# Thread: Gaussian Curvature of a Sphere

1. ## Gaussian Curvature of a Sphere

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?

Thanks for the help!

2. ## Re: Gaussian Curvature of a Sphere

What is the definition of "Gaussian Curvature"?

3. ## Re: Gaussian Curvature of a Sphere

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

4. ## Re: Gaussian Curvature of a Sphere

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:

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

5. ## Re: Gaussian Curvature of a Sphere

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.

6. ## Re: Gaussian Curvature of a Sphere

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.