phinds said:
I think that those of you who are insisting that things in freefall in space follow curved lines are making the mistake of applying Euclidean Geometry in a domain where it is not relevant. Yes, in Euclidean Geometry those paths ARE curved, but so what? That is utterly irrelevant because the geometry of spacetime is not Euclidean, it is pseudo-Riemannian and in that geometry the paths are geodesics, which as has already been pointed out are considered by many to be a logical generalization of "straight line".
Technically, I do not think this is quite the right distinction to make.
It is not the difference between Euclidean and Pseudo-Riemannian that matters. It is the difference between Euclidean and non-Euclidean. Consider special relativity (SR) for a moment. In the flat Minkowski space of SR, geodesics are straight lines. Yet this is a pseudo-Riemannian geometry.
As I know you understand already, it is not that geodesics are curved. It is the space within which they exist that is curved.
It is only when we apply a non-default metric to the space in question that the notion of curvature becomes a meaningful concept. This is perhaps more easily seen if we go back to a two dimensional analogy in a space that is not pseudo-Riemannian -- paper maps of the surface of the Earth.
If we have a flat map of the surface of the Earth this will necessarily be some sort of projection. For instance a Mercator projection. If we look at lines on the map corresponding to straight roads on the Earth, some of those lines will be curved. They will be [sections of] great circle arcs. This is reflected in the distance metric.
This "metric" amounts to a big table of distances. For any pair of points on the Earth's surface metric would tell you how far it is [along the surface] from point A to point B. For instance, 800 miles from New York to Chicago.
If you translate this metric and present it in terms of map coordinates (for instance in terms of latitude and longitude if you are using a Mercator projection) then you will find that it does not match the standard Euclidean metric. The Euclidean metric would be, for instance:$$D = 60 * \sqrt{\Delta\text{ lat}^2+\Delta\text{long}^2}$$where D is in nautical miles and lat and long are in degrees. As should be obvious, this metric matches distances measured with ruler on a flat paper map but does not match distances measured with an odometer on a real earth.
Locally on this flat paper map we will almost always be able to find a metric which fits the Euclidean form (##D^2 = \Delta x^2 + \Delta y^2##) and locally approximates the true metric. We may have to put the local x and y axes at an angle. And we may have to scale them by a constant factor, but we can still obtain something Euclidean-looking. In the case of the Mercator projection we won't have to mess with the angles of the local x and y axes. With some other projections we might need to.
[It is "almost always" because you can get coordinate singularities. In the case of the Mercator projection you get a coordinate singularity at the North and South poles. Other projections tear or irreparably stretch the map in other places]
It is a similar situation when comparing the flat space of special relativity with the curved space of general relativity. Locally you can almost always fix things up so that the Minkowski metric (##D^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 - \Delta t^2##) is approximated well. One may have to scale the axes and put them at odd angles, but you can still recover something Minkowski-looking.
[Again, you can get coordinate singularities -- for instance at the event horizon of a black hole when using Schwarzschild coordinates. This is in addition to true singularities such as at the "center" of a Schwarzschild black hole]
It is possible to apply the metric (as presented in terms of map coordinates) to figure out how to extend a line on the map so that it corresponds to a straight great circle arc on the surface of the Earth. Of course, this line will not be straight on the map. But it will be straight on the surface of the Earth.
Perceived geodesic curvature is all about the projection, not about the geometry.
Note: I have never taken a course covering differential geometry. Pretty much everything think I know has been absorbed over the years here.
