1 The beginnings of relativity

Transcription

1 Physis 46 Fall 26 Susan M. Lea The beginnings of relativity The priniple of relativity was first expressed by Galileo in the 7th entury: If tworeferenefranes moveat onstant relative veloitywith respet to eah other, theresults of any physis experiment onduted in either frame will be the same. Equivalently, thelaws ofphysis areindependent of refereneframe. Galileo used this priniple tounderstand projetilemotion (free fall under gravity ombined with onstant horizontal veloity). But at the beginning of the 2th entury, Einstein notied that there were di ulties in applying this priniple to E&M.. Einstein imagined riding along with an EM wave by moving to a frame travelling at speed with respet to the original frame. He notied that the osillating fields he would apparently see in this frame were inonsistent with Maxwell s equations. 2. Maxwell s equations predit a fixed speed for light: = / p µ ε = 3 8 m/s. But Galilean relativity predits that an observer in a frame moving at speed v with respet to the "lab" frame, and parallel to the diretion of propagation, should measure wave speed v. This is what happens with sound waves, for example. (See LB 6.4.) 3. There is a simple experiment that we use to demonstrate Faraday s Law. Take a oil and push a bar magnet toward it (LB 3.). As the magneti flux through the oil inreases, there is an indued eletri field that drives urrent in the oil. But what happens if we hold the magnet fixed and push the oil toward themagnet? Aordingto the priniple of relativity, we get the same urrent. But in this ase the fore driving the urrent is part magneti, sine theeletrons in the oil now have a non-zero veloity through the magneti field, and part eletri due to indued eletri field. This appears to violate the priniple of relativity. It would appear that we need to throw out either the priniple of relativity or Maxwell s equations. But Einstein realized there was another alternative: we an rethink our ideas of spae and time. He started with the following postulates of Speial Relativity:. The speed of light is onstant and equal to / p µ ε independent of referene frame. 2. The laws of physis are independent of (inertial) referene frame.

2 The wholetheory follows from these twopostulates. We shall disover that some physial quantities are invariant: that is, they have the same value in all referene frames. These are obviously very important quantities, and we will want to fouus attention on them. Invariant quantities inlude the speed of light (postulate ), as well as the mass and harge of a partile. Relative quantities have di erent values in di erent referene frames. They inlude eletri and magneti fields, as well as most distanes and times.. Relativity oftime intervals A proper understanding of relativity requires that we understand the proess of measurement. Physial proesses are defined by a series of events. An event is defined by its spatial oordinates and the time that it ours its time oordinate. Now onsider an experiment to measure time intervals. We design a lok by generating a pulse of light that travels from its soure to a mirror and bak to areeiver o-loated with the soure. The distane from soure to mirror is l, and so the time interval between the two events (a) pulse is generated and (b) pulse is reeived is t = 2l/. Now we plae our lok on a train travelling at speed v with respet to the original, or "lab", frame. The piture of the events as seen from the lab is as follows: The pulse is emitted at time t, and refleted at time t 2, both measured in the lab frame. But the mirror has moved a distane L = v (t 2 t ) during this time, and so the light pulse has to travel a distane s = p l 2 + L 2 to reah the mirror, and thus takes a time t 2 t = s 2

3 The return trip is similar, giving an observed lok interval of t = t 3 t = 2(t 2 t ) = 2s Squaring and simplifying, we get µ " t 2 = µ t 2 2 l 2 + v or where t 2 γ = = l p v 2 / 2 t = γ t () p β 2 ; β = v Gri ths desribes equation () with the phrase "moving loks run slow", but this is inredibly misleading, as the phenomenon has nothingto dowith thelok per se, and I strongly suggest that you forget it immediately. The important issue is: The smallest time interval between two events is measured by an observer who sees both events happen at the same plae in her/his referene frame. All other obervers in other frames measure a longer time interval between the two events. This smallest time interval is alled the proper time interval between the two events. As we shall see, the proper time interval is an invariant, while the oordinate time interval is a relative quantity..2 Length ontration The measurement of lengths is a little more ompliated, but is losely related to the measurement of time intervals. We an see this by setting up an experiment to measure the length of a roket. Engineers on board the roket measure its length in the usual way using stationary meter stiks, and get a result l. (An engineer at eah end of the roket reads the meter stik at the same time.) The roket is moving at speed v past a spae station. How an engineers on board the station measure its length? Simple: they note the time t that the front end passes them, and the time t 2 that the bak end passes, and onlude that l = v (t 2 t ) The engineers on the roket see the spae station fly past them, and also note the times: t when the station passes the nose one and t 2 when the station passes the tail fins. They onlude that (2) t 2 t = l v 3

4 Now whih observers see both events at the same plae? engineers! So their time interval is the shorter one. The spae station and thus and thus t 2 t = γ (t 2 t ) l v = γ l v l = l γ (3) The roket appears shorter to the observers on the spae station, who see the roket moving by them. This e et is alled length ontration. It e ets lengths measured parallel to the relative veloity v. Lengths perpendiular to v are not a eted. To seewhy, onsider atrain moving along rails separated by a distane w. If w were a relative quantity, then we d have a dilemma. Suppose the observers on the train see the separation of the moving traks ontrated to a distane w < w. Then the train s wheels would fall o the trak to the outside as the train aelerates. But if the observers on the ground see the moving train s axle ontrated to a distane w < w, the train s wheels would fall o the trak to the inside. Both an t be true, so we onlude that w = w. Again, I strongly suggest that you avoid Gri th s misleading statement that "moving objets are shortened". Nothing happens to the objets: the proper length of an objet is an invariant, and length ontration is a purely observational e et. i The greatest length of an objet (its proper length) is measured by observers who observe the objet to be at rest..3 Relativity of sumultaneity Along with these e ets we find another related, and perhaps stranger, e et. Whether twoevents are simultaneous depends on theobserver. Tosee why this happens, onsider Einstein s example. A train of proper length l is moving at speed v with respet to the ground ("lab frame"). An observer, Oliver, on the ground sees two lightning bolts hit the ends of the train simultaneously, just as the enter of the train passes him. How does he measure this? Well, the light from the bolts reahes him at the same time t, so he onludes that the bolts hit eah end at time t hit = t l/2 γ Remember: he oberves the train s length to be ontrated, hene the γ in the denominator. Now let s onsider the events as observed by Penny, sitting on the middle of the train. Light from the front bolt reahes Penny at time t after it struk 4

5 the train. During the travel time t, Penny advanes a distane v t toward the bolt strike, and so the distane travelled is not l/2 but l/2 v t, and so and thus Similarly, for the rear bolt we get t = l 2 v t t = l 2( + β) l t 2 = 2( β) Thus Penny observes the front hit before the rear hit. But if Penny observes the front hit at t, and knows that the front end is a distane l/2 from her, she onludes that the bolt struk at t hit, front = t l and if she observes the bak hit at t 2 > t, she onludes that t hit, bak = t 2 l Sine these times are not equal, she onludes that the two bolts did not hit simultaneously! The sequene of events is shown in the drawing below. 5

6 2 The mathematis of Spaetime To desribe an event in spaetime, we need four oordinates: x, y, z and t. We an all t the fourth oordinate, or, more frequently, we use x = t, x = x, x 2 = y, x 3 = z where t is the zeroth oordinate. The values of these oordinates are di erent for di erent observers. But some ombinations are invariant. For example, onsider alight wavethat starts at theorigin (of spaetime! t = x = y = z = ). After a time dt it has travelled a distane dt : q dl = (dx) 2 + (dy) 2 + (dz) 2 = dt and thus the ombination ds 2 = 2 dt 2 dl 2 = (4) and this must hold in any referene frame beause the speed of light is the same in every referene frame (postulate). The quantity ds is alled the di erential spae-time interval between two events on the light ray. It is an invariant. When ds =, as here, the interval is lightlike. Timelike intervals have ds 2 > and spaelike intervals have ds 2 <. These distintions are also invariant. An interval annot be spaelike in one frame and timelike in another. We an draw a diagram, alled a spae-time diagram, in whih we represent the events as points. It is neessary to suppress at least one spae dimension to draw the diagram. Light rays travel at 45 ± in suh a diagram. These diagrams are useful for getting a visualimage of the situation of interest. Allevents that an be reahed from an event E onstitute the future of that event, and they lie on or within an upward-opening one with opening angle 45 ± and apex at point E. Similarly the past is the set of events from whih E an be reahed. These events lie in a downward-opening one with apex at E. 2. The Lorentz transformation Now we ll look at how the oordinates hange as we move from frame to frame. Sine we expet an objet moving at onstant veloity in one frame (whih we ll allunprime) to be moving at some onstant veloity in any other frame (prime), the transformation must be linear. If this were not true, we would have aelerations (and thus fores) in one frame and not in another. This would violate postulate 2. For simpliity, let s put the x axis along the relative veloity v. Thus, sine we have already established that dimensions perpendiular to v are not ontrated, we expet y = y; z = z But t = At+ Bx x = Dt + Ex 6

7 Now we use the invariane of the spae-time interval : ds 2 = (dt ) 2 (dx ) 2 (dy ) 2 (dz ) 2 = (dt) 2 (dx) 2 (dy) 2 (dz) 2 Next insert our assumed linear transformation: (Adt + Bdx) 2 (Ddt + Edx) 2 (dy) 2 (dz) 2 = (dt) 2 (dx) 2 (dy) 2 (dz) 2 This statement must be true for all values of dt, dx, dy, dz. So first let dx = dy = dz =. then A 2 D 2 = (5) Next let dt = dy = dz =. B 2 E 2 = (6) Finally with dy = dz =, we get A 2 D 2 (dt) 2 + B 2 E 2 + dx 2 + (AB DE) 2dtdx = and using the results (5) and (6), this simplifies to Thus B = DE/A and inserting this into (6) we get and then using equation (5) we get AB = DE (7) µ D E 2 2 A = 2 so E 2 A 2 = E = A But if A were negative, time would be running bakward in the prime frame, and that an t be. Similarly, if E were negative, distanes would be inreasing in opposite senses in the two frames this would be inonvenient to say the least. So we require both E and A to be positive, and thus E = A. Then we also have B = D, and our transformation has simplified: t = At + Bx x = Bt + Ax Now onsider an objet at rest at the spatial origin of the prime frame: x = y = z = for all t. As measured in the unprime frame, this objet has oordinates x = vt, y = z = 7

8 And thus and then from (5) we get So finally our transformation is: Lorentz transformation: x = At + Bvt = ) A = Bβ A 2 β 2 = A = p β 2 = γ t = γ(t βx) x = γ(x βt) (8) As β!, γ! and we get bak the Galilean transformation x = x vt In a spae-time diagram for the unprime frame, the prime axes are sloping lines making angle θ with the x and t axes, where tan θ = β The new spae and time axes oinide along a 45 ± line as β!, thus we find that we annot have a frame travelling at a speed greater than with respet to the first. Use this diagram to show that events simultaneous in the unprime frame are not simultaneous in the prime frame, and vie versa.. Now let s revisit time dilation Two events have oordinates (t,x,,) and (t 2,x 2,,) in the unprime frame. In the prime frame, the oordinates are t = γ (t βx ) 8

9 and t 2 = γ (t 2 βx 2 ) If x 2 = x, the time interval between events in the prime frame is t = t 2 t = γ (t 2 t ) = γ t The events our at the same plae in the unprime frame, and so the time interval in the prime frame is longer. But suppose x 2 6= x. Then x = γ (x βt ) x 2 = γ (x 2 βt 2 ) and now let s require that x 2 = x : the events our at the same plae in the prime frame. x βt = x 2 βt 2 ) x 2 x = β(t 2 t ) Then t = t 2 t µt = γ 2 t β (x 2 x ) = γ µ t 2 t β β(t 2 t ) = γ (t 2 t ) β 2 = t 2 t γ Now the interval is shorter in the prime frame, as expeted. 2.2 The veloitytransformation The veloity of a partile measured in the prime frame is µ dx u = dt, dy dt, dz dt Let s look at the x omponent first. have u x = while for the y omponent, we get = u y = dy dt = Using the Lorentz transformation, we γ (dx βt) γ(dt βdx/) = dx/dt β (β/) dx/dt u x v βu x / = u x v β u/ dy γ (dt βdx/) = u y ³ γ β u/ (9) () It is easy to hek that we get bak the right Galilean results as β!. But look at what happens if u x! (and then of ourse u y = u z = ) u x = v β = 9

10 This result is required by postulate. It s a little more interesting if u y!, u x = Then we get u y = γ ; u x = v and j u j = q u y 2 + (u x ) 2 = s 2 γ 2 + v2 = p 2 v 2 + v 2 = 2.3 Metris and 4-vetors We an write the oordinates of an event in spaetime as the omponents of a 4-dimensional vetor: r = (t,x,y,z) Then a di erential displaement d r = (dt,dx,dy,dz) with magnitude equal to the spae-time interval (4): d r d r = 2 dt 2 dx 2 + dy 2 + dz 2 This is almost the usual rule for finding dot produts, exept for a sign hange. The metri for spaetime is g = A and we have d r d r = d r g d r = dr µ g µν dr ν () where we an use matrix multipliation to evaluate the produt on the right. We have disovered the rule for taking dot produts of four vetors. A useful way to remember this is: Dot produt = produt of time omponents - sum of produts of spae omponents Just as the dot produt of a 3-vetor is a salar, so here the dot produt of two 4-vetors is an invariant. Notie that in equation () we wrote the index on the 4-vetor above the symbol, and the index on the metri tensor below the symbol. This is important. In the tensor mathematis of spaetime, it is only permissable to sum over a repeated index, if one of the two indies is up and one is down.

11 The Lorentz transformation an also be done using matrix methods: r = r where = In index notation, we write γ γβ γβ γ r µ = µ νr ν One we have established the orret mathematis for these 4-vetors, we an write all of the physis of SR in a very ompat form. The spae-time interval for a partile moving at veloity v = v^x is A ds 2 = 2 dt 2 v 2 dt 2 = 2 dt 2 and if v =, ds = dt. So the proper time is just dτ = ds = dt γ where ds is timelike. This motivates the definition of the 4-veloity. If we di erentiate the 4-position with respet to the invariant proper time, we get γ 2 (2) v = d r dτ = γd dt (t, r) = γ(, v) (3) where v is the usual 3-veloity measured in a speifi referene frame. Now if we apply the rule we have disovered for taking dot produts of 4-vetors, we get v v = γ 2 2 v 2 = 2 whih is ertainly an invariant! The Lorentz transformation may be used to transform any 4-vetor. Let s do the veloity transformation. Let u be the veloity of a partile in the unprime frame. Then where γ = u = p β 2, γ γβ γβ γ β = v B and γ u = γ u γ u u x γ u u y γ u u z A p u 2/ 2

12 v = v ^x being the relative veloity of the two frames. Thus γγ u ( βu x ) γ u u = B γγ u (u x γ u u y A = B γ u u γ u u A y γ u u z γ uu z So that, from the time omponent, and from the x omponent γ u = γγ u ( βu x /) (4) u x = γγ u (u x β) γ u = u x v ( βu x /) in agreement with (9). I ll leave it to you to hek the perpendiular omponents. One we see how to get the 4-vetors, the rest of physis gets easy. Remembering that the mass of a partile is an invariant, we may form the 4- momentum p = m v = γm (, v) = (γm, p) where the orret relativisti expression for the 3-momentum is p = γm v. (See LB p89 for a proof of this.) The invariant magnitude of the 4-momentum, squared, is p p = γ 2 m 2 2 v 2 = m 2 2 = E 2 2 (5) the square of the rest energy E = m 2 of the partile, divided by 2. This is also an invariant, sine both m and are. The 4-momentum is p = µ E, p where the total energy of the partile is E = γm 2 m 2 = p v 2 / 2 µ = m 2 + v 2 µ µ 3 v ! 4 + = m mv2 + 3 v = rest energy + Newtonian kineti energy + relativisti orretions We an verify this result as follows. First define the 4-aeleration a = d v dτ 2

13 and the 4-fore Then the dot produt F = m a = d p dτ F v = m d v dτ v = m d ³ v v = m 2 dτ 2 d dτ 2 = sine is a onstant, independent of proper time. But if we do it using omponents and oordinate time, we get F v = mγ d dt µ = mγ 2 dγ dt, d p dt [γ (, v)] γ (, v) (, v) = γ 2 µ m 2 dγ dt v F Sine we have already established that the dot produt is zero, we get P = F v = d dt γm 2 = de dt The power expended by the fore equals the rate of hange of the partile s energy. The momentum 4-vetor allows us to ombine two laws of physis onservation of energy and onservation of momentum into one onservation law: onservationof 4-momentum. For an isolated system, (6) p total,before = p total, after If we ombine this with the invariane of p p we an solve some interesting problems quite easily. The Bevatron at Berkeley was designed to produe antimatter through the reation p+ p! p + p+ p+ p where p is a proton and p is an antiproton. To what energy must a proton be aelerated so that this reation will produe antiprotons in a ollision with a stationary proton? In the ollision, both energy and momentum are onserved. In the lab frame a proton with speed v ollides with a stationary proton, and the reation produts arry away the initialmomentum γmv. For the energy required to be a minimum, the reation produts havenointernal energy, that is, eah partile has the same veloity v final. Viewed in the M frame (enter of momentum, in 3

14 relativisti physis), the two protons approah with equal and opposite veloities, and the reation produts are at rest. We ould do the problem using the Lorentz transformation, transforming to the M frame, but there s an easier way. We an alulate the invariant magnitude of the total 4-momentum in the lab frame before the ollision and in the M frame after the ollision. energy momentum invariant E 2 p 2 2 inoming proton, lab frame γm 2 γmv stationary proton, lab frame m 2 Total, lab frame m 2 (γ + ) γmv m 2 2 (γ + ) 2 γ 2 m 2 v 2 2 reation produts, M frame 4m 2 4m 2 2 Thus, setting invariant totals before equal to totals after m 2 2 h (γ + ) 2 γ 2 β 2i = 6 m 2 2 Thus γ 2 β 2 + 2γ + = 6 2(γ + ) = 6 With γ = 7, the kineti energy of the proton is γ = 7 T = γm 2 = 7(.938 Gev) = 5.6 GeV The betatron was designed to produe 6.4 GeV per proton. 2.4 Transformation law for F The 4-fore is F = d p dτ = γ d µ µ E dt, p de = γ dt, F where F is the 3-fore, and, from (6) with β = u/, So Applying the Lorentz transformation de dt = β F F = γ³ β F, F F = F So, with = / p v 2 / 2, and v = v^x ³ γ Fx = γ F x v β F γ F z = γf z ³ = γ F x v β F 4

15 We have already found γ (equation 4), so Fx = v βx ³ F x v β F (7) If F is perpendiular to β in the unprime frame, then F x = F x For the y and z omponents the transformation is F z = v (8) βx F z β x v (9) In thespeialasethat the partile is instantaneously at rest in the unprime frame, then F x = F x, F? = F? 3 Relativisti E&M The transformation laws for eletri and magneti fields are interesting. Suppose we have a uniform sheet of harge with uniform harge density σ lying in the x z planethere is a onstant eletri field E = σ^n/2ε above the sheet. A partile with harge q plaed above the sheet experienes a fore F = q E that aelerates it away from the sheet. If we now move to a referene frame moving with speed v parallel to the sheet, v = v^x, we see things di erently. In this frame there is a surfae urrent density K = σ v, and a magneti field B = µ K ^n/2. The harge also moves with a veloity v in this frame and thus experienes a fore due to both the eletri and magneti fields: ³ E + v B F = q µ µ σ^n = q v µ σ v ^n 2ε 2 = q σ ^n + µ ε v ( v ^n) ^nv 2 ª 2ε = q σ 2ε ^n µ v2 2 = qγ σ 2 ^n 2ε The total harge Q on an area A = l w in the unprime frame is the harge Q on an area A = l w = lw/γ in the prime frame. So σ = Q A = γ Q A = γσ 5

16 Thus F = q σ ^n γ 2ε Sine F is perpendiular to v, we may use relation (9) to ompute F from F : F = q σ ^n γ 2ε The two expressions agree. What this shows is that an eletri field in the unprime frame transforms to a ombination of eletri and magneti fields in the prime frame. In this ase E is perpendiular to v, and ³ E = γ E + v B (2) ³ B = γ B v E/ 2 (2) It turns out that the omponents of E and B parallel to v are unhanged. It is interesting that the omponents of E and B get mixed together in the transformation law. We an see from this that neither E nor B an be parts of a four-vetor, sinethe Lorentz transformation mixes up omponents of asingle vetor but does not ombine omponents of di erent vetors. We need a bigger unit a field tensor. A salar (invariant) is a single number: = 4. A vetor has 4 omponents, so we need 4 numbers. The next objet up the sale is a rank 2 tensor with 4 2 = 6 omponents. We an write these numbers using a 4 4 matrix. We an figure out what these omponents are by working from the potentials. The salar and vetor potentials form a 4-vetor potential: A = µ V, A and the 4-dimensional gradient vetor is = µ t, r so that the 4-divergene of the 4-vetor potential is (using our rule for dotproduts) A = V 2 t + r A = This is the Lorentz gauge ondition, and it is an invariant. This is what makes the Lorentz gauge ondition so useful: if it is true in one frame it is true in all. Now ³ r A B x = = A z x y A y z B x = 2 A A 2 while E x = V x A x t = A A 6

17 This suggests that we formour field tensor from theantisymmetri omponents This leads to F µν = F µν = µ A ν ν A µ The transformation law for this tensor is E x / E y / E z / E x / B z B y E y / B z B x E z / B y B x F = F T A (22) or F µν = µ αf αδ δ ν Verify that this rule reprodues equations (2) and (2). Let s look at the invariants we an form from this tensor. The dot produt of two vetors is A µ g µn A ν = A µ A µ where g is the metri tensor, and A µ = g µν A ν is the so-alled ovariant vetor formed from the vetor A. By analogy, we an find the invariants = F µν g µξ g νλ F ξλ = F µν F µν or, in matrix notation, F µν = gf g T µν and is the sum of the produts of all the elements of the two matries.sine the metri tensor is diagonal, g = g T, so E x / E y / E z / gf g T = B B E x / B z B y E y / B z B x E z / B y B x = F µν = E x E y B E x E y E z E x B z B y E y B z B x E z B y B x E z Ex B z B y E y B z B x E z B y B x Note that the spae omponents of the ovariant vetor have the opposite sign from the spae ³ omponents of the ontravariant (or usual) vetor. Thus the ovariant vetor µ = t, r. A A A (23) 7

18 omparing with (22), we see that the top row and first olumn have hanged sign. Thus = 2 µb 2 E2 In an EM wave, B = E/ and so =. Some fields are primarily magneti in harater, >. On the other hand, if <, the fields are of eletri harater. A seond invariant is found from the dual tensor whih is reated using the 4-d analogue of the Levi-ivita symbol. The result is 2 F µν = ε µναβ F αβ 2 = F µν F µν 2 / E B For an EM wave, this invariant is also zero. If > and 2 = there will be some frame in whih B =. If < and 2 = there is some frame in whih E =. If 2 6=, there is no frame in whih either E or B is zero. For the harged sheet we onsidered above, we have, in the lab frame, = 2 µ σ 2 4ε 2 2 = σ2 2ε 2 2 = µ σ 2 2ε and in the prime frame = 2 2 = " µ 2 (K ) 2 4 (σ ) 2 4ε 2 2 = µ2 (σ v) 2 (σ ) 2 µ 2 2ε µ µ 2 = β 2 2 µ γ 2 σ 2 2 2ε = µ 2ε β 2 γ 2 σ 2 = µ σ 2 2ε and 2 is also zero sine E is perpendiular to B. The harge and urrent form another 4-vetor: In the lab frame, for our sheet, ³ j = ρ, j j = (σδ (y), ) 8

19 Verify that the Lorentz transformation of this 4-vetor orretly gives the harge and urrent in the prime frame. Now we may write harge onservation ompatly as j = = ρ t + r j = Finally we get Maxwell s equations: µ F µν = µ j ν and µ F µν = The Lorentz fore on a partile with harge q and 4-veloity v is Let s verify this qf µν v ν = q = q F µ L = qf µν v ν = qf µν g να v α E x / E y / E z / E x / B z B y E y / B z B x E z / B y B x E x / E y / E z / E x / B z B y E y / B z B x E z / B y B x E x γv x + Ey γv y + Ez = q B E x γ + B z γv y B y γv E y γ B z γv x + B x γv z E z γ + B y γv x B x γv y E ³ β E x + v B ³ x = qγ B E y + v ³ y A E z + v B z γv z A B B γ γv x γv y γv z A B Apart from the fator of γ,the 3-vetor part of this 4-vetor is thenon-relativisti expression for the Lorentz fore, while the zeroth omponent is the power, as usual. 4 Motion of a harged partile in a uniform eletri field In Newtonian mehanis, a harged partile in a uniform field experienes a onstant fore F = q E and aelerates at a onstant rate forever. Its speed γ γv x γv y γv z A 9

20 inreases without limit. But Einstein told us that this an t happen. The partile s speed annot exeed. Let s see how the mathematis enfores this result. The seret is to use proper time. Then the fore law is F µ L = dpµ dτ If we put the x axis along the diretion of the eletri field and the y axis so that the partile s initial veloity is in the x y plane, then the equations are: -omponent -omponent 2-omponent 3-omponent qγ E β = d dτ γqe v x µ E = d dτ (γm) qγe = d dτ p x = d dτ (γmv x) = dp y dτ = d dτ (γmv y) = dp z dτ The last two equations tell us that p y and p z are eah onstant. But sine p z is initially zero, it stays zero and we need not onsider it further. The first two equations may be simplified by di erentiatingagain. First, define = qe/m. Then d 2 dτ 2 (γ) = d dτ ( γv x) = 2 γ The solution for γ is γ = Ae τ + Be τ where Then A + B = γ = γv x = p v 2 / 2 d dτ γ = Ae τ Be τ Suppose the partile is moving perpendiular to E at t = τ =. Then So A = B = γ 2 γ = γ osh τ v x = tanh τ 2

21 Then finally so For short times, τ, we have γmv y = γ mv v y = v osh τ v x = τ and the speed inreases linearly in time, with v y remaining onstant. But for τ large, we find tanh τ!, and so vx! and vy!. The latter result is initially surprising, sine there is no fore in the y diretion. But sine the partile s γ is ontinuously inreasing, and the momentum omponent stays fixed, the veloity omponent must derease. To express the results in terms of oordinate time, we integrate again. The omponent of the 4-veloity is γ = d dτ (t) = γ osh τ Thus t = γ sinh τ where we set the integration onstant to zero so that t = when τ =. Then for τ,, t = γ τ, but for τ À, and t = γ 2 e τ t/γ v x = q + ( t/γ ) 2 v y = v q + ( t/γ ) 2 The graph shows β x (blak) and β y (red) versus t/γ, with β y =.5 at t =..8.6 beta omega t/gamma 2