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#41 |
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#42 |
Penultimate Amazing
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Well, they don't have to agree about what time it is on Earth. They just have to synchronize their own clocks. Like, if the one going away started the journey at 6AM, when it passes the other guy by, he transmits a signal that says, "hey, my clock says it's 12:34 right now", and the other one simply also sets his own clock to 12:34 too, and lets it tick normally from there. When he passes earth by, he and the guy who stayed home exchange similar signals and yep, they're different.
Mind you, when the guy sets the clock to match there is basically a discontinuity (his clock probably showed something else), but that's how you simulate the same clock ticking on both legs of the journey. It has to continue on the trip back from where it reached on the way out. |
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#43 |
Penultimate Amazing
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True. But you still aren't looking at one inertial frame for the traveling twin. Again, that throws off the symmetry you're assuming.
Here's another way to look at it. "Proper time", the time experienced by a traveler moving along a line, is basically just a form of distance measurement. So you're basically just comparing the distance between two points along different lines. In Euclidean geometry, how do we measure distance? For a straight line segment, the length s can be found as s2 = x2+y2+z2 For a curved line, we chop it up into infinitesmal bits and add up the lengths of each bit. So s2 = integral(ds2) = integral(dx2+dy2+dz2) You can also abreviate this as simply ds2 = dx2 + dy2 + dz2 This is called the metric for Euclidean space. One of the consequences is that the shortest path between two points is a straight line. Deviate from that, and you increase the length. But special relativity doesn't happen in Euclidean space. The metric is different: ds2 = d(ct)2 - dx2 - dy2 - dz2 In this metric, for time-like separated points (ie, were ds2 is positive), a straight path is the LONGEST distance between them. Curve the path, and you make it shorter. Furthermore (and this is important), it doesn't matter which reference frame you look at it from, s will always be the same. The different components may change (just like in Euclidean space, if you rotate everything your x's, y's, and z's will all change but your distances won't), but the lengths won't. So let's compare the twins. One twin, the "stationary" one, takes a straight path between the two points in question. One twin, the "traveling" one, takes a curved path. The curved path is shorter than the straight path. That isn't a function of the curvature itself, exactly, but the curvature is still a necessary part of that. Same as Euclidean space, except you're going shorter instead of longer. |
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#44 |
Penultimate Amazing
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I keep telling you how it's solved. And you don't even have to take MY word for it. Here's a Fermi lab physicist saying the same thing: https://www.youtube.com/watch?v=GgvajuvSpF4
One of many. There's no shortage of it getting explained if you just Google or search YouTube for "twins paradox." |
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#45 |
Penultimate Amazing
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Careful here. Instant reversal requires infinite power, but not infinite energy.
An alternate version I like is where the traveling twin goes in a big circle. They're constantly accelerating, but at each point along the trip their relative velocities are always the same. The acceleration makes the traveling twin non-inertial, but here's the kicker: it doesn't matter what the acceleration is, only the velocity. The traveling twin could trace out one big circle (small acceleration), or he could trace out a small circle a whole bunch of times (large acceleration), the final answer won't change. The acceleration is a necessary component to the problem, but you never need to use it in any of the calculations. |
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#46 |
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#47 |
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#48 |
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#49 |
Penultimate Amazing
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Actually, technically so far all you've shown is that you don't know what you're talking about -- as in, not even the nomenclature -- but insist that that shows a problem with everyone ELSE's understanding. That's really not helping anyone. And I mean, not even as illustration of Dunning-Kruger, since we already had people like Pixie Of Key illustrate that far better
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#50 |
Penultimate Amazing
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It's unrealistic for that jump to happen instantaneously, sure. That's a simplification. But the transition can happen in a very short length of time, short enough compared to the overall trip that it's negligible.
Quote:
![]() I made this diagram for triplets. The red one stays home, the green one goes off to the left, the blue one goes off to the right. Diagram 1 shows this from the "home" reference frame. Green and blue are time dilated, red is not. Red experiences more time than green or blue, as measured in the home frame. Now let's look at it from the right moving frame (diagram 2). Red is now time dilated. Green is time dilated for part of the journey, not time dilated for part. But it's time dilated to an even greater degree than red for the portion that is time dilated, and that portion is over half the journey in this frame. So overall green will end up with less total time than red because of that greater time dilation. Same with blue, except that it's the second part of its journey which has the increased time dilation, not the first. Diagram 3 is the other direction. No matter which inertial frame you pick, red ends up with more time than green or blue. But you have to stick with an inertial frame, or Lorenz factors won't suffice. That means you can't pick any frame in which green or blue are straight lines. And it doesn't matter to this whether the turnarounds are instant or not. That rounds the corners of the green and blue, but doesn't fundamentally change anything. Now if you want to know how to handle being in an accelerating reference frame, well, that's quite interesting. The math is really nasty but you can get some of the concepts without getting too into the weeds. But you've got to be able to wrap your head around what's happening in different inertial reference frames first, or you won't have a solid structure to build off of. |
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#51 |
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#52 |
Penultimate Amazing
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I'm not sure what's confusing you there. Regardless of what reference frame you choose, you can plot the {x, y, z, t} coordinates for everyone involved in it. They're ALWAYS in the same reference frame, regardless of what you do. You could put your reference frame on Jupiter, and they'll still both be in it. Those coordinates will be different, based on what frame you're using, but there's no such thing as one being in the reference frame and the other not being in it.
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#53 |
Penultimate Amazing
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#54 |
Muse
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Look at your diagram 1.
All 3 triplets are in the same reference frame at the 'half time'. They have the same simultaneity line. The horizontal line. Please, show me that same simultaneity line of one reference system in the diagrams 2 and 3. Please, remember the triplets are supposed be in the same frame, no relative motion in 2 and 3. Edit: The diagrams 2 and 3 contradict 1 because they do not show acceleration for the left and right triplets. The blue triplet does not have an acceleration on the first leg in 2 and the green triplet does not have an acceleration in 3. |
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#55 |
Penultimate Amazing
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It's not a black hole, it's an event horizon. No, it's not magic, yes, you could describe it as virtual, and no, it's not a problem.
Physicists' preference for inertial reference frames predates special relativity. You encounter similar things in Newtonian mechanics. F=ma works in an inertial reference frame, it doesn't work in a non-inertial reference frame. But you can still do Newtonian mechanics in non-inertial frames. How? Well, you just have to subtract the acceleration from the non-inertiality : F = mareal = m(aapparent - aframe) Rewrite this a bit and you get: maapparent = F + maframe = Fapparent So we can account for our non-inertiality if we know how our frame is accelerating. This leads to so-called "fictitious forces": Fapparent = Freal + Ffictitious Ffictitious = maframe Thes fictitious forces are always proportional to mass, and in many cases (such as rotation), they are also position-dependent. You can do similar stuff in special relativity to account for non-inertiality, but it gets even messier than Newtonian mechanics. When you accelerate, you introduce fictitious forces, but also fictitious Lorenz factors (if you want to work within an accelerating reference frame). And these are position-dependent, because in special relativity, acceleration is a form of 4D rotation. This actually touches on an insight that Einstein had: a uniform gravitational field is the same as acceleration. GR gets significantly messier than special relativity (even special relativity in non-inertial reference frames) because gravity isn't uniform. So you have to make these "fictitious" effects real. But it's still the same jumping off point. |
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#56 |
Muse
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Here is the problem:
![]() The point A in the diagram is a 'virtual black hole'. It is an intersect with all the simultaneity lines of the deceleration and the acceleration shown in the diagram. There is not motion in A, no time flow, nothing - it appears there like a 'black hole'. If we do this exercise in the intergalactic space there is no real black hole at A. The event A is a contradiction to what Minkowski space-time diagram is supposed to show. |
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#57 |
Penultimate Amazing
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But they will disagree about when that simultaneity line is, because they didn't get to that the same way. Look at diagram 2 and 3: that "half time" isn't simultaneous if you pick either the right-moving or the left-moving frame. It's ONLY simultaneous in diagram 1. But diagram 1 is the only diagram in which any triplet is stationary for the whole duration, and only one of the triplets is stationary for it.
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#58 |
Penultimate Amazing
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No it isn't a contradiction.
Give this a try. Go outside, spin around, and watch what happens to the sun (or stars at night). It moves! It spins around you! This massive thing is moving at incredible speed, and experiencing incredible acceleration. And that's just within our solar system. The speed of other galaxies massively exceeds the speed of light, and the forces required to accelerate them in a circle around you are incomprehensibly large. Or are they? You can probably understand that the sun didn't really move. It didn't really experience centrifugal acceleration as you spun. This is just an artifact of you spinning. Inertial frames are different than non-inertial frames. If you take the non-inertial frame to be real, you can reach some very strange conclusions. That's all that this is. ETA: point A is not a black hole. The line delta1 is an event horizon. There is actually a difference. A black hole is surrounded by an event horizon, but it isn't synonymous with one. |
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#59 |
Muse
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#60 |
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#61 |
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#62 |
Penultimate Amazing
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#63 |
Penultimate Amazing
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Acceleration is curvature. Technically speaking, all three diagrams were drawn without any curvature, using simply straight line segments.
Of course, we can posit a scenario in which there is finite acceleration but on a scale too small (ie, too short a time period) to be seen in the diagrams. This is fine, because the acceleration itself doesn't need to enter into the calculations. It can be made an arbitrarily small perturbation to the problem, so we can solve the issues of interest without reference to it. So we actually have multiple possible scenarios for what's happening at t=0. We can have all three tripplets start at rest in diagram 1, and then the green and blue accelerate outwards. We could also have one or both of blue and green start out already in motion. Hell, we could even start with all three moving to the left, and then blue stops and red reverses direction. These are all indistinguishable from each other, if the accelerations occur quickly enough. And all of them can be covered by all three diagrams. There is no contradiction, they all display the same thing to within the specificity possible by these simple diagrams. Any difference you allege between diagram 1 and 2, for example, is either a misinterpretation or can be covered by ambiguity of each diagram individually. |
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#64 |
Penultimate Amazing
Join Date: Mar 2009
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You know some conversation has taken a wrong turn at Albuquerque when you have objections on par with 'but... but... but the horse is not a point.'
![]() Same thing here. We work with a simplified model, because that's all we need. Yes, to actually change direction in zero time, you'd need infinite acceleration, but that doesn't change the fact that it shows how the key to the whole thing is that one of the observers switches frames. You could work with a more sane acceleration in between, but as Ziggurat said, that's basically just rounding the corners on the graph and making the maths a bit more complicated. Now you get to integrate over a curve instead of a triangle, but the basic idea is the same. Also, just like Ziggurat said, having an event horizon doesn't actually mean that a black hole has formed. The light doesn't actually go into some singularity there. And when you stop accelerating, it can reach you again just the same. And it's not even just at relativistic speeds. If I accelerate an old steam locomotive at 1m/s, I have an event horizon behind me just the same. It's too insanely far away to be even noticed, but it exists just the same. So what? And, again, when you stop accelerating, it disappears and the light can reach you again. No information has actually been lost. In particular, if we're talking infinite acceleration for a duration of zero, then that's also exactly how long that event horizon exists: zero. After that, you can see the guy waiting back on Earth, or really any other point in flat space, just as if nothing happened. In essence, that's why you can ignore it in the model. |
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#65 |
Penultimate Amazing
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Additionally, since we're talking virtual even horizons, that exists only in the opposite direction than your acceleration. Since the ship would be accelerating towards the Earth, there is no moment when it would lose sight of Earth because of that event horizon. Because the event horizon is in the other direction. In every possible sense of the word, the twins never cease to be visible in the same frame.
And that is another reason to ignore it from the model: it just happens outside the space that we're interested in. |
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#66 |
Muse
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There is a contradiction.
The green triplet measures acceleration on his accelerometer in the first leg in diagram 1. The green triplet DOES NOT measure any acceleration on his accelerometer in the first leg in diagram 3. These are two distinct and different scenarios. Your diagrams do not describe the same stuff. |
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#67 |
Muse
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#68 |
Muse
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Well, no, to the bold part.
If you go without an acceleration in the solution then we are left with the paradox. The SR does not have a hierarchy of the inertial frames. Once we switch to accelerations there is a difference. The tag scenario is not a solution to twin paradox, only acceleration can help. |
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#69 |
Muse
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![]() Please, look at the reciprocity. There is an inertial moving frame identical with with AP red wordline. This is an additional triplet, let's call him the red triplet. The red triplet is right there flying by when the black triplet accelerates from the stay home triplet. The acceleration away, the 0 to T/4 interval, it is like deceleration for the red triplet frame. The the traveling black triplet decelerates to the event P, T/4 to T/2 interval, this is like acceleration to the red triplet frame. The same thing repeats on the way back. How do we solve this time dilation? |
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#70 |
Penultimate Amazing
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I'm still not sure what the problem is, to be honest, or rather how is it tied to the reference frames.
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#71 |
Penultimate Amazing
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#72 |
Muse
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The SR does not have a solution for the twin paradox.
The acceleration is not a simple solution as well as shown in my previous post. The analysis has to be done carefully. The triplet diagrams show the contradictions. If a hierarchy of the reference frames cannot be achieved we have a problem. The experiments point towards the frame hierarchy to be real. |
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#73 |
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#74 |
Penultimate Amazing
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No, there isn't. There are TWO inertial frames which together cover the red worldline. But it's two different frames, not one, and the fact that it's two is critical. The two frames WILL NOT AGREE about simultaneity for the red world line of the traveling twin and the grey world line for the earthbound twin. The frame for the outbound twin will think that at the turnaround point for the traveling twin, the earthbound twin has already experienced less time. But the frame for the inbound twin will think that at the turnaround point for the traveling twin, the earthbound twin will have already experienced MORE time.
If you cannot understand that my three diagrams all depict the exact same scenario in three different inertial reference frames, then you either don't understand the scenario, or you don't understand Lorentz transformations. It really is that simple. Special relativity has been around for over 100 years. It is a mathematically elegant and fully self-consistent theory. You will not find any actual paradoxes. If you think you found one, it's because you don't understand something. And that's perfectly understandable: special relativity is counter-intuitive. Nobody understands it without study. But you are not smarter than every physicist of the last 100 years. You haven't figured out what nobody else has figured out. |
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#75 |
Penultimate Amazing
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You missed the point entirely. Yes, you need acceleration in the problem. But if you want to plug in actual numbers, you can make the acceleration such that it contributes an arbitrarily small perturbation to the final answer. Therefore we can ignore what the acceleration is for our calculations for simplicity. That a twin accelerates is vital. What the acceleration is, however, is not.
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#76 |
Penultimate Amazing
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Yes it does, and it has been explained to you repeatedly. Just repeating the same falsehood one more time won't make it true. We're not in The Hunting Of The Snark and you're not the Bellman. Something doesn't become true just because you say it three (more) times.
Again, nowhere have you shown that the maths doesn't work or anything else of substance. Your objection has been just, basically, that it's impossible or unrealistic to perform that experiment. But that's a completely unrelated issue than whether SR holds or not. Basically equally I could calculate the thrust force if monkeys were to fly out of my ass. More specifically, adult capuchin monkeys, at 100m/s and a rate of one monkey per second. Is it a realistic or even possible scenario? No, it isn't. But that doesn't mean that the thrust equation is wrong. Err... so? No, seriously. The maths being complicated doesn't make something wrong or anything. No it doesn't. So far it just showed that you don't understand it. No they don't. In fact, not even in GR. So even introducing acceleration into it still won't point in the direction you think. The whole point of why it has "relativity" in the name is that it is, surprisingly enough, relative. GR lets you calculate things in an accelerating frame (or indeed in a gravity well, same thing), but it's still not a hierarchy. If you have one guy on the Earth and one on the moon, the curvature of the space (because of acceleration, which gravity is a part of too) will be different, but one frame (or chart, before someone objects to my misusing terms) is still not in any way better than the other. None of them is inherently the preferred frame. You might prefer one or the other because it makes your maths easier, but essentially neither of them is inherently THE one true frame to use. |
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#77 |
Penultimate Amazing
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#78 |
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#79 |
Muse
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Is there an 'event horizon/black hole' at event A here?
![]() Why would I ask? Because when the front of the train car aligns with the platform frame where x=x'=0 and t=t'=0 and the front train car observer is going to 'ask' his comoving observer at the back what do see outside? Meaning the comoving observer knows to record what is outside at t'=0 and x'=-3.4641cs' ( L0=3.4641cs) The train back observer sees -6.9282cs outside and time -6s. The same is done by the platform observers. The 'back' platform observer records what he sees on the train. The 'back' platform observer located at x=-1.732cs records -3.4641cs' and t'=3s' on the train. How does the train 'shrinks' from the event B to event C with no time flowing at the front of the train car? Is event A a 'black hole'??? The time is moving in both frames at the back but it is not moving at the front? This all goes back to what time dilation is and how it is reciprocal. |
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#80 |
Penultimate Amazing
Join Date: Mar 2009
Posts: 17,729
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Not sure you understand what a black hole is. A black hole has an event horizon. Not every event horizon is a black hole or around a black hole.
Basically just like a bird has wings, but not everything that has wings is a bird. I just saw four wings on a helicopter, for example. Additionally, unless I'm reading your graph wrong, there isn't even an event horizon there. An event horizon ONLY happens when there's acceleration involved. Gravity IS acceleration, or at least equivalent, which is why a black hole has one. An accelerating body would also 'see' one behind it, although it would have to be accelerating really hard, and it's not really there for anyone else. Some guy standing on the ground next to the accelerating train would not actually see any event horizon there. So basically not only there is no black hole there in any case, but if it's a constant speed motion (which seems to be the case in your graph) there isn't an event horizon at all either. |
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