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Announcements • Reading for Monday: 2.1 – 2.5 Today’s class • Velocity transformation • Spacetime interval • HW 3 is posted. Due next Wed. noon. • Exam 1 soon. It will cover Chapters 1 & 2. Start preparations now. I want to fly to a distant star that is 200 lightyears away. a) I can get there if I fly really close to the speed of light b) No way for me to get there before I die, because it is not possible to go faster than the speed of light. Remember time dilation/length contraction!!! I want to fly to a distant star that is 200 light-years away. a) If I fly REALLY close to the speed of light, not only can I make it there, but noone on Earth will notice that I was gone b) When I come back from my trip I will not know anyone No matter how fast I fly, more than 200 years will go by on Earth!! Last class: Lorentz transformation (Relativistic version of Galileo transformation) y y' So we are already nearly done with the transformation laws (x,y,z,t) S x z (x',y',z',t') S' We now can convert: locations: x x’ etc. time: t t’ v x' Lorentz transformation (relativistic) z' But we still have to figure out: velocities: u u’ Lorentz transformation v = 0.5c x" = γ (x − vt) y" = y z" = z v t" = γ (t − 2 x) c u’=0.5c S € S’ u is what we were looking for! (i.e. velocity measured in S) y Suppose a spacecraft travels at speed v=0.5c relative to the Earth. It launches a missile at speed 0.5c relative to the spacecraft in its direction of motion. How fast is the € missile moving relative to Earth? Δx' γ ( Δx − vΔt ) Δx − vΔt Δx /Δt − v ux − v ux' = = = = 2 = 2 2 2 Δt' γ ( Δt − vΔx /c ) Δt − vΔx /c 1 − vΔx /( Δtc ) 1 − ux v /c uy Δy' Δy Δy /Δt uy' = = = = 2 Δt' γ ( Δt − vΔx /c ) γ 1 − vΔx /( Δtc 2 ) γ (1 − ux v /c 2 ) uz Δz' uz' = = Δt' γ (1 − ux v /c 2 ) ( € Velocity transformation: Which coordinates are primed? ) y' u (x,y,z,t) S Earth z S' x z' (x',y',z',t') Spacecraft v x' v = 0.5c Relativistic transformations u’=0.5c S’ S Suppose a spacecraft travels at speed v=0.5c relative to the Earth. It launches a missile at speed 0.5c relative to the spacecraft in its direction of motion. How fast is the missile moving relative to Earth? (NOTE: Remember which coordinates are the primed ones! And: Does your answer make sense?) a) 0.8 c b) 0.5 c c) c Lucy d) 0.25 c The “object” could be light, too! Suppose a spacecraft travels at speed v=0.5c relative to the Earth. It shoots a beam of light out in its direction of motion. How fast is the light moving relative to the Earth? (Get your answer using the formula). a) 1.5c b) 0.5 c c) c d) d e) e e) 0 v ? Lucy v ? ... -3 -2 -1 0 George 1 2 3 ... ... -3 George has a set of synchronized clocks in reference frame S, as shown. Lucy is moving to the right past George, and has (naturally) her own set of synchronized clocks. Lucy passes George at the event (0,0) in both frames. An observer in George’s frame checks the clock marked ‘?’. Compared to George’s clocks, this one reads A) a slightly earlier time B) a slightly later time C) same time -2 -1 0 1 George 2 3 ... The event has coordinates (x = -3, t = 0) for George. In Lucy’s frame, where the ? clock is, the time t’ is , a positive quantity. ‘?’ = slightly later time Time vs. space In the reference frame of the snake. Hatchets are 80 cm apart, but they fall 2.5 ns apart In the reference frame of the boy: Hatchets are 100 cm apart, but they fall simultaneously Spacetime interval Neither space nor time are invariant under Lorentz trans. May be some combination is Remember this? (from 2nd class) The distance between the blue and the red ball is: If the two balls are not moving relative to each other, we found that the distance between them was “invariant” under Galileo transformations... …but not under Lorentz transformations! (Length contraction.) need new definition for distance? Spacetime interval Say we have two events: (x1,y1,z1,t1) and (x2,y2,z2,t2). Define the spacetime interval (sort of the "distance") between two events as: With: Spacetime interval The spacetime interval has the same value in all inertial reference frames! I.e. Δs2 is “invariant” under Lorentz transformations. (Homework #3!) Remember Lucy? Remember Ethel? cΔt’ h Lucy Δy’ = h Δx’ Event 1 – firecracker explodes Event 2 – light reaches detector Geometrical distance between events is cΔt’ Distance between x-coordinates is Δx’ and: (cΔt’)2 = (Δx’)2 + h2 And Lucy got We can write: 0 = (cΔt #) 2 − (Δx#) 2 0 = (cΔt #) 2 − (Δx#) 2 − (h) 2 since Event 1 – firecracker explodes Event 2 – light reaches detector Geometrical distance between events is h. Time between the two events is Δt. And we know that: or: or: (Δs)2 ≡ (cΔt)2 − (Δx)2 − (Δy) € This 2 − Spacetime ct Here is an event in spacetime. (Δz)2 distance is not in our usual space but in Spacetime (h)2 € € Let’s define distance in a new way: that is invariant under Lorentz transformation − x Any light signal that passes through this event has the dashed world lines. These identify the ‘light cone’ of this event. Spacetime ct Spacetime ct The blue area is the future of this event. x The pink is its past. Here is an event in spacetime. A x The yellow area is the “elsewhere” of the event. No physical signal can travel from the event to its elsewhere!