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Transcript
Astronomy 114 Lecture 24: Black Holes & Relativity Martin D. Weinberg [email protected] UMass/Astronomy Department A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—1/15 Announcements PS#5 solutions posted today. PS#6 posted on line (due next Wednesday) Exam next Friday covering: Stars, Stellar Evolution, Stellar remnants A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—2/15 Announcements PS#5 solutions posted today. PS#6 posted on line (due next Wednesday) Exam next Friday covering: Stars, Stellar Evolution, Stellar remnants Today: Black holes Relativity Black Holes, Chap. 24 A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—2/15 Announcements PS#5 solutions posted today. PS#6 posted on line (due next Wednesday) Exam next Friday covering: Stars, Stellar Evolution, Stellar remnants Today: Black holes Relativity Black Holes, Chap. 24 Next up (Wed?): Optics and Telescopes Optics and Telescopes, Chap. 6 A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—2/15 Conservation of Energy Example: potential energy is energy that can be converted to kinetic energy by moving in a force field v=0 v=vf Potential energy is positive to start Kinetic energy is gained as ball rolls down hill Ground level Potential energy is zero at bottom of hill A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—3/15 Conservation of Energy Example: potential energy is energy that can be converted to kinetic energy by moving in a force field v=0 v=vf Ground level v=0 Kinetic energy is gained as ball rolls into well Potential energy is negative at bottom of well v=vf A114: Lecture 24—06 Apr 2007 Potential energy is zero at top of well Read: Ch. 24 Astronomy 114—3/15 Conservation of Energy Example: potential energy is energy that can be converted to kinetic energy by moving in a force field v=0 v=vf Ground level v=0 Potential energy is zero at top of well Kinetic energy is gained as ball rolls into well Potential energy is negative at bottom of well Total energy (kinetic plus potential) is conserved v=vf A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—3/15 Escape velocity (1/2) Need a critical velocity to escape the surface of a gravitating body To escape the gravitational pull of a planet, need an initial velocity that will make your total mechanical energy be unbound Total mechanical energy of a blob (the energy of motion) added to the energy given up by falling in: E = KE + P E 1 2 GM m mv − = 2 R A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—4/15 Escape velocity (2/2) Energy must be greater than zero to be unbound: 1 2 GM m mv − 0 = 2 R s 2GM vesc = R Examples: Earth: vesc = 11.2 km/s Moon: vesc = 2.4 km/s A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—5/15 Schwarzschild Radius For a given velocity and mass, we can compute the radius we need to just escape: 2GM R= 2 vesc A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—6/15 Schwarzschild Radius For a given velocity and mass, we can compute the radius we need to just escape: 2GM R= 2 vesc Nothing moves faster than the speed of light. . . Light cannot escape from a Black Hole if it comes from a radius closer than the Schwarzschild Radius, Rs : 2GM Rs = c2 where M is the mass of the hole. A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—6/15 Sizes of black holes 2GM Rs = c2 A black hole with a mass of 1 M⊙ has Rs =3 km 0.6 M⊙ White Dwarf: radius of 1 Rearth = 6370 km 1.4 M⊙ Neutron Star: radius of 10 km A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—7/15 Implications for the Laws of Physics Need to modify laws of motion to think about velocities near the speed of light (Special Relativity) Newton’s theory gravity implies that the force propagates infinitely fast (action at a distance) Also need to modify law of gravity to incorporate new laws of motion General Relativity: Einstein’s theory of gravitation (1915) First solutions by Karl Schwarzschild (1916) Not taken seriously until the 1960s A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—8/15 New theory of motion Thought experiment. . . Q: Two cars have a head-on collision. Each is moving at 45 mph. What is their relative speed? A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—9/15 New theory of motion Thought experiment. . . Q: Two cars have a head-on collision. Each is moving at 45 mph. What is their relative speed? A: 90 mph A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—9/15 New theory of motion Thought experiment. . . Q: Two cars have a head-on collision. Each is moving at 45 mph. What is their relative speed? A: 90 mph Q: Two spaceships have a head-on collision. Each is moving at 0.9 c. What is their relative speed? A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—9/15 New theory of motion Thought experiment. . . Q: Two cars have a head-on collision. Each is moving at 45 mph. What is their relative speed? A: 90 mph Q: Two spaceships have a head-on collision. Each is moving at 0.9 c. What is their relative speed? A: 1.8 c? No, it is 0.99 c (according to special relativity) A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—9/15 New theory of motion Thought experiment. . . Q: Two cars have a head-on collision. Each is moving at 45 mph. What is their relative speed? A: 90 mph Q: Two spaceships have a head-on collision. Each is moving at 0.9 c. What is their relative speed? A: 1.8 c? No, it is 0.99 c (according to special relativity) Why do we need a new theory? A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—9/15 Special Relativity What is the medium that propagates light? c. 1900: scientists look for differences in the speed of light in different directions Michelson-Morley experiment: no difference Einstein showed that this can be explained assuming the speed of light is constant for all observers A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—10/15 Special Relativity Two principles of Special Relativity: The laws of physics must be the same in all reference frames (True in Galilean relativity, Newtons Laws, too) Speed of light must be constant in all reference frames (New!) A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—10/15 Consequences of Special Relativity (1/2) Consider two observers moving at near the speed of light relative to each other At the instant that they are at the same place, one observer fires a flash bulb or strobe Sphere of light propagates from this point c c A B v=0.8 c c What does each observer see? c A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—11/15 Consequences of Special Relativity (2/2) Each observer must see a sphere of light. How can this be? Stationary observer A sees q B’s length contracts: L = Lo 1 − v 2 /c2 q B’s clock dilates: T = To / 1 − v 2 /c2 q B’s energy increases: E = Eo / 1 − v 2 /c2 From B’s point of view, the same thing happens to A. Both observers A & B see light sphere from the flash bulb. A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—12/15 Photon clock Photon Clock 1.5 m Clock ticks when pulse arrives, sends new pulse Pulse from laser travels 3 meters to receiver t = 3m/3 × 108 m/s = 10−8 s Laser source Photocell detector A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—13/15 Photon clocks in moving frames As seen by spaceship pilot (B) at v = 0.8c A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—14/15 Photon clocks in moving frames v=0.8 c As seen by mission control (A) A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—14/15 More consequences Lengths and clocks Simultaneous events in one frame are not simultaneous in another Pole and barn paradox A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—15/15 More consequences Lengths and clocks Simultaneous events in one frame are not simultaneous in another Pole and barn paradox 80 m pole, 40 m barn with doors at both ends Runner runs through barn at v = 0.5c √ In barn frame: L = 80 1 − 0.52 = 34.9 m √ Runner: barn is L = 40 1 − 0.52 = 17.4 m Barn doors close and open simultaneously in barn’s frame but not in Runner’s frame A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—15/15 More consequences Lengths and clocks Simultaneous events in one frame are not simultaneous in another Pole and barn paradox Twin paradox A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—15/15 More consequences Lengths and clocks Simultaneous events in one frame are not simultaneous in another Pole and barn paradox Twin paradox One twin travels at 0.99c to star 10 ly away Travelling twin sees length to star √ contracted: L = 10 ly 1 − 0.992 = 1.41 ly Time to get there and back: T = 2 × 1.41 ly/c = 2.82 years Twin on Earth sees: 2 × 10 ly/c = 20 years A114: Lecture 24—06 Apr 2007 Read: Ch. 24 Astronomy 114—15/15