On 12/18/2024 1:58 PM, Alan Grayson wrote:


On Wednesday, December 18, 2024 at 2:42:39 PM UTC-7 Brent Meeker wrote:




    On 12/17/2024 11:21 PM, Alan Grayson wrote:


    On Tuesday, December 17, 2024 at 10:16:51 PM UTC-7 Brent Meeker
    wrote:

        On 12/17/2024 7:52 PM, Alan Grayson wrote:

               On Tuesday, December 17, 2024 at 6:57:28 PM UTC-7 Alan
        Grayson wrote:

                      On Tuesday, December 17, 2024 at 2:33:46 PM
        UTC-7 Brent Meeker wrote:

                             On 12/17/2024 9:25 AM, Alan Grayson wrote:


                Yes, you look at it just in terms of lengths, which
                is what I did in the
                first pair of diagrams.  But the relativity of
                simultaneity is another
                way to look at the same problem, which is what I
                showed in my last posting.


            *Another way, but not the only way. AG *


        We seem to be on the same page concerning use of length
        contraction to explain the
        differing results in the frames under consideration. But I
        remain unclear how the
        disagreement of simultaneity can also give the same results.
        For example, suppose
        from the pov of the garage frame, the car fits in the garage
        for sufficient v, with room
        to spare, but the front and rear end EVENTS do not Lorentz
        transform into simultaneous
        events in the car frame. Can't there be other ways for the
        car to fit, using another set
        of events which*are* simultaneous in the car frame? AG

        Sure. If  the car's speed was just right, it would be the
        same length as the garage.  Then in the diagram A and B would
        be at the same time in the garage frame the car would be just
        the right length such that the rear of the car entered the
        garage just as the front exited the garage.  Since we know
        the car is 12 long and the garage is 10 long we can calculate
        the required speed from 10/12 =sqrt{1-v^2} which yields
        v=0.553 if I did the arithmetic right.


    That would be 0.553c. So, if the front and back events in the
    garage frame are simultaneous in the car frame AND in the garage
    frame,
    Nobody said that the events were simultaneous in the car frame. 
    The car is contracted in the car frame.  You keep throwing shit in
    problem just to keep it going.  I'm starting to suspect you're
    just a troll.

    Brent


*My question for you is this; when will you learn to read English? You act like an uneducated prick who can't read basic English. The consensus view in the physics community is that the solution to this problem involves disagreement about simultaneity. I don't see this as correct. For example, that's what Quentin wrote several times, mocking me, and that's what a link claimed, without proof, which someone posted. And even Jesse, if I read him correctly, claims that the result in one frame must be false if there's no simultaneity. So that's why I posed the question to you, an alleged expert who teaches relativity. I am NOT a FUCKIN' troll! The only problem here is that YOU cannot read plain English, despite being educated at a great university. FU, AG*
*I read English just fine and reading what I wrote I see that I did not say the events were simultaneous in the car frame.  You just made that up to keep this ridiculous thread going.  Troll!

Brent
*


    why is it claimed that the solution to the problem, whatever it
    is, depends on disagreements of simultaneous events, when there
    are none? And if we get different results for fitting in the
    garage, where, for example, the car never fits, is there anything
    about this result that implies something contradictory or
    paradoxical? AG

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