What Is the Induced Metric on a Spacelike Hypersurface t=const?

[Xander314]

Homework Statement

Let g_{\mu\nu} be a static metric, \partial_t g_{\mu\nu}=0 where t is coordinate time. Show that the metric induced on a spacelike hypersurface t=\textrm{const} is given by
<br /> \gamma_{ij} = g_{ij} - \frac{g_{ti} g_{tj}}{g_{tt}} .<br />

Homework Equations

Let y^i be the coordinates on the hypersurface and x^\mu the spacetime coordinates. The induced metric on a generic hypersurface defined by the embedding x^\mu = X^\mu(y^i) is given by
<br /> \gamma_{ij} = g_{\mu\nu} \partial_i X^\mu \partial_j X^\nu .<br />

The Attempt at a Solution

I really don't see how this can work. Since it is a hypersurface of constant coordinate time, the embedding is given by X^\mu = (t_0, X^i) so that \partial_i X^\mu = (0,\partial_i X^j). Then it immediately follows that
<br /> \gamma_{ij} = g_{kl} \partial_i X^k \partial_j X^l .<br />
There are no g_{ti} cross terms in my answer, nor is it clear to me that \partial_i X^k=\delta_i{}^k. What am I doing wrong?

 

[WannabeNewton]

What am I doing wrong?

The definition of the induced metric is $\gamma_{ab} = g_{\mu\nu}\frac{\partial x^{\mu}}{\partial Y^a}\frac{\partial x^{\nu}}{\partial Y^b}$ where $x^{\mu}$ are coordinates on space-time and $Y^a$ are coordinates on the hypersurface.
The embedding is $Y^a = Y^a(x^{\mu})$. So you are doing the differentiation incorrectly.

Here is a very simple example to get you started. Let's compute the induced metric on a 2-sphere in $\mathbb{R}^3$. The coordinates on the 2-sphere are $(\theta,\phi)$ and the embedding map is $x = R\sin\theta\cos\phi, y = R\sin\theta\sin\phi, z = R\cos\theta$. Hence \gamma_{\theta\theta} = g_{xx}\frac{\partial x}{\partial \theta}\frac{\partial x}{\partial \theta} + g_{yy}\frac{\partial y}{\partial \theta}\frac{\partial y}{\partial \theta} + g_{zz}\frac{\partial z}{\partial \theta}\frac{\partial z}{\partial \theta} = R^2 \cos^2\theta \cos^2\phi + R^2 \cos^2\theta \sin^2\phi + R^2 \cos^2\theta = R^2 and \gamma_{\phi\phi} = g_{xx}\frac{\partial x}{\partial \phi}\frac{\partial x}{\partial \phi} + g_{yy}\frac{\partial y}{\partial \phi}\frac{\partial y}{\partial \phi} + g_{zz}\frac{\partial z}{\partial \phi}\frac{\partial z}{\partial \phi} = R^2 \sin^2\theta \sin^2 \phi + R^2 \sin^2\theta\cos^2 \phi = R^2 \sin^2 \theta A similar calculation shows that $\gamma_{\theta\phi} = 0$.