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In general relativity, (unlike in special relativity where time-space can be made universal) there is no concept of universal time-space, thus general observers have observations those are highly dependent at the space-time locations of the observers. Two observers who are standing apart in space-time may observe the same phenomena with astonishingly ...


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Since all the velocities are constant working out $\frac{dx^{1}}{dx^{0}}$ is simple division (no calculus required). Note that the entries in your vector in $S$ are the values you're interested in: $$(dx^0, dx^1, 0, 0)=(\gamma(d {x^0}' - v d {x^1}'), \gamma (d {x^1}' - v d {x^0}'), 0, 0)$$. So we have: \begin{equation} \frac{dx^{1}}{dx^{0}} = \frac{d ...


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If you have two vectors you can project one onto the other. It you orthogonally project a spacetime vector onto your unit tangent then the length of that projection is how much time you observe separating the two events. That might be enough to answer your question right there. So to get an average velocity between two events first compute the vector A ...


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This is not exactly the twin paradox, but it's close. First, let's make the problem more precise. Let's assume the train is one light year away from a planet, traveling near light-speed, and at the very beginning of the journey, a person on the planet views the clocks as synchronized. Then a person on the train does not view the clocks as synchronized. This ...


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Further to udrv's answer, and, the technicalities he raises aside, there are two ways to argue the reciprocity relationship that the boost from observer $A$ to $B$ is the boost from $B$ to $A$ but with $v\mapsto-v$. By the detailed arguments in the afterword, we find that the transformations between inertial frames form a group and that group acts linearly ...


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I think udrv's answer hits the nail on the head, but I'll expand a little bit on the intuitive way to look at it. In your example, you have a ship leaving earth flying towards Neptune at relativistic speed. While you might think that the ship is moving fast and the Earth is essentially still (though in truth, the Earth is moving around the sun and around ...


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Your question seems to be: if the observer on Earth sees the spaceship moving at velocity v, how do we know that the observer on the spaceship will see Earth moving at velocity -v? This is known as the "reciprocity principle" problem and it is a good one, in the sense that it raises the following issue: "Does the reciprocity principle follow from the basic ...


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Q: How reasonable it it to conclude that, from a remote observer’s frame, matter falling towards a black hole never crosses the event horizon, because ∆ t → 0 as v → c (according to the Lorentz transform)? This is not reasonable at all because the property of the equivalence principle indeed does say that the infalling object falls into the black hole. The ...


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Can you tell your absolute speed in space? Yes. You just look at the CMBR dipole anisotropy. This tells you how fast you're going relative to the universe, and that's as absolute as it gets. "From the CMB data it is seen that our local group of galaxies (the galactic cluster that includes the Solar System's Milky Way Galaxy) appears to be moving at ...


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The problem with questions like this is that they include many misunderstandings of physics! For example, you say "as one approaches light speed more energy is required to accelerate faster". What you may not be aware of is that in classical mechanics, it's also true that to an observer on the ground, the faster you are going, the more energy you need to ...


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That won't work, and here is why. It's subtle. Say that you are on a ship, leaving the solar system with some technology that is able to thrust you in such a way that you experience a constant acceleration of 1g, as measured on the ship. You can measure this by placing a 1kg weight on a scale. From the point of view of the passengers of the ship (where the ...


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I'm afraid this won't work. If for example you have a rocket motor capable of producing 1g of thrust then it will still produce 1g of thrust no matter how long you accelerate for (assuming you don't run out of fuel). From the perspective of the observer on Earth your acceleration will indeed slow down, but at the same time the Earth observer sees your time ...



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