Modern Physics, Chapter 1: Special Relativity (2024-2025) PDF

Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar Textbook: Concepts of Modern Physics (by Arthur Beiser), 6th ed. 2003. 1.1. Introduction Modern physics began in 1900 with Max Planck's discovery of the role of energy quantized in blackbody radia...

Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar Textbook: Concepts of Modern Physics (by Arthur Beiser), 6th ed. 2003. 1.1. Introduction Modern physics began in 1900 with Max Planck's discovery of the role of energy quantized in blackbody radiation, followed by albert Einstein's discovery of the theory of relativity and quantum theory of light. The theory of relativity is a theory that deals with the motion of objects when their speed is close to the speed of light c=3×108m/sec and says that all motion is relative. In 1905 a young physicist of twenty-six named Albert Einstein show measurement of time and space are affected by motion between an observer and what is being observed. 1. The theory of relativity consists of two rather different theories. The special theory, developed by Einstein and others in 1905, concerns the comparison of measurements made in different frames of reference moving with constant velocity relative to each other (inertial frame). 2. The general theory, also developed by Einstein (around 1916), is concerned with accelerated reference frames and gravity (non-inertial frame). 1.2. Frames of Reference It is the place that someone (observer) observe an event and make the measurements. Any point in a space can be determine using (𝑥, 𝑦, 𝑧, 𝑡) as shown in Figure (1): 1.2.1. Types of Frames i. Inertial frame: any frame of reference that move with constant velocity relative to an inertial frame is itself an inertial frame. It deals with Newton's first law, the law of inertia, states that if an object is at rest it will stay at rest if no force is acting on it, 1 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar and if an object is moving it will keep on moving at constant velocity if no force is acting on it. ii. Non-Inertial frame: The frames in which the law of inertia does NOT hold are those that are accelerating with respect to inertial frames see Figure (2): Figure (2) Note: All inertial frames are equally valid (no universal frame of reference) because all bodies and places in motion, example if something changing its position with respect to us with constant velocity. Is it moving or are we moving. 1.3. Postulates of Special Relativity Special relativity of relativity is based on the following two postulates: 1) The laws of physics are the same in all inertial frames of reference. This postulate follows from the absence of a universal frame of reference. If the laws of physics were different for different observers in relative motion, the observers could find from these differences which of them were “stationary” in space and which were “moving.” But such a distinction does not exist, and the principle of relativity expresses this fact. 2) The speed of light in free space has the same value in all inertial frames of reference 𝑐 = 3 × 108 𝑚⁄𝑠. 2 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar To explain the second postulate, we take the following example: 1.3.1. Michelson-Morley Experiment Michelson and Morley carried out in 1887 an experiment (Figure 3) to measure the velocity of the earth with respect to the ether (a hypothetical medium pervading the universe in which light waves were supposed to occur) and measure the speed of light relative to the earth. Figure (3) A pair of light beams formed by a half-silvered mirror. One light beam is directed to a mirror along a path perpendicular to the ether current, and the other one goes to a mirror along path parallel to the ether current. Both beams end up at the same viewing screen. The clear glass plate ensures that both beams pass through the same thicknesses of air and glass. If the transit times of the two beams are the same they will arrive at the screen in phase and will interfere constructively. An ether current due to earth's motion parallel to one of the beams, however, would cause the beams to have different transit times and the result would be distractive interference at the screen. This is the essence of the experiment. The negative result had two consequences: a) It showed that the ether does not exist and so there is no such thing as "absolute motion" relative to the ether: all motion is relative to a specified frame of reference. b) The speed of light is the same for all observers which is not true of waves that need a material medium in which to occur (such a sound wave or water wave). 1.4. Time Dilation A moving clock ticks more slowly than a clock at rest Measurement of time intervals are affected by relative motion between an observer and what is observed. As a result clock that moves with respect to an observer ticks more slowly than it does without such motion. 3 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar Example: If someone in a moving spacecraft finds that the time interval between two events in the spacecraft is to, we on the ground would find that the same interval has the longer duration 𝑡. The quantity 𝑡0 , which is determined by events that occur at the same place in an observer's frame of reference is called the proper time of the interval between the events. When a witnessed from the ground, the events that mark the beginning and end of the time interval occur at different space, and in consequence the duration (to dilate is to become larger). The derivative of the time dilation: consider two clocks Fig. (4) in each clock a pulse of light is reflected back and forth between two mirrors Lo apart. Whenever the light strikes lower mirror, an electrical signal is produced that marks the recording tape. Each mark corresponds to the tick of an ordinary clock Figure 4: A simple clock. Each “tick” corresponds to a round trip of the light pulse from the lower mirror to the upper one and back.  Two clocks one is at rest in a laboratory on the ground and the other is in a spacecraft that moves at the speed v relative to the ground.  An observer in the lab watches both clocks. Does the observer find that they tick at the same rate? No. Figure (5) shows the lab. clock in operation (clock at rest). The time interval between ticks is the proper time 𝑡0 and the time needed for the light pulse to travel between the mirrors at the speed of light 𝑐 is 𝑡0 ⁄2. Hence 𝑡0 ⁄2 = 𝐿0 ⁄𝑐 and 2𝐿0 𝑡0 = (1.1) 𝑐 Fig. (4) 4 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar Figure (5) shows the moving clock with its mirrors perpendicular to the direction of motion relative to the ground. The time interval between two ticks is 𝑡. Because the clock is moving, the light pulse, as seen from the ground, follows a zigzag path. On its way from the lower mirror to the upper one in the time 𝑡⁄2, the pulse travels a horizontal distance of 𝑣(𝑡/2) and the total distance 𝑐(𝑡/2). Sine 𝐿0 is the vertical distance between the mirrors Fig. (5) 𝑐𝑡 2 2 v𝑡 2 𝑡2 2 ( ) = 𝐿0 + ( ) (𝑐 − v 2 ) = 𝐿20 2 2 4 2 4𝐿20 (2𝐿0 )2 2𝐿0 𝑡 = 2 = 𝑡= (1.2) 𝑐 − v2 v2 v2 𝑐 2 (1 − 2 ) 𝑐 √1 − 2 𝑐 𝑐 ∵ 2𝐿0 ⁄𝑐 is the time interval 𝑡0 between ticks on the clock on the ground, as in Eq. (1.1), and so 𝑡0 ∴ 𝑇𝑖𝑚𝑒 𝑑𝑖𝑙𝑎𝑡𝑖𝑜𝑛 𝑡= = 𝛾𝑡0 (1.3) √1 − v2 𝑐2 Here is a reminder of what the symbols in Eq. (1.3) represent: 𝑡0 = time interval on clock at rest relative to an observer proper time. 𝑡 = time interval on clock in motion relative to an observer. v = speed of relative motion. 𝑐 = speed of light. v2 Because the quantity √1 − < 1 always, for a moving object, 𝑡 > 𝑡0 always. The 𝑐2 moving clock in the spacecraft appears to tick at a slower rate than the stationary one on the ground, as seen by an observer on the ground. Exactly the same analysis holds for measurements of the clock on the ground by the pilot of the spacecraft. To him, the light pulse of the ground clock follows a zigzag path that requires a total time 𝑡 per round trip. 5 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar His own clock, at rest in the spacecraft, ticks at intervals of 𝑡0. He too finds that 𝑡 as given in Eq. (1.3). So the effect is reciprocal: every observer finds that clocks in motion relative to him tick more slowly than clocks at rest relative to him. Example (1): A spacecraft is moving relative to the earth. An observer on the earth finds that, between 1 P.M. and 2 P.M. according to her clock, 3601 s elapse on the spacecraft’s clock. What is the spacecraft’s speed relative to the earth? Solution: Here 𝑡0 = 3600 𝑠 is the proper time interval on the earth and 𝑡 = 3601 𝑠 is the time interval in the moving frame as measured from the earth. We proceed as follows: 𝑡0 v2 𝑡0 2 𝑡= 1− 2 = ( ) 2 𝑐 𝑡 √1 − v 2 𝑐 𝑡0 2 3600 2 8 √ v = 𝑐√1 − ( ) = 3 × 10 1 − ( ) = 0.071 × 108 𝑚⁄𝑠 𝑡 3601 1.5. Doppler Effect The Doppler effect can be described as the effect produced by a moving source of waves in which there is an apparent upward shift in frequency (higher frequency) for observers towards whom the source is approaching and an apparent downward shift in frequency (lower frequency) for observers from whom the source is receding. Doppler effect in sound is given by the following equation: 1 + v⁄ v𝑠 𝜈 = 𝜈0 ( ) (1.4) 1 − V⁄v𝑠 where v𝑠 is speed of sound v is speed of observer (+for motion toward the source, −for motion away from it). V is speed of the source (+for motion toward the observer, − for motion away from him). If the observer is stationary v = 0, and if the source is stationary, V = 0. 1.5.1. Doppler Effect in Light The Doppler effect in light differs from that in sound. We can analyze the Doppler effect in light by considering a light source as a clock that ticks 0 times per second and emits a wave of light with each tick. We will examine the three situations shown in Fig. (6). 6 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar 1. Transverse Doppler effect in light (Fig1-7(1)): Observer moving perpendicular to a line between him and the light source. The proper time between ticks is 𝑡0 = 1⁄𝜈0 , so between one tick and the next the time v 2 𝑡0 1 √1 − 𝑐 2 𝑡= 𝜈(𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒) = = 2 𝑡 𝑡0 √1 − v 2 𝑐 v2 ∴ 𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒 𝐷𝑜𝑝𝑝𝑙𝑒𝑟 𝑒𝑓𝑓𝑒𝑐𝑡 𝑖𝑛 𝑙𝑖𝑔ℎ𝑡 𝜈 = 𝜈0 √1 − (1.5) 𝑐2 The observed frequency 𝜈 < 𝜈0 always. Figure 6: The frequency of the light seen by an observer depends on the direction and speed of the observer’s motion relative to its source 2. Longitudinal Doppler effect in light (Fig. (7-2,3)) observer approaching or receding from the light source. The observed frequency is higher than the source frequency. Again, the same formula holds for motion of the source toward the observer. v 1+ 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝐷𝑜𝑝𝑝𝑙𝑒𝑟 𝑒𝑓𝑓𝑒𝑐𝑡 𝑖𝑛 𝑙𝑖𝑔ℎ𝑡 𝜈 = 𝜈0 √ 𝑐 (1.5) v 1− 𝑐 by adopting the convention that v is + for source and observer approaching each other and − for source and observer receding from each other. Example (2): A driver is caught going through a red light. The driver claims to the judge that the color she actually saw was green (𝜈 = 5.6 × 1014 Hz) and not red (𝜈0 = 4.8 × 1014 Hz) because of the Doppler effect. The judge accepts this explanation and instead fines her for speeding at the rate of $1 for each 𝑘𝑚/ℎ she exceeded the speed limit of 80 𝑘𝑚/ℎ. What was the fine? Solution: Using the Eq. (1.5) 7 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar v v 1+ 𝜈2 1 + 𝑐 𝑐 + v 𝜈2 𝜈 = 𝜈0 √ 𝑐 = = v 𝜈02 1 − v 𝑐 − v 𝜈02 1− 𝑐 𝑐 𝑐𝜈02 + v𝜈02 = 𝑐𝜈 2 − v𝜈 2 v(𝜈 2 + 𝜈02 ) = 𝑐(𝜈 2 − 𝜈02 ) (𝜈 2 − 𝜈02 ) 8 [(5.6)2 − (4.8)2 ] × 1028 ∴v=c 2 = 3 × 10 = 4.59 × 107 𝑚⁄𝑠 (𝜈 + 𝜈02 ) [(5.6)2 + (4.8)2 ] × 1028 = 1.65 × 108 𝑘𝑚⁄ℎ ∵ 1 𝑚/𝑠 = 3.6 𝑘𝑚/ℎ. The fine is therefore $(1.65 × 108 − 80) = $164,999,920. --------------------------------------------------- 1.6. Length Contraction “Faster means shorter” Measurements of lengths as well as of time intervals are affected by relative motion. The length 𝐿 of an object in motion with respect to an observer always appears to the observer to be shorter than its length 𝐿0 when it is at rest with respect to him. This contraction occurs only in the direction of the relative motion. The length 𝐿0 of an object in its rest frame is called its proper length. To understand length contraction, consider a spaceship traveling with a speed v from one star to another and two observers, one on Earth and the other in the spaceship. The observer at rest on Earth measures the distance between the stars to be 𝐿0 , where 𝐿0 is the proper length. According to this observer, the time it takes the spaceship to complete the voyage is 𝑡0 = 𝐿0 ⁄v. What does an observer in the moving spaceship measure for the distance between the stars? Because of time dilation, the space traveler measures a smaller time of travel: 𝑡 = 𝑡0 ⁄γ. The space traveler claims to be at rest and sees the destination star as moving toward the spaceship with speed v. Because the space traveler reaches the star in the shorter time 𝑡, he or she concludes that the distance, 𝐿, between the stars is shorter than 𝐿0. This distance measured by the space traveler is given by 𝑡0 𝐿 = vt = v γ ∵ 𝐿0 = v𝑡0 𝑡0 = 𝐿0 ⁄v 𝐿0 v2 𝐿𝑒𝑛𝑔ℎ𝑡 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 ∴ 𝐿= = 𝐿0 √1 − 2 (1.7) 𝛾 𝑐 8 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar 1.7. Twin Paradox “A longer life, but it will not seem longer” This paradox involves two identical clocks, one of which remains on the earth while the other is taken on a voyage into space at the speed and eventually is brought back. It is customary to replace the clocks with the pair of twins Dick and Jane, a substitution that is perfectly acceptable because the processes of life—heartbeats, respiration, and so on— constitute biological clocks of reasonable regularity. Dick is 20 y old when he takes off on a space voyage at a speed of 0.8𝑐 to a star 20 light- years away. To Jane, who stays behind, the pace of Dick’s life is slower than hers by a factor of v2 (0.8𝑐)2 √1 − = √1 − = 0.6 = 60% 𝑐2 𝑐2 To Jane, Dick’s heart beats only 3 times for every 5 beats of her heart; Dick takes only 3 breaths for every 5 of hers; Dick thinks only 3 thoughts for every 5 of hers. Finally Dick returns after 50 years have gone by according to Jane’s calendar, but to Dick the trip has taken only 30 y. Dick is therefore 50 y old whereas Jane, the twin who stayed home, is 70 y old (Fig. 7). Figure 7: An astronaut who returns from a space voyage will be younger than his or her twin who remains on earth. Speeds close to the speed of light (here 0.8c) are needed for this effect to be conspicuous To look at Dick’s voyage from his perspective, we must take into account that the distance L he covers is shortened to v2 (0.8𝑐)2 𝐿 = 𝐿0 √1 − = (20 𝑙𝑖𝑔ℎ𝑡 − 𝑦𝑒𝑎𝑟𝑠) √1 − = 12 𝑙𝑖𝑔ℎ𝑡 − 𝑦𝑒𝑎𝑟𝑠 𝑐2 𝑐2 9 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar To Dick, time goes by at the usual rate, but his voyage to the star has taken 𝐿⁄v = 15 𝑦 and his return voyage another 15 y, for a total of 30 y. Of course, Dick’s life span has not been extended to him, because regardless of Jane’s 50-y wait, he has spent only 30 y on the roundtrip. Example (3): Dick and Jane each send out a radio signal once a year while Dick is away. How many signals does Dick receive? How many does Jane receive? Solution: On the outward trip, Dick and Jane are being separated at a rate of 0.8𝑐. With the help of the reasoning used to analyze the Doppler effect in Sec. 1.6, we find that each twin receives signals v 1+ 𝑇1 = 𝑡0 √ 𝑐 = (1 𝑦)√1 + 0.8 = 3 𝑦 v 1 − 0.8 1− 𝑐 On the return trip, Dick and Jane are getting closer together at the same rate, and each receives signals more frequently, namely v 1− 𝑇2 = 𝑡0 √ 𝑐 = (1 𝑦)√1 − 0.8 = 1 𝑦 v 1 + 0.8 3 1+ 𝑐 To Dick, the trip to the star takes 15 y, and he receives 15⁄3 = 5 signals from Jane. During the 15 y of the return trip, Dick receives 15⁄(1⁄3) = 45 signals from Jane, for a total of 50 signals. Dick therefore concludes that Jane has aged by 50 y in his absence. Both Dick and Jane agree that Jane is 70 y old at the end of the voyage. 1.8. Relativistic Momentum In classical mechanics, the linear momentum of a body is defined ⃗ = 𝑚v p ⃗ (1.8) It is a useful quantity because it is conserved in a system of particles not acted upon by outside forces. When an event such as a collision or an explosion occurs inside an isolated system, the vector sum of the momenta of its particles before the event is equal to their vector sum afterward. To derive the linear momentum relativistically, Let us consider an elastic collision (that is, a collision in which kinetic energy is conserved) between two particles A and B, as witnessed by observers in the reference frames 𝑆 and 𝑆 ′ which are in uniform relative motion. The properties of A and B are identical when determined in reference frames in which they are at rest. The frames S and 𝑆 ′ are oriented as in Fig. (8), with 𝑆 ′ moving in the +𝑥 direction with respect to 𝑆 at the velocity v ⃗ 10 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar Before the collision, particle 𝐴 had been at rest in frame 𝑆 and particle 𝐵 in frame 𝑆 ′. Then, at the same instant, 𝐴 was thrown in the +𝑦 direction at the speed 𝑉𝐴 while B was thrown in the −𝑦 direction at the speed 𝑉𝐵′ , where 𝑉𝐴 = 𝑉𝐵′ (1.9) Hence the behavior of A as seen from S is exactly the same as the behavior of B as seen from 𝑆 ′. When the two particles collide, 𝐴 rebounds in the −𝑦 direction at the speed 𝑉𝐴 , while 𝐵 rebounds in the +𝑦 ′ direction at the speed 𝑉𝐵′. If the particles are thrown from positions 𝑌 apart, an observer in 𝑆 finds that the 1 collision occurs at 𝑦 = 𝑌 and one in 2 1 𝑆 finds that it occurs at 𝑦 ′ = 𝑦 = 𝑌. ′ 2 The round-trip time 𝑇0 for 𝐴 as measured in frame 𝑆 is therefore 𝑌 𝑇0 = (1.10) 𝑉𝐴 and it is the same for B in 𝑆 ′ Figure 8: An elastic collision as observed in two different frames of reference. The balls are initially Y apart, which is the same distance in both frames since S moves only in the x direction. 𝑌 𝑇0 = (1.11) 𝑉𝐵′ In 𝑆 the speed 𝑉𝐵 is found from 𝑌 𝑉𝐵 = (1.12) 𝑇 where 𝑇 is the time required for 𝐵 to make its round trip as measured in 𝑆. In 𝑆 ′ , however, B’s trip requires the time 𝑇0 , where 11 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar 𝑇0 𝑇= (1.13) 2 √1 − v 2 𝑐 according to our previous results. Although observers in both frames see the same event, they disagree about the length of time the particle thrown from the other frame requires to make the collision and return. Substituting Eq. (1.12) into Eq. (1.11), we have v2 𝑌 √1 − 𝑐2 𝑉𝐵 = (1.14) 𝑇0 𝑌 From Eq. (1.10) 𝑉𝐴 = 𝑇0 If we use the classical definition of momentum, p ⃗ = 𝑚v ⃗ , then in frame 𝑆 𝑌 p𝐴 = 𝑚𝐴 V𝐴 = 𝑚𝐴 ( ) 𝑇0 v2 𝑌 p𝐵 = 𝑚𝐵 V𝐵 = 𝑚𝐵 √1 − 2 ( ) 𝑐 𝑇0 This means that, in this frame, momentum will not be conserved if 𝑚𝐴 = 𝑚𝐵 , where 𝑚𝐴 and 𝑚𝐵 are the masses as measured in 𝑆. However, if 𝑚𝐴 𝑚𝐵 = (1.15) 2 √1 − v 2 𝑐 then momentum will be conserved. If V𝐴 and V𝐵 are very small compared with v, the relative velocity of the two frames. In this case an observer in 𝑆 will see 𝐵 approach 𝐴 with the velocity v, make a glancing collision (since 𝑉𝐵′ ≪ v), and then continue on. In the limit of V𝐴 = 0, if m is the mass in 𝑆 of 𝐴 when 𝐴 is at rest, then 𝑚𝐴 = 𝑚. In the limit of 𝑉𝐵′ = 0, if 𝑚(v) is the mass in 𝑆 of 𝐵, which is moving at the velocity v, then 𝑚𝐵 = 𝑚. Hence Eq. (1.15) becomes 𝑚 𝑚(v) = (1.16) √1 − v2 𝑐2 We can see that if linear momentum is defined as 12 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar 𝑚v ⃗ ⃗ = 𝛾𝑚v p ⃗ = (1.17) 2 √1 − v 2 𝑐 then conservation of momentum is valid in special relativity. When v ≪ 𝑐, Eq. (1.17) becomes just p ⃗ = 𝑚v ⃗ , the classical momentum. 1.9. Relativistic Mass The increase in an object’s momentum over the classical value as being due to an increase in the object mass. Then we would call 𝑚0 = 𝑚 the rest mass of the object and 𝑚 = 𝑚(v) from Eq. (1.16) its relativistic mass, its mass when moving relative to an observer, Figure (9) shows how p varies with v⁄𝑐 for both 𝛾𝑚v and 𝑚v: i. When v ≪ 𝑐, this gives 𝛾𝑚v ≅ 𝑚v. (For v = 0.01𝑐, the difference is only 0.005 percent; for v = 0.1𝑐, it is 0.5 percent, still small). ii. As v → 𝑐, the curve for 𝛾𝑚v rises more and more steeply (for v = 0.9𝑐, the difference is 229 percent). iii. If v = 𝑐, this leads 𝑝 = ∞, which is impossible. We conclude that no material object can travel as fast as light. Figure 9: The momentum of an object moving at the velocity relative to an observer. 1.10. Relativistic Second Law In relativity, Newton’s second law of motion is given by 1 2 −2 𝑑p ⃗ 𝑑 𝑑 v ⃗ = F = (𝛾𝑚v ⃗ ) = 𝑚 [v⃗ (1 − 2 ) ] 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑐 3 1 2 −2 2 −2 1 v 2v⃗ 𝑑v ⃗ v 𝑑v ⃗ = 𝑚 [− ⃗v (1 − 2 ) ∙ (− ) + (1 − 2 ) ] 2 𝑐 𝑐 2 𝑑𝑡 𝑐 𝑑𝑡 v2 v2 v2 1 𝑑v ⃗ + (1 − ) 𝑚𝑎 𝑐2 𝑐2 𝑐2 =𝑚 3 + 1 =𝑚 3 𝑎= 3 = 𝛾𝑚𝑎 (1.18) 𝑑𝑡 v2 2 v2 2 v2 2 v2 2 [ (1 − 𝑐2 ) (1 − ) 𝑐2 ] [ (1 − 𝑐2 ) ] (1 − 𝑐2 ) 13 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar The acceleration of the particle is therefore 3 ⃗ F v2 2 𝑎 = (1 − 2 ) 𝑚 𝑐 Even though the force is constant, the acceleration of the particle decreases as its velocity increases. As v → 𝑐, this leads 𝑎 → 0, so the particle can never reach the speed of light, a conclusion we expect. 1.11. Mass and Energy The most famous relationship Einstein obtained from the postulates of special relativity— how powerful they turn out to be—concerns mass and energy. The work 𝑊 done on a body by a constant force of magnitude F that acts through the distance 𝑠, where 𝐹 is in the same direction as 𝑠, is given by 𝑠 𝑠 𝑊 = ∆𝐾𝐸 = ∫ 𝐹 ∙ 𝑑𝑠 = ∫ 𝐹 𝑑𝑠 0 0 If no other forces act on the object and the object starts from rest, all the work done on it becomes kinetic energy 𝑠 𝐾𝐸 = ∫ 𝐹 𝑑𝑠 0 1 In nonrelativistic physics, 𝐾𝐸 = 𝑚v 2 2 To find the correct relativistic formula for KE we start from the relativistic form of the second law of motion, Eq. (1.18), which gives 𝑠 v v 𝑑 v 𝐾𝐸 = ∫ (𝛾𝑚v) 𝑑𝑠 = 𝑚 ∫ v𝑑(𝛾v) = 𝑚 ∫ v𝑑 𝑑𝑡 2 0 0 0 √1 − v 2 ( 𝑐 ) where, v = 𝑑𝑠⁄𝑑𝑡. Integrating by parts 𝑏 𝑏 ∫ 𝑥𝑑𝑦 = 𝑥𝑦|𝑏𝑎 − ∫ 𝑦𝑑𝑥 𝑎 𝑎 v v 𝑥=v 𝑑𝑥 = 𝑑v, dy = 𝑑 𝑦= v 2 2 √ √1 − v 2 ( 1 − 𝑐2) 𝑐 14 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar v v v2 v ∴ 𝐾𝐸 = 𝑚 || − ∫ 𝑑v 2 2 √1 − v 2 0 √1 − v 2 [ 𝑐 0 𝑐 ] 1 v 2 2 v 1 v 𝑐 2 (1 − 𝑐 2 ) | − v2 v2 2 v2 =𝑚 − ∫ v (1 − 2 ) 𝑑v = 𝑚 − (− ) v 2 𝑐 v 2 2 1 ⁄2 | √1 − 2 0 √1 − 2 [ 𝑐 ] 𝑐 [ 0] 1 v2 v 2 v2 2 v 2 + 𝑐 2 (1 − ) 𝑐2 =𝑚 + 𝑐 2 {(1 − ) − 1} = 𝑚 − 𝑐2 v2 𝑐2 v2 √ √1 − [ 1 − 𝑐2 ] [ 𝑐2 ] 𝑐2 2 𝑚𝑐 2 =𝑚 −𝑐 = − 𝑚𝑐 2 2 2 √1 − v 2 √1 − v 2 [ 𝑐 ] 𝑐 Relativistic kinetic energy, 𝐾𝐸 = 𝛾𝑚𝑐 2 − 𝑚𝑐 2 = (𝛾 − 1)𝑚𝑐 2 (1.19) The Eq. (1.19) may be written 2 2 𝑚𝑐 2 Total energy, 𝐸 = 𝛾𝑚𝑐 = 𝑚𝑐 + 𝐾𝐸 = (1.20) 2 √1 − v 2 𝑐 If we interpret 𝛾𝑚𝑐 2 as the total energy E of the object, we see that when it is at rest and 𝐾𝐸 = 0, it nevertheless possesses the energy 𝑚𝑐 2. Accordingly 𝑚𝑐 2 is called the rest energy 𝐸0 𝐸0 = 𝑚𝑐 2 H.W.: Deduce the classical formula of kinetic energy from relativistic kinetic energy, at v ≪ c. Hint: Use the binomial expansion. Example (4): A stationary body explodes into two fragments each of mass 1.0 𝑘𝑔 that move apart at speeds of 0.6𝑐 relative to the original body. Find the mass of the original body. Solution: The rest energy of the original body must equal the sum of the total energies of the fragments. Hence 2 2 2 𝑚1 𝑐 2 𝑚2 𝑐 2 𝑚1 𝑐 2 𝐸0 = 𝑚𝑐 = 𝛾1 𝑚1 𝑐 + 𝛾2 𝑚2 𝑐 = + =2 2 2 2 √1 − v12 √1 − v22 √1 − v12 𝑐 𝑐 𝑐 15 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar 1 ∴𝑚=2 = 2.5 𝑘𝑔 √1 − (0.6𝑐) 2 𝑐2 1.12. Energy and Momentum From Eq. (1.20) 𝑚𝑐 2 2 𝑚2 𝑐 4 𝐸= 𝐸 = 2 v2 √1 − v 2 1− 2 𝑐 𝑐 From Eq. (1.17) 𝑚v ⃗ 2 𝑚2 v 2 2 2 𝑚2 v 2 𝑐 2 ⃗ = p p = p 𝑐 = 2 v2 v2 √1 − v 2 1− 2 𝑐 1− 2 𝑐 𝑐 Now we subtract p2 𝑐 2 from 𝐸 2 v2 2 4 𝑚 𝑐 −𝑚 v 𝑐 2 4 2 𝑚2 𝑐 4 (1 − ) 𝑐2 𝐸 2 − p2 𝑐 2 = = = 𝑚2 𝑐 4 v2 v2 1− 2 1− 2 𝑐 𝑐 ∴ 𝐸𝑛𝑒𝑟𝑔𝑦 & 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝐸 2 = (𝑚𝑐 2 )2 + p2 𝑐 2 = 𝐸02 + p2 𝑐 2 (1.21) In classical mechanics, a particle must have rest mass in order to have energy and momentum, but in relativistic mechanics this requirement does not hold. 𝐹𝑜𝑟 𝑀𝑎𝑠𝑠𝑙𝑒𝑠𝑠 𝑃𝑎𝑟𝑡𝑖𝑐𝑙𝑒, 𝑚 = 0, 𝐸 = p𝑐 (1.22)  Electronvolts In modern physics the usual unit of energy is the electronvolt (eV), where 1 eV is the energy gained by an electron accelerated through a potential difference of 1 volt. Since 𝑊 = 𝑄𝑉 1 eV = (1.602 × 10−19 𝐶)(1 𝑉) = 1.602 × 10−19 𝐽 The rest energies of atomic and nuclear particles are often expressed in MeV and GeV and the corresponding rest masses in MeV⁄𝑐 2 and GeV⁄𝑐 2. The advantage of the latter units is that the rest energy equivalent to a rest mass for electron, the rest mass, 𝑚 = 9.1 × 10−31 𝑘𝑔, 𝐸0 = 𝑚𝑐 2 = 9.1 × 10−31 × (3 × 108 )2 = 81.9 × 10−15 𝐽 16 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar 81.9 × 10−15 = −19 = 51.12 × 104 𝑒𝑉 = 0.5112 𝑀𝑒𝑉 1.602 × 10 H.W.: An electron (𝑚 = 0.5112 𝑀𝑒𝑉⁄𝑐 2 ) and a photon (𝑚 = 0) both have momenta of 2 𝑀𝑒𝑉/𝑐. Find the total energy of each. Answer: 𝐸𝑒 = 2.064 𝑀𝑒𝑉, 𝐸𝑝ℎ = 2 𝑀𝑒𝑉 1.13. The Lorentz Transformation Suppose we are in an inertial frame of reference 𝑆 and find the coordinates of some events that occurs at the time 𝑡 are 𝑥, 𝑦, 𝑧. An observer located in a different inertial frame 𝑆′ which is moving with respect to 𝑆 at the constant velocity v will find that the same event occurs at the time 𝑡′ and has the coordinates 𝑥′, 𝑦′, 𝑧′. in order to simplify our work we shall assume that v is in the +x direction. How are the measurements 𝑥, 𝑦, 𝑧, 𝑡 related to 𝑥′, 𝑦′, 𝑧′, 𝑡'? Figure 1.22 Frame S moves in the x direction with the speed v relative to frame S 1.13.1. Galilean Transformation Before special relativity, transforming measurements from one inertial system to another seemed obvious. If clocks in both systems are started when the origins of 𝑆 and 𝑆′ coincide, measurements in the x direction made is 𝑆 will be greater than those made in 𝑆′ by the amount t, which is the distance 𝑆′ has moved in the x direction. That is 𝑥 ′ = 𝑥 − v𝑡 (1.23) There is no relative motion in the 𝑦 and 𝑧 directions, and the time is assumed absolutely 𝑦 ′ = 𝑦, 𝑧 ′ = 𝑧, 𝑡′ = 𝑡 (1.24) The two Eqs. (1.23) & (1.24) is known as the Galilean transformation. To convert velocity components measured in the 𝑆 frame to their equivalents in the 𝑆′ frame according to the Galilean transformation, 𝑑𝑥 ′ v𝑥′ = = v𝑥 − v (1.25) 𝑑𝑡 𝑑𝑦 ′ 𝑑𝑧 ′ v𝑦′ = = v𝑦 , v𝑧′ = = v𝑧 (1.26) 𝑑𝑡 𝑑𝑡 17 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar If we measure the speed of light in the 𝑥 direction in the 𝑆 system to be 𝑐, however, in the 𝑆′ system it will be 𝑐′ = 𝑐 − v this violates the second postulate of special relativity. 1.13.2. Lorentz Transformation A reasonable guess about the nature of the correct relationship between 𝑥 and 𝑥 ′ is 𝑥 ′ = 𝛾(𝑥 − v𝑡) (1.27) Here 𝛾 is a factor that does not depend upon either 𝑥 or 𝑡 but may be a function of v. Because the equations of physics must have the same form in both S and ′ , we need only change the sign of v to write the corresponding equation for 𝑥 in terms of 𝑥 ′ and 𝑡 ′ : 𝑥 = 𝛾(𝑥 ′ + v𝑡 ′ ) (1.28) As in the case of the Galilean transformation 𝑦 ′ = 𝑦, 𝑧 ′ = 𝑧, (1.29) The time coordinates 𝑡 and 𝑡 ′ , however, are not equal. We can see this by substituting the value of 𝑥 ′ given by Eq. (1.27) into Eq. (1.28). This gives 𝑥 = 𝛾[𝛾(𝑥 − v𝑡) + v𝑡 ′ ] = 𝛾 2 (𝑥 − v𝑡) + 𝛾v𝑡 ′ = 𝛾 2 𝑥 − 𝛾 2 v𝑡 + 𝛾v𝑡 ′ ′ 𝛾 2 v𝑡 + 𝑥 − 𝛾 2 𝑥 1 − 𝛾2 𝑡 = = 𝛾𝑡 + ( )𝑥 (1.30) 𝛾v 𝛾v Eqs (1.27), (1.28) & (1.30) constitute a coordinate transformation that satisfies the first postulate of special relativity. The second postulate of relativity gives us a way to evaluate 𝛾. At the instant 𝑡 = 0, the origins of the two frames of reference 𝑆 and 𝑆 ′ are in the same place, according to our initial conditions, and 𝑡 ′ = 0 then also. Suppose that a flare is set off at the common origin of 𝑆 and 𝑆 ′ at 𝑡 = 𝑡 ′ = 0, and the observers in each system measure the speed with which the flare’s light spreads out. Both observers must find the same speed 𝑐 (Fig. 1.23), which means that in the 𝑆 frame 𝑥 = 𝑐𝑡 (1.31) in the 𝑆 ′ frame 𝑥 ′ = 𝑐𝑡 ′ (1.32) Substituting Eqs. (1.27) & (1.30) into Eq. (1.32) 1 − 𝛾2 1 − 𝛾2 𝛾(𝑥 − v𝑡) = 𝑐𝛾𝑡 + ( ) 𝑐𝑥 𝑥 (𝛾 − ( ) 𝑐) = c𝛾𝑡 + v𝛾𝑡 𝛾v 𝛾v 18 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar v v c𝛾𝑡 + v𝛾𝑡 𝛾 (1 + ) 1+ 𝑥= = 𝑐𝑡 [ 𝑐 ] = 𝑐𝑡 [ 𝑐 ] (1.33) 1 − 𝛾2 (1 − 𝛾 2 ) 𝑐 1 𝑐 𝛾−( 𝑐 𝛾 (1 − 1 − ( 2 − 1) 𝛾v ) 𝛾2 v ) 𝛾 v Comparing Eq. (1.33) with Eq. (1.31) v 1+ 1 𝑐 𝑐 v 1 𝑐 𝑐 v 𝑐 =1 1− + = 1 + = − 1 𝑐 𝛾2 v v 𝑐 𝛾2 v v 𝑐 1−( − 1) 𝛾 2 v 1 v 𝑐2 − v2 𝑐2 − v2 v2 1 1 = ( )= =1− 2 𝛾2 = 𝛾= (1.34) 𝛾2 𝑐 𝑐v 𝑐2 𝑐 v2 2 1− 2 𝑐 √1 − v 2 𝑐 The complete transformation of measurements of an event made in 𝑆 to the corresponding measurements made in 𝑆′: 𝑥 − v𝑡 𝑥′ = (1.35) 2 √1 − v 2 𝑐 From Eq. (1.30) ′ 1 − 𝛾2 𝑥 𝛾 𝑥 𝑥 𝑥 v2 𝑥 𝑡 = 𝛾𝑡 + ( ) = 𝛾𝑡 + 2 − 𝛾 = 𝛾 [𝑡 − + (1 − 2 ) ] 𝛾 v 𝛾 v v v 𝑐 v v𝑥 𝑥 𝑥 v𝑥 𝑡− 2 = 𝛾 [𝑡 − + − 2 ] = 𝑐 (1.36) v v 𝑐 2 √1 − v 2 𝑐 1.13.3. Inverse Lorentz Transformation To obtain the inverse transformation, primed and unprimed quantities in Eqs. (1.29), (1.35) & (1.36) are exchanged, and v is replaced by −v: v𝑥 ′ ′ 𝑥 + v𝑡 ′ 𝑡′ + 𝑥= , 𝑦 = 𝑦′, 𝑧 = 𝑧′, 𝑡= 𝑐2 (1.37) √1 − v2 2 √1 − v 2 𝑐2 𝑐 1.14. Velocity Addition Suppose something is moving relative to both 𝑆 and 𝑆 ′. An observer in S measures its three velocity components to be 19 Modern Physics, Chapter 1: Special Relativity (2024-2025) Dr. Manar 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑢𝑥 = , 𝑢𝑦 = , 𝑢𝑧 = 𝑑𝑡 𝑑𝑡 𝑑𝑡 while to an observer in 𝑆 ′ they ar 𝑑𝑥 ′ 𝑑𝑦 ′ 𝑑𝑧 ′ 𝑢𝑥′ = , 𝑢𝑦′ = , 𝑢𝑧′ = 𝑑𝑡 𝑑𝑡 𝑑𝑡 By differentiating the inverse Lorentz transformation equations (1.37) for 𝑥, 𝑦, 𝑧, and 𝑡: v𝑑𝑥 ′ ′ 𝑑𝑥 + vd𝑡 ′ 𝑑𝑡 ′ + 𝑑𝑥 = , 𝑑𝑦 = 𝑑𝑦 ′ , 𝑑𝑧 = 𝑑𝑧 ′ , 𝑑𝑡 = 𝑐2 2 2 √1 − v 2 √1 − v 2 𝑐 𝑐 𝑑𝑥 ′ ′ +v ′ ′ 𝑑𝑥 𝑑𝑥 + vd𝑡 d𝑡 𝑢𝑥′ + v 𝑢𝑥 = = = = (1.38) 𝑑𝑡 𝑑𝑡 ′ + v𝑑𝑥 ′ 1 + v 𝑑𝑥 ′ 1 + v 𝑢𝑥′ 𝑐2 𝑐 2 d𝑡 ′ 𝑐2 𝑑𝑦 ′ 𝑑𝑦 𝑑𝑦 ′ 𝑑𝑡 ′ v2 v2 𝑢𝑦′ v2 𝑢𝑦 = = √1 − = √1 − = √1 − (1.39) 𝑑𝑡 𝑑𝑡 ′ + v𝑑𝑥 ′ 𝑐 2 1 + v 𝑑𝑥 ′ 𝑐 2 1 + v 𝑢𝑥′ 𝑐2 ′ 𝑐 2 𝑐 2 𝑐 d𝑡 2 𝑑𝑧 𝑢𝑧′ v2 𝑢𝑧 = = √1 − (1.40) 𝑑𝑡 1 + v 𝑢𝑥′ 𝑐2 𝑐 2 H.W.: If 𝑢𝑥′ = 𝑐, that is, if light is emitted in the moving frame 𝑆 ′ in its direction of motion relative to S. What the speed will measure an observer in frame 𝑆? Answer: 𝑢𝑥 = 𝑐 Example (5): Spacecraft Alpha is moving at 0.9𝑐 with respect to the earth. If spacecraft Beta is to pass Alpha at a relative speed of 0.5𝑐 in the same direction, what speed must Beta have with respect to the earth? Solution: the earth is considered 𝑆, and the Spacecraft Alpha is considered 𝑆 ′ , so The speed of 𝑆 ′ with respect to 𝑆, is v = 0.9𝑐 The spacecraft Beta with respect to 𝑆 ′ , is 𝑢𝑥′ = 0.5𝑐 The spacecraft Beta with respect to 𝑆, 𝑢𝑥 =? Using the Eq. (1.38) 𝑢𝑥′ + v 0.5𝑐 + 0.9𝑐 𝑢𝑥 = v = = 0.97𝑐 1 + 2 𝑢𝑥′ 1 + 0.9c 0.5𝑐 𝑐 𝑐2 20