# Induced Metric on Surface t=const

Tags:
1. May 26, 2015

### Xander314

1. The problem statement, all variables and given/known data
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
$$\gamma_{ij} = g_{ij} - \frac{g_{ti} g_{tj}}{g_{tt}} .$$

2. Relevant 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
$$\gamma_{ij} = g_{\mu\nu} \partial_i X^\mu \partial_j X^\nu .$$

3. 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
$$\gamma_{ij} = g_{kl} \partial_i X^k \partial_j X^l .$$
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?

2. May 31, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Jun 1, 2015

### WannabeNewton

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 $\gamma_{\theta\phi} = 0$.