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    $\begingroup$ 12.19 is approximately 12.56? I'm not sure why such a crummy approximation would be mentioned anywhere. $\endgroup$
    – Gerry Myerson
    Commented Jan 6, 2015 at 15:05
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    $\begingroup$ @GerryMyerson: They define a function they call $Z_k(q)$. That was just $k=1$. To quote Witten's paper in page 32, "...agreement improves rapidly if one increases $k$." For $k=4$, they used the first coefficient of the q-expansion as, $$\log(81026609428)\approx 25.12,\quad 8\pi \approx. 25.13$$ $\endgroup$
    – Tito Piezas III
    Commented Jan 6, 2015 at 16:45
  • $\begingroup$ for Bekenstein-Hawking entropy see en.wikipedia.org/wiki/Black_hole_thermodynamics $\endgroup$
    – jjcale
    Commented Jan 6, 2015 at 19:09
  • $\begingroup$ @GerryMyerson: You are right: it's not a good approximation and, in fact, the original paper does not say it is. Quoting from the paper (p. 32): "It is illuminating to compare the number $196883$ to the Bekenstein-Hawking formula. An exact quantum degeneracy of $196883$ corresponds to an entropy of $\log 196883 \approx 12.19$. By contrast, the Bekenstein- Hawking entropy at $k = 1$ and $L_0 = 1$ is $4π \approx 12.57$. We should not expect perfect agreement, because the Bekenstein-Hawking formula is derived in a semiclassical approximation which is valid for large $k$." $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jan 6, 2015 at 20:47
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    $\begingroup$ Can you clarify what is the question that you want answered? Is it the mathematical derivation of the approximation? Do you want something beyond what is on the subsequent pages of Witten's paper? Or do you want to understand the physics behind the relation? If so, do you have a specific question beyond or about what's in the paper? $\endgroup$
    – Aaron Bergman
    Commented Jan 7, 2015 at 13:43