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First Fundamental Form from tangent vectors

  1. Jul 18, 2013 #1

    Appologies for formatting issues this is the first time I have submitted something to the forum.

    I have a pretty simple problem, I am just going through the derivation of the First Fundamental Form and I think I am missing someting in the derivation.

    If we have a point x = (x1,x2) on the 2D surface ω(x) then the derivative matrix of the surface can be defined as Dω(x) = (∂ω(x)/∂x1,∂ω(x)/∂x2) which is 3x2 matrix.

    | 1 0 |
    | 0 1 |
    | ∂ω/∂x1 ∂ω/∂x2 |

    thus the first column is the partial derivative of the surface with respect to x1 and the second column is the partial derivative with respect to x2 and are thus the non unit tangent vectors.

    The first fundamental form can be defined as I by taking the inner product of the two tangent vectors :
    I = (Dω(x))' Dω(x).

    which can be written as:
    |1+(∂ω/∂x1)^2 (∂ω/∂x1)*(∂ω/∂x2) |
    |(∂ω/∂x1)*(∂ω/∂x2) 1+(∂ω/∂x1)^2 |

    which can be re-written as the metric tensor:

    |E F|
    |F G|

    However looking here:


    they appear to have dropped the 1 in E and G. I was just wondering where I might have gone wrong?

    Many thanks in advance
  2. jcsd
  3. Jul 18, 2013 #2
    Maybe this will help. Suppose you have a non-flat 2D surface immersed in flat 3D space. Let [itex]\vec{s}[/itex] represent a position vector drawn from an arbitrary origin in flat 3D space to an arbitrary point on the 2D surface. Suppose that there is a u-v coordinate grid laid out on the 2D surface. Then [itex]\vec{s}=\vec{s}(u,v)[/itex]. Now suppose we consider two neighboring points in the surface, one at (u,v) and the other at (u+du, v+dv). The differential position vector [itex]\vec{ds}[/itex] between these points lies within the surface (i.e., is tangent to the surface), and is given by:
    where [itex]\vec{a_u}[/itex] and [itex]\vec{a_v}[/itex] are the coordinate basis vectors, defined as
    [tex]\vec{a_u}=\frac{\partial \vec{s}}{\partial u}[/tex]
    [tex]\vec{a_v}=\frac{\partial \vec{s}}{\partial v}[/tex]
    If we form the inner product of [itex]\vec{ds}[/itex] with itself, we obtain:
    where [itex]g_{uu}[/itex] represents the inner product of [itex]\vec{a_u}[/itex] with itself, [itex]g_{uv}[/itex] represents the inner product of [itex]\vec{a_u}[/itex] with[itex]\vec{a_v}[/itex], and [itex]g_{vv}[/itex] represents the inner product of [itex]\vec{a_v}[/itex] with itself.

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