a) What is the duration of the trip according to Ginette? b) What is the duration of the trip according to Tony?

Transcription

1 Ginette stays on Earth while Tony travels towards a star loated 4.6 lightyears away from Earth. The speed of Tony s ship is 80% of the speed of light. fr.depositphotos.om/ /stok-photo-spaeship.html a) What is the duration of the trip aording to Ginette? b) What is the duration of the trip aording to Tony? hdwpapers.om/spaeship_wallpaper_hd-wallpapers.html Disover how to solve this problem in this hapter.

2 The Debate on the Motion of the Earth Signifiant Effets Aording to Prenewtonian Physis Before the 17 th entury, very few people ould aept the idea of the Earth moving around the Sun. Many opposed this theory beause of the effets we were supposed to observe aording to the physis of the time if the Earth is revolving around the Sun. In prenewtonian physis, speed is assoiated with fore. This means that a fore must be applied on an objet in order to move it. To illustrate the onsequenes of suh a physis, let s imagine that someone drops a stone on the ground while the Earth is moving. As long as the stone is touhing the hand of the person, it ould be argued that the hand of the person is exerting a fore on the stone that allows it to move at the same speed as the Earth. However, when the person drops the stone, there would be no longer any horizontal fore ating on the stone. Aording to prenewtonian physis, this means that the stone would no longer be able to move forwards. The stone would, therefore, lose its horizontal speed while the Earth would ontinue its motion. If it is assumed that the duration of the fall of the stone is 0.6 seonds, then we would have the following situation. slid.es/tofergregg/gravity-and-fluid-dynamis/fullsreen#/ Thus, the person on Earth sees the stone fall behind him. Atually, the stone would fall very far behind sine the Earth is moving at nearly 30 km/s. If it took 0.6 seonds for the 018 Version 10-Relativity

3 stone to fall, then the Earth would have moved 18 km while the stone remained at the same x-oordinate. Therefore, the stone would fall 18 km behind the person. As objets do not fall far behind us when we let them fall, this showed that the Earth is not moving (aording to the physis of the time whih assoiates fore and speed). With similar arguments, a series of effets that should be observed if the Earth is moving an be predited. For example, there should be a ontinuous wind on the surfae of the Earth. The air surrounding the Earth annot move with the Earth beause it is hard to imagine how a fore ould push the air to move it. So, the Earth is supposed to be moving in air at rest, whih would give the impression that there is a ontinuous wind in the opposite diretion of the motion of the Earth. It is ertain that the presene of suh a ontinuous 30 km/s wind would be notied. Galileo s Arguments in Favour of the Motion of the Earth Galileo was the first to find onvining arguments to show that the Earth an be moving without generating any dramati effets. The exat reasoning of Galileo will not be followed here beause he used prenewtonian physis with impregnated fores, but the onlusions will be the same. Imagine what happens when Valteri drops a stone from the top of the mast of a ship when the ship is at rest. In this ase, the stone strikes the abin just above the last window. Now, let s look at what happens if Valteri drops a stone from the top of the mast when the ship is moving. Some, intuitively using prenewtonian physis, would say that the stone will fall a little towards the bak of the ship beause the ship is moving forward while the stone is falling. This isn t going to happen. When Valteri let go of the stone, the stone already has a horizontal veloity equal to the veloity of the boat. As no horizontal fore ats on the stone, it always keeps this same horizontal speed while falling. The stone is thus moving forward at the same speed as the boat and the resulting motion is shown in the next figure (where the boat is moving quite fast). 018 Version 10-Relativity 3

4 ww.relativityoflight.om/chapter5.html First of all, it an be noted that the objet hits the boat at exatly the same plae as it did when the boat was at rest: above the last window. Viewed from the boat, the fall is in fat idential to what was seen when the boat was at rest: as the stone falls, is always stays at the same distane from the mast and hit the abin above the last window. Using suh arguments, it is possible to argue that it is possible be on planet moving at a onstant speed without seeing any effets. Let s use the example of the person dropping a stone on the surfae of the Earth to show this. slid.es/tofergregg/gravity-and-fluid-dynamis/fullsreen#/ When the stone is in the hand of the person, the hand exerts no horizontal fore on the stone. As the stone is going at the same speed as the Earth and as this speed is onstant, there is no aeleration and, therefore, no fore. 018 Version 10-Relativity 4

5 When the stone falls, there is still no horizontal fore ating on it. This means that there is no horizontal aeleration and that the horizontal speed of the stone remains the same. The x-omponent of the veloity of the stone is always idential to the speed of the Earth, and the stone advanes at the same rate as the Earth. This means that the stone will hit the ground in front of the person, exatly under the hand of the person. The same thing would happen is the Earth was not moving. In fat, it an be shown that everything that happens on a planet moving at a onstant speed is idential to what happens on a planet at rest. With suh arguments, the idea of a moving Earth beame a bit more popular (but the majority of the sientists of the time still did not aept that the Earth is moving before the disovery of Newton s laws). The Relativity Priniple These arguments are the basis of the priniple of relativity. It was just shown that a person on the ship sees the stone fall exatly in the same way whether the boat is at rest or moving at a onstant speed. This means that there are no new effets that appear when moving at a onstant speed. Everything happens as if the person were at rest. For example, we know how to pour a up of offee in our house (whih is at rest). If, one day, you beome a flight attendant and you must pour offee into a up in a plane moving at a onstant speed, do not hange your habits: pour the offee on the plane in motion exatly the same way as you do it at home. There is no need to pour the offee a bit in front of the up, thinking that the offee will move a little towards the rear of the airraft beause of its motion. Pool (the game) is played exatly the same way on a plane travelling at a onstant speed as in a pool hall. There is no orretion to do beause of the motion of the plane. In this video, a ball is dropped from a moving truk. A board was installed on the side of the truk to give a referene for the motion of the ball. The ball stays in the middle of the board as it falls, exatly as what would happen if the ball was released while the vehile was stopped. All this also means that if an observer is loked inside a plane in motion at a onstant speed without any window, it is impossible for the observer to tell whether the plane is moving or not. If no new effet appears when in motion, everything appears exatly the same way whether the airraft is moving or not. No experiment an show that it is moving. This also means that if you lose your eyes in a ar in motion (if you re not driving of ourse), you won t be able to say whether the ar is moving or not. You surely think that any moron an tell whether the ar is moving or is stopped, and that this is not true. Atually, it is true. You would know that the ar moves beause the bumps on the road will shake the ar a little. However, we re saying here that the motion is undetetable if the 018 Version 10-Relativity 5

6 speed is onstant. If that is the ase, you must imagine that you are travelling on a road without any bumps beause these bumps aelerate the ar in every diretion. So you must imagine that you are travelling on a freshly paved road and that there is no bump whatsoever. Then, it beomes more diffiult to tell if the ar is moving. You still hear the air passing along the ar, but this ould be a wind blowing on the ar at rest... It would be more onvining with a spaeship. Then, there would be no sound from the wind or the road and it would be impossible to tell whether the spaeship is moving or not if your eyes are losed. The final argument that should onvine you that nothing hanges when moving at a onstant speed is the fat that we are a planet moving around the Sun at 9.8 km/s and nothing peuliar happens. Even better, the Sun revolves around the entre of the Galaxy at 40 km/s, and we do not notie this motion. Everything happens as if the Earth were at rest. If there are no new effets appearing at a onstant speed, it is beause the laws of physis are the same for a moving observer. There are no new fores that ome into play, whih means that the laws are exatly the same. This brings us to the priniple of relativity. The Relativity Priniple The laws of physis are the same for all the observers moving at a onstant speed. Desription of a Phenomenon by two Observers The priniple of relativity has an interesting onsequene: a physial phenomenon an be desribed by taking the point of view of two different observers in motion relative to eah other using the same laws of physis. For example, Joe, on a plane, throws a ball in the air and then athes it Version 10-Relativity 6

7 For Joe, the ball moves in a straight line, upwards and then downwards. If Bob, who is on the ground, look at what Joe is doing, he will say that the ball made a paraboli motion. The desriptions are different, but the law for this phenomenon is the same for two observers: the ball falls down with a 9.8 m/s² downwards gravitational aeleration. Is one desription better than the other? Not at all aording to the priniple of relativity. The desription of the motion of the ball aording to Joe is different from the desription of the motion of the ball aording to Bob, but it is just as good. All observers are on the same footing aording to relativity. Joe will say that the x-omponent of the veloity of the ball is zero, and Bob will say that this omponent is not zero. Who is right? Both are right. The initial onditions of the motion an be different from one observer to another, but the laws of physis are the same. This means that even if Bob and Joe do not agree with eah other on the initial speed, they agree on the fat that the aeleration of the ball is 9.8 m/s² downwards. With Galilean transformation, the desription of a phenomenon aording to one observer an be obtained from the desription of the same phenomenon aording to another observer. Some Definitions An event is something that ours at a speifi time and plae. It ould be an explosion, for example. To desribe an event, the position and the time at whih the event ourred must be given. If I tell you about an event that ourred on April 15, 191, at :0 at position N O in the Atlanti Oean, you know (maybe) that I m talking about the sinking of the Titani. So, an event is noted by its position and time oordinates (x, y, z, t). The event doesn t have to be spetaular. Someone sneezing is an event and the position and time of this event an be noted. An observer is someone who notes the position and the time at whih events our. Observers an be moving relative to eah other at a onstant speed. Aording to the priniple of relativity, no observer is better than another. All observers have different views, but they all are equally valid. Eah observer uses an axes system whih is different from the system used by other observers. These are the referene frames. If an observer is moving at a ertain speed, its axes system moves at the same speed as the observer. Although we don t have to neessarily do it, we an onsider that eah observer is always at the origin of his axes system. Eah observer will, therefore, note the position and the time of the events using his frame of referene. Eah observer notes different oordinates, sine the origins are not at the same plae for eah observer. 018 Version 10-Relativity 7

8 There is a differene here between seeing an event and observing it. To illustrate this differene, imagine that a 5 light-years distant star exploded in 01. (The light-year is a unit of distane whih orresponds to the distane travelled by light in one year. Sine the light travels at 300,000 km/s, this distane is 9.46 x m). Then, it takes 5 years for the light from the explosion to reah us, and the explosion is seen in 017. However, when we see the explosion in 017, we an dedue that the explosion atually ourred in 01. So, we see the explosion in 017 but we observe, with a little math, that it has ourred in 01. (Atually, the alulation an be a little more ompliated than that if the soure is moving. Then, we must find the position of the soure when the light oming from it left. The distane of the soure when the light was emitted must be used, not the urrent distane of the soure.) Transformation Laws Two observers observe an event. Bob is at rest, and Joe moves at a onstant speed v towards the right. Eah of these observers notes the position and the time of the event with different axes systems. en.wikibooks.org/wiki/speial_relativity/print_version Bob s axes system is stationary while Joe s axes system follows Joe and, therefore, travels at a onstant speed v towards the right. Obviously, the two observers will not note the same values for the position of the event sine the origins of the axes systems they use are not at the same plae. They will, therefore, disagree on the oordinates of the position of the event. Fortunately, laws to pass from Bob s oordinates to Joe s oordinates (or vis-versa) an be obtained. If Joe tells Bob the oordinates of the event aording to him, then Bob an alulate the oordinates aording to his own axes with these formulas. 018 Version 10-Relativity 8

9 en.wikibooks.org/wiki/speial_relativity/print_version The distane between the two y-axes always inreases sine Joe moves towards the right. Assuming that the two y-axes were at the same plae at t = 0, the distane between the axes is vt. The distane between the y-axis and the event is x, and the distane between the y -axis and the event is x. Similarly, the distane between the x-axes of both observer and the event gives the values of y (y and y ). From the figure, it is easy to find that x = x + vt y = y These are the transformation laws. They are alled Galilean transformations. They an also be inverted to alulate x and y from x and y. The result is Galilean Transformations x = x + vt x = x vt y = y y = y z = z z = z There is a onvention to determine whih of the two observers notes the oordinates with primes. In our situation, Bob sees Joe travelling towards the positive x-axis. If the point of view of Joe is taken, Bob is travelling towards the negative x-axis. en.wikibooks.org/wiki/speial_relativity/print_version The rule is 018 Version 10-Relativity 9

10 Whih Observer Uses Primes? The observer who sees the other travelling towards the negative x-axis uses primes So, Joe uses primes here. (Assuming, of ourse, that the x-axis is pointing towards the right.) Many interesting results an be deduted from these Galilean transformations. For example, suppose that the observers are wathing a moving objet. The speed of the objet is denoted u (beause v is already used to indiate the speed of Joe with respet to Bob). en.wikibooks.org/wiki/speial_relativity/print_version The veloity of an objet orresponds to the rate at whih its position hanges. Therefore, aording to Bob, the omponents of the veloity are u x dx dy = uy = dt dt Aording to Joe, the omponents are dx dy u x = u y = dt dt If the Galilean transformation x = x + vt is used, the result is x ( + vt) dx d x = dt dt dx dx d vt = + dt dt dt u = u + v x ( ) 018 Version 10-Relativity 10

11 Doing the same for the Galilean transformation for the oordinates y and z, the following laws of transformation for the veloities are obtained. Galilean Veloity Transformations ux = u x + v u x = ux v uy = u y u y = uy u = u u = u z z z z One again, the two observers disagree on the veloity of objets. However, this is logial as the next example illustrates. Example Joe, who is on a train, throws a ball towards the front of the train. If the speed of the ball aording to Joe is 30 m/s and if the train is going at 50 m/s aording to Bob (dressed as a girl here), what is the speed of the ball aording to Bob? The speed of the ball aording to Joe is u x = 30 m/s. (The speed of the objet aording to the observers is always denoted with the letter u. As Joe works with the axes x and y, everything that Joe measures has a prime.) The speed between the two observers is v = 50 m/s. (v is always the speed between the two observers.) We are looking for the speed of the ball aording to Bob whih is u x. (The speed of the objet aording to observers is always u. As Bob works with the axes x and y, everything that Bob measure doesn t have a prime.) Aording to the laws of transformation of veloities, the speed is u = u + v x x m = = Version 10-Relativity 11 s m s m s

12 This makes sense. A ball thrown at 30 m/s towards the front of a train moving at 50 m/s has a total speed of 80 m/s for an observer on the ground who is looking at the train. This video illustrates these veloity transformations. It is no big deal if the two observers do not measure the same speed sine Newton s laws are not entred on speeds. Even if there are some quantities that depend on speed in Newton s physis, the priniples of Newtonian physis are not affeted. For example, two observers will disagree on the momentum of the objets beause the speeds are different, but this does not matter sine they will still agree on the onservation of the total momentum of a system, and this is what is important in Newton s physis. For Newton s laws to be respeted, the observers must agree on the aeleration of an objet. Sine the fores must be idential for the two observers, then the aelerations must also be the same for both observers. A differene would mean that Newton s laws are valid only for one of these two observers and that there would be restritions on the use of the Newton s laws. Let s hek this by finding the aeleration transformation laws. The aeleration of the objet, whih is the rate at whih the veloity hanges, is, aording to Bob, a x du dt x = ay = du dt y Aording to Joe, the aeleration is a du du y = = dt dt x x a y The transformation equation for the aelerations is obtained by deriving the transformation equation for the veloity. For the x-omponent, the result is x ( + v) du d u x x = dt dt du d v x du x = + dt dt dt a = a The derivative dv/dt is zero sine Joe s speed is onstant. Proeeding in the same manner for the transformation of the other omponents of the veloity, the following transformation laws are obtained. x ( ) 018 Version 10-Relativity 1

13 Galilean Aeleration Transformations a = a a = a a = a a = a x x x x y y y y a = a a = a z z z z The omponents of the aelerations are the same for both observers. This means that the fores, whih are, of ourse, idential aording to both observers, give the same aeleration. This implies that Newton s laws are valid for both observers: the one at rest and the one moving at a onstant speed. This means that Newtonian mehanis is onsistent with the priniple of relativity. A person who moves at a onstant speed an apply Newton s laws exatly the same way as a person at rest. The Power of the Relativity Priniple The solution of ertain problems an be simplified, and some interesting results an be obtained from the relativity priniple. Here are two examples to illustrate. A Collision Imagine that there is an elasti ollision between a kg ball moving at 1 m/s and a 1 kg ball at rest. The speed of the balls after the ollision is sought. The onservation of kineti energy and momentum ould be used but relativity will be used instead. Let s hange frames of referene to take the point of view of another observer: Josefina. The speed of this observer is arefully hosen to ensure that the total momentum of the system is zero for her. It is then easy to determine the veloities after the ollision beause the total momentum must remain zero after the ollision. 018 Version 10-Relativity 13

14 Finally, we go bak to Matis s frame to get the veloities that we were looking for. By transferring like that from one frame to another, the equations giving the veloities after a ollision, elasti or not, an be obtained quite easily. Momentum Conservation It will now be shown that momentum must also be onserved if the kineti energy is onserved for every observer in an elasti ollision. If the kineti energy is onserved, then (for the observer using primes) 1 m u + 1 m u = 1 m u + 1 m u If the frame of referene is hanged, the veloities hange. The equation then beomes m1 ( u1 v ) + m ( u v ) = m1 ( u3 v ) + m ( u4 v ) m1 ( u1 v ) + m ( u v ) = m1 ( u3 v ) + m ( u4 v ) m1 ( u1 u1 v + v ) + m ( u1 u1 v + v ) = m1 ( u1 u1 v + v ) + m ( u1 u1 v + v ) m1u 1 m1u 1 v + m1v + mu mu v + mv = m1u 3 m1u 3 v + m1v + mu4 mu4 v + mv m1u 1 m1u 1 v + mu mu v = m1u 3 m1u 3 v + mu4 mu4 v If the energy is onserved in every frame, then the following equation must be true This means that 1 m u + 1 m u = 1 m u + 1 m u m u + m u = m u + m u Version 10-Relativity 14

15 Therefore, the mu² in the equation must anel eah other m1u 1 m u v + m u m u v = m u m u v + m u m u v 4 and we are left with m1u 1 v mu v = m1u 3 v mu4 v m1u 1 v + mu v = m1u 3 v + mu4 v ( m1u 1 + mu ) v = ( m1u 3 + mu4 ) v m u + m u = m u + m u This last line is the equation of momentum onservation. This means that if the kineti energy is onserved for every observer for an elasti ollision, then the momentum must also be onserved. Awesome, isn t it? The Aether and Speed of Light In the last deades of the 19 th entury, some felt that the laws of physis had all been disovered and that it only remains to apply them to explain everything that happens in the universe. At this time, Lord Kelvin stated that only two small blak louds remained in the otherwise blue sky of physis, and that these two problems would soon be solved. These two problems were those related to the speed of light, whih will give birth to relativity, and to the hot body radiation, whih will give birth to quantum mehanis. These two small blak louds ompletely revolutionized physis in the end. Here, let s have a look at the first of these problems, i.e. the speed of light problem. At the end of the 19 th entury, everyone was onvined that light was a wave. However, it was believed that light was a mehanial wave. Aording to the beliefs of the time, light was travelling in a medium alled aether found everywhere in the universe. Light was travelling at 300,000 km/s in this medium. But let s look at what two observers would measure if they observe light propagating through the aether. In this example, Bob is at rest in the aether, and Joe moves at 0.1 in the aether. Both are looking at the light emitted by a soure. In Bob s frame, we have 018 Version 10-Relativity 15

16 Aording to Bob, the speed of light is 300,000 km/s in every diretion. Now, let s look at the same situation by taking the point of view of Joe and alulate the speed of both light beams aording to Joe. Beam travelling towards the right. Speed of the beam aording to Bob: u x = Speed of the beam aording to Joe: u x =? Relative speed of Joe and Bob: v = 0.1 Beam travelling towards the left. u = u v = 0.1 = 0.9 Speed of the beam aording to Bob: u x = - Speed of the beam aording to Joe: u x =? Relative speed of Joe and Bob: v = 0.1 x x u = u v = 0.1 = 1.1 So, this is the situation as seen from Joe s frame of referene. x x 018 Version 10-Relativity 16

17 This result is not really surprising. Aording to Joe, the light going towards the right is slowed by the motion of the aether towards the left, somewhat like a fish swimming against the urrent that goes slower than when there is no urrent. The light going towards the left is driven by the motion of the aether towards the left, giving it greater speed, like a fish swimming in the same diretion as a urrent. So, if you are at rest in the aether, as it is the ase for Bob here, the speed of light will appear to be the same in every diretion. If you are moving in the aether, as it is the ase for Joe here, the speed is different depending on the diretion. Notie that the differene between the speeds of light in opposite diretions aording to Joe is = 0., whih is twie the speed of the aether aording to Joe. This speed differene is always equal to twie the speed of the aether. Mihelson-Morley Experiment The previous reasoning led to the oneption of an experiment to measure the speed of the Earth in the aether. If the Earth is at rest in the aether, the speed of light will be the same in every diretion. If the Earth is moving through the aether, the speed will be different depending on the diretion. The differene between the largest measured speed and the smallest measured speed will be equal to twie the speed of the Earth in the aether. Mihelson did this experiment in 1881 and did it again with Morley, with several improvements, in It was so preise that a speed differene as low as km/s speed ould have been deteted. No variation of the speed of light was measured, a sign that the Earth should be motionless in the aether. So far, no problem. 018 Version 10-Relativity 17

18 The problem omes from the fat that the Earth revolves around the Sun at 30 km/s. It is possible for the Earth to be at rest ompared to the aether at one point, but it annot stay at rest for long. Six months later, the diretion of the motion of the Earth is reversed, and the Earth should then be moving through the aether. So, the experiment was done again 6 months later and the same result was obtained: the speed is the same in every diretion, indiating that the Earth was still at rest in the aether. (Maybe you re thinking that this problem an be settled by assuming that the aether revolves around the Sun at the same speed as the Earth, whih would make the Earth always at rest in the aether. However, this solution was exluded by some other observations, suh as stellar aberration, that were showing that the Earth is moving through the aether.) In fat, all the measurements of the speed of light in vauum at this time were always giving the same value, whatever the irumstanes! It was very surprising. This implies that in the following situation, Version 10-Relativity 18

19 all the observers will measure the same speed for the beam of light. The measurement of the speed of the light beam aording to Joe will give a value of, and not 0.5 as predited by the laws of transformation of veloities. The measurement of the speed of the light beam aording to Nik will give a value of, and not 1.5 as predited by the laws of transformation of veloities. If you try to ath up with the light beam, as Joe does here, the beam will always be 300,000 km/s faster than you, whatever your speed. If you go towards the light, as does Nik here, it will always hit you at 300,000 km/s, whatever your speed. Maxwell s Equations The equations of eletromagnetism, obtained by Maxwell in 1865, predit the existene of eletromagneti waves. When the speed of these waves is alulated, the speed of light is obtained. At the time, this strongly suggested that light is an eletromagneti wave and, at the same time, onfirmed the wave theory of light. There was, however, a small onflit between these equations and relativity. The following situation illustrates this onflit. In this situation, what will happen if Joe travels at the speed of light? Aording to the Galilean equations of relativity, the speed of this beam of light aording to Joe is u = u v = = 0 This is this situation in Joe s frame of referene. x x 018 Version 10-Relativity 19

20 Joe would, therefore, see an eletromagneti wave at rest. However, it is impossible to observe suh a wave at rest aording to Maxwell s equations. If these equations are solved, the result show that this wave must absolutely travel at 300,000 km/s aording to both Joe and Bob. Obviously, there is a problem: Maxwell s equations did not omply with the equations of Galilean relativity. There was, therefore, two possible ourses of ation: either Maxwell s equations are wrong or the equations of Galilean relativity are wrong. While some were trying to hange Maxwell s equations to make them omply with the priniple of relativity, a lone man tried hange relativity to make it onsistent with Maxwell s equations. Albert Einstein was entering the physis sene. (Notie that the problems with Maxwell s equations started Einstein s quest into relativity. It seems that Einstein did not even know about the Mihelson-Morley experiment when he worked on the theory of relativity.) In 1905, Einstein introdued hanges to the equations of relativity based on the following postulates: Einstein s Postulates 1) The laws of physis are the same for all the observers moving at a onstant speed. ) The speed of light in a vauum is always 300,000 km/s (atually 99, km/s) for all observers. 018 Version 10-Relativity 0

21 (Fun fat: the postulates of relativity say that the laws of physis and the speed of light are atually not relative quantities, i.e. dependent on the observer!) The first postulate is simply the priniple of relativity as it was known for some enturies. The seond postulate reflets the observations made during the measurements of the speed of light and the solutions of Maxwell s equations. It may seem odd that several observers moving relative to eah other all measure the same speed when they measure the speed of a light beam. This means that we have the following situation. This is in flagrant ontradition with the equations of Galilean relativity. With this seond postulate, Einstein automatially ruled out the idea that light is a mehanial wave in the aether. A material medium meant that observers moving in this medium would measure a different speed of light in different diretions, whih was impossible aording to the seond postulate. The aether then disappeared from physis. Needless to say that it was already a weird onept. It was undetetable, without mass and offering no resistane while being rigid (so that transverse waves an exist). As a result, thousands of hours of work done to interpret the equations of Maxwell by means of motions or swirls of the aether made by many physiists at the end of the 19 th entury ended up in the trash. Fortunately, some physiist, like Hendrik Anton Lorentz, had already reinterpreted Maxwell s equations starting from the idea that there are harged partiles even before the existene of the eletron was onfirmed in Now, we will have a look at the onsequenes of these two postulates. Prepare to be flabbergasted, they are spetaular. 018 Version 10-Relativity 1

22 The usual onept of time is ompletely shattered by Einstein s postulates. To understand why, some preision on how the observers note the time at whih an event ours must be given. When an observer, say Bob, sees an event, he must alulate when the event really ourred, taking into aount the time it took the light to reah him. If he sees a star exploding in 017 and knows that the light from the explosion has taken 500 years to reah him, he will alulate that this explosion has truly ourred in To simplify this proedure, let s imagine that Bob has plaed loks throughout the universe (figure). When an event ours somewhere, the lok at this plae prints a small piee of paper (a timestamp) or sends a signal to indiate the time. The observer then simply has to pik up the timestamps or the signals to know the time of the events. Web%0Physis/Chapter039.htm For this solution to work, however, all the loks must be synhronized, i.e. they must indiate the same time. Of ourse, Bob an synhronize its loks at home and then go plae them at their loation in the universe but it will be shown later that this solution is not good. (When the loks are moved, the frame of referene hanges and this an affet the time indiated.) Bob must, therefore, put eah lok at its respetive loation before starting them. To start them, his entral lok sends a light signal that starts the loks when they reeive it. We re using a light signal beause we an easily know, by the seond postulate, how suh a signal behaves: it always travels at the speed. But Bob isn t stupid. He knows that it will take some time before the light signal reahes a speifi lok. If all the loks are set at midnight and Bob triggers the signal at midnight, a lok that reeives this signal 3 hours later will start at 3 AM. The lok will then indiate midnight at 3 AM and will be 3 hours behind Bob s entral lok. To ompensate for this, Bob thought of not setting all the loks at midnight. For example, he sets the lok that will reeive the signal at 3 AM at 3 AM. When Bob sends its signal at midnight, this lok will start when the signal is reeived at 3 AM. As the lok was initially set at 3 AM, it will be synhronized with Bob s entral lok. 018 Version 10-Relativity

23 en.wikibooks.org/wiki/speial_relativity/print_version Well done Bob, all your loks are synhronized. The following animation also shows you how this synhronization is done. There is, however, a problem: Let s see what happens aording to another observer (Joe). Joe sees Bob, with its loks, moving towards the left at speed v. en.wikibooks.org/wiki/speial_relativity/print_version en.wikibooks.org/wiki/speial_relativity/print_version But then, aording to Joe, the loks won t be synhronized. As loks to the left of Bob are moving towards the left aording to Joe, they are fleeing away from the light signal heading towards them. The light signal from Bob s entral lok, however, does not travel 018 Version 10-Relativity 3

24 faster as its speed remains beause the speed of light is the same for all observers. Therefore, it will take more time for the signal to reah these loks beause they are fleeing away from the signal, and they will start later than if they were at rest. If it takes 8 hours for the signal to ath up with a lok that was due to start at 6 AM, it will indiate 6 AM when Bob s entral lok is indiating 8 AM. It is then hours behind Bob s entral lok. Thus, aording to Joe, the loks on the left are all behind Bob s entral lok, and the farther away these loks are from Bob, the larger is the time differene with Bob s entral lok. For the loks on the right of Bob, the opposite happens: they will start too early. These loks are moving towards the light signal and they will meet this signal sooner than if they were at rest. If it took only four hours for the signal to get to the lok whih was supposed to be triggered at 6 AM, then this lok starts h too soon so it will be ahead of Bob s entral lok by h. The loks on the right are all ahead of Bob s entral lok aording to Joe, and the time differene gets larger as the loks are farther away from Bob. Therefore, the result of the synhronization, aording to Joe, is en.wikibooks.org/wiki/speial_relativity/print_version In onlusion, Bob s loks are not synhronized aording to Joe, whereas they are synhronized aording to Bob! Joe is going to tell Bob that he did not aount for the fat that he and his loks are moving when he made the synhronization. But Bob is going to reply, and this is orret, that it is Joe and not him who is moving, and that his synhronization is perfet. The problem is even bigger than that beause Joe also has to install his loks everywhere in the universe, and these loks are moving at the same speed as Joe. He synhronizes them the same way and, by the same reasoning, we realize that Joe s loks are synhronized aording to Joe but are not aording to Bob. Who is right? Are the loks synhronized or not? In fat, everyone is right. In relativity, the time at whih an event ourred hanges aording to the observers. When the observers see the same explosion, they all alulate when the explosion truly happened taking into aount the time it took the light to reah them. They then arrive at different onlusions. For example, Bob may infer that an event ourred in 1955 while Joe alulate that the same event ourred in To prove it, eah may show the timestamp printed by the lok that eah observer has installed at the plae where the event ourred. Bob will have his timestamp showing 1955 and Joe would have his timestamp showing The razy thing is that they are both right. The time at whih an event ours depends on the observer. 018 Version 10-Relativity 4

25 Time in Relativity If an observer is moving with respet to another, the moments at whih events our is not the same for these two observers. Of ourse, Joe will always be able to say that the information provided by Bob s lok is not good beause Bob s loks are not synhronized aording to Joe. But Bob will then reply that his loks are perfetly synhronized and that it is rather the information given by Joe s lok that isn t worth shit sine Joe s loks are not synhronized aording to Bob... Joe will answer that his loks are perfetly synhronized and that the information is valid. And they are both right! Already, it is obvious that Einstein s relativity will shake up your oneption of time. The absolute time of Newton s physis and Galilean relativity does not exist anymore. In Newton s and Galileo s theories, an event ourring in 199 ours in 199 for all observers. That is what absolute time means: everyone gets the same value. With Einstein s relativity, the event ours at different times for eah referene frame and this time depends on the speed of the observer. Therefore, time beomes relative. To illustrate one more how the notion of time hanges, let s explore a situation where an observer sees two explosions at the same time. Sarah is on Earth while Maria is travelling in a spaeship that goes very fast. There are two stars who are also moving at speed v towards the right. Maria is halfway between the stars. Just at the moment when Maria passes beside the Earth, the stars explode. Let s assume that the distane between the stars and the Earth is 1 light-years at this moment (aording to Sarah) and that both their alendars indiate 016 when they pass one beside the other. fr.depositphotos.om/ /stok-photo-spaeship.html As Sarah is equidistant from eah explosion (1 light-years), the light of eah explosion will take the same time to get to Sarah (1 years) and she is going to see the explosions at the same time (in 08). In fat, Sarah will do the inverse reasoning. Seeing the flashes at the same time (in 08), she will alulate that two explosions ourred simultaneously 018 Version 10-Relativity 5

26 1 years ago (so in 016) and that the two stars were at the same distane from the Earth when they exploded. Knowing this, Sarah an infer what Maria will see. If the two explosions ourred at the same time, the light leaves from the two explosions at the same time. However, it will take more time for the light oming from explosion A to reah Maria beause the light from this explosion and Maria are both going towards the right. The light, therefore, has to ath up with Maria, and it will take more time for the light of explosion A to get to Maria than to arrive at Earth. The opposite happens with explosion B. The light oming from explosion B goes towards the left, and Maria goes towards the right. Maria is going towards this light and it will take less time for the light oming from explosion B to get to Maria than to arrive at Earth. Maria will, therefore, see explosion B before she sees explosion A. So far, no problem. An observer an see two events at the different time even if they ourred at the same time, it depends on the time taken by the light oming from eah event to reah the observer. If, aording to Sarah, Maria sees explosion B before explosion A, it is ertain that this is what Maria sees. This fat annot be relative to the observer. The problem omes from what Maria will infer from this information. Let s take the point of view of Maria in this situation. fr.depositphotos.om/ /stok-photo-spaeship.html In Maria s referene frame, the spaeship is at rest and the Earth is moving towards the left. Knowing that Maria is halfway between the stars, the light oming from the two explosions takes exatly the same time to reah Maria. But as she saw explosion B before explosion A, she is going to onlude that explosion B atually ourred before explosion A. If Sarah s speed is 0,6 aording to Maria, alulations show that explosion B ourred in 007 and explosion A in 05. (This alulation will be made later in this hapter.) Thus, the two explosions ourred at the same time (016) aording to Sarah while explosion B (in 007) ourred before explosion A (05) aording to Maria. The following film shows a similar situation, but with a train. So who is right? Did the explosions our simultaneously in 016 as Sarah says or in 007 (B explosion) and 05 (explosion A) as Maria says? Both are right. In Sarah s referene frame, the two events are simultaneous and in Maria s referene frame, explosion B ourred before explosion A. Eah ould also provide the timestamps oming from their loks to show that she is right. Sarah would have two timestamps indiating 016 while 018 Version 10-Relativity 6

27 Maria would have a timestamp showing 007 and another showing 05. Of ourse, they will both say that the loks of the other are not synhronized and that the timestamps of the other mean nothing. Still, Sarah loks are perfetly synhronized aording to Sarah and Maria loks are perfetly synhronized aording to Maria and these timestamps truly indiate the time at whih eah explosion ourred aording to eah observer. The same phenomenon omes up here: the moment at whih an event ours beomes relative to the observer. Here, the two events were simultaneous aording to Sarah and that the two events were not simultaneous aording to Maria. Even if this is only one example, the following onlusion an be drawn (a formal proof will be done later). Simultaneity If two events are simultaneous for an observer A, they are not simultaneous for all the observers moving with respet to observer A. A onlusion that was nearly drawn by Henri Poinaré in (He was lose to disovering relativity before Einstein.) The formula It will now be shown that the rate at whih time passes is also relative to the observer. To illustrate this, a very peuliar lok is used. This lok works with a ray of light that is ontinuously refleted between two mirrors. Thus, the light hits the top mirror at regular intervals, say 1 milliseond. Whenever the light returns to the top mirror, the lok moves forward by 1 ms. This speial lok is used beause the only thing we know for the moment is that the speed of light is the same for all the observers. By taking a lok working with light, we an easily find out what s happening for the other observers. ommons.wikimedia.org/wiki/file:light-lok.png The time it takes for the light to make a round trip in this lok when it is at rest is d t0 = However, the time for this round trip will be longer if the lok is moving, 018 Version 10-Relativity 7

28 It is easy to understand why this lok is going at a slower rate when it moves. In this ase, the path of the light is illustrated in the figure on the right. The vertial distane between the mirrors remained the same, but the light is now following a zigzag path. Eah of these diagonal lines is longer than the distane between the mirrors so it will take more time to go from one mirror to another sine the ray must travel a greater distane at the same speed. ommons.wikimedia.org/wiki/file:light-lok.png Let s alulate the time it takes for the light to make one round trip for an observer who sees the lok in motion. Let s say that the time is t to make one round trip aording to this observer. During this time, the lok has moved forward a distane v t. Therefore, the length of a diagonal path (the hypotenuse) is d v t + Therefore, the time required to make one round trip is 018 Version 10-Relativity 8

29 hypotenuse t = t = d v t + This equation must be solved for t. With a little algebra, it beomes v t t = d + Sine d t0 = the equation beomes The end result is 0 = + t t v t v t = t + t 0 v t = 1 t 0 t = t 0 1 v This square root omes up often in relativity, and the following symbol is used to represent this quantity. γ Fator γ = 1 1 v Therefore, the time dilation formula is 018 Version 10-Relativity 9

30 Time Dilation t = γ t = 0 t 0 1 v (v is the relative veloity between the observers.) This formula was obtained for the first time by Einstein in It predited that loks in motion are moving at a slower rate than loks at rest, i.e. that time flows at a slower rate when you are moving! In fat, everything that hanges over time is a lok and humans who age are like loks. If loks in motion are going at a slower rate, this means that humans will age less quikly when they are moving than if they remain at rest! This movie gives you the same explanation. Let s go on with a small example. Example Ginette stays on Earth while Tony travels towards a star loated 4.6 light-years away from Earth. The speed of Tony s spaeship is 80% of the speed of light. a) What is the duration of the trip aording to Ginette? The duration of the trip aording to Ginette is fr.depositphotos.om/ /stok-photo-spaeship.html distane 4.6years 4.6years t = = = = 5.75years speed b) What is the duration of the trip aording to Tony? The time measured in the spaeship is t 0 (see explanation below). So the time is 018 Version 10-Relativity 30

31 t = t 1 0 v = 5.75years 1 ( 0.8) = 5.75years = 3.45years Surprising, isn t it? The effet is even more dramati if the speed is inreased to 99.9% of the speed of light. The time aording to Ginette is then years and the time aording to Tony is 0.06 year or approximately two and a half months. If Tony does a round trip to the star and omes bak to Earth, 9.1 years will have elapsed on Earth, while a little less than 5 months will have passed for Tony. Ginette will then be 9.1 years older whereas Tony will be about 5 months older than when he left the Earth. Let s push a bit further with a more sophistiated example. Suppose now that Tony goes to the entre of Galaxy (6,000 light-years away) in a spaeship. To go there, he aelerates at 9.8 m/s² for one half of the journey and then deelerates at 9.8 m/s² for the other half of the journey. These aelerations give an apparent weight idential to the weight on Earth to Tony. One at the entre of the Galaxy, Tony omes bak to Earth. Calulations (more omplex than the ones made in the example sine the speed hanges onstantly) show that the duration of this trip for Ginette is 5,004 years and only 39.6 years for Tony! An astronaut an make this trip during its lifetime. When Tony returns to Earth, he is about 40 years older than when he left but he doesn t reognize anybody on his return sine 5,004 years have elapsed on Earth. It s not even sure that there is someone who remembers he was gone. Maybe apes would have taken ontrol of the Earth by then... Is It Proven? Is this effet proven or is it a mere fantasy of physiists? There are atually a lot of experimental proof that this effet is real. None implies sending astronauts on long trips sine the urrent tehnology is not advaned enough to ahieve suh large speeds with a spaeship. However, in 1971, a lok was installed on a plane while another idential lok remained on the ground. The loks were previously synhronized. After a small trip, the airraft ame bak, and the lok in the plane was slightly behind the other lok (by 14 ns). The time gap between the two was exatly the one predited by relativity. Even if an astronaut annot be sent at speeds lose to the speed of light, suh large speeds an be ahieved for small partiles. This is what partile aelerators are for. In these aelerators, partiles an go almost at the speed of light. However, some partiles have a limited life span. For example, muons have an average lifetime of. µs when they are at rest. (Muons are partiles very similar to eletrons, but they are 07 times more massive.) When muons move at high speeds, they live muh longer. This is exatly what relativity predits: when they move, they age less quikly and therefore live longer for an observer 018 Version 10-Relativity 31

32 who sees the partiles move. The lengthening of the lifetime of a muon is exatly equal to the lengthening obtained with the time dilation formula. Muons also provided one of the first proofs of relativity. When osmi rays hit the atmosphere, they reate, among other things, muons travelling at very high speeds. Even if the muons were travelling at the speed of light, they would travel only 660 m in. µs. However, these muons are formed at an altitude of a few tens of kilometres and they still reah the surfae of the Earth. The lengthening of their lifetime due to time dilation is what allows them to reah the ground. The number of muons reahing the ground is exatly as predited by the theory of relativity. It is possible to think that relativity plays no role in our daily lives and this is not far from the truth. Be aware, however, that loks in satellites would not tik at the same rate as those on Earth if they were to be plaed next to eah other. One in orbit, they tik at the same rate as those on Earth beause of relativisti effets. If engineers had not taken into aount relativisti effets in the design of satellites, the satellite loks would have slowly drifted ompared to the loks on Earth, whih would have led to misalulations. For example, there are loks inside GPS satellites. If relativisti effets had not been taken into aount, the GPS data ould be off by up to 5 km, even if the loks are resynhronized every 1 hours. Whih Observer Measures t 0? In the previous example alulations, it is said that Tony measures t 0 whih is alled the proper time. Why is it Tony? Preisely, proper time is Proper Time t 0 Proper time is the time between two events aording to an observer who observes the two events in the same plae. In most ases, this means that it is the time between the events aording to the observer who is present at both events. In our example, Ginette is not present at the two events that mark the beginning of the trip (departure) and the end of the trip (the arrival). She is present when Tony leaves, but not at the arrival. However, Tony is present at the departure and at the arrival and it is him that measures the proper time. It s not that obvious that Tony measures the proper time if the true definition of proper time is taken. It s lear that Ginette does not observe both events in the same plae: the departure is lose to the Earth and the arrival is near the star whih is 4.6 ly away from the Earth. This is indisputably not in the same plae. It is less straightforward for Tony. In order to see that the two events are in the same plae for Tony, let s look at this situation 018 Version 10-Relativity 3

33 in Tony s referene frame. Aording to Tony, he is at rest while the Earth reedes and the star is oming towards him. fr.depositphotos.om/ /stok-photo-spaeship.html For Tony, the Earth is next to his ship at his departure and the star is next to his ship at arrival. For Tony, these two events are in the same plae: next to his ship. It is, therefore, him that measures the proper time. Common Mistake: Taking the Formula for Time Dilation When None of the Observers Measures t 0 The time between two events is not neessarily t 0 for one of the two observers. It is possible to have two events at different plaes for both observers. In this ase, do not use the time dilation formula. Use instead the Lorentz transformations (that will be seen later in this hapter). If time dilation were the only effet, there would be a problem of logi in relativity. Let s go bak to the example of Ginette and Tony to illustrate this. Aording to Ginette, the situation is as follows. fr.depositphotos.om/ /stok-photo-spaeship.html Aording to previous alulations, the duration of the trip is 5.75 years for Ginette and 3.45 years for Tony. Let s look at this situation in Tony s referene frame to see what happens aording to him. 018 Version 10-Relativity 33

34 fr.depositphotos.om/ /stok-photo-spaeship.html In this referene frame, Tony is stationary and the star is oming towards him. But if the star were 4.6 ly away and is moving towards Tony at 0.8, the time it would take for the star to arrive at Tony would be distane time = speed 4.6y t = 0.8 t = 5.75y But the star is supposed to arrive in 3.45 years. In order to arrive at this value, there is only one solution: the star must be loser than 4.6 ly aording to Tony. If the star arrives in 3.45 years, the initial distane must be L = v t 0 = y =.76ly Thus, the distane between the Earth and the star is 4.6 ly aording to Ginette but only.76 ly aording to Tony. This is alled length ontration. The distane between two objets at rest is denoted L 0. This variable an also represent the length of an objet at rest beause the length of an objet at rest is the distane between its two ends. L 0 is alled the proper length. The distane between two moving objets (at the same speed) is denoted L. This variable an also represent the length of a moving objet. The length ontration formula an be easily found by repeating the reasoning made in the example of Tony and Ginette. Aording to Ginette, Tony s speed is L 0 v = t Aording to Tony, the speed of the star (whih is equal to Ginette s speed) is 018 Version 10-Relativity 34

35 L v = t 0 As Tony s speed aording to Ginette is the same as that Ginette s speed aording to Tony, the following equation must hold. L = L L0 L = t t L = L t0 t t 0 ² t / 1 v 0 ² The end result is Length Contration ² L0 L = L0 1 v ² = γ (v is the relative veloity between the observers.) This formula was obtained by FitzGerald in 1889 and, independently, by Lorentz in 189. (They were then trying to explain the results of the Mihelson-Morley experiment.) Thus, this ontration is sometimes alled the FitzGerald-Lorentz ontration. Let s hek if the formula works. For Tony, the distane between the Earth and the star is L = L 1 v² 0 ² 0.8 = 4.6ly 1 ( ) = 4.6ly =.76ly Bingo, it works 018 Version 10-Relativity 35

36 Example A train is 0 m long at rest. What is its length if it is moving at 0.8? The length is L = L 1 v² 0 ² 0.8 = 0m 1 ( ) = 0m = 1m The only ontrated dimension is in the diretion of the veloity. The dimensions perpendiular to the veloity remain the same. Here, the train has the same height at rest and in motion. Only the length of the train hanges. The passengers of the train do not feel ontrated. To illustrate this idea, let s take the example of Ginette and Tony again. In Ginette s frame, Tony s spaeship moves and the spaeship is shorter than at rest. However, the Earth, Ginette and the star are at rest in this frame and are, therefore, not ontrated. In Tony s frame, the spaeship and Tony are at rest and are, therefore, not ontrated. For Tony, the Earth, Ginette, and the star are moving and are ontrated. Thus, nobody sees himself ontrated sine eah observer is at rest in its own frame. All the observers are observing that the other observers are ontrated. kspark.kaist.a.kr/twin%0paradox/relativity%0fats.htm The same thing happens with time dilation: all observers see themselves at rest and time runs normally for them. All observers are observing that the time passes more slowly for the other observers. 018 Version 10-Relativity 36

37 Example Tony wants to make a trip from the Earth to a star loated 4.6 ly away aording to observers on Earth. What must be the speed of Tony for the trip to last 1 year aording to Tony? The equation to solve an be obtained by taking either point of view: Tony s or Ginette s (whih is on Earth). Aording to Ginette, we have the following situation. The time aording to Ginette is fr.depositphotos.om/ /stok-photo-spaeship.html t = L 0 v Ginette an then alulate the time aording to Tony with v t = t 1 0 L v 0 = 1 v If the trip is to last one year, the following equation must be solved The solution is 4.6y v 1y = 1 v 1 v = v 1 v = v 1 = 4.6 v Version 10-Relativity 37

38 v = 4.6 v 1 = = ² The speed must then be % of the speed of light. Notie that the same equation to solve is obtained by taking the point of view of Tony. In Tony s referene frame, the situation is as follows. fr.depositphotos.om/ /stok-photo-spaeship.html Aording to Tony, the time it takes for the star to reah his spaeship is t = 0 L v L is then found with the length ontration formula. The equation is now L t = 0 ² 0 1 v ² If the trip is to last 1 year, then the equation to solve is v 4.6a 1 1y = v This is the same equation as the one obtained by taking the point of view of Ginette. v² ² Effets at Low Speed In our daily lives, the effets of time dilation and length ontration are rarely taken into aount. But if you want to take these effets into aount at low speeds, the following approximations make the alulations easier. 018 Version 10-Relativity 38

39 Approximations of the γ Fator if v << 1 1 v v 1+ + v + 1 v 1 Example Pimprenelle stays in Quebe City while Amarante goes to Montreal (distane = 50 km) in a high-speed train travelling at 50 km/h. a) What is the differene between the duration of the trip aording to Pimprenelle and Amarante? The duration of the trip aording to Pimprenelle is distane 50km t = = = 1h = 3600s speed 50 km / h The duration of the trip aording to Amarante is t = 0 t 1 v = 3600s 1 v If this value is alulated with a alulator, 3600 s is obtained again beause the speed is too small and the answer is barely smaller than 3600 s. To see the time differene, an approximation is used. differene = t t m ( ) 8 m ( s ) 018 Version 10-Relativity 39 0 = t t 1 11 v v = 3600s 3600s 1 + v = 3600s + s = 3600s s

40 Amarante s wath is, therefore, behind Pimprenelle s wath by only x s after the trip to Montreal. The effet is not very important when the speed is far from the speed of light. A person who travels on a plane for 100 years would live only 1.5 ms longer than a person who had remained at rest on Earth. b) What is the differene between the distane from Montreal to Quebe aording to Pimprenelle and the distane from Montreal to Quebe aording to Amarante? The distane aording to Pimprenelle is 50 km. The distane aording to Amarante is ontrated. It is, therefore, ² L = 50km 1 v ² If this value is alulated with a alulator, 50 km is obtained again beause the speed is too small and the answer is barely smaller than 50 km. To see the differene in distane, an approximation is used. L = L L 0 = 50km 50km 1 m ( ) 8 m ( s ) v² ² v² ( ) 50km 50km 1 v² 50km ² 50km nm Length ontration makes the distane between Montreal and Quebe City only 6.69 nm shorter aording to Amarante. s ² The formula The Doppler effet seen in previous hapters is modified by relativity beause the rate at whih time passes, related to the period of the wave, is hanged by the speed of the soure. 018 Version 10-Relativity 40

41 1) Classial Doppler Shift The formula of the Doppler effet was v v f = v v 0 s f Sine every referene frame is equivalent, the frame where the observer is at rest and the soure is moving is always used. Therefore, the speed of the observer is always v 0 = 0. Sine the speed of light is always, we have v =. The equation then beomes f = v s f The index s for the speed is then dropped sine there is no possibility of onfusion with other speeds. The frequeny and wave period are thus v f = f and T = T v ) Time Dilation However, if the soure is moving, the period is hanged ompared to what we would have if the soure were at rest. This means that T = T 0 ² 1 v ² where T 0 is the period of the soure when it is at rest. Relativisti Doppler Shift Combining these two effets, we have v T = T v T T = 0 ² 1 v ² This beomes 018 Version 10-Relativity 41

42 T = T T = T 0 = 0 v ² v² v ( v)( + v) v + T T0 v Therefore, the period and frequeny shifts are Relativisti Doppler Effet T = T 0 f = f 0 v + v + v v (v is the speed of the soure aording to a motionless observer.) The sign onvention for speed stays the same as it was. The positive diretion is from the soure towards the observer. Thus, the speed of the soure is positive if it is heading towards the observer and negative if it is moving away from the observer. Remember that the point of view of the observer (observer at rest and soure in motion) must always be taken to use this formula. Example A soure at rest emits light with a 600 nm wavelength (orange). What is the wavelength seen by an observer who sees this soure moving towards him at 30% of the speed of light? The frequeny of this wave when the soure is at rest is f 3 10 = = = 5 10 m 8 m s 0 9 λ Hz Therefore, the frequeny reeived by the observer is f = f 0 + v v 018 Version 10-Relativity 4

43 f = 5 10 = Hz Hz = Hz The wavelength is thus 8 m 3 10 s λ = = = 440.3nm 14 f Hz The soure emits an orange light at rest. It is pereived as being violet by the observer who sees the soure moving towards him at 0.3. What the Observers See The Doppler effet formula allows us to alulate what is seen by an observer. Let s take an example to illustrate. fr.depositphotos.om/ /stok-photo-spaeship.html We already know that time passes more slowly for Naomi aording to Ophelia beause of the time dilation. This is the result of the observations of Ophelia. She alulated that for eah seond elapsed on Earth, only 0.6 seonds have elapsed in Naomi s spaeship. This does not mean that this is what Ophelia sees beause the time it takes for the light to reah Ophelia must also be taken into aount. However, this is exatly what the Doppler effet formula does. Suppose you look at a lok in Naomi s ship and that this lok emits a flash of light every seond. This is a periodi phenomenon and the period of this phenomenon will be pereived (so view) as having a period = v T T0 + v Let s do some alulations. 018 Version 10-Relativity 43

44 Example In Naomi s spaeship, there is a lok that emits a flash of light every seond. What is the time between the flashes as seen by Ophelia if Naomi s spaeship moves away from Earth at 0.8? The time between the flashes is = T T0 + v = 1s = 1s = 3s v fr.depositphotos.om/ /stok-photo-spaeship.html ( 0.8) ( 0.8) The speed is negative beause the positive diretion goes from the soure towards the observer so from Naomi to Ophelia. Ophelia sees a flash every 3 seonds. There is a differene between what is observed and what is seen. Ophelia sees a flash every 3 seonds, but she alulates that the flashes are emitted every seonds. It takes more time to see the flashes, beause the spaeship is farther and farther away at eah flash and it will take more and more time for the light to reah Ophelia, thereby inreasing the time between the flashes. At 0.8, the spaeship moves, aording to Ophelia, 4 x 10 8 m between eah flash if there are seonds between them. The light oming from the next flash must then travel this extra distane before arriving on Earth, and this takes seonds. The time between the flashes as seen by Ophelia is, therefore, 3 seonds (1.667s s). It an, therefore, be onluded that Ophelia sees everything that happens in the spaeship at a rate 3 times slower than what is happening on Earth, like a movie playing in slow motion. 018 Version 10-Relativity 44

45 Example In Naomi s spaeship, there is a lok that emits a flash of light every seond. What is the time between the flashes as seen by Ophelia if Naomi s spaeship is moving towards Earth at 0.8? The time between the flashes is = T T0 + v = 1s = 1s 1 = s 3 v fr.depositphotos.om/ /stok-photo-spaeship.html ( 0.8) ( 0.8) Ophelia sees a flash every 1/3 seonds but observes that the flashes are emitted every seonds. It takes less time to see the flashes, beause the spaeship is loser and loser to Earth at eah flash and it will take less and less time for the light to arrive, thereby dereasing the time between the flashes. At 0.8, the spaeship moves, aording to Ophelia, 4 x 10 8 m between eah flash if there are 1,667 seonds between them. The light oming from the next flash has less distane to travel before arriving on Earth, and this removes seonds to the time. The time between the flashes as seen by Ophelia is, therefore, 1/3 seonds (1.667s s). It an, therefore, be onluded that Ophelia sees everything that happens in the spaeship at a rate 3 times faster than what is happening on Earth, like a movie playing in fast forward, even if the time is going at a slower rate in the spaeship! Here s a great video showing what is seen if someone is moving with speed lose to the speed of light Version 10-Relativity 45

46 A more formal approah to relativity will be now used to obtain the transformation laws from one observer to another. With these transformations, all the formulas obtained previously will be proven again but more formally. More omplex ases an also be dealt with using the transformation laws. The Formulas When two observers observe an event, they note the position and the time at whih the event ourred with its own oordinate system. en.wikibooks.org/wiki/speial_relativity/print_version To find out how to swith from one referene frame to another in aordane with Einstein s seond postulate, let s imagine that a bright flash of light is spreading at speed in every diretion. The light is, therefore, reating a sphere with a radius that inreases at the speed of light. As the equation of a sphere is x² + y² + z² = R² and as the radius of the sphere inreases at the speed of light, the equation of this sphere of light, aording to Bob, is ( ) x² + y² + z² = t Joe also observes an inreasingly large light sphere sya1/relativitypriniple.htm growing at speed sine the speed of light is the same for every observer. Therefore, the equation of the sphere aording to Joe must be ( ) x ² + y ² + z ² = t 018 Version 10-Relativity 46

47 With the transformation laws, the equation of the sphere aording to Bob must transform into the equation of the sphere aording to Joe. Let s first show that Galilean transformations fail to do this. Starting from the equation of the sphere aording to Joe, the transformation gives ( ) x ² + y ² + z ² = t ( x vt) + y² + z = ( t) = ( ) x² xvt v t y² z t Obviously, this is not the equation of the sphere aording to Bob. On the other hand, this result is the first lue needed to obtain the orret transformations. There is a ross term with x and t (the seond term) that must be eliminated. It an be eliminated if t = t + fx where f is a value that will be determined in order to make the seond term vanish. With this new transformation law, the transformation of the equation of the sphere is ( ) x ² + y ² + z ² = t ( ) ( x vt) + y² + z = ( t + fx) x xvt + v t + y + z = t + xft + f x ² ² ² ² ² ² ² ² The seond term on the left an now be anelled out by the seond term of right if f = v ² Then, the equation beomes x² xvt = v t y² z ² t² vxt v² + x ² ² This is not quite it. This result an be written as v² x² + v t + y² + z = ² t² + x² ² v² v² x² 1 + y² + z = ² t² 1 ² ² 018 Version 10-Relativity 47

48 ² The orret result would have been obtained if x and t had been divided by 1 v ². Then, the final result is x y z t ² + ² + = ² ² whih is the equation of the sphere aording to Bob. Thus, the transformation laws obtained for x and t are 1 x = x vt 1 v² ² v² ² ( ) 1 vx t = t 1 The full transformation laws, alled the Lorentz transformations, are Lorentz Transformations (1) ( ) γ ( ) x = γ x vt x = x + vt y = y y = y z = z z = z vx vx t = γ t t = γ t + ² ² (v is the speed of an observer aording to another observer.) These transformations laws were obtained in 1898 by Lorentz and in 1899 by Larmor but they gave a muh different interpretation to the time transformation. Poinaré was lose to making a orret interpretation in 1904 but it was Einstein who interpreted orretly, in 1905, that the times of events are not the same for different observers and that there is no absolute time. Note that if, the Galilean transformations are obtained again. Thus, the results obtained with Galilean transformations are still valid, provided that the speed between the observers is small ompared to the speed of light. These laws of transformation are not that useful sine the position x = 0 and the time t = 0 are somewhat arbitrary. It would be better to have the transformation laws of the distane ( x, y, z) and the time ( t) between the events. The x-omponent of the distane and the time between events are, aording to Bob, 018 Version 10-Relativity 48

49 x = x x 1 t = t t 1 The distane between the events aording to Joe is x = x x 1 Using Lorentz transformations, we have ( ) γ ( 1 1) (( 1) ( 1) ) x γ ( x v t) x = γ x vt x vt x = γ x x v t t = An idential formula as the one we had for the Lorentz transformations is obtained, exept that there are now in front of variables. In fat, this is what always happens and the transformations are Lorentz Transformations () ( ) γ ( ) x = γ x v t x = x + v t y = y y = y z = z z = z v x v x t = γ t t = γ t + ² ² (v is the speed of an observer aording to another observer.) This form is a bit more useful beause it does not depend on the arbitrary x = 0 and t = 0. Before using these formulas, remember the following rule: Whih Observer Uses Primes? The observer who sees the other travelling towards the negative x-axis uses primes Here are some examples of the appliation of these formulas. 018 Version 10-Relativity 49

50 Example Greta is in a spaeship moving away from Earth at 0.8. She then sends a beam of light from the bak towards the front of the ship. Aording to Greta, the distane between the light soure and the target is 300 m. Clemene, on Earth, observe Greta s spaeship. theoryoftime.om/wordpress/?p=36 a) How long does it take for the light to get to the target aording to Greta? The time is simply the time it takes for a beam of light to travel 300 m. 300m t = = m s 6 s Greta is the one who sees the other observer moving towards the negative x-axis. That s why Greta notes the time with a prime. b) How long does it take for the light to get to the target aording to Clemene? The situation, aording to Clemene, is The time an be alulated with the Lorentz transformations. theoryoftime.om/wordpress/?p=36 The time aording to Greta is t = 10-6 s, and the distane between the events (departure and arrival of the light) is, aording to Greta, x = 300 m. Therefore, the time aording to Clemene is 018 Version 10-Relativity 50

51 v x t = γ t + ² m = 10 s ² ² = s So, it takes 3 µs for the light to arrive at the target aording to Clemene while it takes only 1 µs aording to Greta. Note that this problem annot be done with the time dilation formula beause none of these observers measures the proper time ( t 0 ) beause none of the observers is present at both events. Better formulation: the two events (departure and arrival of the light) are not in the same plae for any of these two observers, so none measures the proper time. On the other hand, here s how this problem ould have been handled without using Lorentz transformations. For Clemene, the distane between the starting point of the light and the target is smaller beause of length ontration. For her, the length of the spaeship is L = 300m = 180m The light, travelling at, must ath up with a target at 180 m whih moves away at 0.8. Therefore, the time to ath up is 180m t = = s Simultaneity Lorentz transformations will onfirm that two events that are simultaneous for an observer are not simultaneous for the other observers. If two events are simultaneous, then t = 0. The time, aording to another observer, is therefore t v x = γ t v x t = γ whih annot be equal to zero. This shows that the events annot be simultaneous. 018 Version 10-Relativity 51

52 Example Sarah observes that two stars 4 ly one from the other (aording to Sarah) explode simultaneously. What is the time between the explosions aording to Maria? fr.depositphotos.om/ /stok-photo-spaeship.html Aording to Sarah, the time between events is t = 0 (sine the events are simultaneous) and the distane between the events is x = 4 ly. So, the time between the events aording to Maria is v x t = γ t v x = γ y = = 18y There is an 18-year gap between the explosions aording to Maria. The sign of this answer indiates whih event happened first. As we set x = x' x' 1 = 4 ly as positive, it means that event is explosion. As a negative t = t - t 1 was obtained, t must be smaller than t 1 (t < t 1 ). This means that the event ourred before the event 1. So the explosion ourred 18 years before explosion 1 aording to Maria. (This is exatly the same situation we had when we first talk about simultaneity. We then said that Maria observes that the events our in 007 and 05. The gap is effetively 18 years.) Is it possible to find the years of these explosions aording to Maria, knowing that the two explosions ourred in 016 (aording to Sarah), just as Maria was passing next to the Earth? Of ourse, but to alulate them, the position and the time of eah event aording to Sarah must be found first. 018 Version 10-Relativity 5

53 This year 016 orresponds to the time t = 0, sine t = 0 orresponds to the time when the two observers are at the same plae. Thus, the two 016 explosions ourred at t = 0 aording to Sarah. As Sarah is at x = 0 (the observer is always at the origin of its oordinates), one of the explosions ourred at x =-1 ly and the other at x = 1 ly. Therefore, the time of the explosion 1 at x =-1 ly aording to Maria is vx1 t 1 = γ t y = = 9y As t = 0 is 016, t = 9 years is 05. The time of the explosion at x = 1 ly aording to Maria is vx t = γ t y = = 9y As t = 0 is 016, t = -9 years is 007. (Corretly, the observer must not neessarily be at x = 0 of its referene system. The origin an be put anywhere. However, if the position of the origin is hanged, remember that the t = 0 is the moment where the origins of the two axes systems of the observers are at the same plae.) Time Dilation t 0 is the time between two events ourring at the same plae aording to one observer. So for this observer, we have So, for another observer, the time is t = t 0 x = Version 10-Relativity 53

54 v x t = γ t v 0 t = γ t0 t = γ t 0 whih the time dilation formula. Length Contration Measuring the length of an objet onsist of two separate events: measuring the position of an end of the objet and measuring the position of the other end of the objet. The subtration of the two positions gives the length of the objet (L = x - x 1 ). When the objet is in motion, the position of both ends must be measured simultaneously. If the position of the bak end of the objet is measured at one moment and the position of the front end of the objet is measured 1 seond later, the subtration of the two positions does not give the length of the objet in motion beause the position of the front end of the objet has hanged during this time. The position of both ends has to be measured at the same time. Thus, for an observer who sees the objet moving, we must have L = x x1 = x t = 0 On the other hand, the position of the ends an be measured at different times if the objet is at rest. The Lorentz transformations then give whih is the length ontration formula. L = x x = x 0 1 t = whatever 0 ( ) ( 0) x ' = γ x v t L = γ L v L0 L = γ 018 Version 10-Relativity 54

55 The Interval The interval between two events is defined by Interval ( s) = ( t ) ( x) ( y) ( z) In itself, the interval does not mean muh. However, the interval is interesting beause it is a relativisti invariant. This means that all observers obtain to the same value for the interval. Here is the proof that the interval is the same for everyone. If the observer with primes alulates the interval between two events, then Thus, ( s ) = ( t ) ( x ) ( y ) ( z ) v x = γ ( t ) v x t ( x) ( ) ( ) ( ) = γ t γ x v t y z ² ( s) v γ ( x) v x t ( ) ( + v t ) ( y) ( z) ( ) 1 ( ) 1 ( ) ( ) = γ t x y z ( ) ( ) ( ) ( ) t x y z v² v² ² ² = = Relativisti Invariant ( s ) = ( s) There are not a lot of things on whih different observers agree, but they agree on the value of the interval. Note that if two events are in the same plae for an observer, then the time between two events aording to this observer is the proper time. In this ase, the interval is 018 Version 10-Relativity 55

56 ( s) = ( t) ( x) ( y) ( z) ( t ) ( 0) ( 0) ( 0) = 0 ( t ) = 0 This means that there is a link between the interval and the proper time between two events. However, this link is not always there. Aording to the formula. ( s) = ( t) ( x) ( y) ( z) it is possible to obtain a negative value for the interval. For example, this would be the ase for two simultaneous events loated at some distane from eah other. In this ase, it is impossible to alulate the square root to find the proper time between the events. This simply means that there is no proper time between these two events. This happens beause the two events are too far from eah other so that no observer an see them at the same plae. To be present at the two events, the observer would have to travel faster than light and this is impossible (as will be seen later). Thus, the relation is Proper time between two events ( t ) = ( s) 0 If ( s)² > 0 If ( s)² between two events is positive, it is said that the interval is time-like. In this ase, there is a proper time and the order of the events an never be reversed. All observers will see both events in the same order and no observer an observe that these events are simultaneous. If ( s)² between two events is negative, it is said that the interval is spae-like. In this ase, there is no proper time and the order of the events an be reversed. Some observers observe event A before event B and some other will observe event B before event A. Now, two observers measure the speed of an objet. The speed of this objet is u aording to one observer, and u aording to the other observer. en.wikibooks.org/wiki/speial_relativity/print_version 018 Version 10-Relativity 56

57 To know the speed of an objet, the position of the objet at two different instants must be measured. These are two events. With the distane between the events aording to Bob ( x) and the time between the events aording to Bob ( t), the x-omponent of the veloity of the objet aording to Bob is obtained. u x x = t With the distane between the events aording to Joe ( x ) and the time between the events aording to Joe ( t ), the x-omponent of the veloity of the objet aording to Joe is obtained. x u x = t The transformation law of the x-omponent of the veloity of the objet an be obtained with Lorentz transformations. u x = γ u x x = t γ ( x + v t ) ( t + v x / ² ) By dividing the numerator and the denominator by t, the equation beomes u x x + v t = v x 1+ ² t Sine x u x = t the equation is u x u x + v = vu x 1+ ² This is the transformation law for the x-omponent of the veloity. By doing the same thing with the other omponents of the veloity, the transformation laws for the veloities are obtained. 018 Version 10-Relativity 57

58 Veloities Transformations ux v u x + v u x = ux = vux vu x 1 1+ ² ² v² v² uy 1 ² u y 1 ² u y = uy = vux vu x 1 1+ ² ² v² v² uz 1 ² u z 1 ² u z = uz = vux vu x 1 1+ ² ² (v is the speed of an observer aording to another observer.) These formulas were obtained in 1905 by Einstein, weeks before Poinaré! Example In his spaeship, Henry throws a ball. The speed of the ball aording to Henry is 0.6. What is the speed of the ball aording to Pat, who is on Earth? theoryoftime.om/wordpress/?p=36 Aording to Henry, the veloity is u' x = 0.6. Aording to Pat, the veloity of the ball speed is thus u x u x + v = vu x 1+ ² = ( ) = Version 10-Relativity 58

59 So, this is the situation aording to Pat. theoryoftime.om/wordpress/?p=36 (Note that in Galilean relativity, the speed of the ball would have been 1.4 aording to Pat.) It s time to test the theory. What happens if Henri sends a beam of light towards the front? Example In his spaeship, Henry sends a beam of light towards the front of the ship. What is the speed of light aording to Pat, who is on Earth? theoryoftime.om/wordpress/?p=36 Aording to Henry, the veloity is u' x =. Aording to Pat, the speed of light is thus u x u x + v = vu x 1+ ² = = ( ) So, this is the situation aording to Pat. 018 Version 10-Relativity 59

60 This result is in agreement with Einstein s seond postulate. theoryoftime.om/wordpress/?p=36 (Note that in Galilean relativity, the speed of light would have been 1.8 aording to Pat.) Now, Henry will throw the ball upwards. Example In his spaeship, Henry throws a ball upwards. The speed of the ball aording to Henry is 0.6. What is the veloity of the ball aording to Pat, who is on Earth? theoryoftime.om/wordpress/?p=36 Aording to Henry, the omponents of the veloity of the ball are u' x = 0 and u' y = 0.6. The x-omponent of the veloity of the ball aording to Pat is thus u x u x + v = vu x 1+ ² 0.8 = ( 1+ 0) = 0.8 It is just normal to have the same veloity as Henri s spaeship sine the ball is following the motion of the spaeship. 018 Version 10-Relativity 60

61 The y-omponent of the veloity of the ball aording to Pat is u y v² u y 1 ² = vu x 1+ ² = = 0.36 ( 1+ 0) So, the speed of the ball aording to Pat is u = u + u = x y and its diretion is So, this is the situation aording to Pat. u y θ = artan = 4. u x and theoryoftime.om/wordpress/?p=36 Example On Earth, Florene sees Sam s and Alex s spaeships heading towards eah other with a speed of 0.7. What is the speed of Sam s spaeship aording to Alex? fr.depositphotos.om/ /stok-photo-spaeship.html 018 Version 10-Relativity 61

62 Even if there are three observers in this problem, Florene and Alex are the only interesting observers beause we are going to pass from Florene s point of view to Alex s point of view. As the speed between Florene and Alex is 0.7, we have v = 0.7 (v is always the speed between the observers when we pass from the point of view of one observer to the point of view of another. This speed is never negative.) Eah of these observers measures the veloity of the same objet (Sam s spaeship here). The veloity of that objet is u x aording to Florene and u' x aording to Alex (u x and u' x always refer to the veloity of the same objet but aording to different observers). Note that the values of u x and u' x are negative if the objet is moving towards the negative x-axis aording to the observer. This is what is happening here with Sam s spaeship. It is possible for u x and u' x to have different signs. Therefore, The veloity aording to Alex is So, this is the situation aording to Alex. u x = 0.7 u' x = what we re looking for v = 0.7 ux v u = x vu x 1 ² = ( ) = fr.depositphotos.om/ /stok-photo-spaeship.html If Sam s point of view had been taken, Alex s spaeship would be moving towards Sam with a speed of Version 10-Relativity 6

63 With relativity, it is quite easy to find situations that seem paradoxial at first glane. This means that different results are obtained by taking the point of view of different observers whereas different results should have been impossible. Very early in the history of relativity, some have highlighted suh situations. The most famous is probably Paul Langevin s twin paradox of 1911 (Langevin is also famous for his extramarital affair with Marie Curie...). This paradox is a bit too omplex to be addressed here beause there is an observer that hanges speed. However, simpler situations that seem to lead right to a paradox an be examined. Florene and Alex The first ase onerns time dilation. Florene, who is on Earth, has plaed a flag 6 ly from the Earth. When Alex passes next to the flag, Alex and Florene both start a stopwath. and fr.depositphotos.om/ /stok-photo-spaeship.html When Alex will pass near the Earth, Alex and Florene will show eah other their stopwath. Alex and Florene will then be able to ompare the values given by the stopwathes to see whih is ahead and whih is late. Aording to Florene Here is what is going to happen aording to Florene. Between the moment Alex starts her stopwath and the time she arrives at the Earth 10 years will have elapsed aording to Florene (6 ly at 0.6). However, Florene knows that Alex s stopwath is tiking more slowly beause it is moving. Aording to the time dilation formula, the stopwath will advane by only 8 years while 10 years will have elapsed on Earth. Thus, aording to Florene, Alex s stopwath will indiate 8 years at the time of the meeting, while Florene s stopwath, whih is tiking at a normal pae, will indiate 10 years. 018 Version 10-Relativity 63

64 Aording to Alex A paradox seems to appear if the point of view of Alex is taken. Aording to Alex, the situation is as follows. and fr.depositphotos.om/ /stok-photo-spaeship.html Aording to Alex, the distane between the Earth and the flag is 4.8 ly aording to the length ontration formula. Alex starts her stopwath when the flag passes by her ship and she will stop it when the Earth will pass by her. As the Earth is moving at 0.6 and is initially 4.8 ly away from Alex, the Earth will arrive in 8 years. Thus, Alex s stopwath will indiate 8 years at the meeting, just as Florene had predited. However, Florene s stopwath is tiking more slowly (sine it is moving) aording to Alex. As Florene s stopwath is tiking more slowly, it should be late ompared to Alex s stopwath when they will meet. In fat, aording to the time dilation formula, Florene s stopwath will have advane by 8y = t t = 6.4y 0 Thus, Florene s stopwath should indiate 6.4 years at the meeting aording to Alex. The Apparent Paradox There seems to be a paradox beause they annot be both right. What will be seen on Florene s stopwath when they will meet? Are they going to see 10 years as Florene has predited or will they see 4.8 years as Alex has predited? They annot be both right. If someone takes a piture of Florene s stopwath when they will meet, the display of Florene s stopwath an be seen on the photograph. It is impossible to have a result different from what is seen on the photograph. The Solution In reality, there is no paradox when the problem is done orretly. Florene s reasoning is quite orret: Florene s stopwath will indiate 10 years at the meeting. It is the passage from Florene s referene frame to Alex s that was done inorretly. To do it orretly, the positions and the time of the events aording to Florene must be used. 018 Version 10-Relativity 64

65 fr.depositphotos.om/ /stok-photo-spaeship.html The t = 0 is when the observers meet (this is what was assumed when the Lorentz transformations were obtained). Thus, the start time of the stopwathes is t = 10 years sine they start 10 years before the meeting (whih is the t = 0) aording to Florene. As for the x = 0, it is set where Florene is (whih is at the origin of her axes system). Lorentz transformations are then used to obtain the times and positions in Alex s referene frame. Aording to Alex, the oordinates of the meeting (stopwathes stop) are vxm x M = γ ( xm vtm ) t M = γ tm ² x M = ( 0 0) t M = ² x = 0 t = 0 M (These values make sense sine the time of the meeting is, by definition, t = 0 and t = 0. In addition, the meeting is at the plae where Alex is loated at that time, whih is the origin of Alex s axes.) Aording to Alex, the oordinates of the start of Alex s stopwath (Alex s stopwath starts) are vxa x A = γ ( xa vt A ) t A = γ ta ² y x A = ( 6y y) t A = 10y 4 4 ² x = 0ly t = 8y A This means that Alex s stopwath starts 8 years before the meeting (whih is at t = 0). Alex s stopwath should, therefore, indiate 8 years when they will meet aording to Alex, as expeted by Florene and Alex. M A 018 Version 10-Relativity 65

66 Aording to Alex, the oordinates of the start of Florene s stopwath (Florene s stopwath starts) are vxf x F = γ ( xf vtf ) t F = γ tf ² x F = ( y) t F = 10y 4 4 ² x = 7.5ly t = 1.5y F Oh surprise! Florene s stopwath will run for 1.5 years as it will stop at t = 0. But this stopwath is moving at 0.6 and is, therefore, tiking more slowly. The time dilation formula an be used to alulate by how muh this stopwath will advane. 1,5y = t t = 10y 0 F Florene s stopwath should, therefore, indiate 10 years at the meeting aording to Alex. Alex s foreasts are in line with those of Florene. The error that led to the paradox was to think that the starts of the stopwathes were also simultaneously aording to Alex. If the starts are simultaneous aording to Florene, then they annot be simultaneous aording to Alex. This is what the Lorentz transformations are telling us: Alex starts her stopwath at t = 8 years and Florene starts her stopwath at t = 1.5 years. Florene started her stopwath 4.5 years before Alex aording to Alex. It is true that Florene s stopwath is tiking more slowly aording to Alex, but it will still indiate a greater value than Alex s stopwath beause it has started 4.5 years earlier than Alex s stopwath. Therefore, there is no paradox. The transformations also indiate that Florene was 7.5 light-years away from Alex when she started her stopwath (x = 7.5 ly for the start of Florene s hronometer and Alex is always at x = 0). As Florene moves at 0.6 aording to Alex, she will arrive 1.5 years after the start of Florene s stopwath, as expeted. What if eah observer emits light flashes? Now, it will be assumed that Alex s and Florene s stopwathes both emit a flash of light every year after they have been started. Let s ount how many flashes eah observer will see. The number of flash seen by Florene is equal to the time elapsed for Alex and the number of flash seen by Alex is equal to the time elapsed for Florene. Will these results math with the alulations made previously? 018 Version 10-Relativity 66

67 Florene sees Alex oming towards her. Aording to what was said, Florene should see 8 flashes emitted by Alex s stopwath in 10 years. At first glane, it looks like there is a problem. If the Doppler effet formula is used to measure the time between the flashes as seen by Florene, the result is = v + T T0 v = 1y = 1y = 1y 1 4 = 0.5y Florene sees a flash every 6 months. As it takes 10 years for Alex to arrive, Florene would see 0 flashes of light. This means that 0 years have elapsed in the ship. Yet, it was alulated that 8 years have elapsed Of ourse, there is a mistake in the previous alulation. Dates will be used to help understand the following reasoning. Suppose that, aording to Florene, Alex s spaeship passes by the flag in 00 and passes by the Earth in 030. This gives a journey that lasts 10 years aording to Florene, whih is in agreement with what had been said. However, Florene does not see immediately the first flash emitted by Alex s stopwath. As the flag is 6 ly away, Florene will see this flash only 6 years after Alex has passed by the flag. So, she will see the first flash only in 06. As Alex passes by the Earth in 030, Florene will see flashes during only 4 years. During these 4 years, she sees a flash every 6 months, whih gives a total of 8 flashes. Therefore, 8 years have elapsed in Alex s spaeship, as it has been alulated. What will Alex see? She sees Florene travelling towards her at 0.6. This means that she will also see a flash every 6 months. Aording to Alex, Florene s stopwath is running for 1.5 years. However, Alex will not see 5 flashes of light beause the first flash is emitted when Florene is 7.5 ly away. Thus, this flash will be seen with a 7.5-year delay so that there is only 5 years (1.5 years 7.5 years) between the first and the last flash. With a flash every 6 months for 5 years, 10 flashes will be seen. This is in agreement with our previous results, whih showed that 10 years have elapsed for Florene. Note: Florene will see Alex travelling faster than light (atually 1.5 times faster than light, beause she will see her travel 6 light-years in 4 years), but this does not mean that Alex is travelling faster than light. This is not beause an observer sees an objet travelling faster than light that the objet is really travelling faster than light. In fat, Florene knows she saw the beginning of the trip with a delay of 6 years and that the journey of the spaeship atually lasted 10 years. Therefore, she observes that the speed is 6 ly/10y = Version 10-Relativity 67

68 Guns in a Train Now, here is a situation that also seems paradoxial but involving length ontration this time. Gontran is in a 33 m long train (length at rest) while Euzebe is on the ground and wathes the train passing. Gontran has plaed a gun at eah end of the train. When he presses a button, the two guns shoot a bullet at the same time in the ground. Euzebe drew a ruler on the ground to measure the position of the bullet holes. Here is this situation when the train is at rest. praner.physis.louisville.edu/astrowiki/index.php/speial_relativity Now let s alulate the position of the bullet holes on Euzebe s ruler when the train is moving at 0.8. Gontran s instrutions are to fire when the gun loated at the bak of the train is above the 0 of the ruler drawn by Euzebe. Aording to Gontran In Gontran s referene frame, Euzebe and the ground are moving. praner.physis.louisville.edu/astrowiki/index.php/speial_relativity The guns are 33 m apart aording to Gontran. However, the ground, moving at 0.8 aording to Gontran, is ontrated. The ruler, whih was drawn on the ground, is also ontrated so that the gun in front of the train is not above the 33 m mark when it fires. 018 Version 10-Relativity 68

69 Aording to the length ontration formula, the gun is above the 55 m mark when it fires. Therefore, it makes a hole in the ground at the 55 m mark on Euzebe s ruler. Aording to Euzebe This is the same situation as viewed from Euzebe s referene frame. praner.physis.louisville.edu/astrowiki/index.php/speial_relativity Aording to Euzebe, the train is ontrated and the guns are only 19.8 m apart aording to the length ontration formula. When the guns are fired, the gun loated in front of the train is above the 19.8 m mark. This gun will make a hole in the ground at the 19.8 m mark on Euzebe s ruler. The Paradox There is a paradox: the observers do not have the same position on Euzebe s ruler for the hole made by the bullet fired by the gun in front of the train. The hole should at the 55 m mark aording to Gontran and the 19.8 m mark aording to Euzebe. One again, a photo of the hole and of the marks of the ruler drawn by Euzebe on the ground annot be different for eah observer. The Solution There is obviously something that has not been done properly. To find out what went wrong, Lorentz transformations are used again. There are two events here: the firing of eah gun. Aording to Gontran, the guns shoot simultaneously and are 33 m apart. Therefore, x = x x = m 1 33 t = Version 10-Relativity 69

70 It was assumed here that Gontran is moving towards the positive x-axis aording to Euzebe so that Gontran uses primes. Giving a positive value to x' - x' 1, automatially means that the firing of the gun in front of the train is event and that the firing of the gun at the bak of the train is event 1. Aording to Lorentz transformations, the distane and the time between the events aording to Euzebe are v x x = γ ( x + v t ) t = γ t + ² m x = ( 33m ) t = ² = = 7 x 55m t s Thus, the distane between events, so between the bullet holes, is also 55 m aording to Euzebe. There will be a hole at x = 0 and a hole at x = 55 m, as predited by Gontran. The results of the transformation also show the mistake made. For Euzebe, the guns do not fire at the same time! The time t - t 1 = x 10-7 s means that gun (the one in front of the train) fires x 10-7 s after the gun at the bak of the train. During this time, the train, whih is moving at 0.8, has moved 35. m. Here is the solution of what happens aording to Euzebe. First, the gun in the bak fires and make a mark on the ground. praner.physis.louisville.edu/astrowiki/index.php/speial_relativity x 10-7 s later, the gun at the front of the train fires and leave a mark on the ground. Meanwhile, the train has moved forward by 35. m. 018 Version 10-Relativity 70

71 praner.physis.louisville.edu/astrowiki/index.php/speial_relativity The position of the hole on the ruler is, therefore, 35. m m = 55 m aording to Euzebe, as predited by Gontran. There is no paradox sine the preditions of both observers are the same. The Barn Paradox Here is another situation that seems paradoxial when the solution is not done properly. In this example, there is a 15 m long barn and a 0 m long pole. hyperphysis.phy-astr.gsu.edu/hbase/relativ/polebarn.html The pole is too long to enter into the barn unless it goes very fast. Ronnie takes this pole and runs at 0.8. Beause of length ontration, the pole is only 1 m long then. hyperphysis.phy-astr.gsu.edu/hbase/relativ/polebarn.html Barney, who is beside the door of the barn, opens the door to let the pole in and an lose the door sine the pole an fit entirely inside the barn. 018 Version 10-Relativity 71

72 This situation seems paradoxial, however, if Ronnie s point of view is taken. For him, the pole is at rest and the barn is oming towards him at 0.8. hyperphysis.phy-astr.gsu.edu/hbase/relativ/polebarn.html For Ronnie, the pole is 0 m long while the barn is only 9 m long beause of length ontration. The pole annot possibly fit into the barn and Barney annot lose the door! So, an the door be shut or not? If it an be shut aording to one observer, it must shut aording to every observer. Relativity hanges the time at whih an event ours but events annot appear or disappear with a hange of referene frames. To solve this mystery, it will be assumed first that the bak wall of the barn shatters when it is hit by the pole. In this ase, Barney an surely shut the door beause the pole gets out on the other side of the barn. Notie, however, that the door shuts before the wall breaks aording to Barney, while the wall breaks before the door shuts aording to Ronnie. The order of the two events is reversed depending on the observer. This is quite possible in relativity sine the time at whih an event ours hanges from one observer to another. It is possible that this hange reverses the order of the events. Don t worry, this inversion is impossible if an event is the ause of the other! Let s now assume the wall doesn t shatter upon impat. In this ase, the solution is more subtle. It will be shown a little further than nothing an travel faster than light aording to the equations of relativity. This means that when the front of the pole hits the wall and stops, the bak end of the pole does not stop immediately! Before the bak end of the pole stops, the information saying to the bak end has to stop must travel to the bak end of the pole. As this information annot go faster than light, the bak of the pole ontinues to move forwards for a while before stopping. This motion of the bak end of the pole that takes plae while the front end is stopped makes the pole even shorter than the ontrated length. So, this is what is happening aording to Barney. The pole enters the barn, and he shuts the door. Then, the front end of the pole hits the wall while the bak end of the pole ontinues its motion until the length of the pole is only 6.67 m. Then, the information arrives at the bak end of the pole saying that the pole is now at rest. But this pole is 0 m long at rest. So, it grows at the speed of light until it reahes 0 m long, and the door gets smashed. In Ronnie s referene frame, the moving barn hits the right end of the pole at rest. But the left end of the pole doesn t immediately know that there was a ontat and stays at rest. While the right end of the pole is moving towards the left at 0.8, the left end stays still, 018 Version 10-Relativity 7

73 and the pole gets shorter. This shortening of the pole ontinues until the information of the ollision with the barn arrives at the left end of the pole. When the information arrives, the pole is only 4 m long and is smaller than the 9 m long barn. Barney an then shut the door. As the perh is 1 m long when it is moving at 0.8, it extends at the speed of light until it is 1 m long. The pole then smashes the door, beause it is longer than the barn! There is no paradox. Barney an lose the door aording to both observers. Rest your brain a few moments while new onnetions between neurons are formed... Newton s laws were in agreement with Galilean relativity. Changing from one frame to another using Galilean transformations, the values of the momentum and the kineti energy aording to another observer are obtained. With these transformations, the momentum and the kineti energy are onserved for every observer in an elasti ollision. With the hanges made to relativity by Einstein, the results obtained with p = mu and E k = ½mu² are no longer valid (u is used sine this is the symbol used for the speed of an objet in this hapter). With these formulas, it an be shown that the momentum and the kineti energy are not onserved anymore for every observer in an elasti ollision when Lorentz transformations are used to pass from one frame to another. New formulas for the momentum and the kineti energy has to be found to preserve the priniples of onservation of p and E k. Momentum To find the new formula of the momentum, a ollision where the momentum is obviously onserved for an observer is onsidered. Then, this ollision is examined in the referential of another observer to determine how the momentum formula must be hanged so that there is also momentum onservation for this observer. Here is this proof. Then, the following formula for momentum of an objet of mass m moving at speed u is obtained. 018 Version 10-Relativity 73

74 Momentum mu p = = γ mu ² 1 u ² (γ is alulated with u, the speed of the objet.) Relativisti Energy Four-vetors There is a formalism in relativity that was not seen here. This formalism indiates that a vetor with 3 omponents has no plae in relativity. Instead, there are four-vetors, whih are a kind of 4-dimensional vetors. For example, the vetor x giving the distane between two events, whose omponents are ( x, y, z) annot exist in relativity. This vetor is atually part of the four-vetor s (the interval), whih has the following 4 omponents. ( t, x, y, z) (Note that not all vetors beome four-vetors in relativity. For example, the eletri and magneti fields beome the 6 independent omponents of a 4 x 4 matrix.) The Four-Vetor With Momentum The momentum annot have only 3 omponents in relativity. It must be part of a fourvetor with 4 omponents that looks like this. Then, what is this unknown 1 st omponent? (?, px, py, p z ) The last 3 omponents of this four-vetor an be obtained from the last 3 omponents of the interval. These omponents are obtained by multiplying the mass by the derivative of 018 Version 10-Relativity 74

75 the interval with respet to the proper time. Let s illustrate this for the x-omponents. Aording to what was said, the x-omponent of the momentum is p x dx = m dt 0 Using the time dilation formula dt = dt 0 1 v the momentum is p x = 1 1 v dx m dt The speed v in the square root is also the speed of the objet (u). Indeed, the objet must be at rest for the observer who measures the proper time. So, the speed between observers is also the speed of the objet aording to the observer who sees the objet moving. Thus v = u. Also, dx/dt is the speed of the objet. Therefore, p x = 1 1 u mu This is the momentum formula found in the previous setion. This allows us to find the unknown omponent. It is obtained by doing exatly the same manipulations (mass multiplied by the derivative with respet to the proper time) with the first omponent of the interval. As the first omponent is t, the unknown omponent is ( ) d t m dt Using the time dilation formula, the result is Version 10-Relativity 75

76 ( ) d t m dt dt = m dt 0 0 = = = v v u dt m dt m m (The v beame a u for the same reasons as those mentioned for p.) This is the first omponent of the four-vetor that inludes the momentum. What is this omponent? It remains now to find the meaning of this omponent is. We an have a better idea by alulating what this omponent looks like at low speed. The result is 1 u² m 1+ + m 1 ² u 1 m mu m + mu + The last line allows us to understand the result. The ½mu² is the kineti energy of the partile. This omponent is, therefore, linked to energy. We ll all this energy the relativisti energy. Therefore, the four-vetor, alled the energy-momentum four-vetor, is where E is the relativisti energy, whih is E, p x, p y, p z 018 Version 10-Relativity 76

77 Relativisti Energy E = 1 m² = γ m² 1 u (γ is alulated with u, the speed of the objet.) The low-speed approximation shows that the kineti energy in inluded in this energy. But it also shows that there is something more that depends only on the mass of the objet. This is the mass energy. Mass Energy Emass = m² It then beomes obvious that the relativisti energy is the sum of the mass energy and of the kineti energy. Relativisti Energy Relativisti Energy = Mass Energy + Kineti Energy Therefore, the kineti energy an be found. Thus, Kineti Energy (E k ) E = m² + E E = E m² k E = γ m² m² k k Ek ( γ ) = 1 m² (γ is alulated with u, the speed of the objet.) Note that this equation ould have been obtained by starting from the idea that work is equal to the hange in kineti energy. This means that dp dx E = W = F dx = dx dp u dp dt = dt = x k x x x x 018 Version 10-Relativity 77

78 Using the momentum formula in this equation would have led to the same result. All these energy formulas were obtained for the first time by Einstein in (However, it was realized later that the proofs given by Einstein were inomplete. The first orret proof was made by Max von Laue in 1911 and a still more general proof was made by Felix Klein in Is it not ironi? The equation more often assoiated with Einstein was not proven by Einstein...) What is the meaning of E = m²? First, this equation means that mass has energy. This energy an be onverted into other forms of energy. It must be said that it is an astounding amount of energy. The energy of 1 gram of matter is ( m ) s 8 13 E = m² = 0.001kg 3 10 = 9 10 J This value orresponds approximately to the energy released in the atomi explosion of the Hiroshima bomb! However, this onversion is quite diffiult to do. It an be done by making matter and antimatter touh. In this ase, matter and antimatter annihilate eah other and the entire mass is onverted into another form of energy. If a proton and an antiproton touh, there won t be any matter left after the ontat, there will be only light. It must be said that this proess does not often our beause free antimatter does not exist anymore in the universe. Antimatter is known beause some was reated in partile aelerators, and beause some is reated when osmi rays hit the atmosphere. In high-energy ollisions, part of the kineti energy is transformed into mass energy resulting in the reation an equal amount of matter and antimatter. Seond, this equation means that energy has mass. Here are some examples of this Mass of the Kineti Energy A moving objet has a mass greater than an idential objet at rest. This is so beause the moving objet has some kineti energy and this energy has mass. The total mass of the objet is then 018 Version 10-Relativity 78

79 m = m + m tot kineti energy Ek = m + ² ( γ 1 ) m² = m + ² = m + m = γ m ( γ 1) A Completely Inelasti Collision Two kg objets arriving at idential speeds make a ompletely inelasti ollision. After the ollision, the two balls form a 4 kg objet at rest and the kineti energy is onverted into heat. If the mass of this 4 kg objet is measured very preisely, we will disover that its mass is greater than 4 kg beause the heat, whih is a form of energy, has a mass. (In fat, the total mass has not hanged in this ollision. Before the ollision, eah objet has a mass equivalent to the sum of the mass of the objet and the mass of the kineti energy. After the ollision, the mass is equal to the sum of the mass of the two objets glued together and the mass of the heat. These masses (before and after the ollision) are idential.) Mass of the Potential Energy If a brik is lifted, the gravitational energy inreases. As this form of energy also has mass, the mass of the Earth-brik system inreases slightly when the brik is lifted. For the same reason, the mass of the Earth-Moon system would inrease if the Moon were to move away from the Earth. Thus, the Earth and the Moon together have a lower mass than the mass they would have if the Earth and the Moon were separated. Similarly, the mass of the hydrogen atom inreases if the eletron moves away from the nuleus. As eletri energy inreases when two harges are separated, the mass of the system inreases when the harges are pulled apart. The mass of a proton and an eletron far from eah other is, therefore, greater than the mass of the hydrogen atom. However, in most ases, the hange in mass is so small that it is impossible to measure it. It is impossible to measure the mass hange when a brik is lifted or when an eletron is removed from an atom. In fat, there are only a few ases in whih there is a signifiant hange in mass that an be measured. The only instane when this happens that will be seen here is the atomi nuleus. 018 Version 10-Relativity 79

80 Atomi Nuleus There is a deline in mass when an atomi nuleus is formed. If a proton has to be removed from an atomi nuleus, energy must be provided, whih inreases the mass. If all the protons and neutrons in a nuleus are taken out, one after the other, the mass inreases even more. This means that the mass of the atomi nuleus is smaller than the mass of the protons and neutrons when they are far apart from eah other. In this ase, there is a variation of the order of 1%, and this variation is easily measured. This variation of mass will be dealt with in the hapter on nulear physis. The Eletronvolt A small note to omplete this setion. When the energy of a subatomi partile (suh as a proton or an eletron) is alulated, the values obtained are generally very small if alulated in joules. Physiists usually use the eletronvolt to measure the energy of suh small partiles. There is an exat definition of the eletronvolt (that we will see in eletriity and magnetism) but we an be ontent here with this simple following definition. Eletronvolt (ev) 1eV = J Examples Example a) What is the mass energy of an eletron, whose mass is 9.11 x kg? The mass energy is E = m² m ( ) s = kg 3 10 = = 511keV J b) What is the kineti energy of an eletron travelling at 0.8? The kineti energy is 018 Version 10-Relativity 80

81 Ek ( γ ) = 1 m² 1 = 1 m² 1 ( 0,8) 5 = kev 3 = 340keV Note that with ½mu², the kineti energy would have been 164 kev. Example How muh energy must be given to a 100-ton spaeship initially at rest so that it moves at 0.8? The energy is E = E E = γ m² γ m² f ( γ f γ i ) 1 m² 5 = 1 100, = 6 10 f = i i 8 m ( kg )( ) J As all the power stations on Earth produe approximately 6.8 x J eah year (010 figure), the energy that must be given to the spaeship orresponds to 88 years of all the eletriity produed on Earth! Suh a feat will not be ahieved soon! It looks so easy in siene fition movies... s 018 Version 10-Relativity 81

82 Example A 1-gram objet moving at 0.8 makes a perfetly inelasti ollision in one dimension with a 15 g objet at rest. a) What is the speed of the 16 g objet after the ollision? physis.tutorvista.om/momentum/inelasti-ollision.html As in any ollision, momentum is onserved. Before the ollision, the momentum is p before = γ mu 1 = 0.001kg ( ) = 400, 000 After the ollision, the momentum is kgm s pafter = γ mu = γ (0.016 kg) u Sine momentum is onserved, we have p before kgm 400, 000 = γ (0.016 kg) u s = p.5 10 = γ u 7 m s after It only remains to solve this equation for u. 018 Version 10-Relativity 8

83 .5 10 = 7 m s 7 m s.5 10 = u ( ) u ( ) ( ) 7 m.5 10 s = ( ) ( ) 1 ( u ) = 1 u 1 1 u = (With Newton s laws, the answer would have been u = 0.05.) = u u u u b) How muh kineti energy is lost in the ollision? Before the ollision, the kineti energy is E k before ( γ 1) = 13 1 = ( 0.8 ) = 6 10 m J 8 m ( kg )( s ) After the ollision, the kineti energy is E k after ( γ 1) = m 1 = ( ) = J 8 m ( kg )( s ) The hange in kineti energy is thus = 1 13 Ek J 6 10 J = J 018 Version 10-Relativity 83

84 The kineti energy has, therefore, dereased by 5.5 x J in this ollision. (This energy would normally be in the form of heat after an inelasti ollision but the energy here is pratially equal to the energy released by the atomi bomb of Hiroshima. The 16 g objet would surely be vaporized.) Relation Between E and p There is an interesting link between E and p. Let s alulate what is E² (p)². Thus ( ) 4 E p = γ m γ m u ( ) u ( 1 ) = γ u m = γ 4 u 4 u ( 1 ) 1 = 1 = m Link Between Relativisti Energy and Momentum m ( ) ( ) E p = m Note that the term on the right is a relativisti invariant. All the observers get the same value when they alulate E² (p)². (In fat, many relativisti invariants an be obtained with four-vetors. If there are two a, a, a, a and b, b, b, b, then four-vetors ( ) ( ) a1b 1 ab a3b3 a4b4 is neessarily an invariant. For example, if both a and b are the four-vetor ( t, x, y, z), then the following result is obtained. ) ( t) ( x) ( y) ( z) m 4 = onstant 018 Version 10-Relativity 84

85 Example A partile at rest having a 50 MeV mass energy deays into two partiles whih move in opposite diretions. The mass energy of one of these partiles is 15 MeV, and the mass energy of the other partile is 5 MeV. What are the speeds of the two partiles? As in any proess, the relativisti energy is onserved. Sine the mass energy is 50 MeV before the deay and 40 MeV after the deay, the kineti energy of the two partiles after the deay must be 10 MeV. Sine the momentum is zero before the deay, it must also be zero after the deay. So, we have the following equations. E + E = 10MeV k1 k p + p = 0 1 Using the formulas for p and E k, these equations beome MeV + MeV MeV = u1 u 1 ( ) 1 ( ) 1 15MeV 1 5MeV u 1 + u 0 u1 = u 1 ( ) 1 ( ) We then have two equations with two unknowns, whih an be solved. But as it is very diffiult to solve this system, another approah will be used. This approah uses the following equation. ( ) ( ) E p = m Thus, sine the momentum is onserved, we have 018 Version 10-Relativity 85

86 p p p = p 1 = p 1 = p 1 E m = E m Sine E1 + E = 50MeV the equation beomes ( 50 ) E m = MeV E m E m = 500MeV 100MeV E + E m m = 500MeV 100MeV E m Solving for the energy of partile 1, the result is 500MeV m + m E1 = 100MeV Using the values for the mass energy, the energy is E 1 = ( ) ( ) 500MeV 5MeV + 15MeV = 1MeV From there, it an easily be found that 100MeV E = 50MeV E 1 = 9MeV As the relativisti energy is the sum of the mass energy and the kineti energy, it an easily be found that the kineti energy is 6 MeV for partile 1 and 4 MeV for partile. The sum of these kineti energies is indeed 10 MeV, as predited. The value of γ for eah partile an then be found with the relativisti energy formula. 018 Version 10-Relativity 86

87 E = γ m E = γ m MeV = γ 15MeV 9MeV = γ 5MeV 1 1 γ = 1.4 γ = With the γ s, the speeds an be found = 1.16 = 1 1 u1 u ( ) ( ) 1 u = u = If E Is Conserved, Then p Must Also Be Conserved Small interesting note. If the relativisti energy is onserved in a ollision, then it an be proven, using the priniple of relativity, that the momentum is automatially a onserved quantity. The proof an be seen here: Maximum Speed Let s have a look at the graphs of the momentum and of the kineti energy of an objet as funtions of its speed. On eah graph, the values aording to Newton s Physis (in blue) and the values aording to Einstein s physis (in red) are plotted. 018 Version 10-Relativity 87

88 Three things an be noted. 1) The values given by the relativisti formulas are always larger than those given by the non-relativisti formulas. An example of this an be seen in example (340 kev vs. 164 kev). ) The values are virtually idential for small speeds (say, less than 0% of the speed of light). The non-relativisti equations are, therefore, very good approximations at low speeds. 3) The values of p and E k are infinite at u =. This last point shows that it is impossible for an objet to reah the speed of light beause it will take an infinite amount of energy for the objet to reah this speed. The speed of an objet annot exeed the speed of light. Maximum Speed No objet with a non-vanishing mass an reah the speed of light. The media don t seem to be aware of this fat. 018 Version 10-Relativity 88

89 However, this speed an be reahed if the mass is zero. In this ase, the energy, whih is E = γ m may not be infinite even if γ is infinite beause m = 0. But then, the equation gives E = 0 if γ is not infinite. If the energy is zero, then this objet does not exist. Therefore, an objet without any mass an only exist if its speed is equal to the speed of light. The only thing known that has no mass in all urrent theories is the photon, whih is a light partile (we will talk about it in the next hapter). Photon Speed Photons always travel at the speed of light beause their mass is zero. Moreover, for photons, the relation between E and p is ( ) ( ) E p = m E ( p) = 0 sine m is 0. Therefore, Link Between Momentum and Energy for Photons E = p 018 Version 10-Relativity 89

90 Faster Than The fat that massive objets annot exeed the speed of light does not mean that nothing an go faster than the speed of light. Some things an go faster than light, but they are not material things. For example, the bright spot made by a laser pointer on a sreen an go faster than light. By turning the pointer by 5, the spot moves on the sreen. If the sreen is far, the displaement of the spot an be very large. If the sreen were 1 billion km away and the pointer is turned by 5 in 1 seond, the spot would move 87,000,000 km in 1 seond on the sreen. This speed is greater than the speed of light. However, no partile has made this displaement sine the spot is onstantly made of new photons. Also, remember that, sometimes, it is possible to see an objet moving faster than light even if it s travelling slower than light. This phenomenon was enountered with the example of Florene and Alex. In that example, Florene sees Alex travel 6 ly in 4 years, giving her an apparent speed of 1.5. However, when Florene takes into aount the time of arrival of the light, she observes that the speed is atually 0.6. These speeds apparently higher than the speed of light an be observed in the universe. For example, the jets of gas made by the quasar 3C73 seem to travel faster than light as seen from Earth. en.wikipedia.org/wiki/3c_ Version 10-Relativity 90

91 The Relativity Priniple The laws of physis are the same for all the observers moving at a onstant speed. Galilean Transformations x = x + vt x = x vt y = y y = y z = z z = z The observer who sees the other travelling towards the negative x-axis uses primes Galilean Veloity Transformations u = u + v u = u v x x x x u = u u = u y y y y u = u u = u z z z z Galilean Aeleration Transformations a = a a = a x x x x a = a a = a y y y y a = a a = a z z z z Einstein s Postulates 1) The laws of physis are the same for all the observers moving at a onstant speed. ) The speed of light in a vauum is always 300,000 km/s (atually 99, km/s) for all observers. Time in Relativity If an observer is moving with respet to another, the moments at whih events our is not the same for these two observers. Simultaneity If two events are simultaneous for an observer A, they are not simultaneous for all the observers moving with respet to observer A. 018 Version 10-Relativity 91

92 γ Fator γ = 1 1 v Time Dilation t = γ t = 0 t 0 1 v Proper Time t 0 Proper time is the time between two events aording to an observer who observes the two events in the same plae. Length Contration L = L = ² 0 1 v ² L0 γ Approximations of the γ Fator if v << 1 1 v v 1+ + v + 1 v 1 Relativisti Doppler Effet T = T 0 f = f 0 v + v + v v (v is the speed of the soure aording to a motionless observer.) The sign onvention for speed stays the same as it was. The positive diretion is from the soure towards the observer. The speed of the soure is positive if it is heading towards the observer and negative if it moves away from the observer. 018 Version 10-Relativity 9

93 Lorentz Transformations (1) ( ) γ ( ) x = γ x vt x = x + vt y = y y = y z = z z = z vx vx t = γ t t = γ t + ² ² (v is the speed of an observer aording to another observer.) Lorentz Transformations () ( ) γ ( ) x = γ x v t x = x + v t y = y y = y z = z z = z v x v x t = γ t t = γ t + ² ² (v is the speed of an observer aording to another observer.) Whih Observer Uses Primes? The observer who sees the other travelling towards the negative x-axis uses primes. The Interval ( s) = ( t ) ( x) ( y) ( z) ( s ) = ( s) Proper Time Between Two Events ( t ) = ( s) 0 If ( s)² > Version 10-Relativity 93

94 Veloities Transformations ux v u x + v u x = ux = vux vu x 1 1+ ² ² v² v² uy 1 ² u y 1 ² u y = uy = vux vu x 1 1+ ² ² v² v² uz 1 ² u z 1 ² u z = uz = vux vu x 1 1+ ² ² (v is the speed of an observer aording to another observer.) Momentum mu p = = γ mu ² 1 u ² (γ is alulated with u, the speed of the objet.) Relativisti Energy Relativisti Energy = Mass Energy + Kineti Energy 1 E = m = γ m 1 u ² (γ is alulated with u, the speed of the objet.) Mass Energy Emass = m² Kineti Energy (E k ) Ek ( γ ) = 1 m² (γ is alulated with u, the speed of the objet.) Link Between Relativisti Energy and Momentum ( ) ( ) E p = m 018 Version 10-Relativity 94

95 Maximum Speed No objet with a mass an reah the speed of light. Photon Speed Photons always travel at the speed of light beause their mass is zero. Link Between Momentum and Energy for Photons E = p 10.7 Time Dilation 1. Adolf leaves Earth at 80% of the speed of light to go to the star Proyon, 11.4 ly away from Earth. fr.depositphotos.om/ /stok-photo-spaeship.html a) What is the duration of the trip aording to Eva, who has remained on Earth? b) What is the duration of the trip aording to Adolf?. A lok remains on Earth while another lok is moving relative to the Earth at a speed of 0.6. Initially, the two loks are synhronized. The moving lok goes 90 million km away (aording to observers on Earth) and omes bak on Earth at the same speed. What will the differene between the time indiated by the moving lok and the lok whih has remained on Earth be one the moving lok has returned to Earth? 3. Two twins, Otavius and Augustus, leave Earth at the same time, the day of their 0 th birthday, to go on another planet loated 1 ly from Earth. Otavius travels at 0.8, and Augustus travels at 0.6. What will the age of the twins be when Augustus finally arrives on the planet? 018 Version 10-Relativity 95

96 10.8 Length Contration 4. A baseball has a 7. m diameter at rest. What will the width of the ball (L in the figure) be if it moves at 87% of the speed of light? 5. An alien past beside Tom. Tom and the alien are both holding a 1 m long ruler (length at rest) as shown in the figure. Aording to Tom, the ruler held by the alien is only 80 m long. a) What is the speed of the alien aording to Tom? b) What is the length of Tom s ruler aording to the alien? 6. The distane between the Earth and a star is 500 ly aording to an observer on Earth. How fast must a spaeship travel so that the distane between the Earth and the star is only 0 ly aording to the observers in the spaeship? 7. Raoul wants to travel to the star Betelgeuse, loated 643 ly away from the Earth (aording to an observer on Earth). How fast must he go so that the trip lasts only 0 years aording to Raoul? 8. Ahmed looks at Yitzhak s spaeship passing at 0.9. Ahmed noted that, aording to him, the two spaeships have the same length. What is the ratio of the lengths of the spaeships aording to Yitzhak? s3-us-west-.amazonaws.om/oa/dofiles/5466e916f bdd0100/5466e916f bdd0100.html 018 Version 10-Relativity 96

97 9. A muon lives only. µs when it is at rest. Assume that newly reated muons are launhed at one end of a tunnel and are reeived at the other end of the tunnel. The tunnel is 1 km long. a) How fast must a muon go so that it an travel 1 km during his lifetime? b) At this speed, what is the length of the tunnel aording to the muon? 10. Agathe measures the time it takes for a train going at 80% of the speed of light to pass beside a post. She starts her stopwath when the front of the train is beside the pole and she stops it when the bak of the train is beside the pole. renshaw.telein.om/papers/simiee/simiee.stm On the train, Justin also measures the time it takes the train to past the post exatly as Agathe does it. The train is 000 m long when it is at rest. a) What is the length of the train aording to Agathe? b) How long does the train take to past the post aording to Agathe? ) How long does the train take to past the post aording to Justin? 10.9 The Doppler Effet with Light (Take ) 11. A soure on Earth emits a 98.1 MHz eletromagneti wave. What is the frequeny reeived by Terene if he is in a spaeship moving at 95% of the speed of light,... a) towards Earth? b) away from Earth? 1. A star emits a light wave with a 550 nm wavelength (green). What is the wavelength of the wave reeived by an observer who sees the star moving away from him at 30% of the speed of light? 018 Version 10-Relativity 97

98 13. How fast must an observer move towards a red t-shirt (650 nm wavelength) so that the t-shirt looks blue to him (470 nm wavelength)? A soure on Earth emits a 100 GHz eletromagneti wave towards a spaeship heading towards the Earth at 60% of the speed of light. The wave reflets off the spaeship and returns to Earth. Then, what is the frequeny of the refleted waves reeived by the observers on Earth? 15. In the following situation, what is the frequeny reeived by William if Raphael reeives a 00 MHz frequeny and the spaeship emits a 400 MHz frequeny? and fr.depositphotos.om/ /stok-photo-spaeship.html 16. In the following situation, Ursula s spaeship emits a flash of light every 10 seonds (time aording to Ursula). and fr.depositphotos.om/ /stok-photo-spaeship.html a) What is the time between flashes aording to Wilfrid s observations? b) What is the time between flashes as seen by Wilfrid? ) What is the time between flashes aording to Flavien s observations? d) What is the time between flashes as seen by Flavien? 018 Version 10-Relativity 98

99 17. In the following situation, Ursula s ship emits flashes of light at regular intervals. Wilfrid sees a flash every 4 seonds, and Flavien sees a flash every 9 seonds. and fr.depositphotos.om/ /stok-photo-spaeship.html a) What is the time between the flashes aording to Ursula? b) What is the speed of Ursula s ship aording to Flavian and Wilfrid? Lorentz Transformations 18. Aording to Gertrude, the two explosions are simultaneous and 0 ly from eah other. and fr.depositphotos.om/ /stok-photo-spaeship.html a) What is the time between the explosions aording to Sidney? (Speify whih explosion ourred first aording to Sidney.) b) What is the distane between the positions where the explosions ourred aording to Sidney? 19. Aording to Gertrude, explosion ourred years before explosion 1, and the distane between the two positions where the explosions ourred is 0 ly. and fr.depositphotos.om/ /stok-photo-spaeship.html a) What is the time between the explosions aording to Sidney? (Speify whih explosion ourred first aording to Sidney.) b) What is the distane between the positions where the explosions ourred aording to Sidney? 018 Version 10-Relativity 99

100 0. A ray of light passes from the bak to the front of a train. This train is moving at 0.8 relative to the ground. On the edge of the trak, Jean-Marie is looking at the moving train. The following figure illustrates the situation from the point of view of a passenger on the train (alled Rodolphe). hmmthatsfunny.wordpress.om How long does it take for the light to pass from the bak to the front of the train aording to Jean-Marie? 1. A ray of light passes from the front to the bak of a train. This train is moving at 0.8 relative to the ground. On the edge of the trak, Jean-Marie is looking at the moving train. The following figure illustrates the situation from the point of view of a passenger on the train (alled Rodolphe). hmmthatsfunny.wordpress.om How long does it take for the light to pass from the front to the bak of the train aording to Jean-Marie?. Gontran has fixed two guns at eah end of his train ar whih has a length of 33 m at rest. When he presses a button, the two guns shoot simultaneously (aording to Gontran) at the ground. This train is moving at a speed of 0.95 relative to the ground. On the edge of the trak, Euzebe looks at the train pass. The following figure illustrates the situation from Gontran s point of view. hmmthatsfunny.wordpress.om a) What is the distane between the bullet holes aording to Euzebe? b) What is the time between the firing of the two guns aording to Euzebe? (Speify whih gun has fired first.) 018 Version 10-Relativity 100

101 3. In the situation shown in the figure, Gertrude measures that explosion ourred 4 years before explosion 1. What should the speed of Sidney be so that the explosions are simultaneous for him? (Even if the spaeship is drawn as a spaeship moving towards the right, it may have to go towards the left to observe simultaneous explosions.) and fr.depositphotos.om/ /stok-photo-spaeship.html 4. Serge and Tatiana both measure the time it takes for a train going at 60% of the speed of light to go from one post to another. The distane between the poles is 00 m aording to Tatiana. The train is 70 m long aording to Tatiana. They start their stopwath when the front of the train is beside the first pole and they stop them when the bak of the train is beside the seond post. a) What is the time measured by Tatiana? b) What is the time measured by Serge? ) What is the distane between the posts aording to Serge? renshaw.telein.om/papers/simiee/simiee.stm 018 Version 10-Relativity 101

102 5. In the situation shown in the figure, Sidney observes that explosion ourred 8 years before explosion 1 and that the distane between the positions where the explosions ourred is 5 ly. fr.depositphotos.om/ /stok-photo-spaeship.html a) What is the distane between the positions where the explosions ourred aording to Gertrude if explosion 1 ourred 7 years before explosion aording to Gertrude? b) What is the proper time between the events? Veloities Transformations 6. Aline observes Kim s and Mike s spaeship heading towards eah other with a speed of What is the speed of Mike s spaeship aording to Kim? and fr.depositphotos.om/ /stok-photo-spaeship.html 7. Aording to Bertha, Osar is travelling at 0.8 and is following Ivan who is travelling at 0.9. Osar launhes a missile towards Ivan. Aording to Osar, the speed of the missile is and fr.depositphotos.om/ /stok-photo-spaeship.html a) What is the speed of the missile aording to Bertha? b) Will the missile ath Ivan? 018 Version 10-Relativity 10

103 8. Two 46 m long trains (aording to Erin) are heading towards eah other as shown in the figure. What is the length of John s train aording to Claudette? praner.physis.louisville.edu/astrowiki/index.php/speial_relativity 9. Two spaeships are heading towards eah other as shown in the figure. Annabelle s spaeship emits a 100 MHz signal (aording to Annabelle). What is the frequeny reeived by Esteban? and fr.depositphotos.om/ /stok-photo-spaeship.html 30. Pasal leaves Earth at 80% of the speed of light. When he is 0.9 ly away from Earth (distane aording to Arielle, an observer on Earth), Pablo starts to pursue him at 95% of the speed of light. fr.depositphotos.om/ /stok-photo-spaeship.html a) What is Pablo s speed aording to Pasal? b) Who measures the proper time between the instant Pablo leaves the Earth and the instant Pablo athes up with Pasal? How long will it take for Pablo to ath up with Pasal... ) aording to Arielle? d) aording to Pablo? e) aording to Pasal? 018 Version 10-Relativity 103