On Wed, Jun 18, 2025 at 10:54 AM Alan Grayson <[email protected]> wrote:
*>>> Earlier you wrote that free falling in a gravity is like falling, or > moving along a straight line as in a flat Euclidean space, but the SS is > free falling in a gravity field and traveling in a curved path around the > Earth. Can't you just acknowledge your error?* > > > *>> I'm perfectly capable of making an error but I don't know what I said, > or what you think I said, that you're referring to. * > > > *> You were very clear, two messages back, that a body in free fall will > experience straight line motion as in a flat Euclidean space.* > *As I said before, the Equivalence Principle says if you have no contact with anything that is not in your local space, a.k.a. you are at a point, then you have no way of telling if you are moving in a Euclidean straight line through flat space or if you are in a gravitational field moving along a geodesic in curved non-Euclidean spacetime. * *>and I gave the example of the SS orbiting the Earth. AG* *And as I explained in another post that you evidently have not bothered to read: "No force is being applied to the space station but it is not following a Euclidean straight line because it is not in flat Euclidean space, it is in curved 4D non-Euclidean spacetime and is following a geodesic path. In curved 4D non-Euclidean spacetime the shortest path between any two points along the space station's orbit is the space station's orbit itself.* *>> But I do know that the Equivalence Principle says if you have no > contact with anything that is not in your local space then you can't tell > if you're in the gravitational field in curved non-Euclidean spacetime or > if you're accelerating in a straight line in flat Euclidean space.* > > > *> But with sufficiently sensitive instruments one can tell the > difference.* > *The sensitivity of the instrument is not the issue, no matter how sensitive it is if you pick a small enough region of space it will not be able to tell the difference, or alternately if the gravitational field producing object is sufficiently large. In both cases the limit of the difference is zero. Or are you implying that when he proposed the Equivalence Principle Albert Einstein, the greatest physicist in 300 years, was ignorant that a phenomenon called "tidal effects" existed?! * * >>Spacetime has to be non-Euclidean if time is involved because when it > comes to defining a distance the Pythagorean Theorem must be modified, you > need to throw in a minus into the equation, D^2=X^2+Y^2+Z^2 - (cT)^2.* > > > *> This fall far short of an argument. The definition above is certainly > non-Euclidean insofar as the Pythogorean theorem is violated, but how does > this fact imply geodesic motion, specifically from an initial state of > being spatially at rest? AG * > *I'm not sure what you mean because we're talking about relativity so I have to ask "spatially at resf" relative to what? I will say that if you're standing on the Earth's surface then you can NOT be in an initial state * *And intuitively it sort of makes sense that when it comes to distance the > spatial and temporal coordinates should have opposite signs because the > larger the spatial distance between 2 points the harder it would be to > travel between them, but the larger the time you had to make that journey > the easier it would be. * > > *And the speed of light "c" is just a conversion factor between space and > time. * > > *John K Clark See what's on my new list at Extropolis <https://groups.google.com/g/extropolis>* dd0 > > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAJPayv0Uk-DJwupNk4bj4UDQRsbVRbU2o___x7GBYA1iSTvhPg%40mail.gmail.com.
