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Suppose we have a flat 3-D manifold $ ds^2=\delta_{ab}dx^{a}dx^{b},$ which contains a 2-D surface given by a parametric relationship $r^{a}\left(u,v\right)=x^{a}\left(u,v\right).$ Where $ u $ and $ v $ are two independent parameters.

I know that the Area of this surface is the magnitude of a 3-D vector

$$ dA_{a}=\varepsilon_{abc}\dfrac{\partial x^{b}}{\partial u}\dfrac{\partial x^{c}}{\partial v} du dv $$

$$dA^2=\delta^{ef}dA_{e}dA_{f}$$

How can I generalize this formula to an arbitrary 3-D manifold given by $ ds^2=g_{ab}dx^{a}dx^{b}$, where:

$$ dA^2=g^{ab}dA_{b}dA_{a}? $$

That is, is it possible to write

$$ dA_{a}=\kappa_{abc}\dfrac{\partial x^{b}}{\partial u}\dfrac{\partial x^{c}}{\partial v} du dv $$

and, if so, what is $\kappa_{abc}$ ?

Mahlomola Daniel Cwele
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    So you've got a two-manifold $M$ that embeds into a three-manifold $N$ with embedding map $\iota:M\to N$. Then a metric on $N$ induces a metric on $M$ via the pullback $\iota^$, i.e. a natural metric on $M$ is given by $\iota^g$, where $g$ is the metric on $N$. Besides your (typical physicist) notation, this is not a physics question. – Danu Oct 30 '16 at 00:36
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    Actually, this is connected to a much more involved physics problem. This particular step in the logic is what is holding me back. –  Oct 30 '16 at 01:16
  • I recently tried to derive this in $\mathbb{R}^3$ using dot a cross products and gradients. I don't remember the end result. Although more better suited for [Mathematics.SE], I'd like to see what answers people give. – John Alexiou Oct 30 '16 at 02:34
  • It's just $\int_M d^2 \sigma , \sqrt{|\gamma|}$ where ${\sigma^a}$ are your coordinates on the two-manifold $M$, and $\gamma_{ab}$ is the intrinsic metric or pullback (which is not the same as the first fundamental form). – JPhy Oct 30 '16 at 13:27

2 Answers2

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@Solenodon Paradoxus may be has answered your question already. But let me do it in more detail. Suppose we have a 3-dim manifold $\mathfrak{M}_3$ with metric function $g_{\alpha\beta}$. Note that manifolds does not contain any vectors itself. Vectors emerges in tangent spaces. But we will not be so specific. So we will say that $\mathfrak{M}_3$ contains vectors, metric and etc. bearing in mind that this elements belong to its ($\mathfrak{M}_3$'s) tangent space. OK. Let us suppose also we have 2-dim manifold say $\mathfrak{M}_2\in\mathfrak{M}_3$. But $\mathfrak{M}_2$ can be considered individually with its own metric say $G_{\mu\nu}$. So you are asking the following question:

How do I know $G_{\mu\nu}$ if I know $g_{\alpha\beta}$?

Suppose $\mathfrak{M}_2$ represents some surface $M_2$ in your space. Or like you say $\mathfrak{M}_2$ can be parametrized by two variables $u, v$ (in the rest we will call it $u^1$ and $u^2$) in the following way: $$x_1 = x_1(u^1, u^2),\, x_2 = x_2(u^1, u^2),\,x_3 = x_3(u^1, u^2);$$ If we planed to calculate arc length $ds^2$ on our surface $M_2$ we could do it absolutely in the same way as we would do if this arc belonged $\mathfrak{M}_3$ but with the only specialty: $\vec{x}$ is a function of $u$ nd $v$. Let's do this. $$ds^2 = g_{\alpha\beta}(u^1,u^2)dx^\alpha(u^1, u^2) dx^\beta(u^1, u^2) = g_{\alpha\beta}\frac{dx^\alpha}{du^\mu}\frac{dx^\beta}{du^\nu}du^\mu du^\nu$$

But on the other hand

$$ds^2 = G_{\mu\nu}du^\mu du^\nu$$

So we have eventually $$G_{\mu\nu} = g_{\alpha\beta}\frac{dx^\alpha}{du^\mu}\frac{dx^\beta}{du^\nu}$$

LRDPRDX
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  • Nice as it is, This is not what I asked.

    My question is, how can I generalize this formula:

    $$ dA_{a}=\varepsilon_{abc}\dfrac{\partial x^{b}}{\partial u}\dfrac{\partial x^{c}}{\partial v} du dv $$

    To the case where the 3-D metric is arbitrary. If the 3-D metric is $g_{a b}$, I already think that the magnitude would be

    $$ dA^2=g^{ab}dA_{b}dA_{a} $$

    But how do I formulate the vector $dA_{a}$ such that is encodes the arbitrary metric?

    I.e, If

    $$ dA_{a}=\kappa_{abc}\dfrac{\partial x^{b}}{\partial u}\dfrac{\partial x^{c}}{\partial v} du dv $$

    what is $\kappa$ ?

     – Mahlomola Daniel Cwele Nov 01 '16 at 20:32
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Given an embedding of smooth differentiable manifolds $M \rightarrow N$, it is always possible to pull-back covariant tensors from $N$ to $M$. In physicist's notation:

$$ A_{\mu \nu \dots} (x) = \frac{\partial y^{\alpha}}{\partial x^{\mu}} \frac{\partial y^{\beta}}{\partial x^{\nu}} A_{\alpha \beta \dots} (y(x)). $$

To answer your question: pull back the 3D metric to obtain a 2D induced metric on the surface. In your case $\mu, \nu, \dots$ range through $\{1, 2\}$ and $\alpha, \beta, \dots$ range through $\{1, 2, 3\}$. Integrate the volume form of the induced metric over the surface:

$$ \text{Area} = \int d^2 x \sqrt{\det g_{\text{induced}}}. $$

Prof. Legolasov
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