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Notice that one of the observers has to stop and turn back in order to compare watches and hence feel an acceleration. Thus they are not symmetrical in this sense. That's why the observer, who experiences the acceleration would be younger. As Feynman puts this fact in his famous lectures: This [twin paradox] is called a “paradox” only by the people who ...


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Special relativity is define has either points A or B being motionless to a third point, even if neither observer at point A or B can define that third point. Only after comparing stopwatches will the observers know which of them was at relative rest and which of them was in relative motion to the third point. Can't define absolute rest...but the classic ...


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"Unfortunately, due to time dilation the time for mass to fall into the event horizon becomes infinite, so that nothing can cross the horizon within in the life time of the universe." This can not be understood as any objective statement about coordinate-independent physical realities, since it's only true in certain coordinates like Schwarzschild ...


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I can only offer an opinion rather than a peer-reviewed authoritative response. But I hope it's of some use. We can read various descriptions about falling into a black hole, such as this on Baez. There's also Andrew Hamilton's website. I'm afraid to say I think they're wrong. Take a look at The Formation and Growth of Black Holes on mathspages, and you can ...


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No, relativity is about reference frames. The person falling into a black hole will pass through the event horizon, but due to time dilation, no one else can observe it.


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Where is the flaw in my thinking? In your concept of time. It's little more than a cumulative measure of local motion, see A World without Time: The Forgotten Legacy of Godel and Einstein]. Your macroscopic motion relative to some other guy results in you measuring his local motion to be slow, whilst he measures your local motion to be slow. This sounds ...


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To measure time, a duration, you need two moments – when you press "start" and "stop" button on the stopwatch, respectively. But because the two objects are moving relatively to each other, it isn't possible for them to "meet" at both moments. If their locations coincide at the "start" moment, for example, so that their clocks may be compared at this "start" ...


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if A and C are moving at equal velocities 5c then they should meet B at the same time ,If the stationary object B stands at the mid point of the motion path then A will cut the track to B at 5c and C will do the same so suppose that half the distance is 5 kilos then Then both A will meet B and C after 1/c time ..so the velocity each one will see the other ...


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Consider a spherical symmetric mass distribution in space, located around the origin of a coordinate system. One can forumlate a stress-energy-tensor $T^{\mu,\nu}$ for this situation. Solving the Einstein-field-equations for a reference frame, in which that mass distribution is not moving one obtains the Schwarzschild-metric $g_{\mu,\nu}$. One can now try ...



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