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Implications of the statement Acceleration is not relative 
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#109
Feb1513, 07:27 AM

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#110
Feb1513, 07:53 AM

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#111
Feb1513, 08:33 AM

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The equivalence principle is about this notion of gravitational field: there is no difference (other than the variation of [itex]g[/itex] with location) between the effects of a [itex]g[/itex] due to gravity and a [itex]g[/itex] due to the use of accelerated cooridnates. 


#112
Feb1513, 08:40 AM

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[EDIT: stevendaryl made essentially the same point, but faster!] 


#113
Feb1513, 10:02 AM

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#114
Feb1513, 10:35 AM

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My point remains, that the "gravitational field" he refers to by "A gravitational field appears, that is directed towards the negative xaxis. Clock U1 is accelerated in free fall, until it has reached velocity v. An external force acts upon clock U2, preventing it from being set in motion by the gravitational field. When the clock U1 has reached velocity v the gravitational field disappears" is the field from the Christoffel symbols and it is entirely determined by the choice of coordinates. I have not read all of Einsteins writings, but I believe that this is the meaning he usually attributes to the term. If you believe that he refers to something besides the Christoffel symbols then please be explicit about what mathematical term you think he intends and why. 


#115
Feb1513, 12:08 PM

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#116
Feb1513, 05:56 PM

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Before I say what I was looking for, I will accept a good deal of the criticism in recent posts. I have made some overly broad and not well thought out statements, resulting in errors. I realized that as I went through the mental exercise of constructing the spacetime diagram for the twin paradox, rather than accepting it as a finished product. In my previous post, I asked for a SR solution of the problem in which the noninertial rocket is always at rest. I don't recall anyone presenting such a solution. My sense is that my calls for such a solution have been rebuffed. Every SR solution that I have seen has the rocket in motion. Certainly, in every SR spacetime diagram I have seen, the rocket is in motion; it is the reversing motion of the rocket which produces the knee in the rocket's worldline, and thus the difference in proper times. The article discussing acceleration in special relativity also speaks of the accelerating object as in motion. (http://math.ucr.edu/home/baez/physic...eleration.html) I, in my rocket, claim that I am not in motion. I have been sitting in my rocket in the same position throughout the episode. You claim that I will be younger than my twin at the end of the episode. As proof, you present to me a diagram that shows me in motion. I reject it. I categorically deny that the diagram applies to me. The diagram shows me in motion; I have not moved. I do not deny that my experience has been noninertial. I do not deny that the experience of my twin has been inertial. I do not even deny that I have experienced "proper acceleration", because you have told me that I can experience proper acceleration while remaining motionless. I deny that I have been in motion. Therefore, I insist on a solution that does not put me in motion. What would a diagram of such a solution look like? There would have to be a point representing my position and time at the beginning of the episode, another point representing my position and time at the end of the episode, and a (possibly curvilinear) path that connects the two points. Whatever the shape of the path, the value of the position coordinate must not vary from the beginning to the end of the path, because I do not move. It is maintained that SR can be used to solve this problem. I will attempt to draw the spacetime diagram. For convenience, I'll make my time and position axes orthogonal, in the usual manner with time positive toward the top of the page. And as usual I will set my starting position and time at the origin. Because I do not move, my worldline must be parallel to the time axis; in this case it will be coincident with the time axis. I realize that this is the worldline of an inertial object. I am not inertial. It doesn't seem right that my worldline should be inertial, but that's how it must be, because I do not move. Now to draw the earth. The earth is moving in my frame, and inertial in its own frame. I cannot find any worldline that will satisfy both conditions, and also meet me at the end of my worldline. I cannot complete the spacetime diagram. Being unable to represent in a spacetime diagram the perfectly legitimate scenario of myself at rest throughout the episode, and the earth reversing, I conclude that SR is not suitable for solving the problem. I may be wrong in my conclusion, as has been claimed. If so, I would like to see a SR solution which has me in one position throughout the episode. 


#117
Feb1513, 06:10 PM

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#118
Feb1513, 06:14 PM

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#119
Feb1513, 09:14 PM

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Another, more accurate, way to say it is that your coordinate lines are bent. So lines of constant coordinates are not straight lines and straight lines don't have constant coordinates. A similar thing happens, e.g. in polar coordinates. Since the coordinates are curved the equation r=mθ+b does not represent a straight line. 


#120
Feb1613, 07:56 AM

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I have the impression that some of the contributors on this forum are teachers by trade, so what follows may be of interest. If not, no need to read further. I've been trying to figure out why I have had so much trouble learning relativity. In particular, I have never had the experience of repeatedly thinking that I understand a subject, only to discover that I am profoundly wrong. One reason, no doubt, is the bizarre premises that we are called on to accept. However, that was much more of a stumbling block at the beginning than it is now. At this point, I can "suspend disbelief" and treat the problem as an exercise in abstract logic. The "truth" or "reality" of the premises can be evaluated later. So perhaps I'm just not good at abstract logic. Maybe. I'm sure I'm no Einstein, at any rate. But I'm not profoundly stupid, either. So how do I repeatedly find myself in the position of being profoundly wrong? I had an "aha" moment on this a couple of weeks ago, which was reinforced and clarified last night. It has to do with my pattern of learning. The entire subject of relativity is completely handsoff for me. I'll never see, much less operate, a particle accelerator. I learn new things all the time in my work, but in every case I can test my understanding of what ought to happen against what actually happens when I act on my understanding. In addition, many aspects of relativity are hypothetical (handsoff) for everyone, at least in our lifetimes. We'll never travel at relativistic speeds. So none of us have the opportunity to directly test our understanding by experiment. (We have indirect experimental evidence to support what is predicted; that's not the same thing as making the prediction come to pass.) I have made the mistake of thinking that "unable (in practice) to test by experiment" means "completely unable to verify". In my usual method of learning, I form a mental picture of what ought to happen, then I test it by experiment. Because I am unable to test, I have been in the habit of stopping after forming the mental picture. In my work, I may do calculations after forming a mental image and before conducting an experiment. (I nearly always did when I designed machinery; I rarely do now that I work in a larger company and only write software.) The calculations are viewed as a means of avoiding failure in the experiment; they are never seen as a verification of anything. The calculations are never an end in themselves; they are a means of getting to the end, which is a functioning piece of equipment. In relativity, for someone in my situation, the calculations are both the verification and the deliverable. That's what finally penetrated my thick skull last night. I can no more put something on this forum without verifying it by calculation than I can deliver an untested product to a customer. Now maybe we'll see fewer dumb statements by me on the forum. I did find this interesting, for perspective. Errors are never acceptable. But if even professionals have trouble, I should not be surprised if I have trouble, too. From the paper referenced in post #80: 


#121
Feb1613, 08:53 AM

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Okay, if it really will make you happy, I will post such a demonstration. 


#122
Feb1613, 09:54 AM

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Accelerated Rocket in Inertial Coordinates
Let's choose the zero for [itex]x[/itex] and [itex]t[/itex] to make the mathematics as simple as possible. So assume that one twin is at rest at some location [itex]x=L[/itex], in inertial coordinates. The other twin starts off at location [itex]x=L_1[/itex] at time [itex]t = t_1[/itex], travels in the [itex]x[/itex] direction until he reaches [itex]x=L_0[/itex] at time [itex]t=0[/itex], and then travels in the [itex]+x[/itex] direction until he reaches [itex]x=L[/itex] again at time [itex]t=+t_1[/itex]. The time origin is chosen so that his journey is symmetric in time about the point [itex]t=0[/itex]. The mathematical description of the traveling twin's path is: [itex]x(t) = \sqrt{(L_0)^2 + c^2 t^2}[/itex] Time [itex]t_1[/itex] is chosen so that [itex](t_1)^2 = \dfrac{L_1^2  L_0^2}{c^2}[/itex] The velocity of the traveling twin at any time is given by: [itex]v(t) = \dfrac{c^2 t}{\sqrt{(L_0)^2 + c^2 t^2}}[/itex] and the timedilation factor [itex]\sqrt{1\dfrac{v^2}{c^2}}[/itex] is given by: [itex]\sqrt{1\dfrac{v^2}{c^2}} = \dfrac{L_0}{\sqrt{(L_0)^2 + c^2 t^2}}[/itex] The elapsed time for the traveling twin is given by: [itex]\tau = \int \sqrt{1\dfrac{v^2}{c^2}} dt[/itex] I'm not going to do the integral, but you can see that the integrand is less than 1, so the result will certainly be less than [itex]\int dt = 2 t_1[/itex]. So the traveling twin will be younger. I'm going to make another post where I describe this same situation from the point of view of the traveling twin. 


#123
Feb1613, 12:06 PM

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Accelerated Rocket in Accelerated Coordinates
In inertial coordinates, the path of the traveling twin is given by: [itex]x(t) = \sqrt{L_0^2 + c^2 t^2}[/itex] Now, let's switch to a coordinate system in which the traveling twin is at rest, by making the transformation: [itex]X = \sqrt{x^2  c^2 t^2}[/itex] [itex]T = \dfrac{L_0}{c} arctanh(\dfrac{ct}{x})[/itex] where [itex]arctanh[/itex] is the inverse of the hyberbolic tangent function. In terms of these coordinates, the path of the stayathome twin is given by: [itex]X(T) = L_1\ sech(\dfrac{cT}{L_0})[/itex] where [itex]sech[/itex] is the hyperbolic secant function. To see that this is reasonable, you can look at the Taylor series expansion: [itex]sech(\theta) = 1  \frac{1}{2} \theta^2 + \ldots[/itex]. So for small values of [itex]T[/itex], we have: [itex]X(T) = L_1(1  \frac{1}{2} \dfrac{c^2T^2}{L_0^2} + \ldots)[/itex] [itex] = L_1  \frac{1}{2} g_1 T^2 + \ldots[/itex] where [itex]g_1 = \dfrac{c^2 L_1}{L_0^2}[/itex] That is the path of an object that starts off at [itex]X = L_1[/itex] at time [itex]T=0[/itex] and falls at the acceleration rate of [itex]g_1[/itex]. So in these coordinates, the traveling twin is always at the location [itex]X = L_0[/itex], while the stayathome is at [itex]X = L_0[/itex] at some point (at some time prior to [itex]T=0[/itex], rises to [itex]X=L_1[/itex] at time [itex]T=0[/itex], and then falls back to [itex]X=L_0[/itex] at some later time. In terms of the coordinates [itex]X,T[/itex], the elapsed time [itex]d\tau[/itex] for a traveling clock is given by: [itex]d\tau = \sqrt{\dfrac{X^2}{L_0^2}  \dfrac{V^2}{c^2}} dT[/itex] where [itex]V = \dfrac{dX}{dT}[/itex] Note the difference with the formula for an inertial frame: [itex]d\tau[/itex] not only on the velocity, but on the position [itex]X[/itex]. For the traveling twin, who is always at [itex]X = L_0[/itex], his proper time is: [itex]d\tau = dT[/itex] For the stayathome twin, whose position [itex]X[/itex] is always greater than or equal to [itex]L_0[/itex], the first term [itex]\dfrac{X^2}{L_0^2} > 1[/itex]. If you work out the details, you will find that [itex]d\tau = \sqrt{\dfrac{X^2}{L_0^2}  \dfrac{V^2}{c^2}} dT > dT[/itex] So in accelerated coordinates, it's also the case that the stayathome twin ages more than traveling twin. 


#124
Feb1613, 12:16 PM

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In this kind of situation you can have multiple freefall paths between the same pair of events with different proper times; but each freefall path has maximal proper time compared to all nonfreefall paths within the same topological class. (There are other topological classes possible as well; for example, there could be another freefall twin that went around the cylinder twice, etc.) For example, an accelerating "twin" that didn't go around the cylinder would have less elapsed proper time than the freefall stayathome twin (but not necessarily less than the freefall traveling twin); and an accelerating "twin" that *did* go around the cylinder would have less elapsed proper time than the freefall traveling twin. 


#125
Feb1613, 01:38 PM

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The key concepts of relativity, both special and general, are geometrical (Minkowski geometry for SR and pseudoRiemannian geometry for GR). Just like you can take a piece of paper and draw geometrical figures and discuss many things, such as lengths and angles, without ever setting up a coordinate system. The same thing is possible in relativity. The "piece of paper" is spacetime which has the geometrical structure of a manifold. The "geometrical figures" are worldlines, events, vectors, etc. that represent the motion of objects, collisions, energymomentum, etc. In this geometrical approach, the twin scenario is simply a triangle, and the fact that the travelling twin is younger is simply the triangle inequality for Minkowski geometry. In a coordinateindependent sense, the traveler's worldline is bent, and that in turn implies that his worldline is necessarily shorter as a direct consequence of the Minkowski geometry. Now, on top of that underlying geometry, you can optionally add coordinates. Coordinates are simply a mapping between points in the manifold (events in spacetime) and points in R4. The mapping must be smooth and invertible, but little else, so there is considerable freedom in choosing the mapping. It is possible to choose a mapping which maps straight lines in spacetime to straight lines in R4, such mappings are called inertial frames. It is also possible to choose a mapping which maps bent lines in spacetime to straight lines in R4, a noninertial frame. Such a mapping does nothing to alter the underlying geometry. The bent lines are still bent in a coordinateindependent geometrical sense, but because it simplifies the representation in R4 it can still be useful on occasion in order to simplify calculations. Because the mapping is invertible, in many ways it doesn't matter if you are talking about points in the manifold or points in R4. So you talk about things being "at rest" based on R4, and things happening "simultaneously" based on R4, and many other things. However, it is occasionally important to remember the underlying geometry. I hope this helps. You have been provided explanations and references addressing the topic, which clearly didn't "do it" for you. Read the explanations and references and point out the specific things that you don't understand or don't agree with. Then we can help clarify and make some progress. 


#126
Feb1613, 02:56 PM

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So the "piece of paper" must have the property that you can set up a coordinate system on it, that is its defining property if you want to call it a manifold. The fact that "you don't have to" is obviously relying on the fact that it is implicit in the definition, so I always have trouble understanding the insistence on banning charts. 


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