Achieving normal derivative in spheroidal coordinate

[dexturelab]

Homework Statement

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.

Homework Equations

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.

 

[tim_lou]

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

 

[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 textbooks 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.

 

[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.