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Suppose ## \Delta = r^2 + 2GMr + a^2## and ## \rho^2 = r^2 + a^2 \cos ^2 \theta ##. The Kerr metric is

$$

ds^2 = - (1 - \frac{2GMr}{\rho^2})dt^2 - \frac{4GMar\sin^2 \theta}{\rho^2} d t d \phi + \frac{\rho^2}{\Delta} dr^2 + \rho^2 d \theta^2 + \frac{\sin^2 \theta}{\rho^2} \left[ (r^2 + a^2)^2 - a^2 \Delta \sin^2 \theta \right] d \phi^2

$$

I want to determine the area of the inner horizon of the black hole. But first I need to determine the induced metric on the horizon. Suppose that the radius of the innermost horizon is ## r_*##. My tutor stated without proof that the induced metric was

$$

ds^2 = \rho_*^2 d \theta^2 + \frac{\sin ^2 \theta}{\rho_*^2} (r_*^2 + a^2) d \phi^2

$$

How do one compute this induced metric from the Kerr metric?

Thank you for your time.

Kind regards,

Marius

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# A Metric induced on Kerr event horizon

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