Recent content by PAllen

So am I.

You can, but many people have a hard time picturing how this is possible. My suggestion provides a way to picture this that was helpful to me and many others, over the years.

Actually, it is only a possible property of very special spacetimes. It is a feature of geometry that is hard to visualize in a 3x1 spacetime. If you imagine the case of a 1x1 spacetime (1 spatial dimension, 1 time dimension) and closed, then you can say an expanding geometry is like the surface...

I’ve used the following analogy to picture an infinite expanding universe. Consider the universe as an (countably) infinite collection of cubic boxes of e.g. gas (at any time you can put them together mentally to make a continuous whole). Double the side of each box, you still have the same...

I found an explicit statement in this section of MTW (13.6 bottom of p. 331 in my edition) that makes clear that they have the meaning I claim: " In the case of zero acceleration and zero rotation, (..), the observer's proper reference frame reduces to a local Lorentz frame (...) all along his...

The wording “proper reference frame of an accelerated observer” implies, in ordinary English, the notion of proper reference frame of a non accelerated observer. MTW, so far as I see, does not use “proper reference frame” by itself, to imply a mandatory accelerated observer.

Well, MTW calls it the proper reference frame of an accelerated observer, implying that a standard inertial frame is the proper reference frame of a non accelerating observer (in flat spacetime). The discussion in the section of this name (p.327 in my edition) clearly includes zero acceleration...

No, that's not the issue I realized. You have a piece of ##E^2## embedded in ##S^3##, all embedded in ##E^4##. I was thinking that a geodesic of the ##E^2## is a case of a geodesic of ##E^4## contained in ##S^3## that is not a geodesic of ##S^3##. What I ignored was that the ##E^2## in this...

What I was thinking of here was based on the 3-sphere embedded in ##E^4## being able to contain a surface that is metrically ##E^2##. However, such a contained surface, while intrinsically flat, has extrinsic curvature in ##E^4## (as well as within the 3-sphere). As such, its contained geodesics...

I am saying this is not necessarily true. True is that you can find a hypersurface containing it such it is a geodesic of the hypersurface, but you also may find a hypersurface containing it such that it is NOT a geodesic of the hypersurface. I gave an example of this. [edit: turns out that...

I am not sure that is always true. Consider, by analogy, a 3-sphere embedded in Euclidean 4 space. It can contain a geodesic of the 4-space that is not a geodesic of the 3-sphere. I don’t immediately see any reason that couldn’t happen in a Lorentzian 4 space. [edit - see below for more...

Yes.

Actually, if you take any small part of the geodesic, and a small neighborhood around that part, then that geodesic segment is minimal - even though a bigger piece is not minimal.

True, but that is just a recipe for constructing a pure Riemannian manifold, that happens to a submanifold. So it is actually covered in what I said.

No, not longest distance - you can deviate from a non minimal great circle path and make it longer - by any amount with squiggles. In straight Riemannian geometry, for two points and a neighborhood containing them, when sufficiently small, the geodesic connecting them within the neighborhood is...