Surface Normal and Parametric Surface?

[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..

 

[Chestermiller]

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..

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

 

[HallsofIvy]

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?

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), go <b>through</b> the sphere or do they <b>curve</b> 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.)<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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? </div> </div> </blockquote> A surface is, pretty much by definition, <b>two</b> dimensional. That <i>means</i> that we need two numbers, whether parameters or not, to identify a point on a surface.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Thanks.. </div> </div> </blockquote>

 

[Chestermiller]

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), go <b>through</b> the sphere or do they <b>curve</b> 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.)

<br /> <br /> I respectfully disagree. If \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:<br /> 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<br /> where \vec{a_θ} and \vec{a_{\phi}} represent the coordinate basis vectors in the θ and \phi coordinate directions, respectively.<br /> <br /> 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 <i>differential position vectors</i> within the surface.

 

[Calculuser]

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.

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.

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.

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. ??

 

[Chestermiller]

Can I generalize the state so that;

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

This is a line.

In R^{3}, \vec{r} must be at least two variables.

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

In R^{n}, \vec{r} must be at least n-1 variables. ??

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.

 

[Calculuser]

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