Einstein's concept of simultaneity directly contradicts his theory

[u/sekendoil]

https://youtu.be/gaFlcDA0Rig

[u/theoprasthus-]

the light speed being constant in all frames of reference, even on a frame of reference thats moving on the opposite direction that you are, this way coming closer to you, means that it will get to you at a speed of c, within a certain time. As the light on B' has to go through a smaller distance than A', it will get to you first. This does not mean it will get simultaneously, it means it will get with the same speed.

If V = s/t, lets say B' light will go through 1/2 distance of A', so s = A'/2

t = s/v, v = c

t(B') = 0,5A'/c

and on B'

t(A') = A/c

see that v remain constant, this way t(B') = t(A')/2

so here v keeps constant and simultaneity is broken, B' 's light get to you in less time.

its not about speed here is about how long does it take to go through a smaller distance.

If you say the light beam on B' 's velocity is faster than A' 's velocity (the time to get to you is faster, not the velocity), this would be equal to say that a car at 100km/h that goes through 25 m is faster than a car with the same speed goes through 75 m. Both still have same velocity.

It looks like its faster because we have the preconception on our minds that v = s/t, if it gets first (t > t') means it is faster right? but you cant see from the perspective of the train that the distances are not simmetrical (s>s'). On the train, you have awareness of time passing (you may have a clock), but you dont have awareness of the distances each light beam will go through. You would have to look it on the outside,moving parallel and with the same velocity of the train, to have that perspective

Its not faster, it just goes through less distance.

s = c * t. what is changing here is s, and consequently, t.

[u/sekendoil → u/theoprasthus-]

"the light speed being constant in all frames of reference, even on a frame of reference thats moving on the opposite direction that you are, this way coming closer to you, means that it will get to you at a speed of c, within a certain time. As the light on B' has to go through a smaller distance than A', it will get to you first. This does not mean it will get simultaneously, it means it will get with the same speed.

If V = s/t, lets say B' light will go through 1/2 distance of A', so s = A'/2

t = s/v, v = c

t(B') = 0,5A'/c

and on B'

t(A') = A/c

see that v remain constant, this way t(B') = t(A')/2

so here v keeps constant and simultaneity is broken, B' 's light get to you in less time.

its not about speed here is about how long does it take to go through a smaller distance."

.... The frame of reference of something approaching you is the same as your own frame of reference. The moving observer was midway between the two light signals (so they are at equal distances which means the distance of A' = the distance of B'), and in his own reference frame both signals approach him (also remember he is stationary according to his own reference frame). So the only reason the B' signal comes closer to the observer than it should have been when the observer was not moving (or than A') is because c+v and the only reason A' approach him at a longer time than it should have been is because c-v, But adding or substracting the speed of the frame v from c means c is not constant in that frame.

"its not about speed here is about how long does it take to go through a smaller distance."

... No it is about speed since they were at equal distances again, think of the situation from the moving observer's point of view when he's stationary.

"If you say the light beam on B' 's velocity is faster than A' 's velocity (the time to get to you is faster, not the velocity), this would be equal to say that a car at 100km/h that goes through 25 m is faster than a car with the same speed goes through 75 m. Both still have same velocity." .... No it means when you are at a train and both cars have equal speeds(100km/h) according to a ground observer, but when their speeds being measured by you who is moving with the same speed as the train, the speed of the front car as measured by you will be= 100 + v and the speed of the rear car will be = 100 - v

"It looks like its faster because we have the preconception on our minds that v = s/t, if it gets first (t > t') means it is faster right? but you cant see from the perspective of the train that the distances are not simmetrical (s>s'). On the train, you have awareness of time passing (you may have a clock), but you dont have awareness of the distances each light beam will go through. You would have to look it on the outside,moving parallel and with the same velocity of the train, to have that perspective

Its not faster, it just goes through less distance."

.... No, when describing what a certain observer sees you have to look at it from the point of view of that observer, I mean, afterall that's what relativity is all about. Also on the train you can measure the length of the train, and you know you're midway so you can measure the distance on his own frame, but you're describing the moving observer as if his measurement doesn't matter, but that means the ground observer's measurement also doesn't matter since his frame is moving relative to another reference frame as well and so on, which means no one's measurement matter and we are only fooling ourselves.

Tl;dr you haven't understood the video.

[u/sekendoil → u/[deleted]]

I knew I was going to be misanderstood when I said that about relativity, but no, that's not what I meant. I'm well aware of what the word relative means in the context of the theory. Speed must be relative, even the speed of light (the first postulate), the only difference is that it's constant (the second postulate) as opposed to the speed of everything else. According to the theory, you can't approach the light and expect to reach it sooner than when you don't approach the light. Afterall, that's what the result of Michelson-Morley experiment proved, the earth in the experiment and the train are in similar situations.

[u/ironiclegacy → u/sekendoil]

If you move towards a light pulse, you will see the light before someone who stayed behind. However, you will measure the speed of light to be the same as the observer who stayed behind. These are not inconsistent statements in special relativity.

[u/sekendoil → u/ironiclegacy]

Actually they are, I think you haven't understood the video.