i think that what i fail to grasp is WHY ( that is, for what practical reasons ) the light beams will reach the MOVING box doors at the same time, like all of you said. The way i think of it, exactly due to the fact that c is constant and invariant, it will take it more time to reach the front...
Yes i understand what you mean, that the computer screen acts like sort of an "ether" but i think the analogy is not valid. The simulation doesn't happen in the screen, it happens in computer memory and in the cpu ( so it's just mathematics ). The screen is just the visual representation medium...
all of those observers are inherently wrong, either they are on the train, on the platform, on mars, on another solar system or on andromeda. They can see only their relative movements, but that doesn't mean that there is no TOTAL DEFINITE ABSOLUTE movement of each of them in the universe...
no i don't think that this is bugging me. I think the computer simulation example was pretty good. So in the computer simulation of this we have a box which moves in the screen at 10pixel/second to the right. inside the box there is a light emitter ( which also moves at 10 pixel/second ), at...
I see what you mean. What I don't understand is WHY each observer will see the light reaching the doors of his own box, at the SAME TIME. Despite the fact that he DOESN'T know if he's moving, light "knows", because he is actually moving despite the fact that he doesn't know it.
Imagine if we...
that's a good question. i think the answer is this:
they both can conduct the experiment with the two clocks in the front and the back of their respective box. Whoever finds a difference in the measured time can deduce that they are moving. if they don't find a difference, they aren't.
that's the part that i don't understand. the doors are not in reality stationary, they are moving forwards, they just "look" stationary to the inside observer, so in order to make sure he puts two clocks, one in the front one in the back, so because of the fact that the doors indeed are moving (...
hm, but the fact that I ( who am on the train ) think that I'm at rest, doesn't really mean that I'm at rest! The fact that it's difficult for me to tell who's moving, doesn't mean that I'm not moving and that the other guy is. There "must" be some - even theoretical - frame of reference.
I guess I'm getting confused because most probably i have in my mind a notion of "absolute frame of reference" where no-matter what each observer "thinks" (whether he is stationary or moving ), there is some absolute movement.
hm, if one is moving at 0.5c from point M and the other at -0.5c from point M ( minus stands for opposite direction ) then they each will "see" the other moving at -c from them.
yes, the observer within the box cannot fathom whether or not the box is moving (but it's still moving even if he doesn't know it). That's why I thought that by installing two clocks one at the front and one at the back, it would be possible to actually have evidence on whether it is moving or...
I can understand why C is constant and I accept that. What I don't understand is why the time is always 2*L/c. for the front door, since the door is moving away from the light (despite the fact that we don't know it yet), it makes sense that the light beam would need more time to reach it ( from...
yes but we had already established that the two clocks on the train reference (one front door one back door) were already synchronised. we are not talking about two clocks one in the train one in the platforms
i didn't know that they measured it and it was 2*L/c both ways. ok i do accept that, and the first "rational" explanation that comes to mind is: c is not constant. in one way it's c+v in the other it's c-v.
i know of course that C is constant. just saying
yes you are right the roundtrip wouldn't be the same.
But the most unexpected ( for me ) is that they measured it like you said and it's always 2*L/c even when the box is moving. that's weird! I don't understand why that happens. here is a possible explanation: c is not constant ( lol ).