# String theory hilbert space - Gas of free gravitons

I am trying to understand the arguments given in MAGOO in chapter 3.4.1(Hilbert Space of String Theory). The authors give descriptions of the Hilbert space of String Theory when we consider our theory in different energy regimes. There are 4 such regimes. Since each regime is enough material for one question I will spread this in 4 different questions. Let's start here with the first regime where we have

1: Gas of Free Gravitons($$E\ll m_s$$)

In this low energy regime we can approxiamate the states of our Hilbert space by the Fock space of gravitons in $$AdS_5\times S^5$$. Then the states are the stationary wave solutions as shown in section 2.2.2 which are the solutions of the scalar KG-equation in $$AdS_{p+2}$$ $$\phi=e^{i\omega\tau}G(\theta)Y_l(\Omega_p)$$ with $$G(\theta)=(\sin\theta)^l(\cos\theta)^{\lambda_{\pm}} \;_2F_1(a,b,c;\sin\theta)$$ and $$a=\frac{1}{2}(l+\lambda_{\pm}-\omega R)\\ b=\frac{1}{2}(l+\lambda_{\pm}+\omega R)\\ c=l+\frac{1}{2}(p+1)$$

and $$\lambda_\pm=\frac{1}{2}(p+1)\pm\frac{1}{2}\sqrt{(p+1)^2+4(mR)^2}$$ Then they argue that stationary modes are quantised "in the unit set by R" so they view the supergravity particles as confined in a box of size R(the curvature radius). Why are they confined in this box of size R?

I understand that I'm probably missing something in the derivation of this, but the next point is very confusing to me.

They also state the entropy, as a function of density of states $$S(E)\sim(ER)^\frac{9}{10}$$ where did this come from? How is this entropy calculated(estimated?)

For the purposes of thermodynamics, you can treat the propagating modes of the graviton (and superpartners) as independent scalar fields satisfying $$$$\nabla_\mu \nabla^\mu \phi + \xi R \phi = 0.$$$$

And then there's a very quick and dirty way to verify that $$AdS_{d + 1}$$ will behave like a box. Instead of a minimally coupled scalar with $$\xi = 0$$, you can consider a conformally coupled scalar with $$\xi = \frac{d - 1}{4d}$$. This will allow you to perform a Weyl rescaling on the metric to get \begin{align} &ds^2 = -\left ( 1 + \frac{r^2}{R^2} \right ) dt^2 + \left ( 1 + \frac{r^2}{R^2} \right )^{-1} dr^2 + r^2 d\Omega_{d - 1}^2 \\ \mapsto \;\; &ds^2 = -dt^2 + \left ( 1 + \frac{r^2}{R^2} \right )^{-2} dr^2 + \left ( 1 + \frac{r^2}{R^2} \right )^{-1} r^2 d\Omega_{d - 1}^2. \end{align} We can now get an effective volume of global $$AdS_5$$ as \begin{align} V &= \int \sqrt{g} \, dr \, d\Omega_3 \\ &= \mathrm{Vol}(S^3) \int_0^\infty \left ( 1 + \frac{r^2}{R^2} \right )^{-5/2} r^3 dr \\ &= \frac{2}{3} \mathrm{Vol}(S^3) R^4. \end{align} In the present problem there should of course be another piece due to the $$S^5$$.

To proceed with analyzing a gas of free fields, we know that for a fermion, there can be 0 or 1 quanta with momentum $$p$$ so we get a single mode partition function of $$Z(p) = 1 + e^{-\beta |p|}$$. For a boson, there can be arbitrarily many excitations so $$$$Z(p) = 1 + e^{-\beta |p|} + e^{-2\beta |p|} + \dots = (1 - e^{-\beta |p|})^{-1}.$$$$ To get the log of the full partition function, we sum over all modes, leading to \begin{align} \log Z &= \sum_p \log Z(p) \\ &\approx \int | \log (1 \pm e^{-\beta |p|}) | \frac{V d^d p}{(2\pi)^d} \\ &= \frac{V}{(2\pi \beta)^d} \Gamma(d) \mathrm{Vol}(S^{d - 1}) \zeta^{\pm}(d + 1). \end{align} In the last step, we have Taylor expanded the log and integrated term by term. The answer involves the Riemann zeta function $$\zeta^-$$ for bosons and the so called alternating zeta function $$\zeta^+$$ for fermions. Knowing that the free energy is $$$$F = E - TS = -T \log Z,$$$$ the formula for the entropy now follows from $$S = - \frac{\partial F}{\partial T}$$. However, the above is overkill if we aren't interested in the precise prefactor.

The slick way to find that $$S \sim E^{d/(d + 1)}$$ (so $$E^{9/10}$$ in $$9 + 1$$ dimensions) is to use scale invariance. Gravitons are massless and cannot introduce a scale beyond the $$AdS$$ radisu $$R$$. Therefore at temperatures much larger than $$R^{-1}$$ (but still much smaller than the string scale), entropy must be extensive. But entropy is a dimensionless quantity so the only way to achieve this is to have $$$$S \sim V T^d.$$$$ The same goes for internal energy but that needs to have one more power of temperature $$$$E \sim V T^{d + 1}.$$$$ Eliminating $$T$$ then yields $$S$$ in terms of $$E$$. As an aside, a free theory is not the only case where it's possible to determine the prefactor. Another is a 2d conformal field theory where modular invariance leads to the Cardy formula for the density of states.

• Thank you very much! So the derivation of the partition function is not used? Essentially we use dimensional analysis? May 27 at 8:00
• Yes. The last of the four regimes ($E^{3/4}$ for SYM theory in 3 + 1 dimensions) can be argued in the same way. So it is then a nice check that MAGOO recovers this by considering large black holes. May 27 at 12:10

AdS behaves like a box, since the boundaries at infinity can be reached in a finite amount of time. See for instance Does one observe light reach infinity in finite time in AdS spacetime?

Regarding the entropy, you should look into how to derive the entropy of an ideal gas in 10D. See here for a calculation in 3D in the microcanonical ensemble.

• I understand that lightrays can be reflected at finite time, even though AdS can be thought of as an infinite box but why should it have size R? May 25 at 6:34
• It is explained in 2.2.2. By imposing that the energy-momentum flux of a scalar field through the boundary is zero you get 2.41, which show you that the wave are quantised in the unit set by R: $$w = w_0 + 2n/R$$ May 25 at 8:27
• R is also the only scale in AdS, so it has to be expected that the quantisation comes in units or R. May 25 at 8:28
• Thank you for your remarks. Would it be possible to add the derivation of the entropy? May 25 at 9:38