Surface Normal and Parametric Surface?

In summary, the conversation discusses the concept of surface normal vectors and parametrization of surfaces in multivariable calculus. It is clarified that a single vector cannot be considered as a vector field and that the set of all normal vectors to a surface form a vector field. The use of position vectors in the context of surfaces is also discussed, with the suggestion to focus on other methods of representation. It is also mentioned that a surface, by definition, requires two variables to establish position within it. The possibility of generalizing this concept to higher dimensional spaces is also briefly discussed.
  • #1
Calculuser
49
3
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 ([itex]\vec{n}[/itex]) as a vector field ([itex]\vec{F}[/itex]) or just a position vector ([itex]\vec{r}[/itex]) which doesn't make sense to me?

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

Thanks..
 
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  • #2
Calculuser said:
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 ([itex]\vec{n}[/itex]) as a vector field ([itex]\vec{F}[/itex]) or just a position vector ([itex]\vec{r}[/itex]) which doesn't make sense to me?

2)Do we always have to parametrize a surface with two variables ([itex]\vec{r}(u,v)[/itex]) which I haven't seen any of ([itex]\vec{r}(t)[/itex]) 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
 
  • #3
Calculuser said:
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 ([itex]\vec{n}[/itex]) as a vector field ([itex]\vec{F}[/itex]) or just a position vector ([itex]\vec{r}[/itex]) 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 [itex](\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.)

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

Thanks..
 
  • #4
HallsofIvy said:
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 [itex](\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.)

I respectfully disagree. If [itex]\vec{r}(θ,\phi)[/itex] represents a position vector from the origin (i.e., the center of the sphere) to a point on the surface of the sphere, then [itex]\vec{r}(θ,\phi)=r\vec{i_r}(θ,\phi)[/itex], where [itex]\vec{i_r}(θ,\phi)[/itex] is the unit vector in the radial direction. A differential position vector within the surface is then given by:
[itex]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[/itex]
where [itex]\vec{a_θ}[/itex] and [itex]\vec{a_{\phi}}[/itex] represent the coordinate basis vectors in the θ and [itex]\phi[/itex] 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
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 [itex]R^{2}[/itex], [itex]\vec{r}[/itex] must be at least single variable.
In [itex]R^{3}[/itex], [itex]\vec{r}[/itex] must be at least two variables.
In [itex]R^{n}[/itex], [itex]\vec{r}[/itex] must be at least [itex]n-1[/itex] variables. ??
 
  • #6
Calculuser said:
Can I generalize the state so that;

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

This is a line.
In [itex]R^{3}[/itex], [itex]\vec{r}[/itex] 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 [itex]R^{n}[/itex], [itex]\vec{r}[/itex] must be at least [itex]n-1[/itex] 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.
 
  • #7
Okay, I guess I got it. Thanks for help..
 

1. What is a surface normal?

A surface normal is a line that is perpendicular to a surface at a specific point. It is used to determine the orientation or direction of a surface at that point.

2. How is a surface normal calculated?

A surface normal is calculated using the cross product of two tangent vectors on the surface. These tangent vectors can be found using the partial derivatives of a parametric equation for the surface.

3. What is a parametric surface?

A parametric surface is a mathematical representation of a surface in three-dimensional space. It uses two independent variables, usually denoted as u and v, to define points on the surface.

4. How is a parametric surface different from a regular surface?

A regular surface is defined by a single equation, while a parametric surface is defined by multiple equations using two independent variables. This allows for more complex and curved surfaces to be represented.

5. What are some real-world applications of surface normal and parametric surface?

Surface normal and parametric surface are used in computer graphics, computer-aided design (CAD), and computer-aided manufacturing (CAM). They are also used in fields such as engineering, physics, and architecture to model and analyze surfaces and objects.

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