# Surface Normal and Parametric Surface?

1. Jan 3, 2014

### Calculuser

I've been working on Multivariable Calculus with a few books. In Vector Analysis, I've had some parts which made some questions come up in my mind. I have two questions about them.

1)Can we think of surface normal vector ($\vec{n}$) as a vector field ($\vec{F}$) or just a position vector ($\vec{r}$) which doesn't make sense to me?

2)Do we always have to parametrize a surface with two variables ($\vec{r}(u,v)$) which I haven't seen any of ($\vec{r}(t)$) giving us a single variable parameterized surface? Why?

Thanks..

2. Jan 3, 2014

### Staff: Mentor

I really don't understand question 1, but with regard to question 2: A surface is a 2D entity, and requires two independent variables to establish position within the surface.

Chet

3. Jan 4, 2014

### HallsofIvy

We certainly cannot think of a single vector as a vector field since a vector field is a set of vectors. The set of all normal vectors to a given surface form a vector field.

Personally, I would recommend that you forget about "position vectors" entirely. They only make sense, to begin with, in Cartesian coordinates, in a given coordinate system, so are not really general. (Suppose we set up a coordinate system on the surface of a sphere. Do "position vectors", from (0,0) to $(\theta, \phi)[itex], go through the sphere or do they curve around the sphere? I used to worry a lot about that! Of course the answer is that there are NO "position vector" on such a surface.) A surface is, pretty much by definition, two dimensional. That means that we need two numbers, whether parameters or not, to identify a point on a surface. 4. Jan 4, 2014 ### Chestermiller ### Staff: Mentor I respectfully disagree. If [itex]\vec{r}(θ,\phi)$ represents a position vector from the origin (i.e., the center of the sphere) to a point on the surface of the sphere, then $\vec{r}(θ,\phi)=r\vec{i_r}(θ,\phi)$, where $\vec{i_r}(θ,\phi)$ is the unit vector in the radial direction. A differential position vector within the surface is then given by:
$d\vec{r}(θ,\phi)=\frac{\partial \vec{r}(θ,\phi)}{\partial θ}dθ+\frac{\partial \vec{r}(θ,\phi)}{\partial \phi}d\phi=\vec{a_θ}dθ+\vec{a_{\phi}}d\phi$
where $\vec{a_θ}$ and $\vec{a_{\phi}}$ represent the coordinate basis vectors in the θ and $\phi$ coordinate directions, respectively.

In short, even though there are no position vectors within such a surface, position vectors from an arbitrary origin to points on the surface can be used to establish differential position vectors within the surface.

5. Jan 4, 2014

### Calculuser

Anyway I know it doesn't mean a single vector. My question is considering whether its notation refers to a vector field or a position vector. Okay, I've considered it correctly then.

Can I generalize the state so that;

In $R^{2}$, $\vec{r}$ must be at least single variable.
In $R^{3}$, $\vec{r}$ must be at least two variables.
In $R^{n}$, $\vec{r}$ must be at least $n-1$ variables. ??

6. Jan 4, 2014

### Staff: Mentor

In $R^{2}$, $\vec{r}$ must be at least single variable.

This is a line.
You can also use r in this case as a function of a single variable to define a curved line in space.

Even in higher dimensional spaces, r as a function of two variables is still a 2D surface. In this case, r as a function of m=1,...,n-1 variables is an m dimensional "surface." Note that this only works if R^n is a flat space. However, some people (myself included) are comfortable with assuming that the m dimensional curved "surface" is immersed in a higher dimensional flat space. For example, in general relativity, I have no qualms about imagining that curved 4D space-time is immersed in a higher dimensional flat space.

7. Jan 4, 2014

### Calculuser

Okay, I guess I got it. Thanks for help..