Relativity 2

175 Views

Download Presentation
## Relativity 2

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Harrison B. Prosper**Florida State University YSP Relativity 2**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, non-accelerating) frames of reference. • Constancy of the speed of light: The speed of light in vacuum is independent of the motion of the light source.**SpacetimeDiagrams**Events can be represented as points in a spacetimediagram Events with the same time values, such as events A and B, are said to be simultaneous Time A B Space**Event: A place at a given time**Spacetime: The set of all events Earth’s Time Axis 3000 AD now D C B (t,x,y,z) 2500 AD A 2000 AD y x O**tC**C light light tA A Mapping Spacetime – I Starship’s worldline v = speed = BD / OB = x / tB Earth’s worldline B tB c = BD / AB D tD Line of simultaneity “Now” tB = γ tD tD = κ tA O**tC**C tB B D tD tA A O Mapping Spacetime - II 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**Relativistic Doppler Factor Dilation Factor**κ and γ Factors**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?**Line of simultaneity**Δt = tB - tE tE E But events D and E are simultaneous for the starship so tE = tD / γ When is Now ? Events B and D are simultaneous for Earth so tD = tB / γ tB tD B D Line of simultaneity O**When Is Now? - II**“Nows” do not coincide! Δt = tB - tE Line of simultaneity Line of simultaneity Writing distance between B and D as x = BD the temporal discrepancy is given by B tB D tD tE E O 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**P**A O B The Metric dl 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 calledametric dy dx 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 = rdφ CB = dr AB = dl r θ φ y Δφ C x B A**Q**O P The Interval Suppose that O and Q are events. How far apart are they in spacetime? First guess ds2 = (cdt)2 + dl2 ds cdt dl Unfortunately, this does not work! In 1908, Hermann Minkowski showed that the correct expression is ds2 =(cdt)2 – dl2 ds2 is called the interval Hermann Minkowski 1864 -1909**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)2dl>cdt or null ds2 = (cdt)2– dl2 = 0 dl = cdt**C**1. Which is the longest side and which is the shortest side? A B F ct 2. Which path is longer, D to F or D to E to F? 3 E 3 5 6 D units: light-seconds x from Gravity by James B. Hartle**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)**Betty’s Now in 2056**Δt = (0.8) (8y) = 6.4 years 2053.6 Betty’s worldline Betty and Ann 2060 Ann’s Now in 2060 2056 Ann’s Worldline Ann’s Now in 2050 2050 8 light years (ly) β = 0.8 1/γ = 0.6 Example from About Time by Paul Davies**Betty’s Now in 2062**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 Betty’s Now in 2056 Betty’s Now in 2050 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 Ann’s Now in 2050 8 light years Twin Paradox Betty 8 years younger!**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! Betty’s Now in 2050 Betty’s Now in 2056 8 light years 2048 Ann and Betty’s Now in 2047.2 2047.2 A Super-luminal signal sent from A arrives in 2048, preventing signal sent in 2050! Temporal Paradox 2060 Ann’s Now in 2060 Ann’s Worldline 2053.6 8 light years Ann’s Now in 2050 2050**The Two Spaceship Problem**t During the acceleration, do the captains measure a fixed, or changing, distance between the spaceships? Explain! Earth’s worldline Captain Vivian Captain Luke t2 t1 x**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**Proper Distance**By definition: The proper distance is the spatial separation between two simultaneous events. C & D are simultaneous in the Red frame of reference. E & D are simultaneous in the Yellow frame of reference. C D x’ E x A B**t**t' P x' C Q x O A B The Lorentz Transformation Define Δt = tBC+ tCP Δx = xOA + xAB What is the interval between event O and event P? Δx' = xOQ Δt' = tQP tCP = tQP /γ tBC =βΔx/ c xOA = xOQ / γ xAB = vΔt**t**t' P tCP = tQP / γ x' C Q tBC = v Δx/ c2 x O A B xOA = xOQ / γ xAB = v Δt The Lorentz Transformation We obtain the Lorentz transformation (Δx, Δt) → (Δx', Δt') Δx' = γ (Δx – βcΔt) cΔt' = γ(cΔt–βΔx) 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 lyfrom Earth). (Give answer in light-years.)