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Example: relativistic snake Solution: Boy is right Reason: Hatchets do not fall at the same time in snake’s reference frame Let’s assume that the left hatchet falls at tL=tR=t’L=0 (S’-snake, S-boy) Snake’s proper length = 100cm V=0.6c – relativistic snake! – γ=5/4 t Boy drops two hatchets 100cm apart in his reference frame. € From boy’s reference frame: Snake is moving – it is shorter – snake will be unharmed From snake’s reference frame: The hachets are moving – distance between them < my length I will lose my head! Both can’t be right! What is the distance between the places where the two hatchets fall in snake’s frame? 1. From Lorentz transformations (TMZ): 5 x'R = γ (x R − vt R ) = (100cm − 0) = 125cm 4 2. From “relativistic common sense”: € $ vx ' 5 $ 0.6c × 100cm ' t'R = γ & t R − 2R ) = &0 − ) = −2.5ns % ( c ( 4% c2 So the right hatchet fall before the snake’s head gets to it in both reference frames and the snake is unharmed € 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. In snake’s frame: distance between the hatchets in the air: 100cm/γ=80cm; time delay between left and right: 2.5ns. Distance traveled by snake during that time: 0.6c*2.5ns=45cm; 80+45=125cm; Same answer!" Try to check your math this way. Easy to make a mistake!!! Remember time dilation/length contraction!!! 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. 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 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!! So we are already nearly done with the transformation laws We now can convert: locations: x ! x’ etc. time: t ! t’ But we still have to figure out: velocities: u ! u’ Lorentz transformation v = 0.5c x" = γ (x − vt) u’=0.5c y" = y z" = z Velocity transformation: Which coordinates are primed? S’ S u’ is what we were looking for! (i.e. velocity measured in S’) v t" = γ (t − 2 x) c 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 ) ( € y' u (x,y,z,t) S Earth z (x',y',z',t') S' x z' v x' Spacecraft ) € € v = 0.5c Relativistic transformations x" = γ ( x − vt ) vu −v t " = γu("t =− 2 x ) 2 1c− uv / c u−v u' = 1 − uv / c 2 u’=0.5c u= S’ S u= u! + v 1 + u !v / c 2 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 d) 0.25 c e) 0 u! + v 1 + u !v / c 2 u" = u −v 1 − uv / c 2 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