logicman

Why isn't -2X in this drawing? Because these are two separate drawings, which were separately in the video earlier, and for preview purposes they were combined into one. These two drawings combined into one are an illustration of the fact that, according to special theory of relativity, from the point of view of observer A station A is at rest and from the point of view of observer B station B is at rest. But apparently this simple obviousness is too complicated for you. The distances I gave are only meaningless representations of concepts 'far' and 'near'. But apparently this simple obviousness is too complicated for you. And now I will tell you why I made this entry here. Because I want this video to have as many views as possible. And it doesn't matter what idiots, that can't understand the obvious, are clicking on the video link. Let's say this is the case: Station A and station B accelerated in the same way to half the speed of light. Then the observers turn off the engines. Now, from the viewpoint of observer A, clock B shows later time than the clock A, and from the viewpoint of observer B, the clock A shows later time. [It's what you claim.] When stations are flying at constant speed, from the viewpoint of observer A, the clock B is running slower, and from the point of view of the observer B, the clock A is running slower. [It's what special theory of relativity says.] So, according to you, after covering some distance at constant speed by both stations, both clocks will show the same time. Because the relative increase of clocks' running speed would be offset by their relative delay. Now is the second case: In the second case everything is the same, exept that the stations covered at constant speed the distance 1000 times bigger. So in this case, relative time dilation was occuring 1000 times longer than in the first case, but the relative increase in clocks' running speed was the same as in the first case. In other words, in the second case the relative time dilation was 1000 times greater than relative gain of time. So even for Beatrice is obvoius that if in the first case the clocks showed the same time, in the second case they couldn't show the same time. But you claim that in every case both clocks would show the same time. When I told it to my she-goat, she almost died of laughter. Let's say this is the case: Station A and station B accelerated in the same way to half the speed of light. Then the observers turn off the engines. Now, from the viewpoint of observer A, clock B shows later time than the clock A, and from the viewpoint of observer B, the clock A shows later time. [It's what you claim.] When stations are flying at constant speed, from the viewpoint of observer A, the clock B is running slower, and from the point of view of the observer B, the clock A is running slower. [It's what special theory of relativity says.] So, according to you, after covering some distance at constant speed by both stations, both clocks will show the same time. Because the relative increase of clocks' running speed would be offset by their relative delay. Now is the second case: In the second case everything is the same, exept that the stations covered at constant speed the distance 1000 times bigger. So in this case, relative time dilation was occuring 1000 times longer than in the first case, but the relative increase in clocks' running speed was the same as in the first case. In other words, in the second case the relative time dilation was 1000 times greater than relative gain of time. So even for Beatrice is obvoius that if in the first case the clocks showed the same time, in the second case they couldn't show the same time. But you claim that in every case both clocks would show the same time. When I told it to my she-goat, she almost died of laughter.

So in your opinion the situation looks like this: From the point of view of observer A, the hand of clock B moved forward relative to clock A by a value depending on the value of acceleration of station B, and from the point of view of observer B, the hand of clock A moved forward relative to clock B by a value depending on the value of acceleration of station A. And the clocks will be always aligned regardless of whether the stations cover in constant speed the distance separating me from my mother-in-law, or they cover a distance equal to the diameter of the Milky Way? Is that so? And what if the stations traveled the distance separating us from the Andromeda Galaxy? Or the other end of the universe? Before you answer, let me remind you that special theory of relativity says that the clock delay depends on the speed and DISTANCE that the clock has overcome.

For 1000 years Ptolemy's geocentric model was such a well-established theory. So instead of talking relativistic rubbish, try to solve the Kaziuk paradox.

Because this video is the best form of presentation of this paradox. So in your view, from the viewpoint of observer A, the clock at station B was running faster during acceleration, and from the viewpoint of observer B, the clock at station A was running faster during acceleration. And when the stations flew at a constant speed, then from the viewpoint of observer A, the clock at station B was running slower, and from the viewpoint of observer B, the clock at station A was running slower. As a result everything equalized. So when the stations merged, the clocks showed the same time. Is that so? But the stations accelerated very briefly (during acceleration they traveled a distance equal to the distance separating Earth from the Moon), and they flew at a constant speed for a very long time (they traveled a distance equal to the distance separating the Sun from the nearest star). So since from the viewpoint of observer A, clock B was running faster very briefly and was running slower for a very long time, and from viewpoint of observer B, clock A was running faster very briefly and was running slower for a very long time, then how after the stations merged, the clocks can show the same time? And what would the clocks show if the stations accelerated the same and were flying at a constant speed 10 times further?

So in your view, from the viewpoint of observer A, the clock at station B was running faster during acceleration, and from the viewpoint of observer B, the clock at station A was running faster during acceleration. And when the stations flew at a constant speed, then from the viewpoint of observer A, the clock at station B was running slower, and from the viewpoint of observer B, the clock at station A was running slower. As a result everything equalized. So when the stations merged, the clocks showed the same time. Is that so? But the stations accelerated very briefly (during acceleration they traveled a distance equal to the distance separating Earth from the Moon), and they flew at a constant speed for a very long time (they traveled a distance equal to the distance separating the Sun from the nearest star). So since from the viewpoint of observer A, clock B was running faster very briefly and was running slower for a very long time, and from viewpoint of observer B, clock A was running faster very briefly and was running slower for a very long time, then how after the stations merged, the clocks can show the same time? And what would the clocks show if the stations accelerated the same and were flying at a constant speed 10 times further? But they didn't disagree on simultaneity earlier. That's the point. Instead of talking relativistic rubbish, try to solve this paradox. This is not about the general. It's about the Kaziuk paradox.

On YouTube is video Unsolvable Paradox In Einstein's Theory. SPAM LINK DELETED Is it really unsolvable?