vuxx said:
What I am not sure about is when you said 'And each frame will observe the other's clocks to go slow'. Are you saying that both clocks will observed to be slow by each person in each of the frame? But the equation \Delta t = \Delta t_0 / \sqrt{1 - v^2/c^2} will only mean one of the time to be slow so I am quite confused.
Here's what I mean: Two frames, A and B, each with their own clocks. Time dilation says: (1) Frame A will make measurements of the behavior of the B clocks and will deduce that they are slow compared to his A clocks, and (2) Frame B will make measurements of the behavior of the A clocks and will deduce that they are slow compared to his B clocks. Note that this is a specific statement about the behavior of a
single moving clock; it is not a statement about time intervals
deduced from the readings of
multiple moving clocks. Each frame is equally permitted to use the time dilation formula to describe the behavior of the other frame's clocks compared to his own.
As about the 3 SR effects, i have only know the first 2 so far ie. time dilation and length contraction and i think to go deep in these 3 fields will take quite some time. Let me give an example to show you what i don't understand.
I repeat, there is no way to understand SR without considering
all three SR effects working together. The one you left out--the desynchronization of moving clocks--is key!
Lets say I am on a train platform and i observed a train pass by me with speed 0.8c (assuming I am able to see the train). Theres a clock behind me and also a clock in the train. Both of us agree that it is noon at that time.
OK, so far there's a clock on the platform and a clock in the front of the train.
When the train pass by me, i see a time of 12.15pm in the train's clock.
On which clock? The one in the front of the train? (That one's long since past. How do you see it?) Or another one in the back of the train?
The clock behind me will show a time that is longer than 12.15 and the time interval of my frame will correspond to the answer given by the above equation and adding it to 12.00 we have 12.25pm.
Careful. If you are saying this: When the front of the train passed you, that clock at the front of the train showed 12:00. Then when the rear of the train passed you, that
same clock at the front of the train showed 12:15, then you are correct that the clock behind you must show a longer time (say 12:25). It is a simple application of the time dilation formula for a moving clock: the clock at the front of the train.
But this is according to
you, on the ground. The train observers will disagree! The train observers say that when the back of the train passes you, their train clocks will read a time greater than 12:25. You, on the ground, however, see the train clocks (at front and rear of the train) as be out of synchronization -- to you they don't read the same times at all! (That's the desynchronization effect.)
But if there's a person in the train and he sees the clock behind me as the train pass by, what time will he sees? Will the time he sees on the clock behind me be 12.25pm or a time slower than his?
If he's looking at
your clock, he'd better see the same thing you do!

Maybe you are asking: What will he see on
his clock at the back of the train when the back of the train passes you? In that case, his clock at the rear of the train will read greater than 12:25.
I come at this question because if the person behind me uses the equation to solve the time interval from his frame he will get an answer that is slower than 12.25pm.
And that makes sense, since from the view of the train, your clock is a single moving clock, which the train observer sees as going slow.