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Implications of the statement "Acceleration is not relative" |
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| Feb12-13, 10:35 AM | #52 |
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Implications of the statement "Acceleration is not relative"Me neither. It looks completely consistent. Using the fictitious force to keep the 'rocket' twin stationary does not change the physics - viz. the travelling twin is non-inertial some of the time but the other one is always in free-fall. Therefore the travelling twin ages less as she should according to the other frames. |
| Feb12-13, 11:01 AM | #53 |
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Consistency is not the same a uniqueness. There are different, reasonable, choices for simultaneity for the accelerating twin. Each produces a different metric (though they converge near the 'time axis' represented by the accelerating twin), with different statements as to how much of the aging (of the inertial twin) occurs during turnaround (assuming e.g. coast, turn, coast). Further, the most common way this is presented will not work at all for a W shaped traveler trajectory (the simultaneity surfaces will intersect and multiply map the inertial twin world line, preventing you from having any coordinate chart in which to integrate proper time). So then you must use a different set of simultaneity surfaces to handle this case. |
| Feb12-13, 11:17 AM | #54 |
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I understand you are advocating caution, but I was addressing the OP's question about a consistent treament of the twins in which the travelling twin remains stationary ( ie has a vertical worldline). |
| Feb12-13, 11:37 AM | #55 |
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| Feb12-13, 12:09 PM | #56 |
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I have now realised that the OP was bothered because there is no treatment of the twins case with the travelling twin being inertial. As you and others have already pointed out, that is impossible. |
| Feb12-13, 06:00 PM | #57 |
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| Feb12-13, 06:10 PM | #58 |
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The Lorentz transformation works even if the β parameter depends on time, so we have a transformation from inertial to non-inertial coordinates. |
| Feb12-13, 06:15 PM | #59 |
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- The mathematics of SR is simplest in inertial frames, but all phenomena may be analyzed in such frames, including non-inertial motion. - There is no such thing as a global non-inertial frame; non-inertial frames are local. - It is possible, in many ways, to set up coordinates in which a non-inertial world line has constant spatial coordinates of 0. For any such coordinates, you have to transform the Minkowski metric. This transformed metric leads to different formulas for time dilation, light paths, and geodesics. Different choices for such coordinates will produce different answers for coordinate dependent properties, but will produce the same answers as inertial frames for any observations or measurements. |
| Feb12-13, 06:17 PM | #60 |
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If you deal with non-inertial frames within the confines of special relativity, then you have the same problem that Newton had: There is an absolute quality to acceleration; there is a preferred frame. I maintain my position that this does damage to the principle of relativity. |
| Feb12-13, 06:21 PM | #61 |
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| Feb12-13, 06:27 PM | #62 |
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In general relativity as well, acceleration is distinguishable, and there is a precise mathematical difference between a local inertial frame and a local non-inertial frame in GR: in the former, the connection coefficients vanish, in the latter they do not. As for laws taking the same form, this is just a matter of the mathematical way you write them (stevendaryl has explained this before on this thread, I believe). If, in SR, you write laws explicitly using the metric and vector/tensor quantities, as you do in GR, then the laws will take the same form in non-intertial coordinates as they do in inertial coordinates. This is still not GR, because there is no gravity involved, nor is the EFE (the equation defining GR) used. |
| Feb12-13, 06:30 PM | #63 |
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Also, arguing about definitions is not the same as arguing about physics. Can you state a *physical* objection that doesn't depend on a particular definition for what "the principle of relativity" says? |
| Feb12-13, 06:52 PM | #64 |
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However, the discussion about inertial vs non-inertial frames is not relevant to the statement "acceleration is not relative". The statement "acceleration is not relative", as we have mentioned, refers to proper acceleration. Proper acceleration is a property of a worldline, not a property of a reference frame. It doesn't matter what reference frame you use, inertial or not, the proper acceleration is the same in all of them. So, the statement "acceleration is not relative" is about worldlines, not reference frames. I think that you are getting distracted by irrelevancies. The travelling twin has non-zero proper acceleration regardless of what reference frame is used. |
| Feb12-13, 06:56 PM | #65 |
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Once you know how the physics works in inertial frames, then figuring out the physics in any other frame is simply a matter of performing a change of variables to the coordinates (aka coordinate transform). All of the usual math for doing a chang of variables still applies. Thus, even though the postulates only describe physics in inertial frames, you can use them indirectly to derive the physics in non-inertial frames. |
| Feb12-13, 07:56 PM | #66 |
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So let's consider an example, Newton's second law of motion. The relativistic 4D version of this, for a particle of constant mass, is[tex] F^\lambda = m \frac{d^2x^\lambda}{d\tau^2} [/tex]when measured in any inertial (Minkowski) coordinate system. This is pretty simple and almost the same as the non-relativistic version. However in non-inertial coordinates, the equation becomes[tex] F_\lambda = m \sum_{\mu=0}^3 g_{\lambda \mu} \frac{d^2x^\mu}{d\tau^2} + \frac{m}{2} \sum_{\mu=0}^3 \sum_{\nu=0}^3 \left( \frac{\partial g_{\lambda \mu}}{\partial x^\nu} + \frac{\partial g_{\lambda \nu}}{\partial x^\mu} - \frac{\partial g_{\mu \nu}}{\partial x^\lambda} \right) \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} [/tex]You don't need to understand the meaning of this, just observe that it's very complicated. So, yes the laws of physics can be expressed in a form that is the same in all frames, inertial or non-inertial, but such expression is much more complicated than the inertial-frame-versions of the laws. |
| Feb13-13, 01:45 AM | #67 |
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| Feb13-13, 02:17 AM | #68 |
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- http://web.archive.org/web/200608290...ry/gtext3.html It would be good if physics textbooks discussed this topic, but I don't know any that does. |
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