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Physics 25 Chapter 28 Special Relativity Dr Physics 25 Chapter 28 Special Relativity Dr. Alward Video Lecture 1: Mass-Energy Equivalence Video Lecture 2: Relativistic Energies Video Lecture 3: Time Dilation Video Lecture 4: Length Contraction Video Lecture 5: Relative Velocity Mass-Energy Equivalence According to Albert Einstein’s Special Theory of Relativity, energy added to an object causes its mass to increase; similarly, all or part of the mass of an object can undergo a transformation into energy (heat, light, sound, potential energy). This “equivalence” of mass and energy is expressed in the following equation: 2 Δm = Q /c Example: Gain in energy causes a gain in mass, as the example below illustrates: 600 J of light energy Q is absorbed by a 10- kg object. What is the new mass of the object? Δm = Q /c2 = 600 /(3.0 x 108)2 = 6.67 x 10-15 kg m = 10.000 000 000 000 067 kg 1 The most important example of a conversion of mass into energy occurs in the interior of the sun where two different isotopes of hydrogen, 2 3 “deuterium” (1H ) and “tritium” (1H ), combine (“fuse”) to form helium. The mass of the product (the helium nucleus) is less than the mass of the reactants (the hydrogen isotopes) by the amount 3 x 10-29 kg: Δm = -3 x 10-29 kg Q = (Δm) c2 = (-3 x 10-29) (3 x 108)2 = -2.70 x 10-12 J The created energy is electromagnetic energy that’s radiated away. The electron’s “antiparticle” is the positron. If they encounter each other, they will annihilate each other and, in their place will appear electromagnetic energy. The reverse can occur: Under the right conditions, electromagnetic energy can disappear, and in its place will appear the equivalent amount of mass in the form of an electron-positron pair. 2 Total Energy According to Einstein, the total energy of an object moving with speed v is: E = mc2 / (1 - β2)1/2 β = v/c, where c = 3.00 x 108 m/s. Note that when the object is at rest, β = 0, and E = mc2, which is called the “rest-mass energy.” Relativistic Kinetic Energy Subtract the rest portion (mc2) of the energy from the total energy, and what’s left is the motion part of the energy kinetic energy. K = E - mc2 = mc /(1-β2)1/2 - mc2 Note that if v = 0, then K = 0, as expected for an object that’s not moving. We often refer to this kinetic energy as the “relativistic” kinetic energy to alert the reader to the fact that this kinetic energy is the true kinetic energy equation, while the other kinetic energy equation, ½ mv2, is called the “classical” kinetic energy. 3 Example: (a) What is the rest-mass energy of an electron (in MeV)? Electron Mass: 9.11 x 10-31 kg E = mc2 = 9.11 x 10-31 (3.0 x 108)2 = 8.20 x 10-14 J = 8.20 x 10-14 J / 1.6 x 10-19 J/eV = 0.51 x 106 eV = 0.51 MeV (b) What the total energy of an electron moving at 0.9995 c? β = 0.9995 E = mc2 / (1-β2)1/2 = 0.51 / (1 - 0.99952)1/2 = 16.13 MeV (c) What is its relativistic kinetic energy? K = E - mc2 = 16.13 - 0.51 = 15.62 MeV (d) What is the kinetic energy using the “classical” equation? K = ½ mv2 = ½ (9.11 x 10-31)(0.9995 x 3.0 x 108)2 = 4.10 x 10-14 J = 4.10 x 10-14 J/ 1.6 x 10-19 J/eV = 0.26 x 106 eV = 0.26 MeV At very high speeds the classical kinetic energy equation is wildly inaccurate, as we see above, but at lower speeds the classical kinetic energy equation is extremely accurate, as we will see on the next page. 4 Example: Calculate and compare the classical and relativistic kinetic energies (in joules) of a proton traveling at the “non- relativistic” speed, v = 0.01c (β = 0.01). v = 0.01 (3.0 x 108) = 3.0 x 106 m/s Classical: K = ½ mv2 = ½ (1.67 x 10-27) (3.0 x 106)2 = 7.515 x 10-15 J Relativistic: mc2 = 1.67 x 10-27 (3.0 x 108)2 = 1.503 x 10-10 J K = mc2 / (1- β2)1/2 - mc2 = 1.503 x 10-10 (1 - 0.012)-1/2 - 1.503 x 10-10 = 7.516 x 10-15 J Even at speeds as large as 3.0 x 106 m/s, the classical kinetic energy equation is still accurate to within one- hundredth of one percent. 5 Classical versus Relativistic Kinetic Energy Prove: At small speeds v, the classical kinetic energy equation (K = ½ mv2) predicts kinetic energies as accurately as does the relativistic kinetic energy equation. The “binomial theorem”: If x << 1, then (1 - x)n 1 – nx For example, let x = 0.0001: (1 – 0.0001)2 = 0.9998 Compare to 1.0000 – 2 (0.0001) = 0.9998 Proof: The role of x is played by β2, while the role of n is played by -½: (1 - β2)-1/2 = 1 - (-½) β2 = 1 + ½ β2 if β << 1 K = mc2 [ (1 - β2)-1/2 - 1] = mc2 [1 + ½ β2 - 1] = mc2 (½ β2) = mc2 (½) (v2/c2) = ½ mv2 if v << c 6 Example: How much work (in MeV) would need to be done on a proton to increase its speed from βo = 0.60 to β = 0.70? Mass of proton: m = 1.67 x 10-27 kg mc2 = 1.67 x 10-27 (3.0 x 108)2 = 1.50 x 10-10 J = 1.50 x 10-10 J / 1.6 x 10-19 J/eV = 9.39 x 108 eV = 939 x 106 = 939 MeV K = mc2 / (1- β2)1/2 - mc2 2 1/2 = 939 / (1 - 0.70 ) - 939 = 375.9 MeV 2 2 1/2 2 Ko = mc / (1- βo ) - mc = 939 / (1 - 0.602)-1/2 - 939 = 234.8 MeV Use the work-kinetic energy theorem: W = K - Ko = 375.9 - 234.8 = 141.1 MeV 7 Time Dilation The time required for an event to occur depends on the observer’s motion relative to the event. Observers moving relative to an event measure a longer time than do observers stationary relative to the event. --Albert Einstein, 1905 Times measured by observers at rest relative to the event are called the “proper time.” To = “proper time” Times measured by observers moving relative to the event are called the “dilated time.” T = “dilated time” 2 1/2 = To / (1 - β ) Note: The denominator is less than 1.0, so T > To. Note: the word “proper” does not imply it’s the “real” time, and that the times observed by moving observers are somehow flawed because of measurement errors or illusion. The dilated time measured by the moving observer is just as real for that observer as it is for the observer who is at rest relative to the event. 8 Example A: An observer at rest relative to an ice cube says it took eight minutes for the ice to melt. Proper Time: To = 8.0 minutes How long does an observer, moving at a speed of four- tenths the speed of light say it took for the ice to melt? β = 0.40 2 1/2 T = To / (1 - β ) = 8.00 / (1 - 0.402)1/2 = 8.73 minutes Dilated Time: 8.73 minutes Example B: A runner completes a journey in 2.00 minutes, according to her. Does she measure the dilated time, or the proper time? The runner is located wherever the event (the running) is occurring, so she measures the proper time. To = 2.00 minutes Suppose a passenger in a spaceship flying overhead at 0.80 c observes the runner. How long will he say it took the runner to complete her run? T = 2.00 /(1 - 0.802)1/2 = 3.33 minutes 9 Example: A hovering drone in a hypothetical universe observes an automobile drive at speed v = 40 m/s across town in 14 minutes. The automobile driver says the trip took 12.0 minutes; what is the speed of light in that universe? The event is the trip; it is occurring wherever the driver is, so she measures the proper time. The drone’s location relative to the automobile is constantly changing, i.e., the drone moving relative to the event, so it measures the dilated time, T = 14.0 minutes. To = 12.0 minutes T = 14.0 minutes Assume Einstein’s Special Theory holds true in the hypothetical universe. 2 1/2 T = To / (1 - β ) 14.0 = 12.0 / (1 - β2)1/2 β = 0.515 40/c = 0.515 c = 77.67 m/s 10 Muons Muons are unstable particles that live for only 2.2 x 10-6 s before decaying—according to observers stationary relative to the muon. This is the proper time. If muons are moving relative to observers, they live longer—according to the moving observers. For example, muons are created when protons streaming toward Earth from the sun collide with molecules in the upper atmosphere; these muons move at β = 0.998 toward the ground. A hypothetical rider atop the muon would say that the muon after creation would travel the following distance through the atmosphere before decaying: Distance = Speed x Time = (0.998)(3.0 x 108) (2.2 x 10-6) = 659 m Earth observers measure the dilated time: 2 1/2 T = To / (1 - β ) = (2.2 x 10-6) /(1 - 0.9982)1/2 = 3.48 x 10-5 s Earth observers say the muon in fact travels this many meters before dying: (0.998) (3.0 x 108 m/s) (3.48 x 10-5 s) = 10, 419 m 11 Length Contraction 2 1/2 L = Lo ( 1 - β ) Lo = Length measured by an observer at rest relative to the object = “the proper length” L = Length measured by an observer moving relative to the object = “the contracted length” Both lengths are correct for their respective observers.
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