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The gradient vector

  1. Mar 16, 2007 #1
    What is the gradient vector, really? My textbook both states that it is a vector normal to a certain point on a surface, but also that it is a vector that points in the direction with the maximum slope of a surface. I find this slightly ambiguous.
     
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  3. Mar 16, 2007 #2

    berkeman

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    It's confusing you because the gradient is normal to the equipotential surface. 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.
     
  4. Mar 16, 2007 #3

    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)?
     
  5. Mar 16, 2007 #4

    berkeman

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    You will only get the gradient if the surface is an equipotential surface. Maybe I'm not understanding your question. Can you give a specific example of what is causing confusion?
     
  6. Mar 16, 2007 #5
    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? :confused:
     
  7. Mar 16, 2007 #6

    berkeman

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    The gradient of a vector potential field is defined as being the normal to the equipotential surface at a point. The normal to a general surface is not necessarily the gradient.

    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.
     
  8. Mar 16, 2007 #7

    berkeman

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    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,

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

    Where [tex]\phi[/tex] is the scalar electric potential.
     
  9. Mar 16, 2007 #8

    arildno

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    kasse:
    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.
     
  10. Mar 17, 2007 #9

    mathwonk

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    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.)
     
    Last edited: Mar 17, 2007
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