1. Relativity Chapter 1

2. A Brief Overview of Modern Physics 20 th Century revolution: - 1900 Max Planck Basic ideas leading to Quantum theory Basic ideas leading to Quantum theory - 1905 Einstein Special Theory of Relativity Special Theory of Relativity 21 st Century Story is still incomplete Story is still incomplete

3. Basic Problems  Newtonian mechanics fails to describe properly the motion of objects whose speeds approach that of light  Newtonian mechanics is a limited theory –It places no upper limit on speed –It is contrary to modern experimental results –Newtonian mechanics becomes a specialized case of Einstein’s special theory of relativity when speeds are much less than the speed of light

4. Galilean Relativity  To describe a physical event, a frame of reference must be established  There is no absolute inertial frame of reference –This means that the results of an experiment performed in a vehicle moving with uniform velocity will be identical to the results of the same experiment performed in a stationary vehicle

5. Galilean Relativity  Reminders about inertial frames –Objects subjected to no forces will experience no acceleration –Any system moving at constant velocity with respect to an inertial frame must also be in an inertial frame  According to the principle of Galilean relativity, the laws of mechanics are the same in all inertial frames of reference

6. Galilean Relativity  The observer in the truck throws a ball straight up –It appears to move in a vertical path –The law of gravity and equations of motion under uniform acceleration are obeyed

7. Galilean Relativity  There is a stationary observer on the ground –Views the path of the ball thrown to be a parabola –The ball has a velocity to the right equal to the velocity of the truck

8. Galilean Relativity – conclusion  The two observers disagree on the shape of the ball’s path  Both agree that the motion obeys the law of gravity and Newton’s laws of motion  Both agree on how long the ball was in the air Conclusion: There is no preferred frame of reference for describing the laws of mechanics

9. The cornerstone of the theory of special relativity is We shall see that many surprising consequences follow from this innocuous looking statement. The cornerstone of the theory of special relativity is the Principle of Relativity: The Laws of Physics are the same in all inertial frames of reference. We shall see that many surprising consequences follow from this innocuous looking statement. Frames of Reference and Newton's Laws

10. Let us review Newton's mechanics in terms of frames of reference. A "frame of reference" is just a set of coordinates - something you use to measure the things that matter in Newtonian mechanical problems - like positions and velocities, so we also need a clock. A point in space is specified by its three coordinates (x,y,z) and an "event" like, say, a little explosion by a place and time – (x,y,z,t).

11. The "laws of physics" we shall consider are those of Newtonian mechanics, as expressed by Newton's laws of motion, with gravitational forces and also contact forces from objects pushing against each other. _____________________________ For example, knowing the universal gravitational constant from experiment (and the masses involved), it is possible from Newton's second law, force = mass x acceleration, to predict future planetary motions with great accuracy.

12. Suppose we know from experiment that these laws of mechanics are true in one frame of reference. How do they look in another frame, moving with respect to the first frame? To figure out, we have to find how to get from position, velocity and acceleration in one frame to the corresponding quantities in the second frame. Obviously, the two frames must have a constant relative velocity, otherwise the law of inertia won't hold in both of them.

13. Let's choose the coordinates so that this velocity is along the x-axis of both of them.

14. Notice we also throw in a clock with each frame. Now what are the coordinates of the event (x,y,z,t) in S'? It's easy to see t' = t - we synchronized the clocks when O‘ passed O. Also, evidently, y' = y and z' = z, from the figure. We can also see that x = x' +vt. Thus (x,y,z,t) in S corresponds to (x',y',z', t' ) in S', where That's how positions transform - these are known as the Galilean transformations.

15. What about velocities ? The velocity in S' in the x' direction What about velocities ? The velocity in S' in the x' direction This is just the addition of velocities formula

16. How does acceleration transform?

17. the acceleration is the same in both frames. This again is obvious - the acceleration is the rate of change of velocity, and the velocities of the same particle measured in the two frames differ by a constant factor - the relative velocity of the two frames. Since v is constant we have

18. If we now look at the motion under gravitational forces, for example, we get the same law on going to another inertial frame because every term in the above equation stays the same. Note that acceleration is the rate of change of momentum - this is the same in both frames. So, in a collision, if total momentum is conserved in one frame (the sum of individual rates of change of momentum is zero) the same is true in all inertial frames.

19. Maxwell’s Equations of Electromagnetism in Vacuum Gauss’ Law for Electrostatics Gauss’ Law for Magnetism Faraday’s Law of Induction Ampere’s Law

20. 1 2 The Equations of Electromagnetism..monopole.. ?...there’s no magnetic monopole....!! Gauss’s Laws

21. 4 The Equations of Electromagnetism 3.. if you change a magnetic field you induce an electric field........... if you change a magnetic field you induce an electric field......... Faraday’s Law Ampere’s Law.. if you change an electric field you induce a magnetic field........... if you change an electric field you induce a magnetic field.........

22. Electromagnetic Waves Faraday’s law: Faraday’s law: dB/dt electric field Maxwell’s modification of Ampere’s law dE/dt magnetic field These two equations can be solved simultaneously. The result is: E(x, t) = E P sin (kx-  t) B(x, t) = B P sin (kx-  t) ˆ z ˆ j

23. Plane Electromagnetic Waves x EyEy BzBz E(x, t) = E P sin (kx-  t) B(x, t) = B P sin (kx-  t) ˆ z ˆ j c

24. Plane Electromagnetic Waves x EyEy BzBz E(x, t) = E P sin (kx-  t) B(x, t) = B P sin (kx-  t) ˆ z ˆ j c Notes: Waves are in Phase, but fields oriented at 90 0. but fields oriented at 90 0. k=2π/λ. k=2π/λ. Speed of wave is c=ω/k (= fλ) Speed of wave is c=ω/k (= fλ) At all times E=cB

25. It was recognized that the Maxwell equations did not obey the principles of Newtonian relativity. i.e. the equations were not invariant when transformed between the inertial reference frames using the Galilean transformation. Lets consider an example of infinitely long wire with a uniform negative charge density λ per unit length and a point charge q located a distance y 1 above the wire.

26. The observer in S and S ’ see identical electric field at distance y 1 =y 1 ’ from an infinity long wire carrying uniform charge λ per unit length. Observers in both S and S’ measure a force on charge q due to the line of charge.

27. However, the S’ observer measured and additional force due to the magnetic field at y 1 ’ arising from the motion of the wire in the -x’ direction. Thus, the electromagnetic force does not have the same form in different inertial systems, implying that Maxwell’s equations are not invariant under a Galilean transformation.

28. Speed of the Light It was postulated in the nineteenth century that electromagnetic waves, like other waves, propagated in a suitable material media, called the ether. In according with this postulate the ether filed the entire universe including the interior of the matter. It had the inconsistent properties of being extremely rigid (in order to support the stress of the high electromagnetic wave speed), while offering no observable resistance to motion of the planet, which was fully accounted for by Newton’s law of gravitation.

29. Speed of the Light The implication of this postulate is that a light wave, moving with velocity c with respect to the ether, would travel at velocity c’=c +v with respect to a frame of reference moving through the ether at v. This would require that Maxwell’s equations have a different form in the moving frame so as to predict the speed of light to be c’, instead of

30. Conflict Between Mechanics and E&M A.Mechanics Galilean relativity states that it is impossible for an observer to experimentally distinguish between uniform motion in a straight line and absolute rest. Thus, all states of uniform motion are equal.

31. Conflict Between Mechanics and E&M B.E&M Initially- The initial interpretation of the speed of light in Maxwell's theory was this c was the speed of light seen by observers in absolute rest with respect to the ether. In other reference frames, the speed of light would be different from c and could be obtained by the Galilean transformation.

32. Problem- It would now be possible for an observer to distinguish between different states of uniform motion by measuring the speed of light or doing other electricity, magnetism, and optics experiments.

33. Possible Solutions 1. Maxwell's theory of electricity and magnetism was flawed. It was approximately 20 years old while Newton's mechanics was approximately 200 years old. 2. Galilean relativity was incorrect. You can detect absolute motion! 3. Something else was wrong with mechanics (I.e Galilean transformation).

34. Experimental Results Most physicists felt that Maxwell's equations were probably in error. Numerous experiments were performed to detect the motion of the earth through the ether wind. The most famous of these experiments was the Michelson-Morley experiment. Because of the tremendous precision of their interferometer, it was impossible for Michelson and Morley to miss detecting the effect of the earth's motion through the ether unless mechanics was flawed!

35. The Michelson-Morley experiment is a race between light beams. The incoming light beam is split into two beams by a half-silvered mirror. The beams follow perpendicular paths reflecting off full mirrors before recombining back at the half mirror. Time differences are seen in the interference pattern on the screen.

36. Theory We will simplify the calculations by assuming that L 1 = L 2 = L. The time required to complete path 1 (horizontal path) is given by where we have used the Galilean transformation and velocity = distance/time.

38. Since u<<c, we can use the binomial approximation:

39. We can determine the time required to complete path 2 (vertical path) using the distance diagram below: Using the Pythagorean theorem, we have: u(T 2 /2)

41. Again, using the binomial approximation:

42. Thus, the time difference for the two paths is approximately We can now calculate the phase shift in terms of wavelengths as follows:

43. Thus, the phase shift in terms of a fraction of a wavelength is given by

44. Using a sodium light, = 590 nm, and a interferometer with L = 11 m, we have This was a very large shift (20%) and couldn't have been over looked.

45. Result - No shift was ever observed regardless of when the experiment was performed or how the interferometer was orientated!

46. Einstein’s Postulates In 1905 Albert Einstein published a paper on the electrodynamics of moving bodies. In this paper, he postulated that absolute motion can not be detected by any experiment. That is, there is no ether. The reference frame connected with earth is considered to be at rest and the velocity of the light will be the same in any direction. His theory of special relativity can be derived from two postulates:In 1905 Albert Einstein published a paper on the electrodynamics of moving bodies. In this paper, he postulated that absolute motion can not be detected by any experiment. That is, there is no ether. The reference frame connected with earth is considered to be at rest and the velocity of the light will be the same in any direction. His theory of special relativity can be derived from two postulates: Postulate 1:Postulate 1: Absolute uniform motion can not be detected. Postulate 2:Postulate 2: The speed of light is independent of the motion of the source.

47. The Lorentz Transformation Galilean transformations: x = x’ + vt’, y = y’, z = z’, t = t’ The inverse transformations are x’ = x – vt, y’ = y, z’ = z, t’ = t These equations are consistent with experimental observations as long as v is much less than c. They lead to the familiar classical addition law for velocities. If a particle has velocity u x = dx/dt in frame S, its velocity in frame S’ is from here we have

48. The Lorentz Transformation It should be clear that the Galilean transformation is not consistent with Einstein’s postulates of special relativity. If light moved along the x axis with speed u x ’=c in S’, these equations imply that the speed in S is u x =c+v rather than u x =c, which is not consistent with Einstein’s postulates and experiment. The classical transformation equations must therefore be modified.

49. The Lorentz Transformation We assume that the relativistic transformation equation for x is the same as the classical equation except for a constant multiplier on the right side: where γ is a constant that can depend on v and c but not on coordinates. The inverse transformation in this case

50. Lorentz Transformation The transformation described by these equations is called the Lorentz transformation. Lorentz’s equations replace the flawed Galileo transformation equations in relating the measurements of two different observers in uniform motion relative to each other.

51. Lorentz Transformation

52. Addition of Velocities We start by taking the derivative with respect to t' of the first Lorentz transformation equation: From Calculus, we have that:

53. Using the definition of velocity, we have:

54. Differentiating the fourth Lorentz equation with respect to t, we obtain:

55. We now substitute this result into our velocity equation to obtain: where v is velocity of object seen by unprimed observer, v ' is velocity of object seen by primed observer, u is velocity of the primed observer as seen by unprimed observer.

56. EXERSICE: Show that both observers agree on the speed of light. SOLUTION: We determine the speed v' as seen by the primed observer for a beam of light v = c seen by the unprimed observer. Using the velocity addition formula we have:

57. Lorentz-Fitgerald (Length) Contraction Fitzgerald's length contraction is now a direct consequence of the Lorentz transformation. Consider two different observers in relative motion who measure the length of a box as shown below: observer sees stationary observer sees box moving box at u S S’ x1x1

58. At a single instant of time t, the unprimed observer measures the box's length in S as while the unprimed observer measures the box's length in S ’ as Using the Lorentz transformation equation for x', we have that

59. Therefore, we have that

60. Lorentz-Fitzgerald (Length) Contraction where L is the improper length L o is the proper length u is the relative speed between the observers

61. Length Contraction

62. Warning: Your everyday use of English terms can get you in trouble in this material. The term "proper" is not meant to indicate that its the "correct" or "right" answer. Two observers might measure the length of a train as 10 m when they see the train as stationary. If one observer rides in the train as it moves down the track, he will measure the train's length as 10 m while an observer standing by the track sees the train as shorter than 10 m. Both Observers Are CORRECT!!! The question "What is the length of the train?" is meaningless!! You must specify the frame. There is no such thing as absolute distance anymore!

63. Another interesting aspect of relativity is that moving clocks always run slower. This is not an artifact of the clock! It is a consequence of the nature of time itself according to Einstein. All clocks will behave this way include your biological processes! Time Dilation

64. Let us consider a "Light Clock" in which one unit of time corresponds to the time it takes a light pulse to travel between two meters a distanceL apart. We will place the clock on a train traveling at speed u with the unprimed observer. Let us consider a "Light Clock" in which one unit of time corresponds to the time it takes a light pulse to travel between two meters a distance L apart. We will place the clock on a train traveling at speed u with the unprimed observer. Time Dilation L Light Path As Seen By Train Observer Mirror

65. Time Dilation For the unprimed observer on the train, the time it takes the light pulse to make a single tick is given by

66. Time Dilation The primed observer standing by the railroad track see the train and clock pass with a speed ofu. Thus, the path of the light beam appears to follow the diagonal path shown below: The primed observer standing by the railroad track see the train and clock pass with a speed of u. Thus, the path of the light beam appears to follow the diagonal path shown below: L u u t' ct'

67. Time Dilation From the Pythagorean theorem, t' is given by

68. Time Dilation We now plug in our results for the unprimed observer and we find that Thus, our observers do not agree upon time!

69. where T o is the proper time T’ is the improper time T’ is the improper time Time Dilation Equation

70. Time Dilation as a Function of Speed

71. An astronomer on Earth observes a meteoroid in the southern sky approaching the Earth at a speed of 0.800c. At the time of its discovery the meteoroid is 20.0 ly from the Earth. Calculate (a) the time interval required for the meteoroid to reach the Earth as measured by the Earthbound astronomer, (b) this time interval as measured by a tourist on the meteoroid, and (c) the distance to the Earth as measured by the tourist.

72. (a) The 0.8c and the 20 ly are measured in the Earth frame, so in this frame,. (b) We see a clock on the meteoroid moving, so we do not measure proper time; that clock measures proper time.

73. An astronomer on Earth observes a meteoroid in the southern sky approaching the Earth at a speed of 0.800c. At the time of its discovery the meteoroid is 20.0 ly from the Earth. Calculate (a) the time interval required for the meteoroid to reach the Earth as measured by the Earthbound astronomer, (b) this time interval as measured by a tourist on the meteoroid, and (c) the distance to the Earth as measured by the tourist. (c) Method one: We measure the 20 ly on a stick stationary in our frame, so it is proper length. The tourist measures it to be contracted to Method two: The tourist sees the Earth approaching at

74. Classical Doppler Shift Anyone who has watched auto racing on TV is aware of the Doppler shift. As a race car approaches the camera, the sound of its engine increases in pitch (frequency). After the car passes the camera, the pitch of its engine decreases. We could use this pitch to determine the relative speed of the car. This technique is used in many real world applications including ultrasound imaging, Doppler radar, and to determine the motion of stars.

75. I. Moving Observer Assume that we have a stationary audio source that produces sound waves of frequency f and speed v. Source v Observer x y A stationary observer sees the time between each wave as

76. I. Moving Observer If the observer is now moving at a velocity u relative to the source then the speed of the waves as seen by the observer is where the positive sign is when the observer is moving toward the source. The time between waves is now

77. I. Moving Observer Taking the ratio of our two results we get that

78. II. Moving Source We now consider the case in which the source is moving toward the observer. In this case, the wave's speed is unchanged but the distance between wave fronts (wavelength) is reduced. For the source moving away from the observer the wavelength increased. l Source u Observer x y uT v

79. II. Moving Source

80. Thus, the frequency seen by the observer for a moving source is given by

81. Moving Observer Moving Source Note: The motion of the observer and source create different effects. For sound, this difference is explained due to motion relative to the preferred reference frame! This preferred frame is the reference frame stationary to the medium propagating the sound (air)!

82. Relativistic Doppler Effect For light in vacuum the distinction between motion the source and detector can not be done. Therefore the expression we derived for sound can not be correct for light. Consider a source of light moving toward the detector with velocityv, relative to the detector. If source emits N electromagnetic waves in time Δt D (measured in the frame of the detector), the first wave will travel a distance cΔt D and the source will travel the distance v Δt D measures in the frame of detector. Consider a source of light moving toward the detector with velocity v, relative to the detector. If source emits N electromagnetic waves in time Δt D (measured in the frame of the detector), the first wave will travel a distance cΔt D and the source will travel the distance v Δt D measures in the frame of detector.

83. Relativistic Doppler Effect The wavelength of the light will be: The frequency f ’ observed by the detector will therefore be:

84. Relativistic Doppler Effect If the frequency of the source is f 0 it will emit N=f 0 Δt S waves in the time Δt S measured by the source. Then Here Δt S is the proper time interval ( the first wave and the N th wave are emitted at the same place in the source’s reference frame). Therefore times Δt S and Δt D are related as:

85. Relativistic Doppler Effect When the source and detector are moving towards one another: - approaching - approaching - receding - receding

86. Relativistic Doppler Shift Since light has no medium, there should be no difference between moving the source and the observer. The problem with our previous derivation when dealing with light is that we haven't considered that space and time coordinates are different for the source and observer. Thus, we must account for the contraction of space and dilation of time due to motion. After accounting for differences in time and space, we get the following result for both moving source and moving observer:

87. Police radar detects the speed of a car as follows: Microwaves of a precisely known frequency are broadcast toward the car. The moving car reflects the microwaves with a Doppler shift. The reflected waves are received and combined with an attenuated version of the transmitted wave. Beats occur between the two microwave signals. The beat frequency is measured. (a) For an electromagnetic wave reflected back to its source from a mirror approaching at speed v, show that the reflected wave has frequency where f source is the source frequency. (b) When v is much less than c, the beat frequency is much smaller than the transmitted frequency. In this case use the approximation f + f source ≈ 2 f source and show that the beat frequency can be written as f beat = 2v/λ. (c) What beat frequency is measured for a car speed of 30.0 m/s if the microwaves have frequency 10.0 GHz? (d) If the beat frequency measurement is accurate to ±5 Hz, how accurate is the velocity measurement?

88. Let f c be the frequency as seen by the car: (a)Let f c be the frequency as seen by the car: and, if f is the frequency of the reflected wave: Combining gives: (b) Using the above result: The beat frequency is then:

89. (c)(d)

90. A ball is thrown at 20.0 m/s inside a boxcar moving along the tracks at 40.0 m/s. What is the speed of the ball relative to the ground if the ball is thrown (a) forward (b) backward (c) out the side door?

91. (a)(b)(c)

92. A muon formed high in the Earth’s atmosphere travels at speed v = 0.990c for a distance of 4.60 km before it decays into an electron, a neutrino, and an antineutrino. (a) How long does the muon live, as measured in its reference frame? (b) How far does the muon travel, as measured in its frame?

93. For, (a)The muon’s lifetime as measured in the Earth’s rest frame is and the lifetime measured in the muon’s rest frame is.

94. A muon formed high in the Earth’s atmosphere travels at speed v = 0.990c for a distance of 4.60 km before it decays into an electron, a neutrino, and an antineutrino. (a) How long does the muon live, as measured in its reference frame? (b) How far does the muon travel, as measured in its frame? (b)

97. The identical twins Speedo and Goslo join a migration from the Earth to Planet X. It is 20.0 ly away in a reference frame in which both planets are at rest. The twins, of the same age, depart at the same time on different spacecraft. Speedo’s craft travels steadily at 0.950c, and Goslo’s at 0.750c. Calculate the age difference between the twins after Goslo’s spacecraft lands on Planet X. Which twin is the older?

98. In the Earth frame, Speedo’s trip lasts for a time: Speedo’s age advances only by the proper time interval during his trip.

99. The identical twins Speedo and Goslo join a migration from the Earth to Planet X. It is 20.0 ly away in a reference frame in which both planets are at rest. The twins, of the same age, depart at the same time on different spacecraft. Speedo’s craft travels steadily at 0.950c, and Goslo’s at 0.750c. Calculate the age difference between the twins after Goslo’s spacecraft lands on Planet X. Which twin is the older? Similarly for Goslo, While Speedo has landed on Planet X and is waiting for his brother, he ages by: Then ends up older by.

100. An atomic clock moves at 1 000 km/h for 1.00 h as measured by an identical clock on the Earth. How many nanoseconds slow will the moving clock be compared with the Earth clock, at the end of the 1.00-h interval?

101. Solution:

102. Solution:

103. A spacecraft is launched from the surface of the Earth with a velocity of 0.600c at an angle of 50.0° above the horizontal positive x axis. Another spacecraft is moving past, with a velocity of 0.700c in the negative x direction. Determine the magnitude and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft.

104. Solution: Let frame S be the Earth frame of reference. Then As measured from the S’ frame of the second spacecraft:

105. A spacecraft is launched from the surface of the Earth with a velocity of 0.600c at an angle of 50.0° above the horizontal positive x axis. Another spacecraft is moving past, with a velocity of 0.700c in the negative x direction. Determine the magnitude and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft. Solution: As measured from the S’ frame of the second spacecraft: Iu’I =

106. A moving rod is observed to have a length of 2.00 m and to be oriented at an angle of 30.0° with respect to the direction of motion, as shown in Figure. The rod has a speed of 0.995c. (a) What is the proper length of the rod? (b) What is the orientation angle in the proper frame?

107. is a proper length, related to by.

108. We have seen that Einstein’s postulates require important modifications in our ideas of simultaneity and our measurements of time and length. Einstein’s postulates also require modification of our concepts of mass, momentum, and energy We have seen that Einstein’s postulates require important modifications in our ideas of simultaneity and our measurements of time and length. Einstein’s postulates also require modification of our concepts of mass, momentum, and energy.

109. Relativistic Momentum In classical mechanics, the momentum of a particle is defined as a product of its mass and its velocity, mu. In an isolated system of particles, with no net force acting on the system, the total momentum of the system remains the same. However, we can see from a simple though experiment that the quantity Σm i v i is not conserved in isolated system.

110. We consider two observers: observer A in reference frame S and observer B in frame S’, which is moving to the right in the x direction with speed v with respect to the frame S. Each has a ball of the mass m.

111. One observer throws his ball up with a speed u 0 relative to him and the other throws his ball down with a speed u 0 relative to him, so that each ball travel a distance L, makes an elastic collision with the other ball, and returns.

112. Classically, each ball has vertical momentum of magnitude mu 0. Since the vertical components of the momenta are equal and opposite, the total vertical momentum are zero before the collision. The collision merely reverses the momentum of each ball, so the total vertical momentum is zero after the collision.

113. Relativistically, however, the vertical components of the velocities of the two balls as seen by the observers are not equal and opposite. Thus, when they are reversed by the collision, classical momentum is not conserved. As seen by A in frame S, the velocity of his ball is u Ay =+u 0. Since the velocity of ball B in frame S’ is u ’ Bx =0, u By ’=-u 0, the y component of the velocity of ball B in frame S is u By =-u 0 /γ. So, the Σ m i u i is conserved only in the approximation that u<<c.

114. We will define the relativistic momentum p of a particle to have the following properties: 1. In collisions, p is conserved. 2.As u/c approaches 0, p approaches mu. We will show that quantity is conserved in the elastic collision and we take this equation for the definition of the relativistic momentum of a particle.

115. Conservation of the Relativistic Momentum One interpretation of this equation is that the mass of an object increases with the speed. Then the quantity is called the relativistic mass. The mass of a particle when it is at rest in some reference frame is then called its rest mass. m rel = γm rest

116. Conservation of the Relativistic Momentum

117. The speed of ball A in S is u 0, so the y component of its relativistic momentum is: The speed of B in S is more complicated. Its x component is v and its y component is –u 0 /γ.

119. Using this result to compute √1-(u 2 B /c 2 ), we obtain and

120. Conservation of the Relativistic Momentum The y component of the relativistic momentum of ball B as seen in S is therefore

121. Conservation of the Relativistic Momentum Since p By =-p Ay’ the y component of the total momentum of the two balls is zero. If the speed of each ball is reversed by the collision, the total momentum will remain zero and momentum will be conserved. with

122. Relativistic momentum as given by equation versus u/c, where u is speed of the object relative to an observer. The magnitude of momentum p is plotted in units of mc. The fainter dashed line shows the classical momentum mu.

123. Relativistic Energy In classic mechanic: The work done by the net force acting on a particle equals the change in the kinetic energy of the particle. In relativistic mechanic: The net force acting on a particle to accelerate it from rest to some final velocity is equal to the rate of change of the relativistic momentum. The work done by the net force can then be calculated and set equal to the change of kinetic energy.

124. Relativistic Energy As in relativistic mechanic, we will define the kinetic energy as the work done by the net force in accelerating a particle from rest to some final velocity u f. where we have used u=ds/dt.

126. Relativistic Energy Substituting the last result in the equation for kinetic energy we obtain: The second part of this expression, mc 2, is independent of the speed and called the rest energy E o of the particle. E 0 =mc 2 The total relativistic energy:

127. Relativistic Energy The work done by unbalanced force increases the energy from the rest energy to the final energy where is the relativistic mass.

128. Relativistic Energy We can obtain a useful expression for the velocity of a particle by multiplying equation for relativistic momentum by c 2 :

129. Energies in atomic and nuclear physics are usually expressed in units of electron volts (eV) or mega-electron volts (MeV): 1eV = 1.602 x 10 -19 J A convenient unit for the masses of atomic particles is eV/c 2 or MeV/c 2, which is the rest energy of the particle divided by c 2. A convenient unit for the masses of atomic particles is eV/c 2 or MeV/c 2, which is the rest energy of the particle divided by c 2.

130. The expression for kinetic energy does not look much like the classical expression ½(mu 2 ). However when v<<c, we can approximate 1/√1-(u 2 /c 2 ) using the binomial expansion: (1+x) n =1 + nx + n(n-1)x 2 /2 + …….≈ 1 + nx

131. Then Then and when v<<c

132. The equation for relativistic momentum: and equation for the total energy: can be combined to eliminate the speed u: E 2 = p 2 c 2 +(mc 2 ) 2

133. E 2 = p 2 c 2 +(mc 2 ) 2 We received the relation for total energy, momentum, and rest energy. If the energy of a particle is much grater than its rest energy mc 2, the second term on the right side of the last equation can be neglected, giving the useful approximation: If the energy of a particle is much grater than its rest energy mc 2, the second term on the right side of the last equation can be neglected, giving the useful approximation: E ≈ pc(forE>>mc 2 ) E ≈ pc (for E>>mc 2 )

135. An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is 2.50 × 10 –28 kg, and that of the other is 1.67 × 10 –27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?

136. Relativistic momentum of the system of fragments must be conserved. For total momentum to be zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and subscript 1 to the lighter,

137. An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is 2.50 × 10 –28 kg, and that of the other is 1.67 × 10 –27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?

138. An unstable particle with a mass of 3.34 × 10 –27 kg is initially at rest. The particle decays into two fragments that fly off along the x axis with velocity components 0.987c and –0.868c. Find the masses of the fragments. (Suggestion: Conserve both energy and momentum.)

140. This reduces to:

141. A pion at rest (m π = 273m e ) decays to a muon (m μ = 207m e ) and an antineutrino ( ). The reaction is written. Find the kinetic energy of the muon and the energy of the antineutrino in electron volts. (Suggestion: Conserve both energy and momentum.)

144. SUMMARY 1. Einstein’s Postulates: 1. Einstein’s Postulates: Postulate 1: Absolute uniform motion can not be detected. Postulate 1: Absolute uniform motion can not be detected. Postulate 2: The speed of light is independent of the motion of the source. Postulate 2: The speed of light is independent of the motion of the source.

145. SUMMARY 2. The Lorentz Transformation: 2. The Lorentz Transformation:

146. SUMMARY 6. The Velocity Transformation:

147. SUMMARY 3. Time Dilation: 3. Time Dilation: 4. Length Contraction: 4. Length Contraction: 5. The Relativistic Doppler Effect: 5. The Relativistic Doppler Effect: approaching receding

148. SUMMARY 7. Relativistic Momentum: 8. Relativistic Energy: 9. Rest Energy: 10. Kinetic Energy

149. SUMMARY 10. Useful Formulas for Speed, Energy, and Momentum: