Gaussian Curvature of a Sphere

**PStudent175** · December 27th 2012, 05:19 AM

Hi all,

I am trying to find the Gaussian Curvature of the sphere with equation:

X(theta, phi) = (Rcos(theta)sin(phi) , Rsin(theta)sin(phi) , Rcos(theta)

I have found the coefficients of the first fundamental form:

E = R^2*sin(phi)^2

F = 0

G = R^2

But where should I go from here?

My (very unhelpful!) notes just jump to k = 1/R^2

Thanks for the help!

**HallsofIvy** · December 27th 2012, 05:34 AM

What is the definition of "Gaussian Curvature"?

**PStudent175** · December 27th 2012, 05:40 AM

k = eg - f^2 / (EG - F^2) ?

**Deveno** · December 27th 2012, 06:47 AM

the easiest way to get from where you are to where you want to be is to compute the coefficients of the second fundamental form (see here: Second fundamental form - Wikipedia, the free encyclopedia)

to do this, you are first going to have to compute the (unit) normal vector field, given by:

$\mathbf{n} = \frac{\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}}{\left|\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}\right|}$

and then compute the coefficients of the second fundamental form as follows:

$L = \mathbf{n} \cdot \frac{\partial^2X}{\partial \theta^2}$

$M = \mathbf{n} \cdot \frac{\partial^2X}{\partial \theta \partial \phi}$

$N = \mathbf{n} \cdot \frac{\partial^2X}{\partial \phi^2}$

you can then calculate the Gaussian curvature as:

$K = \frac{LN - M^2}{EG - F^2}$ .

**PStudent175** · December 28th 2012, 03:45 AM

Thanks for the help so far Deveno.

I am trying to calculate the unit normal and am unsure where to go from here:

dX/dtheta x dX/dphi = i j k
-Rsin(theta)sin(phi) Rcos(theta)sin(phi) 0
Rcos(theta)cos(phi) Rsin(theta)cos(phi) -Rsin(phi)

Of which the determinant is:

=i(-R^2cos(theta)sin(phi)^2) - j(R^2sin(theta)sin(phi)^2) + k(-R^2sin(theta)^2sin(phi)cos(phi) - R^2cos(theta)^2sin(phi)cos(phi))

I am not sure how to simplify this to get a final answer for N.

Thanks again for your help.

**Deveno** · December 28th 2012, 10:05 AM

for the cross-product i always use the formula:

$(u_1,u_2,u_3) \times (v_1,v_2,v_3) = (u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1)$ .

the way i remember this is:

1st coordinate, leave out "1" subscripts, 2-->3 is positive (the positive cycle goes like this: 1-->2-->3-->1-->2-->3 etc.) (so 3-->2 is negative) so the u₂v₃ term is positive.

2nd coordinate, leave out "2", the positive term is u₃v₁ (see above).

3rd coordinate, well, there's only one possible way to "balance it out" now.

i find this easier to remember than the "determinant formula":

$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\u_1&u_2&u_3\\v_1 &v_2&v_3 \end{vmatrix}$ .

(by the way, you wrote down the wrong "third coordinate" in your parametrization, it should be: $R\cos(\phi)$ not theta. you did get the correct E,F and G, though)

a lengthy calculation shows that:

$\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi} = \begin{bmatrix}(R\cos\theta\sin\phi)(-R\sin\phi) - (0)(R\sin\theta\cos\phi)\\(0)(R\cos\theta\cos\phi) - (-R\sin\theta\sin\phi)(-R\sin\phi)\\ (-R\sin\theta\sin\phi)(R\sin\theta\cos\phi) - (R\cos\theta\sin\phi)(R\cos\theta\cos\phi)) \end{bmatrix}$

$= \begin{bmatrix}-R^2\cos\theta\sin^2\phi\\ -R^2\sin\theta\sin^2\phi\\ -R^2\sin\phi\cos\phi \end{bmatrix}$

which is what you got (you can simplfy your "k" coordinate by taking out the common factor of $-R^2\sin\phi\cos\phi$ ).

this is a normal vector, but it's not a UNIT vector, so we need to "normalize" it (how's that for an odd pun?) by dividing by its norm, which is:

$\left\| \frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi} \right\| = \sqrt{\left(\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}\right) \cdot \left(\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}\right)}$

this is almost as much work as finding the cross-product, first we find that:

$\left(\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}\right) \cdot \left(\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}}\right)$

$= R^4\cos^2\theta\sin^4\phi + R^4\sin^2\theta\sin^4\phi + R^4\sin^2\phi\cos^2\phi$

$= R^4(\sin^4\phi + \sin^2\cos^2\phi) = R^4\sin^2\phi$

taking the square root, we get that the magnitude of our normal vector is $R^2\sin\phi$ so that:

$\mathbf{n} = \begin{bmatrix}-\cos\theta\sin\phi \\ -\sin\theta\sin\phi \\ -\cos\phi \end{bmatrix}$

(a word about the signs, here: our parameterization takes horizontal rays vectors going left-to-right to "lattitude" arcs going counter-clockwise (west-to-east), and vertical rays going down-to-up to "longitude" arcs going from the north pole to the south pole, so the "outward" normal (using a right-hand-rule for the cross-product) in this case points "towards" the center of the sphere (inwards). we could have chosen a different parameterization to make the normal vector n simply be:

$\frac{1}{R}X(\theta,\phi)$

which would have been more intuitively obvious, geometrically (using -θ instead of θ, for example)).

i leave it to you to compute the second derivatives, and thus L,M and N.

Thread: Gaussian Curvature of a Sphere

LinkBack

Thread Tools

Display

Gaussian Curvature of a Sphere

Re: Gaussian Curvature of a Sphere

Re: Gaussian Curvature of a Sphere

Re: Gaussian Curvature of a Sphere

Re: Gaussian Curvature of a Sphere

Re: Gaussian Curvature of a Sphere

Similar Math Help Forum Discussions

Classical question about constant gaussian curvature and geodesics circles.

Given first fundamental form, find gaussian curvature

Question about proof that all surfaces with zero Gaussian curvature are developable

Gaussian curvature

Differential Geometry: Gaussian curvature question

Search Tags