Twin Paradox. Who s correct?

Transcription

1 Twin Paradox Consider the following (attributed to Paul Langevin in 1911). Consider two twins. One twin undertakes a long space journey in a spaceship. She travels at almost the speed of light to a distant star, while the other twin remains on Earth. The traveling twin finally returns to Earth from the distant star, traveling again at almost the speed of light. The Earth-bound twin says, I observe that your velocity is very large relative to mine, so I observe your clock running much slower than mine. Therefore, when you return to Earth, you will be younger than me. The traveling twin says, Well, in my frame, your velocity is very large relative to mine. I observe that your clock on Earth is running much slower than mine. Therefore, when I return to Earth, you will be younger than me. Who s correct?

2 The General Theory of Relativity Gravitational influence of other planets cause the major axis of Mercury s elliptical orbit to precess around the Sun relative to fixed stars at a rate of 574 per century. But, Newton s gravitational force law explains only 43 per century. People tried to explain this by hypothesizing there was another planet within Mercury s orbit (Vulcan) that was perturbing Mercury Mercury Precession is 574 per century

3 The General Theory of Relativity From , Einstein developed a new theory of Gravity. This was not his goal. He was more interested in Accelerated Coordinate Frames. I.e., Accelerated Inertial Consider two objects, separated by a distance r, one of mass m and charge q, and the other of mass M and charge Q. The force due to gravity is: mag = G Mm / r 2 The force due to electrostatics is: mae = (1 / 4πε0 ) Qq / r 2 The mass m on the left-hand sides is an inertial mass. It measures an object s resistance to being accelerated. M and m and Q and q on right-hand side are gravitational and electrostatic charges, govern the strength of their respective forces. Why should an object s resistance to being accelerated be the same as the gravitational charge?

4 The General Theory of Relativity Why should an object s resistance to being accelerated be the same as the gravitational charge? Formally, one can rewrite these formula in terms of their inertial (mi) and gravitational mass (mg). mi ag = G Mg mg / r 2 and mi ae = (1 / 4πε0 ) Qq / r 2 Or, rewriting for the acceleration: ag = G Mg ( mg / mi ) / r 2 and ae = (1 / 4πε0 ) Qq / r 2 / mi Experimentally, physicists find that mg / mi = 1 with an uncertainty of 1 part in 1000 billion (10 12 ). This is remarkable - proportionality of inertial and gravitational masses means all masses experience the same gravitational acceleration. This is referred to as the weak equivalence principle.

5 The General Theory of Relativity

6 The General Theory of Relativity

7 The General Theory of Relativity Einstein (like you) realized that inertial reference frames cannot be defined in the presence of gravity. In 1907, Einstein realized that gravity and acceleration acted exactly the same and that no experiment by an observer can differentiate them. He wrote:... a sudden thought occurred to me: If a person falls freely he will not feel his own weight.... This simple thought made a deep impression on me. It impelled me toward a theory of gravitation. Principle of Equivalence: All local, freely falling nonrotating laboratories are fully equivalent for the performance of all physical experiments. Restriction to nonrotating labs eliminates fictitious forces such as Coriolis and centrifugal. Local, freely falling,nonrotating frames are local inertial reference frames. We can eliminate gravity in a laboratory by entering a state of free-fall. In free fall you can do no (local) experiment to determine that you are experiencing acceleration.

8 The General Theory of Relativity Einstein s new theory is a geometric description of how distances (intervals) are measured in the presence of mass. Near any mass, space and time must be described in a new way. Space surrounding a mass becomes curved through a fourth spatial dimension perpendicular to all the three normal spatial dimensions (xyz).

9 The General Theory of Relativity

10 The General Theory of Relativity Actual Light Path Straight line if no mass Sun Actual Light Path Sun

11 The General Theory of Relativity In 1919, Arthur Stanley Eddington measured the apparent positions of stars close to the Sun during a total eclipse. Einstein s Theory predicted that the positions of stars near the Sun should be shifted from their actual positions by 1.75 o, in excellent agreement with Eddington s measurements! Apparent position of star 1.75 o Earth Sun Actual position of star

12 The General Theory of Relativity Einstein s Theory predicted that the positions of stars near the Sun should be shifted from their actual positions by 1.75 o, in excellent agreement with Eddington s measurements!

13

14

15

16 The General Theory of Relativity Another aspect of GR. Nothing can move faster between two points than light. Light must always follow the shortest distance between two points. Compare to the solid line, a trip on the dashed line would (even at the speed of light): 1. Cover a longer distance than the trip along the solid line; 2. have slower running time (clocks would run slower nearer the mass).

17

18

19 Curvature of Space

20 The General Theory of Relativity Precession of Mercury: Einstein applied his theory for the precession of Mercury. His theory predicted the exact measured precession. This was a 2nd great triumph for his Theory of General Relativity. Mercury Precession is 574 per century

21 The Bending of Light Observer in the box must see photon move in horizontal line across the box. Observer watching from afar sees the box accelerate downward. Photon path must appear curved.

22 L Geometry for the Bending of Light Angle of deflection of photon, ϕ, is small, exaggerated here : Arc of deflection subtends an angle ϕ between the radii OA and OB. If width of lab is L, then photon crosses lab in time t L / c. In this time, lab falls a distance d = (gt 2 ) / 2. ϕ Observer watching from afar sees the box accelerate downward. Photon path must appear curved. Triangles ABC and OBD are similar, thus BC / AC = BD / OD with BC = gt 2 /2 AC = L BD = (1/2) L / cos(ϕ/2) For small angles, cos(ϕ/2) 1 and OD rc

23 Geometry for the Bending of Light Therefore, BC / AC = BD / OD, becomes, (gt 2 /2) / L = L / (2rC) for t = L / c and g=9.8 m/s, near the surface of the Earth the radius of curvature of a photon s path is: rc = c 2 /g = 9.17 x m 1 lyr Angle of deflection is ϕ = L / rc = 1.09 x rad = 2.25 x arcsecond ϕ Observer watching from afar sees the box accelerate downward. Photon path must appear curved. Triangles ABC and OBD are similar, thus BC / AC = BD / OD with BC = gt 2 /2 AC = L BD = (1/2) L / cos(ϕ/2) For small angles, cos(ϕ/2) 1 and OD rc

24 Gravitational Redshift and Time Dilation Consider a photon leaving the floor of a lab at the instant a cable holding the lab is cut. Observer moving with lab measures light s frequency at ν0 at top of lab. From ground, lab falling toward the light, so the light meter has a speed of v=gt=gh/c when the photon gets there. Photon of frequency ν0 Change in frequency is from the doppler shift (v << c, non-relativistic) Δν/ν = v / c = gh/c 2. But, meter sees no frequency change.

25 Gravitational Redshift and Time Dilation Change in frequency is from the doppler shift (v << c, non-relativistic) Consider a photon leaving the floor of a lab at the instant a cable holding the lab is cut. Δν/ν = v / c = gh/c 2. But, meter sees no frequency change. Only explanation is that a gravitational redshift decreases the light frequency of the light as it travels up by Photon of frequency ν0 Δν/ν = -v / c = -gh/c 2. This has been proven by experiments time and again (see Example in book).

26 Gravitational Redshift and Time Dilation Derivation of exact formula (valid for strong and weak gravity) requires more mathematics. Answer is: ν ν0 = [ 1-2GM / (r0c 2 ) ] 1/2 When Gravity weak (x = GM/r0 c 2 << 1), then (1 - x) 1/2 (1 - x / 2), which gives ν ν0 1 - GM / (r 0 c 2 ), for (GM / r0 c 2 ) << 1 Gravitational redshift is then, z = (λ - λ0) / λ0 = ν0 / ν - 1 z = (1-2GM / r0 c 2 ) -1/2-1 GM / r0 c 2. Where the approximation is for the weak case, G M / r0 c 2 << 1.

27 Gravitational Redshift and Time Dilation Imagine a clock that ticks for each Period of a wave of light. The gravitational redshift says these ticks get slower. Δt0 = 1/ν. Time gets slower as spacetime is more curved! Δt0 Δt = ν ν0 = [ 1-2GM / (r0c 2 ) ] 1/2 Example: Sirius B (white dwarf star talked about before) has R=5.5 x 10 6 m and M=2.1 x kg. The radius of curvature at the surface of the star is rc = c 2 / g = c 2 / (GM/R 2 ) = c 2 R 2 / GM = 1.9 x m = 0.13 AU. Note that (GM / R c 2 ) << 1, even for white dwarf stars. Gravitational redshift suffered by a photon emitted from the surface and escaping to infinity is z (GM / Rc 2 ) = 2.8 x 10-4 Excellent agreement with observations of z = (3.0 ± 0.5) x 10-4!

28 Gravitational Redshift and Time Dilation Example: Sirius B (white dwarf star talked about before) has R=5.5x10 6 m and M=2.1 x kg. Compared to time measured from a great distance away, time at the surface of the white dwarf star runs slower. If a distant observer measures a time interval of 1 hr, the recorded at the surface of Sirius B would be less than this: Δt - Δt0 = Δt ( 1 - Δt0/Δt ) = (3600 s) x ( GM/ Rc 2 ) = 1.0 s. The clock on the surface runs 1 second slower for each hour of time that passes for the distant observer.

29 Intervals and Geodesics Spacetime consists of 4 coordinates (x, y, z, t) which specifies an event. Einstein developed the equations for calculating the geometry of spacetime produced by some mass (and energy, they are the same thing!). G = - 8πG c 4 T Einstein s Field Equation T is the Stress-Energy Tensor, evaluates the effect of a given distribution of mass and energy on the curvature of spacetime. G is the Einstein Tensor (Gravity) To learn more about this equation... take a course on General Relativity

30 Intervals and Geodesics Spacetime consists of 4 coordinates (x, y, z, t) which specifies an event. Mathematically, let the distance Δ be a straight line between two points A and B, (Δ ) 2 = (xb - xa) 2 + (yb - ya) 2 + (zb - za) 2 The spacetime interval between events A and B is (Δs) 2 = (cδt) 2 - (Δ ) 2, (Δs) 2 = c 2 (tb - ta) 2 - (xb - xa) 2 - (yb - ya) 2 - (zb - za) 2 Distance light traveled in time tb - ta Distance between A and B (Δs) 2 can be positive, zero, or negative. The sign tells us whether the events A and B are causally connected. (Is there enough time for light to get from A to B?) (Δs) 2 > 0 is a timelike curve - light has enough time to make the trip (Δs) 2 = 0 is a lightlike curve - light has exactly enough time to make the trip (Δs) 2 < 0 is a spacelike curve - light does not have enough time to make the trip

31 timelike curves for Event A are within cone. Event A can influence them or be influenced by them. lightlike curves are on surface of cone. spacelike curves are Elsewhere of A (Δs) 2 can be positive, zero, or negative. The sign tells us whether the events A and B are causally connected. (Is there enough time for light to get from A to B?) (Δs) 2 > 0 is a timelike curve - light has enough time to make the trip (Δs) 2 = 0 is a lightlike curve - light has exactly enough time to make the trip (Δs) 2 < 0 is a spacelike curve - light does not have enough time to make the trip

32 Intervals and Geodesics Spacetime consists of 4 coordinates (x, y, z, t) which specifies an event. Proper Time: time between two events that occur at same location, (Δ ) 2 = 0 τ = Δs / c Proper distance: distance measured between two events in a reference frame where they occur simultaneously (tb = ta): ΔL = - (Δs) 2 For a straight rod connecting two locations, the proper distance is the rest length of the rod. Metrics: metrics are the differential distance formula used to calculate intervals. For flat spacetime, the metric would be: (ds) 2 = (c dt) 2 - (dx) 2 - (dy) 2 - (dz) 2 To determine the interval, you take the line integral of this from point A to B.

33 Intervals and Geodesics In flat spacetime, the interval measured along a straight timelike worldline between two events is a maximum. Other worldlines will have smaller intervals. For massless particles, like photons, all worldlines are 0. Example, the straight line connecting A and B is B Δs(A B) = (ds) 2 A B Δs(A B) = (cdt) 2 - (dx) 2 - (dy) 2 - (dz) 2 A B = c dt = c(tb - ta) A

34 Intervals and Geodesics (a) object at a constant location Three different World Lines. (b) object moving in the y-direction (c) satellite in circular orbit around a planet

35 Intervals and Geodesics Straightest Possible Worldlines are Geodesics. Massless particles (photons) always follow null geodesics. In curved Spacetime, timelike geodesics, (Δs) 2 > 0, has either a maximum or a minimum, it is an extremum. Einstein realized that freely falling objects through spacetime follow geodesics! Three features of General Relativity: 1. Mass acts on spacetime, telling it how to curve. II. Spacetime acts on mass, telling it how to move. III. Any freely falling particle (including photons) follows the straightest possible worldline (geodesic) through spacetime. For massive particles, geodesic has a maximum or minimum interval. For massless particles, the geodesic = 0.

36 Intervals and Geodesics Metrics: recall that the spacetime metric is (ds) 2 = (c dt) 2 - (d ) 2. In spherical coordinates, this becomes, (ds) 2 = (c dt) 2 - (dr) 2 - (r dθ) 2 - (r sinθ dϕ) 2 This is for flat spacetime, that is, spacetime not in the vicinity of a massive object. Important: variables r, θ,ϕ, t are measured by an observer at rest a great distance from any mass (r ). Now, place a sphere of mass M with radius R at the origin O. For an observer at r, spacetime is approx. flat. Light from the sphere will experience time dilation, so (1-2GM / rc 2 ) 1/2 factor should be in Metric. Karl Schwarzschild ( ) In 1916 (two months after Einstein published his theory) Schwarzschild solved Einstein s field equations in the presence of a mass M at the origin. (ds) 2 = (c dt 1-2GM/rc 2 ) 2 - ( ) - (r dθ) 2 - (r sinθ dϕ) 2 dr 1-2GM/rc 2 2

37 Intervals and Geodesics Schwarzschild Metric:. (ds) 2 = (c dt 1-2GM/rc 2 ) 2 - ( ) - (r dθ) 2 - (r sinθ dϕ) 2 Schwarzschild Metric contains all the effects of time dilation (in temporal term). The curvature of space is contained in the radial term. The distance along a radial line (dθ=dϕ=0) is the proper distance (dt=0): dr dl = -(ds) 2 = 1-2GM/rc 2 Thus, the radial distance is longer by a factor of 1 / (1-2GM/rc 2 ) 1/2 compared to flat space. If a clock is at rest at a radial coordinate r then the proper time dτ (dr=dθ=dϕ=0) is related to a time dt measured at an infinite distance. dr 1-2GM/rc 2 dτ = ds / c = dt 1-2GM / r c 2 Note that dτ < dt always. 2

38 Black Holes In 1783 John Michell ( ) considered Newton s theory of light and gravity. He speculated that if the Sun were 500x more massive, then light from its surface would not have enough velocity to escape the Sun s Gravity. Using Newton s Force Law (vescape) 2 = 2GM/r. For vescape=c, R = 2GM/c 2. For M=mass of the Sun, R = 2.95 km. In the 18th century this was too small to be more than a curiosity. In 1939, J. Robert Oppenheimer and Harland Snyder considered the gravitationally collapse of a star that had exhausted its nuclear fuel. J. Robert Oppenheimer ( )

39 Black Holes (ds) 2 = (c dt 1-2GM/rc 2 ) 2 - ( ) - (r dθ) 2 - (r sinθ dϕ) 2 What happens as 2GM/rc 2 1? At this point space becomes infinitely curved. This is the Schwarzschild Radius. Note that at the Schwarzschild Radius, the temporal term in the Metric goes to 0. the Proper time, dτ = 0. Is light frozen in time? (Does time stop?) dr 1-2GM/rc 2 RS = 2 GM / c 2 Consider the apparent speed of light, the coordinate speed of light, the rate at which the spatial coordiantes of a photon change. Start with lightlike curve: (ds) 2 = 0, above. Inserting dθ=dϕ=0. dr c dt 1-2GM/rc 2 = ( ) 1-2GM/rc 2 dr/dt = c( 1-2GM/rc 2 ) = c( 1 - RS / r) When r>>rs, then dr/dt=c. But at r=rs, dr/dt=0. This is a barrier (the Event Horizon). At center of a Black Hole is a singularity. The Event Horizon prevents the singularity from being observed. 2

40 What is a Black Hole? For really massive stars (>18 solar masses), nothing stops the core s gravitational collapse. It forms a Black Hole, an object so powerful that not even light can escape.

41 Black Holes Imagine a trip into a Black Hole. Starting at a safe distance (r >> RS) we fire a radio photon at a Black Hole, which is reflected (somehow) from the Event Horizon. How long does the trip take? Integrate the formula for the coordinate speed of light. dr/dt = c( 1-2GM/rc 2 ) r2 - RS Δt = dr / (dr/dt) = dr / c(1 - RS / r ) = (r2 - r1)/c + (RS / c) ln( ) r1 - RS valid for r2 > r1. Inserting r1 = RS, Δt=infinity! Because the trip is symmetric, the same results applies if the the photon starts at infinity. The photon never makes it to the Event Horizon from the point of view of the observer at infinity. A photon fired at the Event Horizon takes an infinite time to get there.

42 Black Holes Now consider a brave astronaut, who agrees to fall into a black hole. The astronaut starts at r=infinity and falls toward a 10 M black hole, where RS = 30 km. What happens from the astronaut s point of view?

43 Black Holes Now consider a brave astronaut, who agrees to fall into a black hole. The astronaut starts at r=infinity and falls toward a 10 M black hole, where RS = 30 km. What happens from the astronaut s point of view? Progression of radial coordinate from outside observers view Progression of radial coordinate from astronaut s view

44 Black Holes Now consider a brave astronaut, who agrees to fall into a black hole. The astronaut starts at r=infinity and falls toward a 10 M black hole, where RS = 30 km. What happens from the astronaut s point of view? Physically, tidal forces become extreme and stretch and squeeze the poor astronaut. Once inside event horizon, r < RS. For dr = dθ=dϕ=0, the interval is: (ds) 2 = (c dt) 2 ( 1 - RS / r) < 0 This is a spacelike curve, which is not permitted for massive particles. At the singularlity all worldlines converge.

45 Black Holes Can Photons orbit a Black Hole? Classically (Newton): v 2 /r = GM/r 2 Set v=c, and solve for r, yields: R = GM / c 2 This is smaller than RS = 2GM / c 2.

46 Black Holes Can Photon s orbit a Black Hole? For General Relativity, consider the the coordinate speed of light in the ϕ-direction. Start with the Schwarzchild solution: (ds) 2 = (c dt 1-2GM/rc 2 ) 2 - ( ) - (r dθ) 2 - (r sinθ dϕ) 2 dr 1-2GM/rc 2 Set dr=dθ=0, and for the speed of light, ds=0: 2 Which yields:

47 Black Holes Lets equate this to the centripetal velocity, v = (GM/r) 1/2, for stable circular orbits: And solve for the radius where this occurs: This is the photon sphere where photons follow circular orbits around the Black Hole. If you looked into the photon sphere, you would literally see the back of your head...

48

49

50

51

52

53

54 Black Hole Candidates Evidence that the center of our Milky Way Galaxy contains a black hole with a mass of several million solar masses (we ll discuss this more in 2 weeks). First candidate for a Black Hole was in a binary star system, Cygnus X-1 (X-ray source near bright star η Cygni). In the best(?) candidate, V 404 Cyg underwent an X-ray outburst in This object is orbited by a K0 Giant star with a radial velocity of 211 km/s with a period of P=6.473 days. Kepler s Laws give a mass for the unseen companion of 12 M, which would make it much, much too massive to be a neutron star (maximum neutron mass is ~3 solar masses). Other evidence in Active Galactic Nuclei (Black Holes in the centers of other galaxies with masses up to one billion solar masses, much more on this later). The relativistic effects of the black hole can be seen on the emission lines of gas very near the black hole, which gives the mass.

55 Do Black Holes Exist? Most Popular Black Hole candidate is Cygnus X-1.

56 Black Hole Candidates

57

58

59

60 Wormholes normal space light ray path light ray path light ray path Wormhole

61 How to make a time machine: Postulated by Prof. Kip Thorne (Caltech) Similar to Twin Paradox of Special Relativity 1. Kip and his wife make a wormhole and hold hands through it. 2. His wife takes the wormhole on a rocket and travels at 0.6 c for 10 hrs, then stops and returns to Earth taking 10 more hrs. 3. Special Relativity says that in her frame the trip took 20 hrs x ( ) 1/2 = 16 hrs So her clock will be 4 hrs behind Kip s clock. 4. Since they were holding hands through the wormhole the whole time, if anything Kip sends through the wormhole will emerge 4 hrs in the past!

62 Chandra X-ray Observatory Hubble Space Telescope Remember Pulsars? Pulsar = rapidly rotating Neutron Star. Pulsar at the center of the Crab nebula pulses 30 times per second!

63 Chandra X-ray Observatory Hubble Space Telescope Remember Pulsars? Pulsar = rapidly rotating Neutron Star. Pulsar at the center of the Crab nebula pulses 30 times per second!

64 Pulsars = Test of General Relativity In the 1960s, Russell Hulse and Joseph Taylor discovered that the pulsar PSR was a binary system of two neutron stars (or a neutron star and a white dwarf). The orbital separation of the system is a little larger than the Sun s diameter. A 30-year study of this system showed that the orbital period is speeding up as a result of gravitational radiation (gravity waves!). The two stars should merge in ~300 Myr. Theoretical : dp/dt = -( / ) x Measured: dp/dt = -( / ) x Excellent! Hulse and Taylor received the Nobel prize in 1993 for this work.

65 Orbiting Compact Objects emit Gravitational Waves

66 Orbiting Compact Objects emit Gravitational Waves

67 LIGO: Laser Interferometer Gravitational Observatory

68 Hawking Radiation Semiclassical result merging Quantum Mechanics and General Relativity Virtual Particles are able to exist for a time duration given by Heisenberg s uncertainty principle: ΔE Δt ħ, or Δt ħ/e e - e + annihilation Stephen Hawking (b. 1942) creation γ γ annihilation

69 Hawking Radiation Virtual Particles are able to exist for a time duration given by Heisenberg s uncertainty principle: ΔE Δt ħ, or Δt ħ/e Consider what happens to virtual particles produced near the Event Horizon. Hawking worked out that blackholes radiate these particles with spectrum of a black body with a temperature given by:

70 Hawking Radiation Black hole evaporation! As black holes emit energy, they lose mass (Einstein), so as they accrete virtual particles, black holes lose mass. Temperature of Black Hole Radiation is: Schwarzchild radius is: Area of a Black hole is: r s = 2GM c 2 A =4πr 2 s =4π 2GM c 2 2 = 16πG2 M 2 c 4 Luminosity of a perfect black body is: You can rewrite the Stefan- Boltzman constant as: Which gives a luminosity of: L = 16πG2 M 2 c 4 π 2 k 4 B 60 3 c 2 σ = L = AσT 4 π 2 k 4 B 60 3 c 2 c 3 8πGMk B 4 = c πG 2 M 2

71 Hawking Radiation Black hole evaporation! As black holes emit energy, they lose mass (Einstein), so as they accrete virtual particles, black holes lose mass. L = 16πG2 M 2 c 4 π 2 k 4 B 60 3 c 2 c 3 8πGMk B 4 = c πG 2 M 2 Consider that the luminosity is the change in energy (change in mass) with time: L = de dt = dm dt c2 So, Hawking predicts that black holes should evaporate. For a 1 solar mass Black Hole, the Luminosity minuscule: L = W. You can work out how long it takes a black hole to evaporate: The evaporation time increases as the cube of the mass! For a 1 solar mass black hole is tev = 2.1 x yrs. Much, much longer than the age of the Universe. For tev = 1 s, the mass is 2.3 x 10 5 kg (about the size of a large military cargo plane) and the released energy would be 2 x J = 5 x 10 6 megatons of TNT = 200,000 of the most powerful nuclear bombs made.