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Time Dilation. The faster you travel the longer I have to wait for you to return? |
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| Dec9-12, 05:41 AM | #18 |
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Time Dilation. The faster you travel the longer I have to wait for you to return?:http://arxiv.org/pdf/1201.5611v1.pdf |
| Dec9-12, 11:39 AM | #19 |
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Ghwellsjr's spacetime diagrams would not be satisfying because the diagram couldn't show the symmetry breaking mechanism. This is a case where a word could be worth a thousand pictures. Even if the analysis was mathematical, what would be useful is if the equation with the symmetry breaking mechanism was presented. At least one would get an idea of how the symmetry breaking occurred. The problem is that there is no physical hypothesis that distinguishes between trajectories that are geodesics and trajectories that are not geodesics. One can always find a set of variables in which an observer is traveling a geodesic. After all, the physical laws are independent of the path of the observer. "Everything" is relative, or so people have been told. What many people are asking is not whether there is mathematics that distinguish one twin from the other. They are asking for a physical hypothesis that distinguishes between the two observers. They want to know the "symmetry breaking" feature in the calculation. "Physical intuition" is not sufficient for distinguishing between a geodesic and a nongeodesic. Many books on science for laymen say straight out that it is the rockets that break the symmetry. I am just generalizing what these books in "mainstream science" say. Instead of "rockets", I say "external force". What you seem to be saying is that the rockets have nothing to do with the twin on earth aging faster than the traveling twin. What is asked for is a physical hypothesis, not a mathematical proof. The mathematical proof is certainly worth a discussion on its own, but it is irrelevant here. There are many problems in physics where a symmetry is broken. The question of what interaction physically breaks a symmetry is often a valid scientific question, regardless of scientific field. |
| Dec9-12, 01:27 PM | #20 |
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The fact is, twin that traveled has actually taken a shorter path. How? Geodesics are local minima. In fact, they don't even have to be minima, merely extrema, but I don't know if that's ever relevant in GR. At any rate, there can exist paths that are shorter. For a simple analogy, think of an object in a glass of water placed in such a way that you can see the object both through the wall of the glass and through the surface. It looks like there are two copies of the object, one distorted more than the other. I'm sure you've seen this. In optics, light takes the "shortest" path as well. Fact that you can see two images of the same object tells you that there are two "shortest" paths between the object and your eye. Again, the path only needs to be locally shortest. Meaning that any small perturbation of the path has to increase the length. Anyways, back to the twins. I'm not entirely sure whether it's fully equivalent to engine thrust, but imagine that the twin that traveled used gravity of a massive object, like a black hole, to turn around. That way, his trajectory is also a geodesic. It is also locally shortest. However, the question of which path is shorter remains. We know the answer, of course. Special Relativity tells us that the twin that stayed put aged more. That means, his path was longer. |
| Dec9-12, 01:46 PM | #21 |
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The reason the traveling twin's path is shorter is the Lorentzian analog of the triangle inequality: two sides of a triangle are *shorter* than the third, if all three sides are geodesic segments. Physically, this corresponds to the fact that the motion of the traveling twin, who travels along two sides of the triangle, can't be geodesic everywhere; there must be at least one event where he undergoes non-geodesic motion. Only the stay-at-home twin, who moves along only one side of the triangle, can have a path that is geodesic everywhere, so his path is necessarily the longest. If, OTOH, you want to have a scenario where the curvature of spacetime around the massive object is the primary effect, it is easy to construct one where the "stay at home" twin ages less. Just have the stay at home twin be in orbit about the massive object, and have the traveling twin be launched directly upward in such a way that he is pulled back down by the massive object's gravity to meet the stay at home twin after some integral number of orbits. Here both twins are traveling along geodesics, so it is clearer what the difference is between a local maximum and a global one. |
| Dec9-12, 01:59 PM | #22 |
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Physically, what "the path of the traveling twin cannot be geodesic everywhere" means is that the traveling twin has to turn around somehow. If spacetime is flat (i.e., gravity is negligible), the only way he can do that is by firing rockets, getting pushed by a laser, etc.--i.e., he has to experience a force, even if only for an instant (in the idealized version of the scenario that you've been discussing, where the turnaround is instantaneous). This means that an accelerometer carried with the traveling twin will read nonzero, even if only for an instant. The stay at home twin's accelerometer will read zero the entire time. |
| Dec9-12, 02:05 PM | #23 |
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| Dec9-12, 04:09 PM | #24 |
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In that very limited sense, the symmetry is broken by the rockets firing, because that is when everyone can work out who will be the younger twin. But, physically, there is nothing special about the rockets firing. They don't "cause the other twin to age". They just enable the twins to meet up again, which they can't in inertial frames. The analogy that I made earlier, of driving in a straight line from A-to-B compared to driving A-to-C-to-B is perfectly valid. It isn't the act of turning at C that makes the path lengths different, it is the decision to take a different route through space to the same destination. And if you (a third-party) are ignorant of the destination, you can't tell until one car turns which one is going to, so you cannot predict which will have the higher mileage. That's exactly what ghwellsjr's spacetime diagrams are showing you. If you hide the top half of them, the first two graphs just look like mirror images of each other. It's only at the point that the rocket fires and one twin turns around that they start to look like different graphs. |
| Dec9-12, 04:10 PM | #25 |
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Your comment about the accelerometer is very pertinent. I believe that you have proven my point. What breaks the symmetry is a type of acceleration, just as I said. However, it can't be any type of acceleration. A kinematic acceleration, meaning an acceleration not associated with an interaction, can't break the symmetry. The physical quantity that determines where the path is not geodesic is whatever the accelerometer is measuring. I submit that the accelerometer is measuring the total force on the accelerometer applied by the surroundings. The accelerometer is measuring the external force on it by other particles. The observer who is close to the accelerometer and stationary relative to the accelerometer is also under the influence of a force. There may be an accelerometer that is stationary and close to the observer on earth. That accelerometer does not detect an external force. The quantity that the accelerometer is measuring is what I call the dynamic acceleration. I haven't found a reference that uses that phrase. However, I do see scientists refer to the external force. I think the external force is what breaks the symmetry in the twin paradox. The external force usually stated in the twin paradox is caused by the rockets. |
| Dec9-12, 04:24 PM | #26 |
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| Dec9-12, 06:05 PM | #27 |
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Maybe I was saying it wrong. I started out using the word acceleration. Maybe this isn't the right word either. Whatever physical interaction defines the path of the hypothetical observer is what breaks the symmetry. It doesn't have to be a canonical force. In the case of quantum systems, there may not be a canonical force. Maybe it is a measurement, or a decoherence, or some other type of interaction. I was not claiming that the rockets cause the aging. I was claiming that the rockets caused the difference in ages. The question was what caused the asymmetry. I claim that the force of the rockets cause the asymmetry. What physical thing do you think is the cause of the asymmetry? A space time diagram by itself is not enough. If you must present a diagram, circle the part of the diagram which has the physical cause of the asymmetry. The concrete example in the OP's question involved twins, rockets, and aging. Later it turned in burritos, sidewalks and deliveries. The question was originally about the cause of asymmetry, not about twins or burritos. |
| Dec10-12, 03:39 AM | #28 |
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Your concern over asymmetry or a broken symmetry is a red herring. It only seems relevant in the classic Twin Paradox because one twin remains inertial. But in general, we can have any number of observers starting at one location and traveling at different speeds in different directions and ending up in the same location (but not necessarily where they started) where there is no symmetry in the scenario at all and yet we can still analyze the scenario using any arbitrary IRF and determine how each one ages with respect to each other. A spacetime diagram is simply a way to present a whole lot of data in an easily graspable way. It is part of the discipline of the Theory of Special Relativity. But you don't need to analyze scenarios like this using Special Relativity. You can do it simply with a Relativistic Doppler Analysis which shows physically what each person actually observes and measures. But you have to discipline yourself and not ask about physical causes beyond what can actually be measured and observed so I doubt that that would be satisfying to you either. Warp was the one that turned this thread into yet another Twin Paradox issue and I corrected him on some false notions and I'm still trying to correct false notions. |
| Dec10-12, 05:53 AM | #29 |
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| Dec10-12, 08:12 AM | #30 |
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Time dilation not observable? Where do you get this? This is simply wrong. In another thread I recently tried to explain and show you that time dilation is due to relativity of simultaneity. http://www.physicsforums.com/showpos...2&postcount=38 The reason you do not understand this is probably because you are too much stuck to your calculator. (And I'm afraid at the end we will have to discuss what time-coordinates are, the meaning of space-like simultaneous events, etc etc .... and this is a no go in this forum. No wonder you battle to grasp the essence of SR...) |
| Dec10-12, 08:32 AM | #31 |
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| Dec10-12, 08:58 AM | #32 |
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The essence of SR is that time is relative to the coordinate system or reference frame that you arbitrarily select. So is space. So is time dilation. So is simultaneity. Select a different coordinate system and all these characteristics change to different values. But what doesn't change are all the measurements and observations that each observer in the scenario makes. Each coordinate system preserves those measurements and observations. Maybe another way to say this is that the measurements and observations made at each event remain the same, even though the coordinates of each event take on different values in each reference frame. |
| Dec10-12, 10:21 AM | #33 |
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Let me put it to you this way. Say you have a forest full of trees... You can give me thousands of different coordinate systems with enless data lists of observations, from all over the place, and all plotted out in a different diagram. But as long as you do not tell me about the forest itself, I do not get it. This is what happens in SR discussions: data list talks. And where's the forest? (I'm glad that at least Bobc2 knows what the forest is in SR... and names it: 4D block universe) |
| Dec10-12, 10:35 AM | #34 |
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