Recent content by PAllen

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...

Let’s keep in mind that the shadow would have to be 300 meters for a one microsecond delay between front and back. I don’t think the eye/brain could discern this. Meanwhile, Penrose-Terrell rotation relies on the existence of steady illumination such that at all times you are seeing light from...

Consider the scenario for a slow moving object (close to a screen, far away flash bulb). What do you see? How is the fast moving object any different, given the notion of a flash bulb?

If you make the distant light source a flash bulb (at rest in screen frame), the fixes simultaneity. Then, you will definitely see a length contracted shadow (assuming you have a very fast eye). And if the screen is a flat piece of film, it will photograph the length contracted shadow.

Regarding post #21, the formula given for area minimizing ##\theta## simplifies to ##\tan\theta=c/d##. The minimum area just becomes ##2cd##, and you have ##a=d## and ##b=c##.

Ok, if we're bringing this back, I'll share my results. In reference to the diagram in the original post, let c be the segment labeled 4, d be the segment labeled 2. Let A be rectangle area. Let ##\theta## be the right most angle in the diagram. Then the following relations hold (among others)...

So, FW transport along any curve will preserve all of these properties. Parallel transport will only preserve them along a geodesic. For an arbitrary curve, parallel transport will preserve mutual orthogonality, but the timelike vector will no longer be tangent to the curve.

But that operation is typically called the exponential map (and is a completely rigorous construct): https://en.wikipedia.org/wiki/Exponential_map_(Riemannian_geometry) I wonder if this is what @cianfa72 is thinking of.