There are several significant flaws in this story, which would impact, to name a very similar example, the “twin paradox”. Transferring this example to the twins, the leaving twin would have to post-process his brother’s clock, setting it back conveniently to reach some analogue result as in this example. The key point is that the start, both with the twins as in this example, represents a COMMON frame of reference, which one is leaving. There is no point in suggesting that the partner is moving whilst it should be very clear who the moving part is – both in the twin paradox as here (again) between George and Gracie. The example means stretching relativity beyond the limits.

My comment refers to 22.11, where Professor Brian wrote 100’ = (v)(L0/2)/c^2.
Following 8.1 it should be 100’ = (v) (L/2)/(c^2-v^2), and with 200’ = L/v,
results in v = 0.7c; γ = 1.41; L = 141’ and L0 = 200’

My comment refers to 22.11, where Professor Brian wrote 100ns = (v)(L0/2)/c^2.
Following 8.1 it should be 100ns = (v) (L/2)/(c^2-v^2), and with 200ns = L/v,
results in v = 0.7c; γ = 1.41; L = 141’ and L0 = 200’

Hi Emilio, professor Green uses the formula as derived in module 21.

Thank you very much Marius, now I realize that data of 100ns of Train’s front clock refers to the coordinate (Δt’), while I was mistakenly confusing with (Δt). Now Professor Brian explanation is clear to me!

You are welcome! I often confuse the time too.

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?

Hi, Marius, I can’t follow your writing and question. Please specify the Problem’s number you are referring to.

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)……

Good morning, Marius
Try to use the Space Time axes George (x,t) and Gracie (x’,t’) or in other words, World Lines (t,t’) and Now Lines (x,x’). I find much better understanding using them.
For better explanation, while George remains at the Start line, he has a friend (Peter) at the Finish Line. Both of them have clocks in sync.
When Gracie’s World Line (x’=0) crosses the Peter’s World Line (x=156 lt.min), the Now Line of Gracie is (t’=65 min) and the Now Line of Peter and George is (t=169min).
The “Problem” starts when Gracie makes numbers, without using the Space Time diagram:
The Now Line of Gracie (t’=65 min) crosses the World Line of George (x=0) at (t=25 min). Thinking twice now, she sees her Now Line (t’=0) while crossing the George’s World Line (x=0) at (t=0), crosses the Peter’s World Line (x=156lt.min) at (t=144 min) and solve the problem making 25+144 =169 min!
But if she were using the Space Time diagram, she would see that her Now Line (t’=65min) also crosses the World Line of Peter (x=156 lt.min) at (t = 169 min)

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.

Space Time is good thought to enjoy the daily walks. I still remember my bike walks in Eindhoven in the summer 1968, when working as a Student Volontair in Philips!

Given a certain frame of reference, and the “stationary” observer described earlier, if a second observer accompanied the “moving” clock, each of the observers would perceive the other’s clock as ticking at a slower rate than their own local clock, due to them both perceiving the other to be the one that is in motion.

In frame 3 you state that gracie’s clock starts at 144, but start moment should all be zero because there is no relative motion yet.