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Hi Emilio, thanks for your reaction. Today I was making a walk and I suddenly realized that Gracie’s clock at the startplace has moved to the finishplace after t’=65 min. So that’s why that clock must read 65, it’s the same clock that read 0 at the startplace at the moment of the start! Quite obvious but I missed it.
I will look into worldlines and now lines, I never used them before. Hi Emilio, it’s not a problem or excercise but it is the demo in the video in 22.12. Professor Green puts the 65 at Gracie’s clock at the finish (George’s perspective). I would think it has to be placed at Gracie’s clock at the start.
I would think that Gracie’s clock at the moment of the start and at the startplace in George’s perspective reads 0. So Gracie’s clock at the moment of the start and at the finishplace because of asynchronious clocks should read : (156* 12/13C) / C2 = -144 (Clocks must lag behind)…… Some of the clocks haven’t been filled in.
1 Gracie’s clock at the finishplace at the moment of the start in George’s perspective: -144 because of the asynchronisity and in George’s perspective this clock is in the direction of the movement.
2 Gracie’s clock at the startplace at the moment of the finish in George’s perspective: 0+65=65 because of the time dilation.
This would make Gracie’s clock at the finishplace at the moment of the finish in George’s perspective: -144 + 65 =-79. That’s not the same as the 65 that professor Green shows us. What goes wrong here? The only conclusion I can draw is that 5,9 km is wrong. When you use the time passed on Earth (2,0113346 s * 10^-5) to calculate the distance travelled according to an observer on Earth you get 2,0113346 s * 10^-5 * 299,792*0,994 = 5,9936413 km instead of 5,9 km!! Using this 5,9936413 you find 0,655585 km as the Lorentz contracted distance. And this gives you the speed v = 0,655585 / 2,2 * 10^-6 = 297,993 km/s. This is 297,993/299792 =0,994C !
This indeed is a riddle! I agree with your calculations. Also when you calculate the time that passed on earth (2,2 * 10^-6 * gamma = 2,0113346 * 10^-5 s) and use this to calculate the speed you also get v = 5,9/2,0113346 = 293,338 km/s = 0,978C
I don’t see where this goes wrong because the only thing we use is 5,9 km, v=0,994C and lifetime muon in rest = 2,2*10^-6. All the rest is derived using gamma……. If the motion is perpendicular to the plane of the triangle the contraction will have no influence on the shape of the triangle. Therefore they will agree on the slope. If the motion is not perpendicular to the plane of the triangle they will not agree on the slope due to the Lorentz contraction.