If you intersect two 2-spheres, you get a circle.

If you intersect two 3-spheres, do you get a sphere?



eta: or maybe to be more accurate, would the intersection of two 3-spheres, be a 2-sphere, within both 3-spheres?

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a 0-sphere is a pair of endpoints of an interval (−r, r) of the real line
a 1-sphere is a circle of radius r
a 2-sphere is an ordinary sphere
a 3-sphere is a sphere in 4-dimensional Euclidean space

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If I set the budget, we'd have Ares and more. Unfortunately, I don't set the budget, and Ares is just too expensive and too far out for us to accomplish our goals within the budget we were given.

If we halt the ISS, all versions of Ares, and transport Orion and Altair aboard DIRECTv3's Jupiter family of Shuttle-Derived Launch Vehicles, we just might make it back to the Moon by 2020.

YES.

Why is this ball I am holding called a 2 sphere ?.
It occupies 3 dimensions so should be called a 3 sphere.
The hypersphere which occupies 4 dimensions should be the 4 sphere.
Is it an American thing like billion when they mean milliard or month,day,year
for the date or electric switches upside down or fahrenheit or gallons.

a ball isn't a 2-sphere.

It is the surface which is a 2-sphere.

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It is the surface which is a 2-sphere.
Thanks.
That explains the number.
Strange that a 3D object is described by its surface which is one level less important than the volume.

Here is the intersection between two 2-spheres.

click on the thumbnail for a higher resolution view.

that's the intersection of two 3D balls, but a 2-sphere is just the surface is it not?
So the intersection would be a circle, or a 1-sphere.

I think the rule is: the intersection of two n-spheres, gives you a (n-1)sphere, that exists in both n-spheres, but I might be wrong.

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It seems one of the surest ways of analyzing higher dimensional objects is to use a coordinate system and turn it all into algebra--equations with four variables are not much harder than equations with three variables.

A sphere residing in n dimensions (and having an n-dimensional surface) has a surface whose equation is:

r^2 = (x_1-a_1)^2+(x_2-a_2)^2 + ... + (x_n-a_n)^2

where r is the radius, x_1, x_2, ..., x_n are the coordinate positions, and the ^2 is just squaring. (a_1,a_2,...,a_n) is the coordinates of the center of the sphere.

You can find the intersection of two 3-spheres by solving a pair of simultaneous equations. To simplify, let one sphere have radius 1 and center at (0,0,0,0). Take the other sphere to be centered at (0,0,0,2) and have radius 2. All other cases will work the same (if the spheres intersect in more than one point--the other cases are, they are tangent to each other, or they do not intersect at all).

x^2+y^2+z^2+w^2=1

and

x^2+y^2+z^2+(w-2)^2=4

If we subtract the first equation from the second, we get:

(w-2)^2 - w^2 = 3

or

w^2-4w+4-w^22=3, or -4w=-1, or w=1/4

Plugging this back into the first equation shows that the system is equivalent to the following new system:

x^2+y^2+z^2 = 15/16
and
w=1/4

which is the equation of a circle of radius sqrt(15)/4 and whose center is at (0,0,0,1/4), and is parallel to the xyz-plane

Similar reasoning shows you eliminate one variable by solving two equations for n-spheres, effectively giving you an n-1 sphere

(or the degenerate cases: no solution corresdponding to a "sphere" with imaginary radius in the algebra, or tangent at one point, a "sphere" of radius 0)

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Todd (Bowie, MD, US, North America, Earth, Sol System, Vega region, Local Bubble, Orion arm, Milky Way Galaxy, Local Group, Virgo A Cluster, Virgo supercluster, the universe in which spock is clean shaven)

Quidquid latine dictum sit, altum sonatur.

personal page: http://blog.astrosketches.info

Nothing to do with spheres, but I just figured out Frog march's signature picture. I like it!

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If they can get you asking the wrong questions, they don't have to worry about the answers.

so is this the equasion for the intersection of your two 3-spheres?


and that's the equasion for a sphere, isn't it, of radius=sqr(15/16)?

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yes---sqrt(15/16) = sqrt(15)/4.

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Todd (Bowie, MD, US, North America, Earth, Sol System, Vega region, Local Bubble, Orion arm, Milky Way Galaxy, Local Group, Virgo A Cluster, Virgo supercluster, the universe in which spock is clean shaven)

Quidquid latine dictum sit, altum sonatur.

personal page: http://blog.astrosketches.info

It is an equation for a 2-sphere in 4 dimensions. Note that the w coordinate is fixed and if you know two of the coordinates among x,y,z you can find the third from the equation. So the sphere is a 2-dimensinal surface, embedded in the original 4 dimensions, which has the same geometry as the ordinary 2-sphere embedded in 3 dimensions.

You started with two spheres in dimension 4 -- a 3-dimensional hypersurface. Then you intersect them to get a sphere of one lower dimension, but the embedding space remains the same. So you wind up with an equation that describes an ordinary 2-sphere, as it lives in 4-space.

neat, DrRocket


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thanks tdvance.

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tdvance did the hard part.

As I venture out of that dark place I often hide in... I see 'tdvance's' post., and all that maths... As has already been mentioned here, that in different parts of the English speaking world we use the same words for very different meanings. Now you must agree ( you might not ). That if I were to hand that text to a engineer,. He would not build me a ball, or is that better described as a sphere or globe... A ah ! Back to my telescope shed.