On Thu, Sep 18, 2025 at 11:15 AM Alan Grayson <[email protected]> wrote:
*> Please elaborate on the role of the Christoffel symbols that enclode the > spacetime curvative in a gravitational field. What does "encode" mean in > this context? AG* > In ordinary calculus the derivative is just how something changes with respect to something else, but in curved space there’s an extra complication, the coordinate grid itself bends. Christoffel symbols keep track of how the coordinate axes are changing as you move around. Mathematically they look like Γ*i**jk, where **i*, *j*, *k* = 1, 2, ..., *n * and each entry of this 4×4×4 array is a real number because spacetime has 4 coordinates. So there are 64 slots. Fortunately symmetry simplifies that a little, there are only 40 independent numbers. *This is not a tensor but but it tells you how to take the derivative of a tensor in curved space so the result is also a tensor and thus looks the same to everyone. * *> Can a tensor be defined as a mathematical entitity which is invariant > under a coordinate transformation.* > *Yes.* *> What is the form of the entity that guarantees that invariance?* > *I'm not sure I understand the question but momentum and velocity and acceleration are all tensors, but angular momentum is not, it's just a "pseudo tensor". That's because to calculate the angular momentum you need to produce a cross product, and that causes an additional handedness and is why things look different in a mirror. Angular momentum is relative to the choice of origin. If you pick a point for the origin of your coordinate system other than the sun, as astronomers customarily do, then the angular momentum of Jupiter would be different. * *> Brent says, I think, that proper accelation will not be felt. True or > not? AG* *If you are in freefall in a gravitational field then proper acceleration will not be felt because it is zero. If you ARE firing a rocket in a gravitational field then you are NOT in freefall, you are NOT on the straightest possible line through 4D Spacetime, the proper acceleration is NOT zero, and that will be felt. * * John K Clark See what's on my new list at Extropolis <https://groups.google.com/g/extropolis>* 9vq -- 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/CAJPayv20ZEpDYdoJ2AtGx1rf34RmyvDjNv4F28aMiGorcv8-WQ%40mail.gmail.com.
