The first ever photograph of light as both a particle and wave

March 2, 2015

(Phys.org)—Light behaves both as a particle and as a wave. Since the days of Einstein, scientists have been trying to directly observe both of these aspects of light at the same time. Now, scientists at EPFL have succeeded in capturing the first-ever snapshot of this dual behavior.

Quantum mechanics tells us that light can behave simultaneously as a particle or a wave. However, there has never been an experiment able to capture both natures of light at the same time; the closest we have come is seeing either wave or particle, but always at different times. Taking a radically different experimental approach, EPFL scientists have now been able to take the first ever snapshot of light behaving both as a wave and as a particle. The breakthrough work is published in Nature Communications.

When UV light hits a metal surface, it causes an emission ofelectrons. Albert Einstein explained this "photoelectric" effect by proposing that light – thought to only be a wave – is also a stream of particles. Even though a variety of experiments have successfully observed both the particle- and wave-like behaviors of light, they have never been able to observe both at the same time.

A research team led by Fabrizio Carbone at EPFL has now carried out an experiment with a clever twist: using electrons to image light. The researchers have captured, for the first time ever, a single snapshot of light behaving simultaneously as both a wave and a stream of particles.

The experiment is set up like this: A pulse of laser light is fired at a tiny metallic nanowire. The laser adds energy to the charged particles in the nanowire, causing them to vibrate. Light travels along this tiny wire in two possible directions, like cars on a highway. When waves traveling in opposite directions meet each other they form a new wave that looks like it is standing in place. Here, this standing wave becomes the source of light for the experiment, radiating around the nanowire.

 

This is where the experiment's trick comes in: The scientists shot a stream of electrons close to the nanowire, using them to image the standing wave of light. As the electrons interacted with the confined light on the nanowire, they either sped up or slowed down. Using the ultrafast microscope to image the position where this change in speed occurred, Carbone's team could now visualize the standing wave, which acts as a fingerprint of the wave-nature of light.

Credit: Fabrizio Carbone/EPFL

While this phenomenon shows the wave-like nature of light, it simultaneously demonstrated its particle aspect as well. As the electrons pass close to the standing wave of light, they "hit" the light's particles, the photons. As mentioned above, this affects their speed, making them move faster or slower. This change in speed appears as an exchange of energy "packets" (quanta) between electrons and photons. The very occurrence of these energy packets shows that the light on the nanowire behaves as a particle.

"This experiment demonstrates that, for the first time ever, we can film quantum mechanics – and its paradoxical nature – directly," says Fabrizio Carbone. In addition, the importance of this pioneering work can extend beyond fundamental science and to future technologies. As Carbone explains: "Being able to image and control quantum phenomena at the nanometer scale like this opens up a new route towards quantum computing."

Chap20 special relativity_理学_高等教育_教育专区。Chapter 20 The Special Theory of Relativity Key wards antiparticle 反粒子 electron-positron annihila

Chapter 20 The Special Theory of Relativity Key wards antiparticle 反粒子 electron-positron annihilation 正负电子湮灭 the principle of relativity 相对性原理 the principle of the constancy of the speed of light 光速不变原理 time dilation proper time length contraction proper length (rest length) 度，静长 时间膨胀 本征时，原时，固有时 长度收缩 本征长度，原长，固有长 Lorentz transformation 洛伦兹变换 space-time coordinates of an event 事件的时空坐标 the relativity of simultaneity 同时的相对性 relativistic momentum and energy 相对论的动量和能量 relation of mass and energy relation of mass and velocity 质能关系 质速关系 relation of energy and momentum 能量动量关系 What do you know about the Theory of relativity? Can you say something about the theory of relativity? Near the end of the 19th century, many physicists thought that all the important laws of physics had been discovered and that there was little left for them to do other than work out the remaining details. However, the discovery of radioactivity by Becquerel (贝克勒尔, 法国物理学家) in 1896, the theoretical papers of Planck (普朗克,德国物理学家, 量子论确立者) in 1897 and Einstein in 1905, and the work of Rutherford (卢瑟福,英国 物理学家,化学家), Millikan (密立根, 美国物理学家), Bohr (波尔 丹 麦 物 理 学 家 ), de Broglie ( 德 布 罗 意 , 法 国 物 理 学 家 ), Schr?dinger (薛定谔, 奥地利理论物理学家), Heisenberg (海森堡, 德国物理学家), and others in the early twentieth century led to two completely new theories: relativity and quantum mechanics. These theories revolutionized the world of science and became the foundation for new technologies that have changed the face of civilization. 20-1 Troubles with classical physics Classical physics had many triumphs, but there are some difficulties. 1(a). Troubles with our ideas about time Pions (π + or π –) created at rest decay (to other particles) with an average lifetime of only 26.0 ns. While in one particular experiment, pions created in motion at a speed of v =0.913c traveled in the laboratory an average distance of D=17.4 m before decaying→decaying in a time given by D/v=63.7ns. i.e. trest= 26.0 ns, tmotion≠ trest time dilation (时间膨胀) 1(b). Troubles with our ideas about length The distance between the markers =17.4m. In pion frame, speed of the frame u=0.913c, lifetime=26ns, D'=7.1m. Lmotion≠ Lrest Length contraction (长度收缩) Newtonian physics: Time and space are both universal coordinates having identical value for all observers. 2. Troubles with our ideas about speed If the speed of ball is faster than light… Person A throws a ball toward person B at a speed faster than light, the light signal carrying the view of A throwing the ball travels to observer O at speed c,….. The light signal from B (catching the ball) arrives before the light signal from A→ violations of cause and effect. 3. Troubles with our ideas about energy Electron-positron (正电子) annihilation: radiation appears after annihilation, the radiation energy is absorbed by the walls of the container, ΔEint >0, but ΔWext=0, Q=0, no energy enters or leaves the system boundary. Where is the energy from? It appears that the law of conservation of energy is violated in this process. 4. Troubles with our ideas about light A paradox (佯谬): If you move at the speed of light parallel to a light beam traveling in empty space, you would observe “static” electric and magnetic field patterns →violates the theory of electromagnetism. Two choices to resolve this paradox:either electromagnetic theory was wrong or else the classical kinematics that permits an observer to travel along a light beam was wrong. Einstein put his faith in electromagnetic theory and sought an alternative to the kinematics of Galileo and Newton→The theory of relativity. The obstacle which the classical mechanics faced is due to that the view of classical space-time has some problems. The relativity has totally changed the view of classical space-time. What’s the old and classical view of space-time? Our ideas about space and time Newton’s ideas: absolute space and absolute time Galilean transformation S' S r. R r' p r = r′ + R v = v '+ u a = a′ absolute space absolute time u Galilean principle of relativity: The laws of mechanics are the same in all inertial reference frame. 1）The absoluteness of simultaneity---If one event is happen simultaneous in one RF, it is also happen simultaneous in other RF. Y O Y' O’ x1 x2 X' X 2）The measurement of time interval is absolutely. K K' The measurement of elapsed time of one process is the same for different reference frames. 3) The measurement of space interval is absolutely. The length measured in K’ Y Y' u x '1 x1 x'2 X' x2 l'= x ? x ' 2 The length measured in K ' 1 X l = x 2 ? x1 l ' = x ? x = x2 ? x1 = l ' 2 ' 1 ( t 2 = t1 ) 4）Transform of velocity Y o Y' o’ u u' u' y ' = uy u' x ' = ux ? v X z' z X'Galileo’s transform of velocity u' = u uC = c + v > c c v reason： X' X The absoluteness of the view of space -time! x' = x ? vt y' = y ' z =z t' = t 2-2LT及经典时空观.exe These concepts are the reflection of life in human’s mind under low velocity. In the sixties of the 19th century, Maxwell proposed a set of equations to explained the phenomena of electromagnetics: ?2E ?2E ?2E 1 ?2E + + = 2 2 2 ?x ?y ?z C 2 ?t2 ?2H ?2H ?2H + + 2 2 ?x ?y ?z2 C = 1 1 ?2H = C 2 ?t2 ε 0μ0 * The Michelson-Morley Experiment Comparing with the fact that the mechanical wave propagates in the elastic medium, it was therefore natural to expect that some kind of medium supports the propagation of light and other electromagnetic waves. This proposed medium was called the ether (以太). ≈ 3 .0 × 1 0 8 m /s Attempt looking for “ether” Light c A，v B，v According to classical transformation of velocity: A: c + v B: c?v The experimental result is that light speed is c no matter which observer measured. The ideas about space and time must be altered. Relativity ---The theory on the relation between the view of space-time and substance. It consists of two rather different theories, the special theory and the general theory. (1) Special relativity: The theory on the the view of spacetime for IRF. (2) General relativity: The theory on the the view of spacetime for general RF and gravity. 20-2 The postulates (假定) of special relativity The principle of relativity: The laws of physics are the same in all inertial reference frames. The principle of the constancy of the speed of the light: The speed of light in free space has the same value c in all inertial reference frames. Postulate 1 is merely an extension of the newtonian principle of relativity to include all types of physical measurements (not just those that are mechanical). For example, two IRFs, K and K', with constant relative velocity: K K' Mechanical law d (m v ) d ( m ' v ') F = F '= dt dt ' Electrics law K ⊕ K' Q1 R Q2 ⊕ Q'1 ⊕ R’ Q'2 ⊕ Q1Q 2 ? F = K r 2 R Magnetics law Q '1 Q ' 2 ? F '= K ' r' 2 R' I I' Postulate 2 describes a common property of all waves. Classical physics places no upper limit on the speed that an object may attain; relativity does impose such a limiting speed, which, by the first postulate, must be the same for all frames of reference. The two postulates taken together have another consequence: they imply that it is impossible to accelerate a particle to a speed greater than c, no matter how much kinetic energy we give it! The electrons were accelerated by a large voltage difference, … Note: 1） The validity condition of the principle of the constancy of the speed of light: a) Inertial reference frame; b) free space (the velocity of light in medium =c/n). 2）Do not think that the special relativity is the direct result of Michelson-Morley experiment, it is the summarization of many experiments nearly in half century; surely, Michelson-Morley experiment has important effect on confirming the special relativity. 20-3 Consequences of Einstein’s the postulates 1. The relativity of time For S', 2 L0 Δt0 = c (20-1) Δt0----proper time, measured by an observer (S’ in this case) relative to whom the clock is at rest. How about Δt ?---- measured by an observer (S in this case) relative to whom the clock is in motion . For S, 2L Δt = c 2 L02 + (uΔt / 2)2 = c Δ t0 (20-2) (20-3) Then Δt = 1 ? u 2 / c2 Δt≥ Δt0 ------time dilation (膨胀): all observers in motion relative to the clock measure longer intervals. The time intervals are not the same in two frames. In Section 20-1, for pion, Δt0=26.0ns, u=0.913c, Δt = Δt0 1? u / c 2 2 = 26.0ns 1 ? 0.913 2 =63.7ns in agreement with the measured value. 2. The relativity of length For S', L0---proper length, measured by an observer who is at rest with respect to the object being measured. For S, L ---measured by an observer who is in motion with respect to the object being measured. a flash emitted at position A→ arrives in mirror at position B, time △t1, distance traveled by light c△t1; in another respect, the light traveled a distance L+u△t1 , ∴c△t1=L+u△t1 (u is the velocity of train). In return trip, the flash from mirror at position B→detector at position C, time △t2, distance traveled by light in this time interval is c△t2; in another respect, the light traveled a distance L – u△t2, ∴ c△t2= L–u△t2, L L 2L 1 Δ t = Δ t1 + Δ t 2 = + = c?u c+u c 1 ? u2 / c2 comparing with Δt = Δ t0 1 ? u2 / c2 2 L0 = c 2 1 1 ? u2 / c2 2 L = L0 1 ? u / c L≤L0 ------length contraction (收缩): all observers in motion relative to the object being measured measure shorter length. The space intervals are not the same in two frames. Note: length contraction is only for dimensions along the direction of motion; length measurements transverse to the direction of motion are unaffected by the relative motion. In Section 20-1, L0=17.4m, u=0.913c, L = (17.4m ) 1 ? (0.913) 2 = 7.1m Under ordinary circumstances, u<< c and the effects of length contraction are far too small to be observed. Example (sample 20-1 p456) For muons (produced by cosmic rays collide with air molecules), Δt0=2.2μs. If the muons flight a distance of 100km in the reference frame of the Earth, find u=? Solution 1: In Earth’s reference frame: L0= 100km (In Earth’s frame), Δt0=2.2μs (In muons’frame), L0 1 0 0 k m 100 × 103 Δt = = = ( i n E a r t h 's f r a m e ) u u u Δ t0 100 × 103 2 .2 × 1 0 -6 S in ce Δ t = ∴ = → u = 0 .9 9 9 9 7 8 c 2 2 2 2 u 1? u /c 1? u /c Solution 2: In muon’s frame : Since L = L0 1? u2 / c2 L= uΔt0= u (2.2 ×10-6s)= 2.2 ×10-6u ∴2.2 ×10?6 u=(100km) 1? u2 / c2 → u = 0.999978c Note: A time dilation in one RF can be observed as a length contraction in another. 3. The relativistic addition of velocities A particle is emitted by P at speed v0, when the particle reaches F, it triggers the emission of a flash of light that travels to the detector D. The time interval (particle at P→F, then light from F→D) (a). measured by S' who is at rest with respect to the device: Δt0=L0/v0+L0/c (b). measured by S who is in motion (to the left for u>0 with respect to the device): forward: vΔt1=L+uΔt1 (similarly to fig.20.06) backward: cΔt2=L–uΔt2 (v is the speed of particle relative to S and L is contracted length of L0 according to S, Δt1 and Δt2 are times in S). Δt=Δt1+Δt2= L/(v–u)+L/(c+u) Δt = Δ t0 1 ? u2 / c2 and L = L 0 1 ? u 2 / c 2 1 1 L L 2 2 + ) Δ t = Δ t1 + Δ t 2 = + = L0 1 ? u / c ( v?u c+u v?u c+u Δ t0 L0 L0 ) / 1 ? u2 / c2 =( + v c 1 ? u2 / c2 v0 + u v '+ u or v = →v= uv 0 uv ' 1+ 2 1+ 2 c c Here v is the speed of particle relative to S, v0 (or v') is the speed of particle relative to S', while u is the speed of S' relative to S. 20-4 The Lorentz transform The set of equations of Maxwell should have the same form in different IRF.-----The principle of relativity. 1）The coordinate axis is Supposing two IRFs: K, K′ parallel each other; P： x '. y '.z ' t ' 2）K′ is moving along the Y Y' x. y. z.t x axis relative to K with O' constant velocity u ; O Z' X' X 3）Take the timing start u Z point when O and O′ is superposition (重合) . If an event is happen in point P, its coordinates of spacetime are x ' y ' z ' t ', x y z t Transformation→ the relation between them. Note:1) Time and space must be uniform→transform is linear i.e., in the same RF, the time interval and space interval of one event is independent of when and where it happens. Y For example, L1=L2 the same length！ O Z A L1 Y O Z B A L2 B X Another example, one object falls from a height of H. Δ t1 = Δ t2 H X We could prove it using apagogic (反证法的) method: If the transformation for x coordinator is x=ax′ 2, here a is constant, a stick is putting on K′ O' O K' x x ' 1 ' 1 ' 2 ' 2 X' X K' O' x ' 1 x ' 2 X' X x = 0, x = 1m ' x 1' = 1 m , x 2 = 2 m Length: l ' = x ' ? x ' = 1m 2 1 Length: l' = x ? x =1 m ' 2 ' 1 K l = x 2 ? x1 K l = x 2 ? x1 = ax ' 2 2 = ax ' 2 2 ? ax '2 1 ? ax = a (m ) '2 1 = 4 a ? a = 3a (m ) 2）The new transform should be return to Galieo Transform (GT) under the limit of low speed. 牛顿定理与伽利略变换毕竟是低速状态下客观事物 的反映，因此新变换必须在低速下，即v/c<<1 时，回到 伽利略变换（渐进性要求） We will see how the new transform looks like: for the transform of x coordinate, it becomes “GT” under low speed→ x′ =x–ut (u is the speed of S? relative to S ), it’s linear transform，so the new transform should only differ a constant factor independent of x′, t′, x, t with it (say, k). x ' = k (x ? ut) (1) k→1 (under low speed), it becomes “GT”: x?=x–ut . The reverse transform should also be linear. x = k '( x ' + u t ') (2) Any RF has equality of position, there is no difference for transformation from S to S’or from S? to S → k? = k ? x ' = k ( x ? ut) ? ? x = k ( x ' + u t ') (3 ) (4 ) Take the timing start point when O and O′ is superposition (重合), at the superposition point, a light signal is emitted along x′, x axis, reaches point x at time t (point x′ at time t′ ) y K o z Z' Y' u K' X' x z y K o O' Y' K' i x '.t ' X' x O' x .t (5 ) (6 ) From postulate 2 ? x ' = ct ' ? ? x = ct y K o z Z' Y' u K' X' x z O' y K o Y' O' K' X' x i x '.t ' ? x ' = k ( x ? ut ) (3) ? ? x = k ( x '+ ut ') (4) (3)×(4), (5)×(6) ： 2 x .t ? x ' = ct ' (5) ? (6) ? x = ct (7 ) x x ' = k ( x ? u t ) ( x ' + u t ') xx ' = c tt ' 2 (8) 2 Since（7）=（8） ∴ k ( x ? u t )( x ' + u t ') = c tt ' 2 (9 ) ? x ' = k ( x ? u t ) (3 ) ? ? x = k ( x '+ u t ') (4 ) 2 ? x ' = ct ' ? ? x = ct 2 (5 ) (6 ) k ( x ? u t )( x '+ u t ') = c tt ' Inserting（5）to（9） (9 ) (10) (1 1) k ( ct ? ut )( ct '+ ut ') = c tt ' 2 2 i.e.：k tt '( c ? u )( c + u ) = c tt ' 2 2 c 1 k = 2 = 2 2 1 ? (u / c ) c ?u 2 2 (12) (13) Let： β =u/c ∴k = 1 1? β 2 ? x ' = k ( x ? ut ) (3) ? ? x = k ( x '+ ut ') (4) k= 1 1? β 2 (13) ( β = u / c) Since there are no relative motion in y and z direction ; Inserting (13) to (3)： x'= x= x ? ut 1? β 2 ; ….(14) y'= y z'= z (15) (16) x '+ ut ' 1? β 2 ; For time transformation: x' = x ? ut 1? β 2 ; x= x '+ ut ' 1? β 2 2 ; ….(14) x '+ ut ' x' = x ? ut 1? β 2 = 1? β ? ut 2 1? β ; x ' 1? β = 2 x '+ ut ' 1? β 2 ? ut ; x ' 1? β = 2 x '+ ut ' 1? β 2 ? ut ; 2 ….(15) ∴t = ( x '+ ut ' 1? β 2 ? x ' 1 ? β ) / u; 2 u x' u t '+ 2 2 x ' + u t ' ? x '(1 ? β ) c ; /u = = 2 2 1? β u 1? β ux ux ' t? 2 t '+ 2 c ; c ; or: t ' = ∴t = 2 2 1? β 1? β In summary: x'= x ? ut 1? β 2 ; x= x '+ ut ' 1? β 2 ; y'= y z'= z ux t? 2 c ; t'= 2 1? β -------Lorentz transformation ( “LT”) y = y' z = z' ux ' t '+ 2 c ; t= 2 1? β Denoting： γ = 1 1? β 2 ; u t = γ (t '+ 2 x ') c c Note：1） << c, then β → 0, γ →1 LT → GT u u< c, otherwise 1 ? β 2 is imaginary number, and the space-time coordinates are also imaginary numbers---no meaning. x ' = γ ( x ? ut ) y' = y z'= z u t ' = γ (t ? 2 x ) x = γ ( x '+ ut ') y = y' z = z' 3）“LT” means the relation between x, y, z, t and x′, y′ , z′, t′ for the same event in two IRFs (for different event, the relation is not exist). 4）When using “LT” , the space-time benchmark (基准) should consistent (一致的). i）The clock and ruler should be rest relative to its IRF. Otherwise there is no common standard to talk about the space-time coordinates, as the space-time coordinates are dependent on the motion. O’ O X X’ ii）The unit of space and time should have common standard. For example, they all choose some crystal’s elastic variation period or some atom’s half life as the unit. 5）“LT” plays a very important role in special theory of relativity. It reflects the fundamentality difference with “GT” through mathematical language, it’s the key to open the unacquainted world with high speed. 6) “LT” can be solved for Δx and Δt , with the result Δx = γ (Δx' +u Δt' ) , Δt=γ (Δt' +u Δx'/c2 ) We can see: in general Δx ≠ Δx' , Δt ≠ Δt' New ideas of space-time! 20-5 Measuring the space-time coordinates How do we measure the space-time coordinates of an event? Space coordinate: Lay out three rods which are calibrated in that frame, each along one of the three coordinate axes. Time coordinates: Lay out many clocks along the coordinate axes to record the time when the event occur. These clocks must be synchronized. Every inertial observers must define its own coordinate system and synchronize its clocks. Supposing observer S has a large team of assistants, each is given a clock and a measuring rod of a certain length. Each is sent to a post with a rod of some length L and a clock present (预调) at t=L/c. When all the assistants are at their posts, S sets off a flash of light at the origin and simultaneously starts the clock at the origin, which is present to zero. As the light signal reaches the other clocks, each is started in turn at the preset reading. All clocks in the entire system are thus perfectly synchronized (同步的). Note: This calibration holds only for observer S. All other observers(such as S?) must carry out a similar procedure to define a coordinate system and synchronize its clocks. 20-6 The transformation of velocity A particle is moving from P1 → P2, For S: (x1, y1, z1, t1) → (x2, y2, z2, t2); For S? : (x′1, y′1, z′1, t′1) → (x′2, y′2, z′2, t′2) ? x ' = γ (x ? ? From “LT” ? ? t ' = γ (t ? ? Δx ' γ ( Δx ? u Δ t ) ' vx = ' = Δt γ (Δt ? uΔx / c 2 ) u t ), y ' = y, z ' = z, u x) 2 c vx ? u ( Δx / Δ t ? u ) = = 2 (1 ? uΔx / Δtc ) 1 ? uvx / c 2 vx ? u vx = , 2 1 ? uvx / c ' vy = ' vy γ (1 ? uvx / c ) 2 , vz vz = γ (1 ? uvx / c 2 ) ' If vx=c, vy= vz=0, then we have vx′ =c, vy′= vz′=0, i.e. the speed of light is indeed the same in all frames. vx' = vx ? u, vy' = vy , When u<<c, they becomes “GT”: vz ' = vz or v′ + u x vx = 1 + uv′ / c 2 x vy = vy ' γ (1 + uvx '/ c 2 ) vz ' vz = γ (1 + uvx '/ c 2 ) For example, vx ' = 0, v y ' = c, vz ' = 0, then vx = u , v y = c = c 1 ? u 2 / c 2 γ v = v +v +v = c = v' 2 2 x 2 y 2 z 2 2 Example: A supersonic plane moves with speed 1000 m/s (about 3 times the speed of sound) along the x axis relative to you. Another plane moves along the x axis at speed 500 m/s relative to the first plane. How fast is the second plane moving relative to you? Solution: According to the classical formula for combining velocities, the speed of the second plane relative to you is 1000m/s +500m/s =1500m/s. If we assume that you are at rest in the S reference frame and that the first plane is at rest in the S' frame, which is moving at u =1000 m/s relative to S, the second plane has velocity vx'=500 m/s in S'. The correction term for vx in the denominator of inverse velocity transformation is then u v x ' (1 0 0 0 )(5 0 0 ) ?12 c 2 = (3 × 1 0 ) 8 2 ≈ 5 × 10 This correction term is so small that the classical and relativistic results are essentially the same. Example: How fast is the second plane moving relative to you? if in the above example the first plane moves with speed u= 0.8c relative to you and the second plane moves with the same speed 0.8c relative to the first plane. Solution: uvx ' = (0.8c )(0.8c ) = 0.64 → v = 0.8c + 0.8c = 0.98c This is quite different from the classically expected result of 0.8c +0.8c =1.6c. It can be shown that if the speed of an object is less than c in one frame, it is less than c in all other frames moving relative to that frame with a speed less than c. We will see later (section 20-9) that it takes an infinite amount of energy to accelerate a particle to the speed of light. The speed of light c is thus an upper, unattainable limit for the speed of a particle having mass. (Massless particles, such as photons, always move at the speed of light.) c2 c2 x 1 + 0.64 20-7 Consequences of the Lorentz transformation 1. The relativity of time (1) Time dilation: Supposing clock C' is at rest in the frame of S'. x = γ ( x '+ ut '), y = y ', z = z ', t = γ (t '+ u2 x ') c u → Δ t = γ ( Δ t '+ 2 Δ x ') If two events take place at the same location x0' (in S')→ Δx' =0, Δt' =Δt0 (Δt0 is called proper time). So we have Δ t = γΔ t ' = γΔ t0 = Δ t0 1? u / c 2 2 c ---time dilation. Note:The time dilation effect is completely symmetric (in IRF). Each observer believes that the other’s clock is running slower than the ones at rest in the reference frame of the observer. -----Moving clocks run slow Clock.swf Clock-1.swf Now, two events take place at the same location x0′ (in S′) → Δx′ =0, how about in S ? Δx =–γ uΔt′ =– γ uΔt0≠0 , i.e., the events do not occur at the same place in S . In the similar way, when two events that occur at the same place in S → do not occur at the same place in S'. (2) The relativity of simultaneity Take Δt=0, Δx≠ 0, → Δt′=0? from Δx′ =γ (Δx – u Δt ) , Δt′=γ (Δt – u Δx/c2 ) Δt′≠0 → If two events are simultaneous and occur at different place in S, then they are not simultaneous in S′. Δt′ is positive for Δx negative and is negative for Δx positive. Note that this occurs only when two events at different locations in S?. Similarly, take Δt′=0, Δx′≠ 0 → Δt≠0 (3) The Doppler shift Change in the apparent frequency of a wave as observer and source move toward or away from each other. 1 ? u2 / c2 f ' = f0 1 ? ( u / c ) cos θ 在 R 参照系中看，光源 S 的速度为u， 式中 u---S、R 相对速度的绝对值， θ ---u 和 R 到 S 的连线方向(指向 S 的 “视线”方向)间的 夹角 Source and receiver separating, cosθ <0; source and receiver approaching cosθ >0. (4) The twin paradox (clock paradox) Fred and Ethel are identical twins. Ethel travels at high speed to a planet beyond the solar system and returns while Fred remains at home. When they are together again, which twin is older, or are they the same age? The problem is a paradox because of the seemingly symmetric roles played by the twins with the asymmetric result in their aging. There are two ways to resolve the paradox: a. The method of general relativity--- Because the velocity direction of Ethel’s reference is changed when she returns, so the RF is not inertial RF, the special relativity is no longer valid, it should be using the method of general relativity. b. The method of special relativity--- The twins' roles are not really in symmetry.Taking the launching on and return as two IRF, and using special relativity, especially the concept “simultaneity ” correctly. Let planet P and Fred on earth be at rest in S, a distance Lp apart, LP = 8 light-years and u = 0.8c. Then γ = 5/3. It is easy to analyze the problem from Fred's point of view on earth. According to Fred's clock, Ethel travels in S' for a time LP/u = 10 y and in S" for an equal time. Thus Fred is 20 y older when Ethel returns. The time interval in S' between Ethel’ leaving earth and his arriving at the planet is shorter because it is proper time. The time it takes to reach the planet by Ethel ' Δt 10 y clock is Δt ' = = = 6y γ 5/3 Since the same time is required for the return trip, Ethel will have recorded 12 y for the round trip and will be 8y younger (=20–12) than Fred upon her return. The difficulty with the analysis from the point of view of Ethel is that she does not remain in an inertial frame. However, we can get some insight into the problem by having the twins send regular signals to each other so that they can record the other's age continuously. If they arrange to send a signal once a year, each can determine the age of the other merely by counting the signals received. The arrival frequency of the signals will not be 1 per year because of the Doppler shift. The frequency observed is given by 1? u2 / c2 1? 0.64 1 f '= for the case in which the twins are receding(后退) from each other. When they are approaching (u→ –u), f ' = 3f0. Consider the situation first from the point of view of Ethel. During the 6 y it takes her to reach the planet (remember that the distance is contracted in her frame, L' = LP 1+ u / c f0 = f0 = f 0 1+ 0.8 3 γ = 8 light-years = 4.8 light-years 5/3 she receives signals at the rate of 1/3 signal per year, and so she receives 2 signals. As soon as she turns around and starts back to earth, she begins to receive signals at the rate of 3 signals per year. In the 6 y it takes her to return she receives 18 signals, giving a total of 20 for the trip. She accordingly expects her twin to have aged 20 years. From Fred's point of view. He receives signals at the rate of ? signal per year not only for the 10y it takes Ethel to reach the planet but also for the time it takes for the last signal sent by Ethel before she turns around to get back to earth. (He cannot know that Ethel has turned around until the signals begin reaching him with increased frequency.) Since the planet is 8 light-years away, there is an additional 8y of receiving signals at the rate of ? signal per year. During the first 18y, Fred receives 6 signals. In the final 2y before Ethel arrives, Fred receives 6 signals, or 3 per year. (The first signal sent after Ethel turns around takes 8y to reach earth (Lp=8 light-years), whereas Ethel, traveling at 0.8c, takes 10y to return and therefore arrives just 2y after Fred begins to receive signals at the faster rate.) Thus Fred expects Ethel to have aged 12y. In this analysis, the asymmetry of the twins' roles is apparent. 按Fred的观点：他不但在Ethel到达星球的10年期 间，而且在Ethel到达星球时发出的最后一个信号(记为 信号A)到达地球前的时间内都以每年1/3个信号的速率 接收信号(直到接收信号的速率变大之前，他不知道 Ethel已开始返回地球)。因为星球离地球8光年，信号A 到达地球时间为8年，Ethel开始返回后发出的信号被 Fred接收的时间都比信号A晚。所以对Fred而言，共有 18年以每年1/3个信号的速率接收信号-----接收到6个 Ethel发出的信号。剩下2年内， Fred以每年3个信号的 速率接收信号-----接收到6个Ethel发出的信号。共收到 12个Ethel发出的信号，所以Fred认为Ethel度过了12年。 Example: Astronauts in a spaceship traveling away from the Earth at u=0.6c sign off from space control, saying that they are going to nap for 1 hour and then call back. How long does their nap last as measured on Earth? Solution: Since the astronauts go to sleep and wake up at the same place in their reference frame, the time interval for their nap of 1 hour as measured by them is proper time. In the earth's reference frame, they move a considerable distance between these two events. The time interval measured in the earth's frame (using two clocks located at those events) is longer by the factor γ. With u = 0.6c, we have 1 γ= = = 1.25 2 2 0.64 1? u / c 1 Δt=γ Δt0=1.25×1=1.25 h The nap thus lasts for 1.25 hours as measured on earth. 2. The relativity of length To determine the length of a moving object must record the coordinates of the two ends of the object simultaneously! According to S′, Δx′=x2′–x1′=L0 (relative to S? the rod is at rest ) L0, the length as measured in its rest frame of the object, is called the rest length (or proper length). According to S, Δx=x2 –x1, because Δx′ =γ (Δx–uΔt) Δt=0→ L=Δx= Δx/γ = L0 1 ? u2 / c2 ----length contraction Note that L≈ L0, if u<<c. (1) The ladder paradox:The length paradox involves a long ladder traveling near the speed of light and being contained within a smaller garage. Example: A stick that has a proper length of 1 m moves in a direction along its length with speed u relative to you. The length of the stick as measured by you is 0.914 m. What is the speed u? Solution: The length of the stick measured in a frame in which it is moving with speed u is related to its proper length by L=Δx= Δx/γ L0 1m 1 2 2 Then u=0.406c An interesting example of time dilation or length contraction is afforded (给予) by the appearance of muons as secondary radiation from cosmic rays. Since muons are created (from the decay of pions) high in the atmosphere, usually several thousand meters above sea level, few muons should reach sea level. 1? u / c u2 u2 2 1 ? 2 = (0.914) = 0.835 → 2 = 1 ? 0.835 = 0.165 c c L 2 2 γ= = 0.914m = = 1.094 → 1 ? u / c = 0.914 The mean lifetime is about 2μ s for muons at rest. A typical muon moving with speed 0.998c would travel only about 600 m in 2 μ s. However, the lifetime of the muon measured in the earth's reference frame is increased by the factor, which is 15 for this particular speed. The mean lifetime measured in the earth's reference frame is therefore 30 μ s, and a muon with speed 0.998c travels about 9000 m in this time. From the muon's point of view, it lives only 2 μ s, but the atmosphere is rushing past it with a speed of 0.998c. The distance of 9000 m in the earth's frame is thus contracted to only 600 m in the muon's frame as indicated in Figure. 20-8 Relativistic momentum In S frame: vxi1=v, vxi2=?v, vyi1= vyi2=0; vxf1= 0, vxf2=0, vyf1 = v, vyf2= ?v. Pxi=mv+m(?v)=0, Pyi=0 Pxf=0, Pyf=mv+m(?v)=0 In S? frame(which moves relative to S with speed u=?v): LT: vx ? u vx = 1 ? uv x / c 2 ' vy = ' ' vy γ (1 ? uv x / c 2 ) vz vz = γ (1 ? uv x / c 2 ) 2v v xi 1 = , v xi 2 ' = 0, v y i 1 ' = v y i 2 ' = 0, 1 + v2 / c2 v ' ' ' vx f 1 = vx f 2 = v, v y f 1 = = v 1? v2 / c2 , ' γ v yi2 = ' ?v γ = ? v 1 ? v2 / c2 2v 2 mv Pxi ' = m × + m×0 = ; Pyi ' = 0 2 2 2 2 1+ v / c 1+ v / c Pxf ' = mv + mv = 2 mv ; Pyf ' = mv 1 ? v 2 / c 2 + m ( ? v 1 ? v 2 / c 2 ) = 0 S’ will conclude that momentum is not conserved, if ----violating Einstein’s first postulate. 2. Relativistic momentum P = mv Note: not mv/ 1 ? v2 / c2 m/ 1?v2 / c2 P = mv 1 ? u 2 / c2 is sometime referred to as the relativistic mass of the particle; m is then called its rest mass (sometimes is represented by m0). This new definition restores conservation of momentum in the collisions involving particles moving at high speeds. In the S frame, Pxi=Pyi=0, Pxf =Pyf =0 In the S’ frame, Pxi ' = 1? ( mvxi1 ' vxi1 '2 + vyi1 '2 2 + ) 1? ( mvxi 2 ' vxi 2 '2 + vyi 2 '2 c 2 = ) 2mv , Pyi ' = 0 2 2 (1 ? v / c ) 4v 2 (c 2 ? v 2 )2 (1 ? v 2 / c 2 )2 Since, 1 ? ( ) = 1? = 2 2 2= 2 2 2 2 2 c (1 + v / c ) c (c + v ) (1 + v2 / c2 )2 ???????????????????????????????????????? Pxf ' = 1? ( Since, 1 ? ( mvxf 1 ' vxf 1 ' + vyf 1 ' 2 2 2 c vxi1 '2 + vyi1 '2 + ) 1? ( mvxf 2 ' vxf 2 '2 + vyf 2 '2 c 2 c vxf 1 '2 + vyf 1 '2 c2 ) 2mv = , Pyf ' = 0 2 2 (1 ? v / c ) v 2 (2 ? v 2 / c 2 ) (c 2 ? 2v 2 + v 4 / c 2 ) ) = 1? = = (1 ? v 2 / c 2 )2 c2 c2 20-9 Relativistic energy 1. The definition of energy must be changed if the workenergy theorem is to hold for particles at high speeds. For the elastic collision in fig.20-20, In S’ frame 2 mv 2 Ki ' = ; K f ' = mv 2 (2 ? v 2 / c 2 ) (1 + v 2 / c 2 ) 2 Ki′≠ Kf ′, inelastic in S′ but Ki=Kf , elastic in S →violates the relativity postulate. v v dp v v mv ) 2. Relativistic K = ∫v = 0 Σ Fds = ∫0 ds = ∫0 vdp = ∫0 vd ( 2 2 dt 1? v / c energy 2 v ?3 / 2 dv ∵d( = m (1 ? 2 ) 2 2 c 1? v / c v v mv v 2 ?3 / 2 vdv ∴ K = ∫ vd ( ) = ∫ mv (1 ? 2 ) 2 2 0 0 c 1? v / c mc 2 1 ? mc 2 = mc 2 ( ? 1) or K = 1 ? v2 / c2 1 ? v2 / c2 mv K = mc2 1 ? v2 / c2 ? mc 2 → relativistic kinetic energy 1 The expression for kinetic energy consists of two terms. The first term depends on the speed of the particle. The second, mc2, is independent of the speed. The quantity mc2 is called the rest energy of the particle E0= mc2. The total relativistic energy E is then defined to be the sum of the kinetic energy and the rest energy: E = K + mc 2 = mc 2 1 ? v2 / c2 → relativistic energy (for a free particle) 1 v2 K = mc2 ( ? 1) ≈ m c 2 (1 + ? 1) 2 2 2 2c 1? v / c 1 = mv2 2 In practical applications, the momentum or energy of a particle is often known rather than the speed. From P = mv 1? v / c 2 2 and E = mc 2 2 1? v / c 2 → E 2 = (pc) 2 + (mc 2 ) 2 Relation for total energy, momentum, and rest energy This expression replaces E=p2/2m (=mv 2/2) Note: 1) For a massless particle, such as photon, E=pc 2) In many interactions involving fundamental particles , the total energy E is conserved, but the rest mass and rest energy are not conserved; rest energy can be converted to kinetic energy and vice versa. Example: An electron with rest energy 0.511 MeV moves with speed v = 0.8c. Find its total energy, kinetic energy, and momentum. Solution: We first calculate the factor 1/ 1?v2 / c2 =1/ 1?0.82 =1.67 . The total energy is then E= mc 2 1? v / c 2 2 = 1.67(0.511Mev) = 0.853Mev The kinetic energy is the total energy minus the rest energy: K = E – mc2 = 0.853 MeV – 0.511MeV=0.342MeV The magnitude of the momentum is 2 1.33mc = (1.67) m (0.8c ) = p= 2 2 c 1? v / c 1.33(0.511Mev) = = 0.680Mev / c c mv The unit MeV/c is a convenient unit for momentum. Example: A particle of rest mass 2 MeV/c2 and kinetic energy 3 MeV collides with a stationary particle of rest mass 4 MeV/c2 After the collision, the two particles stick together. Find (a) the initial momentum of the system, (b) the final velocity of the two-particle system, and (c) the rest mass of the two-particle system. Solution: (a) Since the moving particle has kinetic energy of 3 MeV and rest energy of 2 MeV, its total energy is El=5 MeV , its momentum 2 2 2 2 2 pc = E1 ? (mc ) = (5MeV) ? (2MeV) = 4.58MeV or p = 4.58 MeV/c , Since the other particle is rest, this is the total momentum of the system. (b) Ef = Ei = El + E2 = 5 MeV + 4 MeV = 9 MeV From P = mv 1 ? v2 / c2 and E = mc 2 1 ? v2 / c2 → v pc 4.58MeV = = = 0.509 c E 9MeV (c) E2 = (pc)2 + (mc2)2→(9 MeV)2 = (4.58 MeV)2 + (mc2)2 mc 2 = 81 ? 21MeV = 7.75MeV → m = 7.75 MeV/c 2 It is instructive to check our answers by computing the initial and final kinetic energies. Ki=3 MeV. Kf =E – Mc2 =9 MeV–7.75 MeV=1.25 MeV energy ΔK=Kf –Ki =1.25MeV–3MeV= –1.75MeV-- loss in kinetic Since the initial rest energy is 2 MeV + 4 MeV = 6 MeV and the final rest energy is mc2 = 7.75 MeV, the gain in rest energy is 7.75 MeV–6 MeV = 1.75 MeV. E=K+E0 remains constant.