The story talks about 2 clocks and 2 pictures. What's happening if we take one photo that "sees" both ends of the pole? What if the photo is taken by Claude, who's travelling as fast as the pole, but from a distance (so he can sees both ends of the pole). My uderstanding is that he'll see all the pole inside the barn, but the time on the right hand side of his photo won't be the same as the time on his left hand side. This is... wierd? Is my undertstanding correct?
In my opinion it is impossible, even if Alice runs at speeds close to the speed of light. If two equal distances are in relative motion to each other at speed v, the distances overlap. In my opinion, if we denote with t the elapsed time in the frame of the barn and if we denote with t_1 the elapsed time in the frame of the pole, it is t = t_1. (and two equal distances overlap, no problem) The length of the pole contracts in the barn frame and the length of the barn contracts in the frame of the pole, the two lengths are equal! (suppose they are equal to d) d = gamma * v * t and d = gamma * v * t_1. When the two distances overlap it is t = t_1 = d / (gamma * v), for each value of d. If the length of the barn is shorter, in my opinion it is impossible! If Alice runs without the pole, then the situation is different. (there are no two distances in relative motion between them) In this case x = v * t and t_1 < t. (Alice's clock slows down compared to Bob's clock)
The barn paradox has been solved by this publication which proves that both ends of a moving pole will match both ends of the barn at the same time. vixra.org/abs/1902.0452
This guy delivers the best explanation of SR on the Internet. Watch all his videos on the subject on Coursera / Stanford. Well done!
Excellent series of explanations. 1st time I fully understood the twin paradox. D*V/C2 did the thing. Surprised to see not more likes.
The story talks about 2 clocks and 2 pictures. What's happening if we take one photo that "sees" both ends of the pole? What if the photo is taken by Claude, who's travelling as fast as the pole, but from a distance (so he can sees both ends of the pole). My uderstanding is that he'll see all the pole inside the barn, but the time on the right hand side of his photo won't be the same as the time on his left hand side. This is... wierd?
Is my undertstanding correct?
In my opinion it is impossible, even if Alice runs at speeds close to the speed of light.
If two equal distances are in relative motion to each other at speed v, the distances overlap. In my opinion, if we denote with t the elapsed time in the frame of the barn and if we denote with t_1 the elapsed time in the frame of the pole, it is t = t_1. (and two equal distances overlap, no problem)
The length of the pole contracts in the barn frame and the length of the barn contracts in the frame of the pole, the two lengths are equal! (suppose they are equal to d)
d = gamma * v * t and d = gamma * v * t_1.
When the two distances overlap it is t = t_1 = d / (gamma * v), for each value of d.
If the length of the barn is shorter, in my opinion it is impossible!
If Alice runs without the pole, then the situation is different. (there are no two distances in relative motion between them) In this case x = v * t and t_1 < t. (Alice's clock slows down compared to Bob's clock)
When I was in college, had a hard time understanding this.
I don't understand 😅
The barn paradox has been solved by this publication which proves that both ends of a moving pole will match both ends of the barn at the same time.
vixra.org/abs/1902.0452
faulty