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Metric of a surface

  1. Feb 11, 2010 #1
    So say I want to find the metric of a surface, which is obviously 2 dimensions, which is immersed in a 3 dimensional space.

    The surface is described by 3 coordinates (x,y,z) in Cartesian, but can be expressed using 2 (r, phi) in polar.

    Does this mean you'd end up with a 2 dimensional or 3 dimensional metric?

  2. jcsd
  3. Feb 12, 2010 #2
    Since you say the metric OF the surface, this is a 2d metric.

    Consider an embedding mapping f of your 2D surface in 3D:

    f: (u^1,u^2) -> (x^1,x^2,x^3)

    where u^1 = r, u^2 = phi, x ^1 = x, x^2 = y, x^3 = z. This would simply be the function that for each (r,phi) gives you the corresponding triple (x,y,z). Also we define components of f by

    f^1 (u^1,u^2) = x^1
    f^2 (u^1,u^2) = x^2
    f^3 (u^1,u^2) = x^3

    Then the "induced metric" on your surface can be expressed as:

    h_ab = sum_ij df^i/du^a df^j/du^b g_ij

    where g_ij is the metric on your 3D space (presumably just the Kronecker delta if your 3D space is Euklidean?), and sum_ij means a sum of all values of i and j. The indices a,b only have the possible values 1,2 since the u-coordinates only have two components, r and phi.

    This metric simply reflects the "restriction" of the 3D metric to the 2d surface.

    Ref: http://en.wikipedia.org/wiki/Induced_metric

  4. Feb 13, 2010 #3
    Hi Ryan(onomous),

    Consider f(x,y) = z, where x=r\cos\phi and y=r\sin\phi,
    we calculate the metric via the 1st fundamental form on the surface:

    g_{11} = E = x_{,r}x_{,r}+y_{,r}y_{,r}+z_{,r}z_{,r},
    g_{12} = F = x_{,r}x_{,\phi}+y_{,r}y_{,\phi}+z_{,r}z_{,\phi},
    g_{22} = G = x_{,\phi}x_{,\phi}+y_{,\phi}y_{,\phi}+z_{,\phi}z_{,\phi}.

    The line element takes the form

    ds^2 = g_{a b} dx^a dx^b = E dr^2 + 2F dr d\phi + G d\phi^2.

    That should do it.
    Last edited: Feb 13, 2010
  5. Feb 14, 2010 #4


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    Science Advisor

    You can do this in two ways; either use the explicit pullback from the space R3 to the surface, or "the easy way".

    The easy way is the following. Take for example a sphere S2 in a space R3. The space R3 has as line element

    ds^2 = dx^2 + dy^2 + dz^2

    Our sphere S2 is parametrized in R3 as

    x^2 + y^2 + z^2 = R^2
    where R is the radius of the sphere. This constraint cuts down the number of independent coordinates from 3 to two. Now, for instance, you can "take the differential of this equation" and write

    2x*dx + 2y*dy + 2z*dz = 0

    because R is a constant. If you now rewrite this in terms of, for example, dx = ... and plug this into the expression for the line element of R3, you have found the "induced line element on the sphere by R3". Hope this helps :)
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