# Achieving normal derivative in spheroidal coordinate

1. Jan 14, 2009

### dexturelab

1. The problem statement, all variables and given/known data

Hi PhysicsForums,
I am calculating something related to the spheroidal membrane and want to ask you a question.

I consider a oblate spheroid (Oblate spheroidal coordinates can also be considered as a limiting case of ellipsoidal coordinates in which the two largest semi-axes are equal in length.)

In spheroidal coordinate, the relationship to Cartesian coordinates is
$$x=a\sqrt((1+u^2) (1-v^2))\cos(\phi)$$
$$y=a\sqrt((1+u^2) (1-v^2))\sin(\phi)$$
$$z=a u v$$

Now, I want to know how to achieve the normal derivative to the surface of a spheroid (u = const), in terms of the derivatives of u, v and $$\phi$$.

2. Relevant equations

3. The attempt at a solution
I firstly think that the normal derivative in this case is the partial derivative to u. Because in limit cases (v =1,-1) this lead to the derivative to z.

Thank you very much.

2. Jan 14, 2009

### tim_lou

writing in Cartesian coordinate, the normal derivative is just
$$\hat{n}\cdot \nabla$$

However, I'm pretty sure you want to work in Oblate spheroidal coordinate, just apply the chain rule and you should be fine.

More intuitively, you could consider the normal derivative as a unit tangent vector (in terms of manifold) acted on a function.
Simply write down the tangent vector and then normalize it using the metric.

3. Jan 14, 2009

### dexturelab

Thank you tim_lou,
In fact, I am totally an amateur in this type of calculation.

So, can you please recommend me any basic text books or simple introductories.

For example, if I have the equation of a membrane in xyz coordinate, what can I do to get the normal vector $$\vec(n)$$.
Moreover, the chain rule seems to be rather stenuous.
Is there any different ways. I need some types in practical ways.
Is there any function in Mathematica to do that?

Thank you again.

Last edited: Jan 14, 2009
4. Jan 15, 2009

### tim_lou

to find the normal vector in xyz coordinates, take the gradient...

to be more precise, suppose you want to find the normal vector to a surface described by the set:
A={(x,y,z)| f(x,y,z)=c}

then the normal is in the direction of n=∇f(x,y,z)
to see why that is, take a curve, (x(t), y(t), z(t)) such that the curve lies in A.
Hence, f(x(t), y(t), z(t))=c and take derivative with respect to t,
∇f(x,y,z)· (x'(t), y'(t), z'(t))=0. We see that n must be always orthogonal to the derivative or any curve lying in A so we conclude n=∇f(x,y,z)

Any tangent vector (b,c,d) attached to a point in R3 can be considered as a tangent vector on the manifold, so,
(b,c,d)=b∂_x + c∂_y + d∂_z
so when you find n·∇=b∂_x + c∂_y + d∂_z, you want to express it in the basis
{∂_a, ∂_u, ∂_v}. Changing basis is a linear algebra problem and those two basis are related via Jacobian.

things simplify when the {∂_a, ∂_u, ∂_v} are orthogonal, in which case, the gradient becomes
f ∂_a + g ∂_u + h ∂_v where f,g,h are functions.