On Sun, Dec 15, 2024 at 3:09 PM Alan Grayson <[email protected]> wrote:
> > > On Sunday, December 15, 2024 at 12:54:08 PM UTC-7 Quentin Anciaux wrote: > > > > Le dim. 15 déc. 2024, 20:45, Alan Grayson <[email protected]> a écrit : > > > > On Sunday, December 15, 2024 at 11:24:36 AM UTC-7 Quentin Anciaux wrote: > > Le dim. 15 déc. 2024, 18:48, Alan Grayson <[email protected]> a écrit : > > On Sunday, December 15, 2024 at 6:45:57 AM UTC-7 Quentin Anciaux wrote: > > Le dim. 15 déc. 2024, 14:31, Alan Grayson <[email protected]> a écrit : > > On Sunday, December 15, 2024 at 6:20:47 AM UTC-7 Alan Grayson wrote: > > On Sunday, December 15, 2024 at 5:41:54 AM UTC-7 John Clark wrote: > > On Sat, Dec 14, 2024 at 11:01 PM Alan Grayson <[email protected]> wrote: > > * > What bothers me is the disagreement between frames about fitness or > not, and why the alleged lack of simultaneity resolves the apparent > contradiction. AG * > > > *In this thought experiment I think even you would agree that no matter > how fast or slow the car is going there will always be times when the front > of the car is in the garage, and times when the back of the car is in the > garage; so the question of the day is " Is there any frame of reference in > which those two events occur SIMULTANEOUSLY?" Einstein's answer is "yes", > provided the car is moving fast enough. And as proof that Einstein's answer > was correct, in this thought experiment, above a certain speed, there is NO > frame of reference in which there is a car shaped hole in the back door of > the garage. * > > *John K Clark See what's on my new list at Extropolis > <https://groups.google.com/g/extropolis>* > > > No. Above a certain speed, since the car is length contracted from the pov > of the garage frame, the car will fit in the garage, and waiting some time, > then being in a different reference frame, it could hit the back door. AG > > > But the real question is this; if from the pov of the car frame, there is > a v such that the car never fits in the garage for this v and greater, why > is it claimed that lack of simultaneity between frames solves the problem, > since we're only considering simultaneity in garage frame where the car > fits? AG > > eua > > > It's crazy you're still on this... wtf is your definition of fits in which > doesn't involve simultaneously having rear and front of the car in the > garage? > > Quentin > > > I never denied that "fitting" requires simultaneity. > > > > Then wtf are you trolling about ? > > > *Earlier you asked if I agreed that fitting requires simultaneity, and I > affirmed that. My problem all along, which apparently went over your head, > is how the two frames can give diametrically opposite results. All you were > capable of doing was to repeat some slogan without offering any genuine > explanation. * > > > Bullshit, everyone here explained it was about ordering of events, it's > nit hand waving, it's just your trollest and unwilling attitude to > understand, or just you as cosmin wanting to troll because you enjoy it. > > > *Obviously, you don't understand but indulge the illusion that you do. > Fact is I don't fully understand the time ordering of events, such as if > two events are simultaneous in one frame, and not in another, will the time > order necessarily be reversed in that frame and all others, and if so, why?* > Just to answer this question, if two events A and B have a spacelike separation (neither event lies in the past or future light cone of the other one), then there will be one frame where they are simultaneous, other frames where A happens before B, and other frames where A happens before B. It may be easier to understand why this is true intuitively if you learn about how to draw spacetime diagrams where you have two observers #1 and #2 in relative motion, and you draw a diagram where the vertical axis is the T coordinate of observer #1 and the horizontal axis is the X coordinate of observer #1, and then you draw the X' and T' axes of observer #2 in that diagram, where both axes look like slanted lines offset from the vertical and horizontal, as in Fig. 5 near the top of the page at https://en.wikibooks.org/wiki/Special_Relativity/Simultaneity,_time_dilation_and_length_contraction . Since the X' axis is a set of points that all have the same T' coordinate for observer #2, the X' axis will show you what observer #2's surface of simultaneity looks like from the perspective of observer #1's frame. Meanwhile the T' axis would be the worldline of an object at rest in observer #2's frame (since it the T' axis has a constant X' coordinate) whose worldline passed through the origin. The X' axis will always make the same angle with a light ray as the T' axis (compare the x' and ct' axes with the yellow light ray in the diagram at https://commons.wikimedia.org/wiki/File:Minkowski_diagram_-_3_systems.svg as well as the x'' and ct'' axes for a third observer with a higher velocity), so if you know how to draw the worldline of an object at a given speed v in observer #1's frame, that's enough to give you an idea of what the surface of simultaneity for that object (its X' axis) would look like when plotted in observer #1's frame, and how it would "tilt" continuously as you tilt the worldline/T' axis (considering objects at higher velocity) relative to observer #1's frame--see the little animation of this tilting starting at 13:09 in the video at https://www.youtube.com/watch?v=5bSy18w8Dh0&t=789s Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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