(July 22, 2018 at 8:18 pm)Jehanne Wrote:(July 22, 2018 at 6:26 pm)polymath257 Wrote: Like I said, it takes some practice to get a 'feel' for it. I've spent *way* too much time learning how to visualize what is going on in 4D and in 3D manifolds.
One issue here is terminology. The sphere (surface only) in 3D is a 2D manifold. The sphere in 4D is a 3D manifold. The dimension of a manifold is determined *internally*, not via the embedding.
So, just like the 2D sphere in 3D, the 3D sphere in 4D has 'geodesics' that are 'great circles'. But there are also 'great spheres', that are the intersection of hyperplanes with the sphere.
Initially, it can be useful to use time as a fourth dimension for visualization purposes. So, a 3D sphere in 4D would start as a single point at some time (the south pole), expand up to the 'diameter sphere' and then shrink again to a single point (the north pole) and disappear. The full sequence is the sphere in 4D and is shown by successive cross sections. The problem with this visualization technique is figuring out what things look like when doing a cross section from a different direction.
To get back to my OP, though, let's say that NASA launched a space probe into orbit between Mercury and Venus, such that the eccentricity of that orbit was zero, that is, a circle. Are you saying that the measured radius to the center of the Sun times 2 times pi would give the circumference of the orbit?
The simple answer is that C=2*pi*r would fail in this situation. I'd have to do the detailed calculation to see by how much (and the internal density of the sun is relevant for this), but my estimate is that the answers would differ in the 8th or 9th decimal place. That is the typical scale on which general relativity is relevant in the solar system.
The more detailed answer needs to deal with some technicalities. The first is that measured distances depend on relative motion. So, the 'radius' of the orbit as measured by someone at rest to the sun will be different than the 'radius' as measured by someone going past at half the speed of light. In fact, the orbit can be circular in the first frame and NOT circular in the second.
So, the first simplifying assumption is that we measure in a frame that is at rest with respect to the (center of) the sun. We assume that in this frame the orbit is circular, in the sense that the same distance is kept from the center of the sun at all times.
Next, we measure the circumference of the orbit also from the frame where the f the sun is at rest and NOT from the frame of the moving Earth (which would give a different answer that is harder to compute). Also, we trace out the orbit and determine the circumference at an instant of time.
These are assumptions made to guarantee your question even makes sense. Other interpretations can be found and those will also give answers where C and 2*pi*r differ, but by different amounts.
From my back of the envelope calculations, the measured circumference will be slightly *larger* than the result of 2*pi*r. I'll try to do a more detailed calculation later, if you wish.
