# Induced metric on a one-sheet hyperboloid

I am trying to find the induced metric on a one-sheet hyperboloid. Suppose we use cylindrical coordinates $(r, \theta, z)$ for the ambient space in which the hyperboloid is embedded. The hyperboloid itself then consists of all points such that

$$r^2 = R^2 + z^2$$

where $R$ is the minimum radius, at the throat. On the hyperboloid itself, we define the following coordinates $(\rho, \varphi)$:

\begin{align} r &= \sqrt{R^2 + \rho^2} \\ \theta &= \varphi \\ z &= \rho \end{align}

We can now find the induced metric using the formula

$$g_{ab} = g_{\mu\nu} \partial_a X^\mu \partial_b X^\nu$$

where $a$ and $b$ describe the indices of the coordinates of the submanifold (in this case, the hyperboloid), while the $X^\mu$ and $X^\nu$ encode the embedding into the ambient space.

First, the nonzero components of the ambient metric tensor in cylindrical coordinates are

\begin{align} g_{rr} &= 1 \\ g_{\theta \theta} &= r^2 \\ g_{zz} &= 1 \\ \end{align}

Thus we have

\begin{align} g_{ab} &= g_{rr} \partial_a X^r \partial_b X^r + g_{\theta \theta} \partial_a X^\theta \partial_b X^\theta + g_{zz} \partial_a X^z \partial_b X^z \\ &= \partial_a X^r \partial_b X^r + r^2 \partial_a X^\theta \partial_b X^\theta + \partial_a X^z \partial_b X^z \end{align}

Hence the nonzero components of the metric tensor on the hyperboloid are

\begin{align} g_{\rho \rho} &= \partial_\rho X^r \partial_\rho X^r + r^2 \partial_\rho X^\theta \partial_\rho X^\theta + \partial_\rho X^z \partial_\rho X^z \\ &= \left(\partial_\rho \sqrt{R^2 + \rho^2}\right)^2 + (\partial_\rho \rho)^2 \\ &= \left(\frac{\rho}{\sqrt{R^2 + \rho^2}}\right)^2 + 1^2 \\ &= \frac{\rho^2}{R^2 + \rho^2} + 1 \end{align}

and

\begin{align} g_{\varphi \varphi} &= \partial_\varphi X^r \partial_\varphi X^r + r^2 \partial_\varphi X^\theta \partial_\varphi X^\theta + \partial_\varphi X^z \partial_\varphi X^z \\ &= 0 + r^2 \cdot 1^2 + 0 \\ &= r^2 \end{align}

The cross-terms $g_{\rho \varphi} = g_{\varphi \rho}$ vanish. Is this derivation correct?

## 1 Answer

Yes, your derivation is correct.

Incidentally, you can avoid expressing the Euclidean metric in cylindrical coordinates: Parametrize the hyperboloid by $$\Phi(\rho, \varphi) = \left(\sqrt{R^{2} + \rho^{2}} \cos\varphi, \sqrt{R^{2} + \rho^{2}} \sin\varphi, \rho\right).$$ The partials are \begin{align*} \Phi_{\rho}(\rho, \varphi) &= \left(\frac{\rho}{\sqrt{R^{2} + \rho^{2}}} \cos\varphi, \frac{\rho}{\sqrt{R^{2} + \rho^{2}}} \sin\varphi, 1\right), \\ \Phi_{\varphi}(\rho, \varphi) &= \left(-\sqrt{R^{2} + \rho^{2}} \sin\varphi, \sqrt{R^{2} + \rho^{2}} \cos\varphi, 0\right), \end{align*} so pulling back the Euclidean metric gives the components \begin{align*} g_{\rho\rho} &= \Phi_{\rho} \cdot\Phi_{\rho} = \frac{\rho^{2}}{R^{2} + \rho^{2}} + 1, \\ g_{\rho\varphi} &= \Phi_{\rho} \cdot\Phi_{\varphi} = 0, \\ g_{\varphi\varphi} &= \Phi_{\varphi} \cdot\Phi_{\varphi} = \rho^{2}. \end{align*}