No perfect circles in space...

[Jehanne]

A dimple is fine in 2D; but, how do we imagine such in 3D?

I can't imagine a dimple on a flat surface in anything other than 3D - dimpling the a flat surface changes two dimensions to three, doesn't it?

Boru

True.  But, we live in a 3D world.  If a satellite is in orbit about the Sun (say, by NASA), and the eccentricity of its orbit is 0, then, its orbit is a circle.  Why does the diameter of its orbit times pi not equal the circumference?  And, how can that be visualized?

[Anomalocaris]

As an example, take a flat surface, make a dimple on it.   Trace an enclosed line, exactly equidistant from some other point on the surface as measured along the surface, and passing through the dimple.   This line forms a perfect circle from the perspective of the surface.

I'm not being intentionally difficult, I'm seriously trying to grasp this.

On a flat surface, the circumference of a circle with a diameter of 10 cm would be 31.4 (ish)cm.

Ok, now we have a flat surface with a dimple.  Let's say the dimple extends 1cm below (or above, doesn't really matter) the rest of the surface. We choose our centre point 5 cm from the dimple, giving us a radius of 5 and a diameter of 10.  But the line marking the circumference of the circle would, when passing through the dimple, go down 1 cm then up 1 cm, making the circumference of this circle 33.4 cm, so C = pi x d wouldn't apply, making this not a perfect circle. 

What am I missing?

Boru

If you define the circle as circumference = pi X D, then it won’t be a circle.  But that relation is normally taken to be the property of a circle inscribed on a flat plane, not the definition what circle is.

But we normally define the circle as a locus of points all of whom are exactly the same distance from a single center point, and forming an enclosed circumference.   Using such a definition the circle on a dimpled surface is a true circle, but it would not have the property of  circumference = pi X D.  For such a circle, it is precisely as you say, the circumference, if measured by tracing along the dimpled surface, would go down and then up, so that is precisely the reason why its circumference would be longer than the circumference of a circle of exactly the same diameter but inscribed on a perfectly flat plane.

---

I can't imagine a dimple on a flat surface in anything other than 3D - dimpling the a flat surface changes two dimensions to three, doesn't it?

Boru

True.  But, we live in a 3D world.  If a satellite is in orbit about the Sun (say, by NASA), and the eccentricity of its orbit is 0, then, its orbit is a circle.  Why does the diameter of its orbit times pi not equal the circumference?  And, how can that be visualized?

I don’t know if this is the crux of the issue.  But I would like to point out the orbit of the earth is not quite circular, but instead is a ellipse with a small non-zero eccentricity.  I believe even without any perturbation from any third source of gravity, Relativistic effects would make the earth not repeat the exact same complete elliptical orbit trace time after time.  Instead the earth would behave as if each each elliptical orbit ends just a tiny fraction too early, and next orbit starts slightly early, so the major axis of the ellipse gradually rotate through space, making one complete revolution every few hundred million years.  

I think, (but do not know for certain) that is what the quoted text meant when it said earth’s orbit is 10km shorter than pi X 2 X distance from earth to the sun projected onto 3D space.

[Jehanne]

The perihelion point of our orbit, as with the inner planets out to Mars (and beyond, technically), does precess over time.

[polymath257]

As I said, a geodesic is a "path"; it can, of course, be a circle.  It does not have to be, and I did not intend to imply such.

P.S.  An "arc"  can be a straight line, just with zero curvature.

Not quite. The definition of a geodesic on a surface is that it is a path that minimizes distances along the path. So, for a flat plane, a geodesic is a straight line. On a sphere, it is a great circle, not any other path or circle.

---

I'm not being intentionally difficult, I'm seriously trying to grasp this.

On a flat surface, the circumference of a circle with a diameter of 10 cm would be 31.4 (ish)cm.

Ok, now we have a flat surface with a dimple.  Let's say the dimple extends 1cm below (or above, doesn't really matter) the rest of the surface. We choose our centre point 5 cm from the dimple, giving us a radius of 5 and a diameter of 10.  But the line marking the circumference of the circle would, when passing through the dimple, go down 1 cm then up 1 cm, making the circumference of this circle 33.4 cm, so C = pi x d wouldn't apply, making this not a perfect circle. 

What am I missing?

Boru

A dimple is fine in 2D; but, how do we imagine such in 3D?

Through practice, mostly. Start by trying to imagine a sphere in 4 dimensions: it provides an example of a curved *three* dimensional manifold.

I might recommend the book 'The Shape of Space' as an introduction:

https://www.amazon.com/Shape-Space-Chapm...e+of+space

---

A dimple is fine in 2D; but, how do we imagine such in 3D?

I can't imagine a dimple on a flat surface in anything other than 3D - dimpling the a flat surface changes two dimensions to three, doesn't it?

Boru

No. It is still a 2 dimensional surface. Only two coordinates *on the surface* are required to determine a point.

[Jehanne]

A dimple is fine in 2D; but, how do we imagine such in 3D?

Through practice, mostly. Start by trying to imagine a sphere in 4 dimensions: it provides an example of a curved *three* dimensional manifold.

I might recommend the book 'The Shape of Space' as an introduction:

https://www.amazon.com/Shape-Space-Chapm...e+of+space

---

Like seeing "dead people" (from the movie)?  I can't imagine that 4th dimension, although, the math (tensors) is completely reasonable and valid.

[polymath257]

Through practice, mostly. Start by trying to imagine a sphere in 4 dimensions: it provides an example of a curved *three* dimensional manifold.

I might recommend the book 'The Shape of Space' as an introduction:

https://www.amazon.com/Shape-Space-Chapm...e+of+space

---

Like seeing "dead people" (from the movie)?  I can't imagine that 4th dimension, although, the math (tensors) is completely reasonable and valid.

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.

[Jehanne]

Like seeing "dead people" (from the movie)?  I can't imagine that 4th dimension, although, the math (tensors) is completely reasonable and valid.

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?

[Anomalocaris]

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?

I suspect that would depend on how you measure the radius and circumference.

[polymath257]

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.

[Jehanne]

Please do; if you read my OP, the authors of 21st Century Astronomy (all PhDs) say that the distance would be shorter.