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How to draw a 2D space in 3D Euclidean space by metric tensor

  1. Oct 24, 2014 #1
    Suppose, I know the metric tensor of a 2D space. for example, the metric tensor of a sphere of radius R,
    gij = ##\begin{pmatrix} R^2 & 0 \\ 0 & R^2\cdot sin^2\theta \end{pmatrix}##
    ,and I just know the metric tensor, but don't know that it is of a sphere.
    Now I want to draw a 2D space(surface) which will have the metric tensor gij and will be embedded on 3D Euclidean space.
    Last edited: Oct 24, 2014
  2. jcsd
  3. Oct 24, 2014 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    Your question does not always have an answer! Not every 2d manifold is isometrically embeddable into 3d Euclidean space.

    When your question does have an answer, it might not be unique. For example, the metric

    $$ds^2 = dx^2 + dy^2$$
    could be an infinite plane, or it could be a cylinder, or it could be a non-circular cylinder, or it could be an infinite plane with a plane wave wiggle on it of arbitrary profile.

    In general, you have a whole arbitrary function's worth of non-uniqueness for isometric embeddings into R^3. The reason for this is that surfaces embedded in R^3 have two curvatures: an intrinsic one (Gauss curvature) and an extrinsic one (mean curvature). As the terminology would imply, only the Gauss curvature is determined by the 2d metric tensor; the mean curvature is a property of the choice of embedding.

    In general, you should write

    $$g_{ij} = \partial_i \vec s(u, v) \cdot \partial_j \vec s(u,v)$$
    where ##g_{ij}## is your 2d metric tensor, ##u, v## are your 2d coordinates, and ##\vec s(u,v)## is a vector (in R^3) describing your surface, as a function of ##u,v##. Your task is to solve these equations for ##\vec s##. You have a coupled set of nonlinear differential equations. It is not always easy to find solutions.
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