are simultaneous in one inertial system may not be simultaneous in another (the man with a flash-lam in the railway carriage). We can return to the question of simultaneity now in terms of the Lorentz transformation. Suose we have two events, which in system S are secified by x A x y A z A A and y z A ct A ct 8

16 and in system S are secified by x A y A z A ct A A and x y z ct A t A be- According to the inverse Lorentz transformation, eq.s(4.8), the time interval t tween the events in S is given by: c(t t A )= ((ct ct A)+(x x A)v/c). Now suose the events are simultaneous in system S, sothatt t A =. Then in system S the time interval is: t t A = (x x A)v/c 2. This means that the events are simultaneous in S only if x A = x. Otherwise, the order in which the events occur deends on the sign of x x A and on the sign of the relative velocity v. 9

17 4.3.5 The interval between events The time interval between two events and the satial distance between events deend on the inertial system in which the events are observed. Nevertheless, four-vectors rovide a way of characterising the interval between two events that does not deend on the inertial system. onsider two coordinate four-vectors and the di erence between them: x A x x A x y A y z A A z A = y A y z A z A = q ct A ct ct A ct Since the Lorentz transformation is linear, the di erence of two four-vectors is also a fourvector. Thus the squared magnitude of q: q 2 = x 2 + y 2 + z 2 c 2 t 2, where x = x A x, y = y A y, z = z A z and t = t A t,isinvariant. Thus for two given events, the quantity q 2 is the same in all inertial frames. Note that q 2 can be ositive, negative or zero: q 2 =:thenthedistancebetweentheevents x 2 + y 2 + z 2 is equal to c t, so that a light signal can ass from one event to the other. q 2 < :inthiscase, x 2 + y 2 + z 2 is less than c t, so that a article can travel from one event to the other at a seed less than the velocity of light. Put another way, there is an inertial system in which x = y = z =,sothatboth events occur at the same osition. q 2 > :inthiscase, x 2 + y 2 + z 2 is greater than c t, soitisimossiblefor asignaltoassfromoneeventtotheotheratlessthanthevelocityoflight.inthis case, there is an inertial system in which the two events are simultaneous. If q 2 = x 2 + y 2 + z 2 c 2 t 2 <, the interval between the two events is called time-like. The events always occur in the same time order in every inertial frame. If q 2 = x 2 + y 2 + z 2 c 2 t 2 >, the interval between the two events is called sacelike. The time order in which the events occur deends on inertial system, and there is an inertial system in which the events are simultaneous. Note that for anything travelling at the seed of light, the interval between all oints on its ath is zero.

18 4.3.6 Addition of velocities Suose an object is moving with a certain seed in a certain direction as observed from inertial system S, whatisitsseedanddirectionasobservedfroms? As usual, the artesian axes of S are arallel to the artesian axes of S, ands moves at seed v along the ositive x-direction relative to S. To describe seed and direction, we need to consider the velocity vector. So in system S, let the velocity vector of the moving body be u =(u x,u y,u z ). In system S,thevelocity vector of the moving body is u =(u x,u y,u z). Our task is to find formulae exressing u x, u y and u z in terms of u x, u y and u z. We can derive the required formulae from the Lorentz transformation by thinking in terms of events. Suose that, as observed from S, attimet A,thebodyisat(x A,y A,z A ), and at time t it is at (x,y,z ). These are two events. The velocity with which the body moves in S has artesian comonents given by: x A u x = x, u y = y y A, u z = z z A. (4.9) t t A t t A t t A Now let these same events A and be secified in system S (x,y,z,t ). Then the velocity vector in S is given by: by (x A,y A,z A,t A )and u x = x t x A t A, u y = y ya t t A, u z = z za t t A. (4.2) Now use the Lorentz transformation (eq. 4.4) to write rimed quantities in terms of unrimed quantities. To relate u x to u x,notethat: x x A = (x x A v (t t A )) t t v A = t t A c (x 2 a A ), where, as usual, =/ v 2 /c 2.Dividingx x A by t t A,weget: u x = x t x A t A = x x A v (t t A ) t t A (v/c 2 )(x x A ) (4.2) Now divide to and bottom by t t A : Doing the same thing for u y : u x = u x v u x v/c 2. (4.22) y u y = y ya = t t A (t t A (v/c 2 )(x x A )) = u y u x v/c. (4.23) 2 y A

19 Similarly, for u z: u z = u z u x v/c 2. (4.24) The corresonding exressions for (u x,u y,u z )intermsof(u x,u y,u z)canbefoundby swaing u x and u x, u y and u y and u z and u z,andchangingthesignofv: u x = u x + v (4.25) +u xv/c 2 u y u y = (4.26) +u xv/c 2 u z = u z +u xv/c 2. (4.27) These formulae can also be derived by the multilying of the two Lorentz matrices; see the roblem sheets. Note that the three comonents of velocity do not transform like the satial comonents of a four-vector Examles of addition of velocities: Suose a saceshi travels directly away from the Earth with a seed v =.5c. Itfiresa sace-robe directly ahead of it (i.e. away from the Earth). The seed of the sace-robe is u =.5c as seen from the saceshi. What is the seed of the sace-robe as seen from the Earth? If we used common sense, instead of relativity, we would just add the velocities, and conclude that the sace-robe moves at the seed of light as seen from the Earth. ut this is incorrect! To aly the formulae for relativistic addition of velocities, let system S be fixed on the Earth, system S be fixed on the saceshi. Let the x-axis be in the direction in which the saceshi travels. Then v is the seed of the saceshi relative to the Earth, u x is the seed of the sace-robe as seen from the saceshi, and u x is the seed of the sace-robe as seen from Earth. With v =.5c, u x =.5c, usingeq.4.25wehave: u x = c + 4 =.8c. If instead of firing a sace-robe with seed u =.5c, the saceshi had fired a ulse of light (obviously in this case u = c), the seed of the ulse of light relative to Earth would be: u x =.5c +.5 = c. This is what we exect, and simly confirms that the seed of light is the same for all observers. 2

20 4.3.8 Relativistic Doler e ect The Doler e ect is the e ect in which the aarent frequency of a wave-motion is altered by the motion of the source, or the observer, or both. It is familiar from everyday life. The itch of the sound from a moving vehicle, e.g. the siren of a fire-engine, is raised when it aroaches you, and is lowered when it recedes from you. The same e ect haens with light sources. I derive here the relativistic formula for the shifts of frequency and wavelength of a monochromatic light source. onsider a light source that emits a regular sequence of light ulses. In the rest-frame of the source, let the time interval between ulses be T. The light source now moves away from the observer along the observer s line of sight with seed v. I will obtain a formula for the time interval T between the arrival of consecutive ulses at the observer. Note first that the sequence of light ulses is like a clock. ecause of time dilation (section ), the clock aears to run slow in the observer s inertial frame, so that the time interval T between the emission of consecutive ulses in the observer s frame is: T = T. (4.28) This interval T is, of course, not the time interval T between the arrival of ulses at the observer it is the time interval between the emission of ulses. To obtain T, consider two consecutive ulses. The first ulse is emitted when the source is at some distance d from the observer, so it takes a time d/c to arrive. The second ulse is emitted at time T later, by which time the distance from the source to the observer has increased to d + vt (note, d, v and T are all as seen in the observer s frame). So the time interval between the arrival of these two ulses at the observer is: T = T + d + vt c = T + v. c d + c = T v c Noting that =/ ( v 2 /c 2 )=/ ( v/c)( + v/c), we can also write this as: s +v/c T = T v/c. (4.29) Now it is easy to go from this formula for the interval between the arrival of ulses to a formula for the Doler shift of the frequency of a monochromatic light source. A monochromatic source gives out a sinusoidal wave, and in the rest-frame of the source there is a time interval T between the emission of consecutive crests of the wave. The frequency of the source in the rest-frame is f =/T.Tocalculatethefrequencyf of light received by the observer, we say f =/T,whereT is the time interval between the arrival of consecutive wave crests at the observer. ut the formula relating T and T for wave crests must the same as the formula relating T and T for light ulses. So from eq. (4.29), the frequency f seen by the observer is given by: s f = = v/c T T +v/c. 3

21 Hence: f = f s v/c +v/c. (4.3) Please note: The signs of v aearing in this formula deend on whether the source is aroaching or receding from the observer. In the version written above, v is the seed with which the source recedes. Some books (e.g. Klener and Kolenkow) give a formula for f in terms of f with oosite signs of v, butthisisbecausetheyassumethatthesource is aroaching. The formula for f in terms of f is easily converted to a formula for wavelengths. Suose is the wavelength of the light source as observed in its rest frame. Then = c/f.if is the wavelength seen by the observer, then = c/f, soforasourcerecedingwithseed v, wehave: s +v/c = v/c. (4.3) For a receding source, the frequency is lowered, and the wavelength is lengthened: we get a red shift. For an aroaching source,frequency is raised,wavelength is shortened:a blue shift. What haens for a source moving erendicular to the line between it and the observer? In this case, unlike the non-relativistic case, there is still a frequency shift due to the time dilation in eq This shift is f = f /. 4

22 4.4 Mass, Energy and E = mc 2 Momentum and energy are absolutely central to hysics, because they are conserved quantities. The total energy and the momentum of an isolated system are constant, and do not change with time. In modern hysics, conservation laws are a cornerstone of fundamental theories, because they are intimately linked to symmetries. The conservation of energy and momentum is maintained and generalised in secial relativity. However, since in relativity we have seen that the addition of velocities is modified, it is reasonable to assume that there will be imlications for the formulae for momentum and kinetic energy Why mass must deend on seed Momentum is a vector. In Newtonian hysics, the momentum of a article is its mass m times the velocity vector v: = mv. According to Newton s laws of motion, the three artesian comonents of the total momentum of an isolated system of interacting articles are all conserved. For examle, if two articles collide, the total momentum vector before the collision is equal to the total momentum after the collision. This is exactly true in the rest frame of the isolated system and by the rinciles of secial relativity it must therefore be true in any inertial frame, regardless of the relative velocity of the frame with resect to any observer. We will use this fact to see how mass must deend on seed. onsider the collision of two articles which are identical to each other. The first figure shows the collision as observed in the inertial frame in which the centre of mass of the two articles is at rest. Particle A Particle ollision in centre-of-mass inertial frame. The collision is elastic, which means that it is reversible in time. If, after the collision, the velocities of the articles are reversed, exactly the same collision will occur in reverse. Take the x-axis as being horizontal in the diagram, i.e. the line of symmetry running through the centre of mass, such that the trajectory of A is the mirror image of the trajectory of (but in reverse time-order) when we reflect in the x-axis. The y-axis is the symmetry axis at right-angles assing through the centre of mass. y conservation of momentum, the x-comonents of momentum of the two articles must be equal and oosite both before and after the collision, as must the y-comonents. 5

23 y symmetry, the x-comonent of velocity of each article before the collision is equal to its x-comonent after the collision. This has to be true because (a) the articles are identical, and there is no reason for one article to seed u while the other slows down and (b) because the collision is reversible (symmetric in time) so we cannot have both articles going either faster or slower than they were before the collision. y similar arguments, the y-comonents of velocity of each article before and after the collision are equal in magnitude, but have oosite signs. Now go to the inertial frame which moves along the ositive x-axis at a seed such that in this frame the x-comonent of velocity of A before and after the collision is zero. all this frame S (see second figure). Particle A u u v u v Particle ollision in S-frame. u In the S-frame, let the x-comonent of velocity of article before and after the collision be called v. eforeandafterthecollision,articlea is moving arallel to the y-axis with y-comonent of velocity that I call u and u (equal and oosite, because they are equal and oosite in the centre-of-mass frame). The y-comonents of velocity of before and after the collision are called u and u. Using the formula for velocities under Lorentz transformations, and the symmetries of the roblem, we can get a formula for u in terms of u as follows: onsider only article, and look at it in the S -frame (see the third icture): u v Particle A u v u Particle u ollision in S -frame. that is, the inertial frame moving along the x-axis to the left relative to the S-frame at seed v. In the S -frame, the x-comonent of velocity of before and after the collision is zero (u x =). Furthermore,they-comonent of the velocity of in the S -frame must be u before and u after the collision, i.e. equal magnitude but oosite direction to the 6

24 y-comonent of the velocity of A in the S-frame. This must be true by symmetry. Hence, u is the y-comonent of velocity in the S-frame (i.e. u y = u )ofaarticlewhich,inthes frame, is moving along the y-axis with velocity u y = u. Now use the relativistic formula relating y-comonent of velocities in di erent frames: u y = u y +u xv/c 2. In the resent case, u x reresents the x-comonent of velocity of article in the S -frame, which is zero, so we have: u y = u = u = u v2 /c 2. (4.32) The crucial oint to areciate here is that u and u are not the same. They-comonents of the velocities of A and are equal in the centre-of-mass frame, but they are not the same in the S-frame or the S -frame, because of the relativistic relations between velocities in di erent inertial frames. If u 6= u, this oses an extraordinary roblem, if we believe that momentum is conserved. The roblem can only be resolved by ostulating that the mass of a article deends on its seed. Here is the roblem: the y-comonent of total momentum before the collision is m A u + m u,andafterthecollisionism A u m u, so that the momentum changes sign in the collision. Momentum can only be conserved if m A u m u =. utsince u 6= u,wemusthavem A 6= m,eventhoughthearticlesareidenticalineveryway. Faced with this roblem, Einstein decided that conservation of momentum is such a fundamental rincile that it must be maintained at all costs. In order to maintain it, he had to say that mass deends on seed Deendence of mass on seed Now we derive a formula exressing how mass deends on seed. Note that we assume that mass deends only on the magnitude of velocity, and not on the individual artesian comonents of the velocity vector. (This must be true if we resect the rincile of relativity that there are no referred inertial frames, and hence no referred directions). Work in frame S. Since the seed of A before and after the collision is u, we write m A = m(u ). The seed of before and after the collision, denoted by w, isw = u 2 + v 2, so we write m = m(w). Then the requirement m A u m u =canbeexressedas: u m(u )=u m(w). (4.33) Putting the formula from eq.(4.32) into eq.(4.33), we have: where I have cancelled a factor u on both sides. m(u )=m(w) v 2 /c 2, (4.34) We can now deduce from eq.(4.34) how m must deend on seed. To do this, we consider what haens when u becomes very small. On the left of eq.(4.34), we just get m,the 7

25 mass of the article when it is at rest. On the right, we note that w! v when u!. This gives: m m(v) = v2 /c = m. (4.35) 2 [You should ask: with m(v) given by eq.(4.35), is eq.(4.34) exactly satisfied even when u is not small? The answer is yes: you might like to try verifying this.] Eq.(4.35) shows that the mass of a article increases with seed, and (if m 6=)tendsto infinity as the seed aroaches the seed of light Total energy of a moving body In Newtonian hysics, a force F does work on a body, causes it to accelerate or decelerate, and changes its kinetic energy. In one dimension, if the body moves through distance x, theworkdoneisf x,andthisisequaltothechangeofkineticenergy K. If the body is moving with seed v, then x = v t, sotherateofchangeofkineticenergyis dk/dt = Fdx/dt = Fv. ut the force is the rate of change of momentum: F = d/dt. The momentum is = mv. In Newtonian hysics, m is constant, so that d/dt = mdv/dt. Putting these facts together, we have: dk dt = v d dv = mv dt dt = d dt 2 mv2. (4.36) This shows that the kinetic energy is equal to 2 mv2,ifwerequirethatk =whenv =. Now we assume that we can aly the same argument in the context of relativity. It is still true that dk dt = v d dt = v d (mv), (4.37) dt but in order to evaluate d(mv)/dt we have to remember that m deends on v. Using the formula m(v) =m / v 2 /c 2 (eq. 4.35), we have: d(mv) dt = d m v dt v2 /c " 2 v 2 /2 = m + v2 c 2 c 2 # v 2 3/2 dv c 2 dt = m v 2 c 2 3/2 dv dt. Hence, from eq. (4.37): dk dt = m v v 2 c 2 3/2 dv dt = d m c 2 dt v2 /c = d 2 dt 8 mc 2. (4.38)

26 Now we integrate this equation to give: K = mc 2 + A, where A is an arbitrary constant. As in the Newtonian case, A is determined by requiring that K =whenv =. utwhenv =,m = m,therestmass,sothat: K = mc 2 m c 2 (4.39) This looks totally di erent from the Newtonian exression K = 2 mv2,butinfactreroduces this formula if v/c is very small: K = mc 2 m c 2 = m c 2 v 2 /2 m c 2 c 2 Eq.(4.39) can be rewritten as: = 2 m v 2 +termsof order(v/c) 2. mc 2 = m c 2 + K. (4.4) Einstein roosed a radical interretation of eq He said the total energy E in a body is mc 2. This means that even when a body is stationary it contains a rest energy m c 2. When the body moves, its energy E = mc 2 increases because its mass m increases. The increase of energy is the kinetic energy, reresented by the di erence mc 2 m c 2. This is the origin of the equation E = mc 2. This formula reresents the total energy of a body, both the kinetic energy and an intrinsic energy that exists simly because the body has a mass. Put another way, we can say that energy and mass are the same thing, aart from a factor of c 2 (which is just a matter of the units used to secify E and m): Whatever has mass has energy, and whatever has energy has mass Internal energy and mass The imlications of E = mc 2 are very remarkable. The equation imlies that any form of energy at all is associated with mass. For examle, if you comress a sring (e.g. when you wind u a clock), energy is stored in the sring. According to E = mc 2,acomressed sring has a larger rest-mass than a relaxed sring. If a comressed sring and a relaxed sring move at the same seed v, thecomressedsringhasagreatermomentum = mv. When you heat u an object, its mass increases. When you charge u a battery, you increase its mass, not because you add material to it (which you don t) but because you add energy to it. If hydrogen and oxygen combine to form water, the number of rotons, neutrons and electrons stays exactly the same, but the total mass decreases, because heat was given o. If two deuterons combine to form a helium nucleus, the number of rotons and neutrons stays the same, but the total mass decreases. 9

27 Now let s be scetical! Suose you don t believe that a comressed sring has a bigger mass! I will convince you by discussing another collision roblem. This time, we have an inelastic collision between two identical bodies A and, each having rest mass m.intheinertialframeinwhichthecentreofmassisstationary,theyaroach each other with equal and oosite velocities v and v along the x-axis. When they collide, they stick together, and the combined body A is stationary, by symmetry. Let s suose that the kinetic energy lost goes into comressing srings inside the bodies. According to Einstein, the total energy before the collision is E =2mc 2 = 2m c 2 v2 /c 2. Since energy is conserved, Einstein says that this E must be equal to m A c2,wherem A is the mass of the combined object. This requires that m A is not equal to 2m ;insteadit is equal to m 2m A = v2 /c. 2 Einstein claims that it is greater than 2m because the energy stored in the srings has a mass. Now we ll test this claim by considering the collision in the inertial frame in which body A is initially at rest. In this frame, the initial velocity of body is u = 2v +v 2 /c 2, (4.4) where we have used the usual formula for relativistic addition of velocities. velocity of the combined body A in this frame is v. From this, the initial momentum, as seen in this inertial frame, is: The final um(u) = m u2 /c 2v 2 +v 2 /c. 2 The momentum after is: m A v2 /c 2 v, where m A,asbefore,istherest-energyofthecombinedbody. Momentumconservation therefore demands that: m A v2 /c 2 = 2m u2 /c 2 ( + v 2 /c 2 ). If we now insert the formula for u (eq.4.4) into the right-hand side, after a little algebra, we obtain: m 2m A = v2 /c. 2 2

28 This means that Einstein must be right! The rest mass of the combined article cannot be 2m.Ifitwas,momentumwouldnotbeconserved. This confirms that a comressed sring must have a greater rest mass than a relaxed sring, and the contribution of the stored energy to the mass is correctly given by E = mc 2.ut if the energy stored in a sring has mass, then all other forms of energy must have a mass too. In the collision roblem just considered, we could just as well have suosed that the kinetic energy lost goes into heating the bodies, charging u batteries, breaking u water into hydrogen and oxygen or breaking alha-articles into deuterons Relativistic Energies Relativistic e ects are most imortant when articles travel at seeds comarable with the seed of light, or when their interaction energies are comarable with their rest energies. The rest energy m c 2 of an electron is aroximately.5 MeV ( MeV = million electron volts). This is enormously greater than the energies that characterise chemical reactions (usually a few ev). The energy region of MeV and uwards is the domain of nuclear and article hysics, and is recisely the region where the total energy of the article can be much bigger than its rest energy. I will illustrate the alication of relativistic ideas using some worked examles from nuclear and article hysics. Other areas of hysics where such high energies are imortant include accelerator hysics and cosmology. As an illustration of the very large interaction energies that lay a role in nuclear hysics, consider the fusion reaction in which two deuterons combine to form a helium-4 nucleus: 2 2 H! 4 He. The deuteron 2 H, a heavy isotoe of hydrogen, consists of a roton and a neutron bound together by the strong nuclear force. The helium-4 nucleus 4 He consists of two rotons and two neutrons bound together. The rest-mass of 4 He is 4.26 u. (The atomic-mass unit u is defined so that the mass of the 2 atomisexactly2u.)themassofthedeuteron 2 H is 2.4 u. You can see from that the mass of 4 He is slightly less than twice that of 2 H: =.256. The discreancy results from the mass equivalent of the energy released in the reaction 2 2 H! 4 He. From the formula E = mc 2,theenergyreleasedis: E = mc 2 = (3 8 ) 2 = J=23.8MeV. Thus we obtain an energy that is enormously bigger than the energies of chemical reactions Energy exressed in terms of momentum In Newtonian hysics, one often writes the kinetic energy as K = 2 /2m. Inrelativitytoo, it is often convenient to exress the energy in terms of the momentum. Let us start from 2