Visualizing the space and structure described by a metric

[Whitehole]

I need help to visualize the geometry involved here,

How can I visualize the last paragraph? Why is the surface of fixed r now an ellipsoid? Also for r = 0, it is already a disk? I've tried searching for the geometry of these but I can't find any image of the geometry that I can just stare at and understand already what the paragraph meant. Anybody can give me insights?

 

[andrewkirk]

Bear in mind that these machinations do not change the space, which remains 3D Euclidean. All they change is the coordinate system, so that the surfaces defined by fixing a coordinate change.
In spherical coordinates in flat space the surfaces of fixed coordinates are:
- spheres centred on the origin for fixed $r$
- cones around the z axis with tip at the origin, for fixed $\theta$
- half planes emanating from the z axis, for fixed $\phi$

The change above alters the first of these so that the spheres become ellipsoids. You can see that must be the case because of the equations:
$$x=f\sin\theta\cos\phi$$
$$y=f\sin\theta\sin\phi$$
$$z=r\sin\theta\cos\phi$$
If $f=r$ these are the equations of a sphere of radius $r$. If $a\neq 0$ than $f=\sqrt{a^2+r^2}>r$ so the horizontal dimensions $x,y$ are expanded while keeping the vertical dimension $z$ constant. That squashes the sphere into an oblate spheroid.

 

[Whitehole]

Bear in mind that these machinations do not change the space, which remains 3D Euclidean. All they change is the coordinate system, so that the surfaces defined by fixing a coordinate change.
In spherical coordinates in flat space the surfaces of fixed coordinates are:
- spheres centred on the origin for fixed $r$
- cones around the z axis with tip at the origin, for fixed $\theta$
- half planes emanating from the z axis, for fixed $\phi$

The change above alters the first of these so that the spheres become ellipsoids. You can see that must be the case because of the equations:
$$x=f\sin\theta\cos\phi$$
$$y=f\sin\theta\sin\phi$$
$$z=r\sin\theta\cos\phi$$
If $f=r$ these are the equations of a sphere of radius $r$. If $a\neq 0$ than $f=\sqrt{a^2+r^2}>r$ so the horizontal dimensions $x,y$ are expanded while keeping the vertical dimension $z$ constant. That squashes the sphere into an oblate spheroid.

How can $a$ only alter the horizontal dimensions, $r$ is a 3-d coordinate here so it should affect the z component if it were to change right? Also, is the term squashed sphere accurate? Because squashed means that the z coordinate is decreasing and eventually flat, correct me if I'm wrong.

 

[andrewkirk]

How can a only alter the horizontal dimensions, r is a 3-d coordinate here so it should affect the z component if it were to change right?

No. I think you are confusing yourself by using the term '3D coordinate'. Look at the formulas for x, y and z. You can see that only the x and y formulas use f, which depends on a. z uses r instead of f, and hence has no dependence on a.

squashed means that the z coordinate is decreasing and eventually flat, correct me if I'm wrong.

No, squashed doesn't mean having zero height. It just means having less height than it had previously.

 

[Whitehole]

No. I think you are confusing yourself by using the term '3D coordinate'. Look at the formulas for x, y and z. You can see that only the x and y formulas use f, which depends on a. z uses r instead of f, and hence has no dependence on a.

No, squashed doesn't mean having zero height. It just means having less height than it had previously.

Now I got it, I didn't carefully saw the "fixed r". Also, yes, specifically squashed means having less height than it had previously. For the last part, how can $θ$ play the role of the radial variable if it is already flat? $φ$ sweeps out the disk from 0 to 2π. But $θ$ . . .

 

[andrewkirk]

Think about what it looks like when you view an ant crawling about on the Northern hemisphere of an origin-centred sphere from above it on the z axis, eg viewing the unit sphere from the point (0,0,10).
Changes in the ant's $\theta$, keeping $r$ and $\phi$ constant, will look like increases or decreases in radius.
Squashing the sphere down to absolutely flat, which is what the set r=0 is, looks the same as viewing a 'fully inflated' sphere from above.

 

[Whitehole]

Think about what it looks like when you view an ant crawling about on the Northern hemisphere of an origin-centred sphere from above it on the z axis, eg viewing the unit sphere from the point (0,0,10).
Changes in the ant's $\theta$, keeping $r$ and $\phi$ constant, will look like increases or decreases in radius.
Squashing the sphere down to absolutely flat, which is what the set r=0 is, looks the same as viewing a 'fully inflated' sphere from above.

Very nice example. Thanks a lot for helping me visualize all this!