I have this $D=4$ space with the following global metric $$ dS^2_4 = L^2\Big(\frac{y^2-h^2}{(y^2-1)(y^2-b^2)}dy^2 + \frac{y^2-h^2}{h^2}d\theta^2 + (y^2-1)\cos^2\theta d\phi_1^2 + \frac{y^2-b^2}{b^2}\sin^2\theta d\phi_2^2 \Big) $$ with $$ h^2(\theta) = \sin^2\theta + b^2\cos^2\theta $$ and, with this substitution (1) \begin{equation} \cosh^2 R = \frac{y^2}{b^2}h^2, \qquad cos^2 \psi = \frac{b^2(y^2-1)\cos^2\theta}{y^2h^2-b^2}. \end{equation} I should obtain this easier form $$ dS^2_4 = L^2\Big(dR^2 + \sinh^2 R(d\psi^2 + \cos^2\psi d\phi_1^2 + \sin^2\psi d\phi_2^2) \Big). $$ My question is how to show that indeed this substitution gives the last metric. I have tried to derive the differentials from the substitution formula (1) but I found it quite troublesome and I couldn't do it, is there a clever way instead of reverting on using Mathematica?


1 Answer 1


We are given the components of the metric $g_{\alpha\beta}$ in a particular set of coordinates: $\{x^\gamma\} = \{y, \theta, \phi_1, \phi_2\}$. For a different set of coordinates, $\{x^{\bar{\sigma}}\}=\{R, \psi, \phi_1, \phi_2\}$, the metric has components $g_{\bar{\mu}\bar{\nu}}$ given by the transformation law between components of $0 \choose 2$ tensors.

$$g_{\bar{\mu}\bar{\nu}} = \frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^\beta}{\partial x^{\bar{\nu}}}g_{\alpha\beta}$$

You can get these partial derivatives by taking derivatives of the "substitution" in the OP. This isn't a particularly clever way of getting the new metric components, but it's the most algorithmic.

For example, differentiating both sides of the first coordinate transformation with respect to $y$

$$2\cosh{R}\sinh{R}\frac{\partial R}{\partial y} = \frac{2yh^2}{b^2} \;\;\;\;\Rightarrow \;\;\;\;\frac{\partial R}{\partial y} = \frac{2yh^2}{b^2}\frac{1}{\sinh{2R}}$$

Similarly we may obtain $\partial R/\partial \theta$, $\partial \psi/\partial y$, and $\partial \psi/\partial \theta$ and construct a matrix with elements $\partial x^{\bar{\mu}}/\partial x^\alpha$. Recall

$$\frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^{\bar{\mu}}}{\partial x^\beta} = \delta^\alpha_\beta$$

So the partial derivatives of $\{y,\theta\}$ with respect to $\{R,\psi\}$ may be obtained by inverting this matrix. Careful here we're mixing two different coordinate systems when doing these calculations as a shorthand, but eventually we should convert everything to the new coordinate system.

  • $\begingroup$ I have tried to do this, but first I should write $y(R,\psi)$ and $\theta(R,\psi)$, which is not easy. Is there any other way? $\endgroup$ Commented Sep 15, 2023 at 14:54
  • $\begingroup$ Deriving $y(R,\psi)$ and $\theta(R,\psi)$ from the "substitution" in the OP is not an easy task, or am I missing something? $\endgroup$ Commented Sep 15, 2023 at 15:42
  • $\begingroup$ @LorenzoZappa I've updated my answer to address this $\endgroup$
    – Aiden
    Commented Sep 15, 2023 at 20:30

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