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RELATIVITY 2 Harrison B. Prosper Florida State University YSP Topics Part 1 Recap Mapping Spacetime When Is Now? Part 2 Distances in Spacetime Paradoxes Summary Recap Einstein’s theory is based on two postulates: Principle of relativity: The laws of physics are the same in all inertial (that is, nonaccelerating) frames of reference. Constancy of the speed of light: The speed of light in vacuum is independent of the motion of the light source. Spacetime Diagrams Events can be represented as points in a spacetime diagram Time A B Space Events with the same time values, such as events A and B, are said to be simultaneous Event: A place at a given time Spacetime: The set of all events Earth’s Time Axis 3000 AD C now D B (t,x,y,z) 2500 AD A x O y 2000 AD MAPPING SPACETIME 6 Mapping Spacetime – I C tC Starship’s worldline v = speed = BD / OB = x / tB Earth’s worldline B tB c = BD / AB tA A D tD tB = γ tD tD = κ tA O Mapping Spacetime - II tC C tB B tA A O D tD κ-factor Relates elapsed times from a common event O to two events A and D that can be connected by a light ray. γ-factor Relates elapsed times from a common event O to two events B and D that are judged simultaneous in one of the frames. κ and γ Factors 1 1 Relativistic Doppler Factor Dilation Factor v c 1 1 2 κ and γ Factors 1 1 Relativistic Doppler Factor Problem 1: derive the formula for κ. Problem 2: light of 500 nm wavelength is emitted by a starship, but received on Earth at a wavelength of 600 nm. What is the relative speed between the Earth and the starship? When is Now ? Δt = tB - tE B E O tB tE Events B and D are simultaneous for Earth so tD = t B / γ tD D But events D and E are simultaneous for the starship so tE = t D / γ When Is Now? - II “Nows” do not coincide! Δt = tB - tE tB B D tE E O tD Writing distance between B and D as x = BD the temporal discrepancy is given by v x x t c c c Problem 3: derive Δt Problem 4: estimate Δt between the Milky Way and Andromeda, assuming a relative speed between the galaxies of 120 km/s Worlds in Collision T. J. Cox and Abraham Loeb DISTANCES IN SPACETIME 14 The Metric P A dl dy O dx B The distance between points O and P is given by: OB2 + BP2 = OP2 = OA2 + AP2 OP2 is said to be invariant. The formula dl2 = dx2 + dy2 for computing dl2 is called a metric In 3-D, this becomes dl2 = dx2 + dy2 + dz2 The Metric z The metric in spherical polar coordinates (r, θ, φ) Consider the spatial plane θ = 90o AC CB AB r θ φ Δφ x A = r dφ = dr = dl dl 2 dr 2 r 2 d 2 y C B The Interval Q Suppose that O and Q are events. How far apart are they in spacetime? First guess ds2 = (cdt)2 + dl2 ds dl Unfortunately, this does not work! O In 1908, Hermann Minkowski showed that the correct expression is ds2 = (cdt)2 – dl2 ds2 is called the interval Hermann Minkowski 1864 -1909 cdt P The Interval In general, the interval ds2 between any two events is either timelike ds2 = (cdt)2 – dl2 cdt > dl or spacelike ds2 = dl2 – (cdt)2 dl > cdt or null ds2 = (cdt)2 – dl2 = 0 dl = cdt 1. Which is the longest side and which is the shortest side? C ct F 3 B A 5 6 D x from Gravity by James B. Hartle 2. Which path is longer, D to F or D to E to F? 3 E units: light-seconds PARADOXES The Pole and the Barn Paradox A 20 m pole is carried so fast that it contracts to 10 m in the frame of reference of a 10 m long barn with an open front door. Consequently, the pole can fit within the barn for an instant, whereupon the back door is swung open. But in the pole’s frame of reference, the barn is only 5 m long, so the pole cannot possibly fit in the barn! Resolve the paradox (hint: draw a spacetime diagram) 21 Betty and Ann 2060 Ann’s Now in 2060 x t c Δt = (0.8) (8y) = 6.4 years 2053.6 2050 Ann’s Now in 2050 8 light years (ly) β 1/γ Example from About Time by Paul Davies = 0.8 = 0.6 Twin Paradox Ann’s Now in 2070 2070 E-mail sent by Betty in 2056, Betty’s time, received by Ann in 2068, Ann’s time. 2068 2060 Ann’s Now in 2060 E-mail sent by Ann in 2052, Ann’s time, received by Betty in 2056, Betty’s time. 2053.6 2052 2050 8 light years Ann’s Now in 2050 Temporal Paradox 2060 Ann’s Now in 2060 Super-luminal signal sent to star A in 2050, arriving in 2056 according to Betty. But for Ann, signal sent in 2050 arrives in 2047! 2053.6 8 light years 8 light years 2050 Ann’s Now in 2050 2048 A 2047.2 Ann and Betty’s Now in 2047.2 Super-luminal signal sent from A arrives in 2048, preventing signal sent in 2050! How To Make A Time Machine! t t2 Earth’s worldline The Two Spaceship Problem During the acceleration, do the captains measure a fixed, or changing, distance between the spaceships? Explain! Captain Vivian Captain Luke t1 x 26 Summary There is no absolute “now”. Each of us has our own “now” determined by how we move about Super-luminal travel, within a simply-connected spacetime, would lead to temporal paradoxes The time between events depends on the path taken through spacetime, with an inertial (non-accelerating) path yielding the longest time APPENDIX Proper Distance By definition: C D The proper distance is the spatial separation between two simultaneous events. x’ E C & D are simultaneous in the Red frame of reference. x A B E & D are simultaneous in the Yellow frame of reference. The Lorentz Transformation t Define t' What is the interval between event O and event P? P Δt Δx = tBC + tCP = xOA + xAB Δx' Δt' = xOQ = tQP tCP = tQP /γ x' C Q O A tBC = β Δx / c B xOA = xOQ / γ xAB = v Δt x The Lorentz Transformation t We obtain the Lorentz transformation (Δx, Δt) → (Δx', Δt') t' Δx' cΔt' P tCP = tQP / γ Q x' C tBC = v Δx/ c2 O A B xOA = xOQ / γ x xAB = v Δt = γ (Δx – β cΔt) = γ (cΔt – β Δx) The interval from O to P is OP2 = (cΔt)2 – (Δx)2 = (cΔt')2 – (Δx’)2 Problem 5 Compute the spacetime distance (ds) between the following events: 1. event 1: solar flare on Sun (in Earth’s) now. event 2: a rainstorm here, 7 (Earth) minutes later. (Give answer in light-minutes.) 2. event 1: the fall of Alexandria in 640 AD event 2: Tycho’s supernova seen in 1572 AD (the star was then 7,500 ly from Earth). (Give answer in light-years.)