(July 21, 2018 at 12:46 pm)polymath257 Wrote:

(July 21, 2018 at 12:37 pm)Jehanne Wrote: That's very true, but the point of my OP is that one cannot visualize the warping of space/time.

I think this is more a matter of practice than inability.

Most people don't spend any time really attempting to visualize four dimensional objects, let alone curved four dimensional manifolds. For that matter, even curved *three* dimensional manifolds are beyond what most people spend any time thinking about. That doesn't make them impossible to imagine. It just means you have to work a bit harder to do so.

But to answer the OP: yes, I find it quite easy to imagine 'circles' (meaning the set of points equidistant from some center point) to not have the 'circumference' (meaning the arc length of the boundary curve' equal to pi times the 'diameter' (twice the defining distance).

Among other possibilities is changing the definition of 'distance'. The standard Euclidean distance (based on the Pythagorean equation) is not the only possible 'metric', even in a 'flat' plane. The taxicab metric is another, perfectly usable measure of 'distance' and the 'circles' for the taxicab metric are actually 'squares' in two dimensions, 'octagons' in three dimensions, etc.

Other types of 'metric' can easily be used, and are used, when discussing general relativity. Again, no fundamental issue is found, but you have to keep track of the definitions a bit more closely.

In a 3d world/model, say, on a sphere, how do you imagine a geodesic (a circle) where C != 2r * pi?

P.S. A geodesic is, of course, just a path on a surface, in which case, it can be an arc.

No perfect squares in hip town!

Perfect schmerfect. I encounter little to no perfection my life so I don't really care much.

Wait, don't tell me,........god is a perfect circle.

(July 21, 2018 at 4:13 pm)Jörmungandr Wrote: No perfect squares in hip town!

Wow, I never realised that you can see the background dancers boobs through their shirts!

(July 21, 2018 at 1:54 pm)Jehanne Wrote:

(July 21, 2018 at 12:46 pm)polymath257 Wrote: I think this is more a matter of practice than inability.

Most people don't spend any time really attempting to visualize four dimensional objects, let alone curved four dimensional manifolds. For that matter, even curved *three* dimensional manifolds are beyond what most people spend any time thinking about. That doesn't make them impossible to imagine. It just means you have to work a bit harder to do so.

But to answer the OP: yes, I find it quite easy to imagine 'circles' (meaning the set of points equidistant from some center point) to not have the 'circumference' (meaning the arc length of the boundary curve' equal to pi times the 'diameter' (twice the defining distance).

Among other possibilities is changing the definition of 'distance'. The standard Euclidean distance (based on the Pythagorean equation) is not the only possible 'metric', even in a 'flat' plane. The taxicab metric is another, perfectly usable measure of 'distance' and the 'circles' for the taxicab metric are actually 'squares' in two dimensions, 'octagons' in three dimensions, etc.

Other types of 'metric' can easily be used, and are used, when discussing general relativity. Again, no fundamental issue is found, but you have to keep track of the definitions a bit more closely.

In a 3d world/model, say, on a sphere, how do you imagine a geodesic (a circle) where C != 2r * pi?

P.S.  A geodesic is, of course, just a path on a surface, in which case, it can be an arc.

First, you seem to be confused about definitions. A circle with a radius at a point P is the collection of points that are some fixed distance from P. A geodesic is a 'great circle', but not all circles are geodesics. Every latitude line is a 'circle'. And, in fact, a geodesic is closer to being a 'line': they form the paths of least distance *on the sphere*.

But, in fact, NO circle on the sphere has C=2*pi*r if the radius is measured *on the sphere*. For example, a geodesic has C=4r if both the geodesic and the radius are measured on the sphere. For example, the distance around the equator is 4 times the distance from the north pole down to the equator.


And this is how things *should* be done on manifolds: what is important is the *intrinsic* distances, not the *extrinsic* ones that have to do with a specific embedding.

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.

There's another issue, which is that we're only ever dealing with our observations and our filtered information, rather than reality itself. So we have no fricking clue what's really going on, we can just assume (without loss of functionality) that it's something like what we observe.

What I'm really saying is that what we imagine to be a "circle in space" may be utter nonsense in relation to reality, even if it relates well to our observations.

This is really just my philosophical mental masturbation. Carry on

Quote: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

(July 22, 2018 at 5:04 am)BrianSoddingBoru4 Wrote:

Quote: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

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

(July 22, 2018 at 7:49 am)Jehanne Wrote:

(July 22, 2018 at 5:04 am)BrianSoddingBoru4 Wrote: 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?

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