- #1

- 366

- 0

- Thread starter kasse
- Start date

- #1

- 366

- 0

- #2

berkeman

Mentor

- 59,012

- 9,109

Think of skiing down the fall line of a slope -- you are following the gravitational gradient on the slope when you are following the fall line.

- #3

- 366

- 0

equipotentialsurface. It does indeed point in the direction of maximum change in the potential, which is normal to the equipotential surface.

Think of skiing down the fall line of a slope -- you are following the gravitational gradient on the slope when you are following the fall line.

But how is it then that we get the normal vector if we compute the gradient at a certain point of a surface f(x,y,z)?

- #4

berkeman

Mentor

- 59,012

- 9,109

- #5

- 366

- 0

Let's say we have the function f(x,y,z)=(2x-3y+5z)^5. If we want to find a plane tangent to the surface xyz=x^2-2y^2+z^3=14 at P(5,-2,3)

If we compute the gradient, we don't get a vector tangent to the surface, but normal to it. Is that because the surface is equipotential? Well, I don't know what that means. How can I know whether a surface is equipotential?

- #6

berkeman

Mentor

- 59,012

- 9,109

To get an equipotential surface, you would need to know the source of the potential field (like gravity, or electric field, or magnetic field, etc.), and the shape of the geometry of any sources (like the earth, or coaxial capacitor plates, or an open coil, etc.). Given the source, you can calculate the field and the equipotential surfaces.

- #7

berkeman

Mentor

- 59,012

- 9,109

Actually, I misspoke there. The potential field is generally a scalar field, and to get the force field, you can use the gradient operator. Like in electrostatics problems,

To get an equipotential surface, you would need to know the source of the potential field (like gravity, or electric field, or magnetic field, etc.), and the shape of the geometry of any sources (like the earth, or coaxial capacitor plates, or an open coil, etc.). Given the source, you can calculate the field and the equipotential surfaces.

[tex]E = - \nabla \phi[/tex]

Where [tex]\phi[/tex] is the scalar electric potential.

- #8

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 132

Let us consider the equation:

[tex]f(x,y,z)=0[/tex]

This equation determines a SURFACE, [tex]\vec{S}(u,v)=(x(u,v),y(u,v),z(u,v))[/tex], where u and v are variables parametrizing the surface.

In particular, this means that the two local TANGENT vectors to the surface are given by the vectors [itex]\frac{\partial\vec{S}}{\partial{u}},\frac{\partial\vec{S}}{\partial{v}}[/tex]

Furthermore, for all (u,v), we have the folllowing IDENTITY:

[tex]f(\vec{S}(u,v))=0[/tex]

Since it is an identity, we may, for example, diffferentiate with respect to u, yielding, with aid of the chain rule:

[tex]\nabla{f}\cdot\frac{\partial\vec{S}}{\partial{u}}=0[/tex]

This tells us that the gradient of f is normal to a tangent vector to the surface; i.e, since this holds for the other tangent vector as well, we see that the gradient must be parallell to the surface normal.

- #9

mathwonk

Science Advisor

Homework Helper

2020 Award

- 11,098

- 1,301

Expand your function in a Taylor series at the given point. Then the gradient vector is the vector of coefficients of the linear term.

I.e. it is the vector such that the best linear approximation to the change in your function near that point is dotting with the gradient. Since the function does not change at all along the level surface, this gradient should dot to zero with the tangent vector to the level surface, i.e. it should be perpendicular to the level surface through the given point.

equivalently the gradient points in the direction of greatest change of the function near the given point, i.e,. as far away as possible from all directions tangent to the level surface.

e.g. the gradient of the function X^2 + Y^2 + Z^2 at (1,1,1) is perpendicular to the sphere of radius r = sqrt(3), i.e, it is parallel to the radius from (0,0,0) to (1,1,1).

(Does this seem right? I make a lot of mistakes in explicit examples.)

I.e. it is the vector such that the best linear approximation to the change in your function near that point is dotting with the gradient. Since the function does not change at all along the level surface, this gradient should dot to zero with the tangent vector to the level surface, i.e. it should be perpendicular to the level surface through the given point.

equivalently the gradient points in the direction of greatest change of the function near the given point, i.e,. as far away as possible from all directions tangent to the level surface.

e.g. the gradient of the function X^2 + Y^2 + Z^2 at (1,1,1) is perpendicular to the sphere of radius r = sqrt(3), i.e, it is parallel to the radius from (0,0,0) to (1,1,1).

(Does this seem right? I make a lot of mistakes in explicit examples.)

Last edited:

- Last Post

- Replies
- 4

- Views
- 695

- Last Post

- Replies
- 7

- Views
- 6K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 13

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 11

- Views
- 40K

- Replies
- 1

- Views
- 693

- Last Post

- Replies
- 8

- Views
- 3K

- Replies
- 4

- Views
- 6K

- Last Post

- Replies
- 3

- Views
- 433