Compute Induced Metric on Kerr Event Horizon

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Discussion Overview

The discussion centers on computing the induced metric on the inner horizon of a Kerr black hole, exploring the mathematical formulation and implications of the Kerr metric. Participants engage with theoretical aspects of general relativity and black hole physics.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • Marius presents the Kerr metric and seeks to compute the induced metric on the inner horizon, referencing a statement from a tutor regarding the induced metric's form.
  • Some participants note that on the horizon, the differentials ##dt## and ##dr## are zero, indicating constants for those variables.
  • One participant corrects Marius's expression for ##\Delta##, suggesting it should be ##\Delta = r^2 - 2GMr + a^2## instead of the original formulation.
  • Another participant provides the equation for the surface area of the inner and outer horizons in the Kerr metric, indicating a relationship between the radii and the area.

Areas of Agreement / Disagreement

There is no consensus on the correct formulation of ##\Delta##, as one participant challenges Marius's initial expression. The discussion remains unresolved regarding the computation of the induced metric and the implications of the corrections presented.

Contextual Notes

Participants express uncertainty regarding the assumptions made in the derivation of the induced metric and the implications of the corrections to the equation for ##\Delta##.

Jonsson
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Hello there,

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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On the horizon ##t## and ##r## are constants, so ##dt=0## and ##dr=0##.
 
martinbn said:
On the horizon ##t## and ##r## are constants, so ##dt=0## and ##dr=0##.

Also ##\Delta = 0## on the horizon.
 
That's right. Got it now. Thanks
 
Jonsson said:
Suppose ##\Delta = r^2 + 2GMr + a^2##

This isn't quite correct. It should be ##\Delta = r^2 - 2GMr + a^2##.
 
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You're probably working through things yourself but if you want a check, the equation for the surface area of both the inner and outer horizon in Kerr metric is-
A_{\pm}=\pm4\pi\left(r^2_{\pm}+a^2\right)
where
r_{\pm}=M\pm\sqrt{M^2-a^2}
where \pm denotes the outer (+) and inner (-) horizon.

One source-
https://redirect.viglink.com/?format=go&jsonp=vglnk_149222179481113&key=6afc78eea2339e9c047ab6748b0d37e7&libId=j1illvfl010009we000DApptnkylp&loc=https%3A%2F%2Fwww.physicsforums.com%2Fthreads%2Fsecond-law-of-thermodynamics-and-the-macroscopic-world.833987%2F%23post-5238097&v=1&out=http%3A%2F%2Fwww.researchgate.net%2Fpublication%2F230923684_Entropy_of_Kerr-Newman_Black_Hole_Continuously_Goes_to_Zero_when_the_Hole_Changes_from_Nonextreme_Case_to_Extreme_Case&ref=https%3A%2F%2Fwww.physicsforums.com%2Fsearch%2F5021084%2F%3Fq%3Dkerr%26o%3Drelevance%26c%5Buser%5D%5B0%5D%3D68677&title=Second%20law%20of%20thermodynamics%20and%20the%20macroscopic%20world%20%7C%20Physics%20Forums%20-%20The%20Fusion%20of%20Science%20and%20Community&txt=Negative%20Temperature%20of%20Inner%20Horizon%20and%20Planck%20Absolute%20Entropy%20of%20a%20Kerr-Newman%20Black%20Hole by Liu Bo Liu & Wen-Biao
 
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