Absorption by D-Branes and Two-Point Functions

The D3-brane is where the two languages of string theory begin to sound almost indistinguishable. In one language, a low-energy closed string hits a stack of branes and turns into open-string excitations. In the other, the same closed-string field is a wave moving in the curved supergravity geometry sourced by the branes. The equality of the two absorption probabilities is one of the cleanest dynamical checks behind the emergence of holography.

The process we study is deliberately simple. Take $N$ coincident D3-branes in type IIB string theory and send in a massless bulk scalar with frequency $\omega$. The scalar can be the dilaton, or any other minimally coupled scalar in the same supergravity multiplet. At low energy, $\omega R\ll 1$, where $R$ is the D3-brane throat radius, the absorption cross-section is

$$\sigma_{\rm abs} = {\pi^4\over 8}\,\omega^3 R^8 = {\kappa_{10}^2 N^2\over 32\pi}\,\omega^3.$$

This equation is worth staring at. The left expression is geometric: it knows about a wave tunneling through the D3-brane throat. The right expression is microscopic: it knows about $N^2$ adjoint degrees of freedom living on the branes. The power $\omega^3$ is also not accidental; it is the spectral-density scaling of a four-dimensional operator of dimension $\Delta=4$.

The same absorption process has two descriptions. In the open-string description a bulk scalar couples to a worldvolume operator and produces brane excitations. In the closed-string description it is a radial wave in the D3-brane geometry, with absorption equal to the flux entering the throat.

The D3-brane background and the scalar wave equation

For an extremal stack of $N$ D3-branes, the ten-dimensional Einstein-frame metric is

$$ds^2 = H(r)^{-1/2}\eta_{\mu\nu}dx^\mu dx^\nu + H(r)^{1/2}\left(dr^2+r^2d\Omega_5^2\right), \qquad \mu,\nu=0,1,2,3,$$

with

$$H(r)=1+{R^4\over r^4}, \qquad R^4=4\pi g_sN\alpha'^2.$$

Equivalently, using the convention

$$2\kappa_{10}^2=(2\pi)^7g_s^2\alpha'^4,$$

one may write

$$R^4={\kappa_{10}N\over 2\pi^{5/2}}.$$

The dilaton is constant for the D3-brane, so the Einstein-frame and string-frame metrics differ only by a constant normalization. This is one reason the D3-brane is so clean.

Let $\phi$ be a minimally coupled scalar, independent of the $S^5$ angles and with no momentum along the brane. We take

$$\phi(t,r)=e^{-i\omega t}\phi(r).$$

The scalar equation is

$${1\over\sqrt{-g}}\partial_M\left(\sqrt{-g}\,g^{MN}\partial_N\phi\right)=0.$$

For the metric above,

$$\sqrt{-g}\,g^{rr}=r^5\sqrt{g_{S^5}}, \qquad g^{tt}=-H^{1/2},$$

and the radial equation becomes

$${1\over r^5}{d\over dr}\left(r^5{d\phi\over dr}\right) +\omega^2\left(1+{R^4\over r^4}\right)\phi=0.$$

Introduce the dimensionless radial coordinate

$$\rho=\omega r.$$

Then

$${1\over \rho^5}{d\over d\rho}\left(\rho^5{d\phi\over d\rho}\right) + \left(1+{(\omega R)^4\over \rho^4}\right)\phi=0.$$

This is the central equation. The term $1$ describes propagation in the asymptotically flat region. The term $(\omega R)^4/\rho^4$ is the effect of the D3-brane throat. At low energy, $\omega R\ll 1$, the wave mostly reflects, but a small fraction tunnels into the throat and is absorbed.

It is useful to remove the first-derivative term by setting

$$\phi(\rho)=\rho^{-5/2}\psi(\rho).$$

The radial equation becomes a one-dimensional Schrödinger-type equation,

$$\left[ -{d^2\over d\rho^2} +{15\over 4\rho^2} -1 -{(\omega R)^4\over \rho^4} \right]\psi(\rho)=0.$$

The term $15/(4\rho^2)$ is the $s$-wave centrifugal barrier in six transverse dimensions. The negative inverse-fourth-power term pulls the wave into the throat. Their competition produces the barrier shown in the figure.

The low-energy matched-asymptotics computation

The small parameter is

$$\epsilon=\omega R\ll 1.$$

The radial equation has two natural approximations.

In the outer region,

$$\rho\gg \epsilon^2,$$

the throat term $\epsilon^4/\rho^4$ is small. The equation reduces to the flat six-dimensional radial wave equation,

$${1\over \rho^5}{d\over d\rho}\left(\rho^5{d\phi\over d\rho}\right)+\phi=0,$$

whose solutions are

$$\phi_{\rm out}(\rho) = \rho^{-2}\left[ A J_2(\rho)+B N_2(\rho) \right],$$

where $J_2$ and $N_2$ are Bessel functions.

In the inner region, it is better to use

$$z={\epsilon^2\over \rho}={\omega R^2\over r}.$$

Near the horizon $r\to0$, one has $z\to\infty$, and the physical boundary condition is purely ingoing flux. The inner solution is written in terms of Hankel functions,

$$\phi_{\rm in}(z)\propto z^2H^{(1)}_2(z),$$

up to a normalization convention.

The two approximations overlap because

$$\epsilon^2\ll \rho\ll 1$$

is a nonempty region when $\epsilon\ll1$. In this region both the inner and outer solutions may be expanded in powers of $\rho$, and the coefficients are matched. The result is the $s$-wave absorption probability

$$\mathcal P_0 = {\pi^2\over 256}(\omega R)^8 + O\!\left((\omega R)^{12}\right).$$

For $\omega R\ll1$, the radial equation has an inner throat region, an outer asymptotically flat region, and a parametrically large overlap region. Matching Bessel expansions fixes the small absorption probability.

The cross-section is obtained from the usual partial-wave relation in the six transverse spatial dimensions. For the $s$-wave,

$$\sigma_0 = {(2\pi)^5\over \omega^5\operatorname{Vol}(S^5)}\,\mathcal P_0.$$

Since

$$\operatorname{Vol}(S^5)=\pi^3,$$

we get

$$\sigma_{\rm abs} = {(2\pi)^5\over \omega^5\pi^3} \,{\pi^2\over 256}(\omega R)^8 = {\pi^4\over 8}\,\omega^3R^8.$$

Using

$$R^4={\kappa_{10}N\over 2\pi^{5/2}},$$

this becomes

$$\sigma_{\rm abs} = {\kappa_{10}^2N^2\over32\pi}\,\omega^3.$$

This is the answer from the supergravity throat.

The brane calculation: closed string into open strings

Now compute the same quantity using the low-energy worldvolume theory on the D3-branes. The massless open-string theory is $U(N)$ $\mathcal N=4$ super-Yang—Mills. At energies small compared to the string scale, the leading coupling of the dilaton to the brane follows from the DBI action.

Schematically,

$$S_{\rm D3} = -T_3\int d^4x\,e^{-\Phi} \operatorname{Tr} \sqrt{-\det\left(\eta_{\mu\nu}+2\pi\alpha'F_{\mu\nu}+\cdots\right)}.$$

Expanding in the dilaton fluctuation and in open-string fields gives an interaction of the form

$$S_{\rm int} = \int d^4x\,\varphi(x)\,\mathcal O_\varphi(x),$$

where $\varphi$ is the canonically normalized bulk scalar restricted to the brane and

$$\mathcal O_\varphi \sim \operatorname{Tr} \left( F_{\mu\nu}F^{\mu\nu} +2D_\mu X^iD^\mu X^i +\text{fermions} +\text{interactions} \right).$$

At the lowest order in open-string perturbation theory, the incoming dilaton can create two massless open-string quanta. If one displays only the gauge-boson part, the process is

$$\varphi\longrightarrow A_\mu^a+A_\nu^b,$$

with adjoint color indices $a,b$. Summing over the $U(N)$ adjoint species gives the characteristic factor $N^2$ at large $N$.

The resulting absorption cross-section is

$$\sigma_{\rm brane} = {\kappa_{10}^2N^2\over32\pi}\,\omega^3.$$

This precisely equals the supergravity result. The equality is striking because the two computations are organized very differently:

Description	Degrees of freedom	Computation
D-brane worldvolume	open strings, low-energy $\mathcal N=4$ SYM	production of brane excitations
Supergravity throat	closed-string fields in the D3 geometry	tunneling and horizon flux
Shared answer	$N^2$ adjoint degrees of freedom	$\sigma_{\rm abs}\propto N^2\omega^3$

At this stage, one should not yet say that the full AdS/CFT dictionary has been derived. But the moral is already visible: the brane theory contains precisely the degrees of freedom needed to reproduce classical absorption by the geometry.

Absorption as an optical-theorem statement

The previous calculation can be made more conceptual. If a bulk field $\phi_0$ couples to a worldvolume operator $\mathcal O$ as

$$S_{\rm int} = \int d^4x\,\phi_0(x)\mathcal O(x),$$

then the absorption probability is controlled by the spectral density of $\mathcal O$. Define the retarded Green function

$$G_R(p) = -i\int d^4x\,e^{ip\cdot x}\theta(t) \left\langle [\mathcal O(x),\mathcal O(0)] \right\rangle.$$

The spectral density is

$$\rho_{\mathcal O}(p) = -2\,\operatorname{Im}G_R(p).$$

For a homogeneous incoming wave with spatial momentum $\vec p=0$, the optical theorem gives the schematic relation

$$\sigma_{\rm abs}(\omega) = {1\over 2\omega}\rho_{\mathcal O}(\omega,\vec0),$$

with the normalization of $\mathcal O$ fixed by the coupling above. Equivalently, in terms of the Euclidean two-point function $\Pi(p)=\langle\mathcal O(p)\mathcal O(-p)\rangle$, absorption is the discontinuity across the Lorentzian branch cut:

$$\sigma_{\rm abs}(\omega) = {1\over 2i\omega} \operatorname{Disc}\Pi(p)\bigg|_{p^2=\omega^2+i0}.$$

The absorbed flux is the imaginary part of the forward amplitude. In the brane theory this is the discontinuity of the two-point function of the operator sourced by the bulk field.

This explains the power of $\omega$ without solving the wave equation. In four-dimensional conformal theory, an operator of dimension $\Delta$ has

$$\langle \mathcal O(x)\mathcal O(0)\rangle \sim {1\over |x|^{2\Delta}}.$$

For the dilaton operator on the D3-brane worldvolume, $\Delta=4$, so

$$\langle \mathcal O_\varphi(x)\mathcal O_\varphi(0)\rangle \sim {N^2\over |x|^8}.$$

The Fourier transform in four dimensions behaves as

$$\Pi(p) \sim N^2p^4\log(-p^2),$$

up to contact terms. The discontinuity of $\log(-p^2)$ at timelike momentum is a constant, so

$$\operatorname{Disc}\Pi(p)\big|_{p^2=\omega^2} \sim N^2\omega^4.$$

Dividing by the incoming flux factor $2\omega$ gives

$$\sigma_{\rm abs}\sim N^2\omega^3,$$

exactly as in the D3-brane result.

The operator dictionary visible from absorption is:

Bulk fluctuation	Worldvolume operator
dilaton $\varphi$	$\operatorname{Tr}F_{\mu\nu}F^{\mu\nu}+\cdots$
R—R axion $C_0$	$\operatorname{Tr}F_{\mu\nu}\widetilde F^{\mu\nu}+\cdots$
graviton $h_{\mu\nu}$ along the brane	stress tensor $T_{\mu\nu}$
transverse metric fluctuations	scalar bilinears and higher Kaluza—Klein descendants

The ellipses are important. The full $\mathcal N=4$ operators are supersymmetric completions, not just the displayed bosonic terms. For example, the dilaton couples to the Lagrangian density of the worldvolume theory, while the graviton couples to the conserved stress tensor. Conservation and supersymmetry strongly constrain their two-point functions.

The $h_{xy}$ example

A particularly transparent example is a graviton polarized along two brane directions,

$$h_{xy}(t,r), \qquad x,y\in\{1,2,3\}.$$

From the brane point of view, it couples to

$$T_{xy}.$$

Thus

$$S_{\rm int} = {1\over2}\int d^4x\,h_{xy}(x)T^{xy}(x).$$

The absorption cross-section is controlled by

$$\langle T_{xy}(\omega)T_{xy}(-\omega)\rangle.$$

In a four-dimensional CFT, the stress-tensor two-point function is fixed up to the central charge. For $\mathcal N=4$ $SU(N)$ SYM at large $N$,

$$C_T\propto N^2.$$

Therefore the stress-tensor spectral density has the same large-$N$ scaling as the area of the D3-brane throat. This is a small but very clean preview of a major theme: the gravitational coupling is measuring the number of field-theory degrees of freedom.

The near-horizon viewpoint

The D3-brane geometry contains two regions. At large $r$ it is asymptotically flat. At small $r$,

$$H(r)\simeq {R^4\over r^4},$$

and the metric becomes

$$ds^2 = {r^2\over R^2}\eta_{\mu\nu}dx^\mu dx^\nu + {R^2\over r^2}dr^2 + R^2d\Omega_5^2.$$

Introducing

$$z={R^2\over r},$$

this is

$$ds^2 = {R^2\over z^2} \left( dz^2+\eta_{\mu\nu}dx^\mu dx^\nu \right) + R^2d\Omega_5^2.$$

That is $AdS_5\times S^5$. The minimally coupled scalar action in the near-horizon region reduces to

$$S = {1\over 2\kappa_{10}^2} \int d^{10}x\,\sqrt{-g}\, {1\over2}g^{MN}\partial_M\phi\,\partial_N\phi$$

or, after integrating over the unit $S^5$,

$$S = {\pi^3R^8\over4\kappa_{10}^2} \int d^4x\int {dz\over z^3} \left[ (\partial_z\phi)^2 + \eta^{\mu\nu}\partial_\mu\phi\,\partial_\nu\phi \right].$$

One must regulate the integral near the AdS boundary, say $z>\epsilon$. The boundary value of $\phi$ acts as a source for the operator $\mathcal O_\varphi$ in the brane theory. This is the beginning of the GKPW prescription, which we will develop next.

The lesson of the absorption calculation is therefore sharper than a mere agreement of numbers. A bulk wave entering the D3-brane throat is encoded by the spectral density of a brane operator. Geometry is already being translated into correlation functions.

Exercises

Exercise 1: Derive the scalar radial equation

Starting from the extremal D3-brane metric,

$$ds^2 = H^{-1/2}\eta_{\mu\nu}dx^\mu dx^\nu + H^{1/2}\left(dr^2+r^2d\Omega_5^2\right),$$

derive

$${1\over r^5}{d\over dr}\left(r^5{d\phi\over dr}\right)+\omega^2H(r)\phi=0$$

for an $s$-wave scalar $\phi(t,r)=e^{-i\omega t}\phi(r)$.

Solution

The scalar equation is

$${1\over\sqrt{-g}}\partial_M\left(\sqrt{-g}\,g^{MN}\partial_N\phi\right)=0.$$

For the D3-brane metric,

$$g^{rr}=H^{-1/2}, \qquad g^{tt}=-H^{1/2}.$$

The determinant factor is

$$\sqrt{-g}=H^{1/2}r^5\sqrt{g_{S^5}},$$

so

$$\sqrt{-g}\,g^{rr}=r^5\sqrt{g_{S^5}}.$$

For an $S^5$-independent scalar with no brane momentum,

$${1\over H^{1/2}r^5}{d\over dr}\left(r^5{d\phi\over dr}\right) -H^{1/2}\partial_t^2\phi=0.$$

Using $\partial_t^2\phi=-\omega^2\phi$ and multiplying by $H^{1/2}$ gives

$${1\over r^5}{d\over dr}\left(r^5{d\phi\over dr}\right)+\omega^2H(r)\phi=0.$$

Exercise 2: Put the radial equation in Schrödinger form

Set $\rho=\omega r$ and $\phi(\rho)=\rho^{-5/2}\psi(\rho)$. Show that

$$\left[ -{d^2\over d\rho^2} +{15\over4\rho^2} -1 -{(\omega R)^4\over\rho^4} \right]\psi=0.$$

Solution

The dimensionless equation is

$$\phi''+{5\over\rho}\phi' +\left(1+{\epsilon^4\over\rho^4}\right)\phi=0, \qquad \epsilon=\omega R.$$

Let $\phi=\rho^{-5/2}\psi$. Then

$$\phi' = -{5\over2}\rho^{-7/2}\psi+\rho^{-5/2}\psi',$$

and

$$\phi'' = {35\over4}\rho^{-9/2}\psi -5\rho^{-7/2}\psi' +\rho^{-5/2}\psi''.$$

Therefore

$$\phi''+{5\over\rho}\phi' = \rho^{-5/2}\psi'' -{15\over4}\rho^{-9/2}\psi.$$

Multiplying the radial equation by $\rho^{5/2}$ gives

$$\psi'' + \left( 1+{\epsilon^4\over\rho^4} -{15\over4\rho^2} \right)\psi=0.$$

Multiplying by $-1$ yields

$$\left[ -{d^2\over d\rho^2} +{15\over4\rho^2} -1 -{\epsilon^4\over\rho^4} \right]\psi=0.$$

Exercise 3: Convert absorption probability to cross-section

Use

$$\mathcal P_0={\pi^2\over256}(\omega R)^8, \qquad \operatorname{Vol}(S^5)=\pi^3,$$

and the six-dimensional $s$-wave formula

$$\sigma_0={(2\pi)^5\over \omega^5\operatorname{Vol}(S^5)}\mathcal P_0$$

to derive

$$\sigma_0={\pi^4\over8}\omega^3R^8.$$

Solution

Substitute the probability:

$$\sigma_0 = {(2\pi)^5\over\omega^5\pi^3} {\pi^2\over256} (\omega R)^8.$$

Since $(2\pi)^5=32\pi^5$,

$${(2\pi)^5\over\pi^3}{\pi^2\over256} = {32\pi^5\over\pi^3}{\pi^2\over256} = {\pi^4\over8}.$$

The powers of $\omega$ give

$${\omega^8\over\omega^5}=\omega^3.$$

Thus

$$\sigma_0={\pi^4\over8}\omega^3R^8.$$

Exercise 4: Match the supergravity and brane answers

Given

$$R^4={\kappa_{10}N\over2\pi^{5/2}},$$

show that

$${\pi^4\over8}\omega^3R^8 = {\kappa_{10}^2N^2\over32\pi}\omega^3.$$

Solution

Square the relation for $R^4$:

$$R^8 = {\kappa_{10}^2N^2\over4\pi^5}.$$

Then

$${\pi^4\over8}\omega^3R^8 = {\pi^4\over8}\omega^3 {\kappa_{10}^2N^2\over4\pi^5} = {\kappa_{10}^2N^2\over32\pi}\omega^3.$$

This is the same normalization obtained from the leading D3-brane worldvolume absorption calculation.

Exercise 5: Explain the $\omega^3$ scaling from conformal invariance

Assume a four-dimensional CFT operator $\mathcal O$ has dimension $\Delta=4$ and

$$\langle\mathcal O(x)\mathcal O(0)\rangle\sim {C\over x^8}.$$

Explain why absorption through this operator scales as $\sigma_{\rm abs}\sim C\omega^3$.

Solution

The Fourier transform of a power-law two-point function in four dimensions obeys

$$\int d^4x\,e^{ip\cdot x}{1\over (x^2)^\Delta} \propto (p^2)^{\Delta-2}$$

away from integer singularities. For $\Delta=4$, the transform has the logarithmic form

$$\Pi(p)\sim C\,p^4\log(-p^2)$$

up to contact terms. The discontinuity across the timelike branch cut is proportional to $C\,p^4$. Setting $\vec p=0$ gives $p^2=\omega^2$, hence

$$\operatorname{Disc}\Pi\sim C\omega^4.$$

The optical theorem divides by the incoming flux factor $2\omega$, so

$$\sigma_{\rm abs}\sim C\omega^3.$$

For the D3-brane theory, $C\sim N^2$, giving $\sigma_{\rm abs}\sim N^2\omega^3$.

Exercise 6: Identify the operator sourced by a graviton fluctuation

A graviton polarized along the brane directions, $h_{xy}$, is turned on as a weak external source. Which operator does it couple to, and why does its two-point function scale as $N^2$?

Solution

A metric perturbation couples universally to the stress tensor:

$$S_{\rm int} = {1\over2}\int d^4x\,h_{\mu\nu}T^{\mu\nu}.$$

Therefore $h_{xy}$ sources $T_{xy}$. In a four-dimensional CFT, the stress-tensor two-point function is fixed up to the central charge coefficient $C_T$:

$$\langle T_{\mu\nu}(x)T_{\rho\sigma}(0)\rangle = {C_T\over x^8} \times \text{fixed tensor structure}.$$

For $\mathcal N=4$ $SU(N)$ SYM, the number of adjoint degrees of freedom is of order $N^2$, and correspondingly

$$C_T\propto N^2.$$

Thus the absorption cross-section for a graviton polarized along the brane has the same large-$N$ scaling as the dilaton result.

Exercise 7: Why is the matching region essential?

The inner and outer solutions are both approximate. Explain why the existence of the region

$$(\omega R)^2\ll \rho\ll1$$

is the reason the low-energy absorption probability can be computed analytically.

Solution

The outer solution is valid when the throat term is small:

$${(\omega R)^4\over\rho^4}\ll1,$$

which means

$$\rho\gg(\omega R)^2.$$

The inner solution is valid in the low-energy near-throat approximation, which requires $\rho\ll1$. Both approximations are simultaneously valid when

$$(\omega R)^2\ll\rho\ll1.$$

This interval exists only if

$$\omega R\ll1.$$

In that overlap region, both solutions reduce to elementary power series in $\rho$. Matching those series fixes the ratio between the ingoing horizon flux and the incoming flux at infinity. That ratio is the absorption probability.