curvature
Author
Could someone give me an explanation of what gaussian and mean curvature on a surface represent? I''d also like to know how to find the curvature analytically. I know how to find the curvature of a curve.
Author
Yeah, I already tried there. I don''t understand alot of the notation, which is why I came here, to a message board, for a simple explanation, and a technical definition that I can understand with 1st year calculus.
Well, they''re not simple concepts and not really first year calculus but I can try 
First imagine a curve in 2 dimensions, then pick any point p on that curve. You can approximate the curve in the immediate neighbourhood of p with a circle. If the curve isn''t curving very much (ie looks more like a line) then it will be a big circle. If it is very curved then it will be a small circle.
The curvature at p is then defined to be the reciprocal of the radius of the ''approximating circle''.
In 3 dimensions it gets a little more complicated. First of all you need to find the normal to the point p. Then each plane parallel to the normal and passing through p you find the curvature at p for the 2d curve formed by the intersection of the surface with the plane. After doing this for each plane you will have a maximum curvature cmax and a minimum curvature cmin. The Gaussian curvature is the product of cmax and b>cmin. The mean curvature is the average of these two values.
First imagine a curve in 2 dimensions, then pick any point p on that curve. You can approximate the curve in the immediate neighbourhood of p with a circle. If the curve isn''t curving very much (ie looks more like a line) then it will be a big circle. If it is very curved then it will be a small circle.
The curvature at p is then defined to be the reciprocal of the radius of the ''approximating circle''.
In 3 dimensions it gets a little more complicated. First of all you need to find the normal to the point p. Then each plane parallel to the normal and passing through p you find the curvature at p for the 2d curve formed by the intersection of the surface with the plane. After doing this for each plane you will have a maximum curvature cmax and a minimum curvature cmin. The Gaussian curvature is the product of cmax and b>cmin. The mean curvature is the average of these two values.
That was easy enough to follow. I imagine actually finding that max and min takes a bit of creativity, but the idea itself seems simple enough.
Author
"The Gaussian curvature is the product of cmax and b>cmin"
Gaussian curvature is the only thing from your post that I didn''t already know... what is b>cmin? Is it a typo? Or is b>cmin something besides cmin. And if gaussian curvature is cmax times cmin, what is the purpose of it? Also, is there an average curvature that is the result of an integral, as opposed a simple geometric mean?
Gaussian curvature is the only thing from your post that I didn''t already know... what is b>cmin? Is it a typo? Or is b>cmin something besides cmin. And if gaussian curvature is cmax times cmin, what is the purpose of it? Also, is there an average curvature that is the result of an integral, as opposed a simple geometric mean?
The b> is two thirds of a bold tag. It should have been cmin
One of the reasons it''s defined as a product is so you can tell if cmax and cmin have different signs.
I haven''t ever used the average curvature so don''t know any closed forms for it. The Gaussian curvature has several analytic expressions, none of which would be easy to explain on a web forum. General relativity courses are probably the best place to learn about it.
One of the reasons it''s defined as a product is so you can tell if cmax and cmin have different signs.
I haven''t ever used the average curvature so don''t know any closed forms for it. The Gaussian curvature has several analytic expressions, none of which would be easy to explain on a web forum. General relativity courses are probably the best place to learn about it.
Author
Well, you''ve told me everything I wanted to know. I assumed that curvature of a surface was just a composite of the curvature of some curves, but I wasn''t sure, and I had no idea how exactly it was defined. It doesn''t seem like finding the max and min curvature would be terribly difficult. Thanks.
By the way, could you recommend a book or a website that would comprehensively explain surface curvature to me? I have read most of my calc 3 book, so I know a lot about vector calculus, and I don''t need any help with curvature of a curve. I am just wondering about finding curvature of surfaces. Thanks again.
By the way, could you recommend a book or a website that would comprehensively explain surface curvature to me? I have read most of my calc 3 book, so I know a lot about vector calculus, and I don''t need any help with curvature of a curve. I am just wondering about finding curvature of surfaces. Thanks again.
Actually take a look at this:
http://www.math.uncc.edu/~droyster/courses/fall98/math4080/classnotes/gaussianformula.pdf
in particular Theorem 4.2 and Lemma 4.4 (don''t worry about the proofs) They give closed form expressions for the gaussian curvature for surfaces in 3 dimensions parameterised by u and v,.
The metric symbols E,F,G are given by
E = (dx/du)2 + (dy/du)2 + (dz/du)2
F = (dx/du)*(dx/dv) + (dy/du)*(dy/dv) + (dz/du)*(dz*dv)
G = (dx/dv)2 + (dy/dv)2 + (dz/dv)2
and for a line segment ds on the surface we have
ds2 = E du2 + 2F du dv + G dv2
Theorem 4.2 refers to the case where F = 0, Lemma 4.4 is the general case.
Some notation you may or may not have seen: fu is the partial derivative ∂f/∂u . fuu is then ∂2f / ∂u2 and fuv = ∂2f / ∂u ∂v and so on.
You should be able to use this to calculate the guassian curvature of any parameterized surface in ℜ3, and will definately keep your vector calculus muscles exercised
I was lucky enough to have good lecture notes for this and don''t know of any good books. Flick through any books on Differential Geometry in your library and you might find one that suits.
http://www.math.uncc.edu/~droyster/courses/fall98/math4080/classnotes/gaussianformula.pdf
in particular Theorem 4.2 and Lemma 4.4 (don''t worry about the proofs) They give closed form expressions for the gaussian curvature for surfaces in 3 dimensions parameterised by u and v,.
The metric symbols E,F,G are given by
E = (dx/du)2 + (dy/du)2 + (dz/du)2
F = (dx/du)*(dx/dv) + (dy/du)*(dy/dv) + (dz/du)*(dz*dv)
G = (dx/dv)2 + (dy/dv)2 + (dz/dv)2
and for a line segment ds on the surface we have
ds2 = E du2 + 2F du dv + G dv2
Theorem 4.2 refers to the case where F = 0, Lemma 4.4 is the general case.
Some notation you may or may not have seen: fu is the partial derivative ∂f/∂u . fuu is then ∂2f / ∂u2 and fuv = ∂2f / ∂u ∂v and so on.
You should be able to use this to calculate the guassian curvature of any parameterized surface in ℜ3, and will definately keep your vector calculus muscles exercised
I was lucky enough to have good lecture notes for this and don''t know of any good books. Flick through any books on Differential Geometry in your library and you might find one that suits.
Author
Hmm... well, that was a brainful of equations. I can understand it easily enough, but now I''m interested in these metrics, too. Do you know where I can learn about them?
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