Absorption by D-Branes and Two-Point Functions

A dilaton incident on a stack of D3-branes can transfer its energy into excitations living on the branes. The same process has a gravitational description: part of the incident wave travels down the throat of the D3-brane geometry. Comparing these descriptions gives a quantitative test of the connection between gauge theory and gravity.

We will calculate the leading low-energy cross-section in both descriptions, including its numerical coefficient:

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

Here $\omega$ is the incident frequency, $R$ is the throat radius, and $\kappa_{10}$ is the physical ten-dimensional gravitational coupling. The gravitational calculation determines a transmitted flux; the worldvolume calculation counts the states into which the incident quantum can decay. Their relation to a two-point function will then explain why this comparison works across different coupling regimes.

This page assumes familiarity with the D3-brane solution, the quadratic expansion of the Dirac–Born–Infeld action, and relativistic phase space. We use the volume conventions, including its dilaton normalization, and supply the Bessel-function limits needed below. The low-energy limit here is $\omega R\ll1$, for an extremal brane at zero temperature and an incident wave with no momentum along the brane.

A dilaton wave in the D3-brane geometry

Classical supergravity requires $\lambda\gg1$ to suppress curvature corrections and $g_s\ll1$ to suppress string loops. The free worldvolume calculation instead applies at $\lambda\ll1$. Keeping only massless string excitations also requires $\omega\sqrt{\alpha'}\ll1$; at weak coupling this does not follow from $\omega R\ll1$ alone. The two calculations therefore have different perturbative regimes. Their agreement will require a protected observable; low incident energy alone does not make the two coupling expansions simultaneously valid.

For worldvolume indices $a,b=0,1,2,3$, the extremal Einstein-frame metric is

$$\begin{aligned} ds^2&=H^{-1/2}\eta_{ab}dx^a dx^b +H^{1/2}\bigl(dr^2+r^2d\Omega_5^2\bigr),\\ H(r)&=1+\frac{R^4}{r^4}. \end{aligned}$$

The background dilaton is constant. The fluctuation has the quadratic action

$$S_\phi^{(2)}=-\frac1{4\kappa_{10}^2}\int d^{10}x\sqrt{-g}\, g^{MN}\partial_M\phi\,\partial_N\phi.$$

Varying it gives $\nabla^2\phi=0$: the linear dilaton fluctuation is a minimally coupled scalar in this background. We take the $S^5$ s-wave, with no worldvolume spatial momentum:

$$\phi(t,r)=e^{-i\omega t}\phi(r), \qquad \omega>0.$$

The determinant and inverse metric give

$$\sqrt{-g}=H^{1/2}r^5\sqrt{\gamma_5}, \qquad \sqrt{-g}\,g^{rr}=r^5\sqrt{\gamma_5}, \qquad g^{tt}=-H^{1/2},$$

where $\gamma_5$ is the determinant of the unit-sphere metric. Multiplying the wave equation by $H^{1/2}$ therefore gives

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

Introduce

$$\rho=\omega r, \qquad \epsilon=\omega R\ll1.$$

The dimensionless radial equation is

$$\phi''+\frac5\rho\phi' +\left(1+\frac{\epsilon^4}{\rho^4}\right)\phi=0,$$

where primes now mean derivatives with respect to $\rho$. Removing the first derivative with $\psi=\rho^{5/2}\phi$ gives a useful scattering form:

$$-\psi''+V_{\rm eff}(\rho)\psi=\psi, \qquad V_{\rm eff}(\rho)=\frac{15}{4\rho^2}-\frac{\epsilon^4}{\rho^4}.$$

The effective energy is $1$. The positive term creates a barrier, while the negative term opens a propagation region deep in the throat. Even the s-wave has this barrier: the $15/(4\rho^2)$ term comes from the radial measure in six transverse dimensions, rather than from nonzero angular momentum.

The radial equation has effective energy $E=1$ and potential $V_{\rm eff}=15/(4\rho^2)-\epsilon^4/\rho^4$. The example uses $\epsilon=0.4$ to display the barrier; the analytic calculation below takes $\epsilon\ll1$. Increasing $\rho$ leads toward the asymptotically flat region, while decreasing $\rho$ leads into the throat.

Matching the throat wave to the incident wave

The full radial equation is not a Bessel equation. At small $\epsilon$, however, two Bessel approximations have a common domain. Matching their amplitudes in that domain determines how much of a wave sent from infinity reaches the throat. This method follows the original D3-brane absorption calculation, section 2.

The inner solution and its boundary condition

Use the inverted radial coordinate

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

The exact equation becomes

$$\frac{d^2\phi}{dz^2}-\frac3z\frac{d\phi}{dz} +\left(1+\frac{\epsilon^4}{z^4}\right)\phi=0.$$

Setting $\phi=z^{3/2}f$ gives

$$f''+\left(1-\frac{15}{4z^2}+\frac{\epsilon^4}{z^4}\right)f=0.$$

For $z\gg\epsilon^2$, the last term is small compared with the retained inverse-square term when $z\ll1$, and small compared with $1$ when $z\gtrsim1$. Thus it can be dropped in the leading inner approximation. In the original coordinate this domain is $\rho\ll1$. The resulting solutions are $z^2J_2(z)$ and $z^2Y_2(z)$.

At the extremal horizon $r\to0$, impose an ingoing wave: energy travels down the throat, with no wave emerging from it. To fix the Hankel-function choice, the near-horizon tortoise coordinate satisfies

$$r_*\simeq-\frac{R^2}{r}=-\frac z\omega.$$

An ingoing mode has phase $e^{-i\omega(t+r_*)}=e^{-i\omega t+iz}$. Since $H_2^{(1)}(z)$ behaves as $e^{iz}/\sqrt z$ at large $z$, a convenient normalization is

$$\phi_{\rm in}=i z^2H_2^{(1)}(z).$$

The prefactor $i$ is a freely chosen overall phase. It makes the small-$z$ limit real:

$$J_2(z)=\frac{z^2}{8}+O(z^4), \qquad Y_2(z)=-\frac4{\pi z^2}-\frac1\pi+O(z^2\log z),$$

and consequently $\phi_{\rm in}\to4/\pi$ for $z\to0$ within the inner approximation.

The outer solution and the overlap

In the equation for $\psi$, the term $\epsilon^4/\rho^4$ is negligible for $\rho\gg\epsilon^2$: compare it with $1/\rho^2$ at small $\rho$, and with $1$ at large $\rho$. The leading outer solution is therefore

$$\phi_{\rm out}=\rho^{-2}\bigl[A J_2(\rho)+B Y_2(\rho)\bigr].$$

The two approximations overlap when

$$\epsilon^2\ll\rho\ll1.$$

Both $\rho$ and $z=\epsilon^2/\rho$ are then small. Notice that this overlap is broader than a comparison of $1$ with $\epsilon^4/\rho^4$ would suggest. Such a comparison would give the incompatible conditions $\rho\ll\epsilon$ and $\rho\gg\epsilon$. The inverse-square term in the transformed equations is what permits matching.

For $\epsilon\ll1$, the inner approximation applies at $\rho\ll1$ and the outer approximation at $\rho\gg\epsilon^2$. In their common domain, both Bessel arguments are small. The logarithmic radial arrangement is schematic; the inequalities denote parametrically separated scales.

In the overlap, $\rho^{-2}J_2(\rho)\to1/8$, whereas $\rho^{-2}Y_2(\rho)$ contains a $\rho^{-4}$ term. Matching the leading constant gives

$$\frac A8=\frac4\pi, \qquad A=\frac{32}{\pi}$$

at leading order. Thus

$$\phi_{\rm out}\simeq\frac{32}{\pi}\,\rho^{-2}J_2(\rho).$$

This leading approximation sets $B$ to zero only for the purpose of finding the leading incident amplitude. The exact solution has a small imbalance between its incoming and outgoing components. A pure real $J_2$ wave has zero net flux, so its leading form alone cannot resolve that imbalance. We instead evaluate the transmitted flux directly from the ingoing throat solution and divide by the incident part of the outer solution.

The flux ratio

The radial equation implies conservation of

$$\mathcal F_\rho =\frac{\rho^5}{2i}\bigl(\phi^*\phi'-\phi\phi'^*\bigr) =\rho^5\operatorname{Im}(\phi^*\phi').$$

A positive sign means motion toward increasing $\rho$. Overall factors from the sphere volume, the bulk action, and converting $r$ to $\rho$ cancel in a flux ratio.

The Bessel Wronskian is

$$J_2(x)Y_2'(x)-Y_2(x)J_2'(x)=\frac2{\pi x}.$$

For $\phi_{\rm in}=iz^2H_2^{(1)}(z)$ it gives

$$\operatorname{Im}\left(\phi_{\rm in}^*\partial_z\phi_{\rm in}\right) =\frac{2z^3}{\pi}.$$

Using $\partial_\rho=-(z/\rho)\partial_z$, the transmitted flux is

$$\mathcal F_{\rm throat} =-\frac2\pi\rho^4z^4 =-\frac{2\epsilon^8}{\pi}.$$

At infinity,

$$J_2(\rho)=\frac12\left[H_2^{(1)}(\rho)+H_2^{(2)}(\rho)\right].$$

With time dependence $e^{-i\omega t}$, $H^{(2)}$ is the incoming component there. Its coefficient is $A/2$, so the same Wronskian gives

$$\mathcal F_{\rm incident}=-\frac{|A|^2}{2\pi}.$$

The probability of absorption is therefore

$$\mathcal P =\frac{|\mathcal F_{\rm throat}|}{|\mathcal F_{\rm incident}|} =\frac{\pi^2}{256}\,\epsilon^8\,[1+o(1)].$$

Here $o(1)$ tends to zero as $\epsilon\to0$. Leading matching does not determine the next term. In fact, improving the matching produces logarithmic corrections, so an unqualified power-series remainder would be misleading; see Gubser, Hashimoto, Klebanov, and Krasnitz, section 3.

From a partial-wave probability to a cross-section

For an s-wave in $d$ transverse spatial dimensions, plane-wave normalization gives

$$\sigma_0=\frac{(2\pi)^{d-1}}{\omega^{d-1}\Omega_{d-1}}\,\mathcal P,$$

where $\Omega_{d-1}$ is the area of the unit sphere. One way to obtain this factor is to project the unit-amplitude plane wave onto its angular average:

$$\frac1{\Omega_{d-1}}\int d\Omega_{d-1}\,e^{i\rho\cos\theta} =\frac{(2\pi)^{d/2}}{\Omega_{d-1}}\, \rho^{1-d/2}J_{d/2-1}(\rho).$$

Splitting this Bessel function into incoming and outgoing waves gives the incident s-wave flux relative to the original plane-wave flux. For $d=6$, $\Omega_5=\pi^3$, and substitution yields

$$\begin{aligned} \sigma_{\rm abs} &=\frac{(2\pi)^5}{\pi^3\omega^5}\, \frac{\pi^2(\omega R)^8}{256}\,[1+o(1)]\\ &=\frac{\pi^4}{8}\,\omega^3R^8\,[1+o(1)]. \end{aligned}$$

The result has dimension length to the fifth power, appropriate to scattering in six transverse spatial dimensions. For the infinite brane it is understood per unit three-dimensional worldvolume. Expressing $R$ through $N$ gives the coefficient stated at the beginning of the page.

The worldvolume calculation with normalized states

We now calculate the same coefficient using the D3-brane fields at weak coupling. The leading process converts one incident dilaton into two massless gauge bosons with opposite momenta along the brane. Both the field normalization and the number of physical polarizations matter.

Why the leading dilaton vertex contains gauge fields

For one brane, with the other bulk fields set to zero, start from the string-frame action

$$S_{\rm DBI} =-\mu_3\int d^4x\,e^{-\Phi} \sqrt{-\det\bigl(P[G^{\rm S}]+2\pi\alpha' F\bigr)}.$$

Here $P$ denotes pullback to the worldvolume. Since $G^{\rm S}=e^{\phi/2}g$, factoring $e^{\phi/2}$ from the four-dimensional matrix supplies $e^\phi$ from the square root of its determinant. This cancels $e^{-\phi}$ outside it. The Einstein-frame action is thus

$$S_{\rm DBI} =-T_3\int d^4x\, \sqrt{-\det\bigl(P[g]+2\pi\alpha'e^{-\phi/2}F\bigr)}.$$

For a flat brane with transverse displacement fields $Y^i$,

$$\begin{aligned} S_{\rm DBI}=-T_3\int d^4x\,\Bigl[ 1&+\frac12\partial_aY^i\partial^aY^i\\ &+\frac{(2\pi\alpha')^2}{4}e^{-\phi}F_{ab}F^{ab} +\cdots\Bigr]. \end{aligned}$$

The scalar kinetic term has no direct linear dilaton coupling in this frame. Adding a term $\phi(\partial Y)^2$ by multiplying the entire quadratic action by $e^{-\phi}$ would count a coupling that the frame transformation has canceled. At the same leading on-shell order there is no dilaton-to-two-fermion contribution. The explicit string and worldvolume amplitudes are given in Gubser, Klebanov, and Tseytlin, section 3.

For $N$ branes, canonically normalize the $N^2$ real adjoint gauge fields. Denote their linearized field strengths by $f^A_{ab}$, with $A=1,\ldots,N^2$. Their quadratic action and the relevant interaction become

$$\begin{aligned} \mathcal L_{\rm gauge}&=-\frac14\sum_A f^A_{ab}f^{Aab},\\ \mathcal L_{\rm int}&=\frac{\sqrt2\,\kappa_{10}}4\, \varphi\sum_A f^A_{ab}f^{Aab}. \end{aligned}$$

The canonical component basis removes trace-normalization ambiguity from the amplitude below. The positive sign of the interaction follows by expanding $-e^{-\phi}f^2/4$. Nonabelian interactions are omitted only in this leading weak-coupling calculation.

Polarizations and colors

The incident particle has worldvolume momentum $q^a=(\omega,\mathbf0)$. Let the final massless momenta be $p_1$ and $p_2$, and their transverse polarization vectors be $e_1$ and $e_2$. Differentiating the interaction with respect to the two identical gauge fields gives, up to an overall phase,

$$\begin{aligned} \mathcal M^{AB} =\sqrt2\,\kappa_{10}\delta^{AB}\Bigl[ &(p_1\cdot p_2)(e_1\cdot e_2)\\ &-(p_1\cdot e_2)(p_2\cdot e_1)\Bigr]. \end{aligned}$$

This expression vanishes when either polarization is replaced by its momentum, as required by gauge invariance. In the center-of-mass frame,

$$p_1^a=\left(\frac\omega2,\mathbf k\right), \quad p_2^a=\left(\frac\omega2,-\mathbf k\right), \quad |\mathbf k|=\frac\omega2, \quad p_1\cdot p_2=-\frac{\omega^2}{2}.$$

Choose two real spatial polarizations perpendicular to $\mathbf k$. The mixed momentum-polarization contractions vanish. Of the four polarization pairs, the two parallel pairs each have $|\mathcal M|^2=\kappa_{10}^2\omega^4/2$ for an allowed color. The two perpendicular pairs give zero. Summing over polarizations and colors, with no average over the scalar initial state, gives

$$\sum_{A,B,\,\mathrm{pol}}|\mathcal M^{AB}|^2 =N^2\kappa_{10}^2\omega^4.$$

The factor $N^2$ follows from $\sum_{A,B}\delta^{AB}\delta^{AB}=N^2$, rather than from $N^4$ independent color pairs.

Phase space and the identical-particle factor

The integrated phase space for two massless particles with total energy $\omega$ is

$$\begin{aligned} \int d\Pi_2 &=\int\frac{d^3p_1}{(2\pi)^3\,2E_1} \frac{d^3p_2}{(2\pi)^3\,2E_2} (2\pi)^4\delta^{(4)}(q-p_1-p_2)\\ &=\frac1{4(2\pi)^2}\int d\Omega\int_0^\infty dk\, \delta(\omega-2k) =\frac1{8\pi}. \end{aligned}$$

The momentum delta function first sets $\mathbf p_2=-\mathbf p_1$; the two energy denominators cancel the $k^2$ in the radial measure. The remaining energy delta function supplies $1/2$.

For a canonically normalized incident bulk quantum, dividing the transition rate per unit brane volume by its incident number flux supplies $1/(2\omega)$. The final particles are identical bosons, so the phase-space sum also contains $1/2!$. Consequently,

$$\begin{aligned} \sigma_{\rm abs} &=\frac1{2\omega}\,\frac1{2!} \int d\Pi_2\sum_{A,B,\,\mathrm{pol}}|\mathcal M^{AB}|^2\\ &=\frac1{2\omega}\,\frac12\,\frac1{8\pi}\, N^2\kappa_{10}^2\omega^4\\ &=\frac{\kappa_{10}^2N^2}{32\pi}\,\omega^3. \end{aligned}$$

Every numerical factor now has an origin: canonical normalization fixes the vertex, the two physical polarizations fix its squared sum, the adjoint index gives $N^2$, and the phase space and identical-particle factor fix the denominator. This agrees with the gravitational result after using $R^8=\kappa_{10}^2N^2/(4\pi^5)$.

Absorption measures a spectral function

The inclusive absorption rate can be calculated without resolving each final state individually. Define the operator sourced by the dimensionless dilaton through

$$\mathcal O_\phi(x) =\left.\frac{\delta S_{\rm D3}}{\delta\phi(x)}\right|_{\phi=0}, \qquad S_{\rm int}=\sqrt2\,\kappa_{10}\int d^4x\, \varphi(x)\mathcal O_\phi(x).$$

In the free on-shell two-particle calculation its relevant term is

$$\mathcal O_\phi=\frac14\sum_A f^A_{ab}f^{Aab}.$$

At finite coupling, the source variation defines the full supersymmetric operator, including interaction terms. The quadratic expression alone does not define a complete interacting composite operator. Terms proportional to equations of motion and local contact terms must also be distinguished from the part that creates physical states.

Using the shared Fourier and retarded-response conventions, insert a complete set of states into the spectral density. At positive frequency in the vacuum this gives

$$\rho_{\mathcal O}(\omega,\mathbf q) =\sum_f(2\pi)^4\delta^{(4)}(q-p_f) \bigl|\langle f|\mathcal O(0)|0\rangle\bigr|^2.$$

The sum includes the appropriate phase-space measures and identical-particle factors. It is nonnegative. With a coupling $g\varphi\mathcal O$, the same state sum appears in the absorption rate, so

$$\sigma_{\rm abs}(\omega) =\frac{g^2}{2\omega}\rho_{\mathcal O}(\omega,\mathbf0).$$

For the D3 dilaton, $g=\sqrt2\kappa_{10}$. The pair-production calculation therefore determines

$$\rho_{\mathcal O_\phi}(\omega,\mathbf0) =\frac{N^2}{32\pi}\omega^4, \qquad \omega>0.$$

This is the optical theorem in a form useful for a brane: the imaginary part of a forward response equals the sum over states that can absorb the incident energy.

The timelike cut and the operator dimension

For the momentum invariant in this calculation, write

$$p^2=-\omega^2+\mathbf q^2, \qquad s=-p^2=\omega^2-\mathbf q^2.$$

Physical absorption occurs on the timelike cut $s>0$, or $p^2<0$. A dimension-four scalar operator in four dimensions has a separated-point correlator proportional to $1/(x^2)^4$. Its momentum-space nonlocal part scales as $(p^2)^2\log p^2$, rather than as a pure polynomial. With the shared retarded and source conventions, the normalized answer can be written

$$G_R(\omega,\mathbf q) =\frac{N^2}{64\pi^2}(p^2)^2 \log\!\left[\frac{\mathbf q^2-(\omega+i0)^2}{\mu^2}\right] +\text{local terms}.$$

For $\omega>0$ and $s>0$, the logarithm is $\log(s/\mu^2)-i\pi$. Therefore $-2\operatorname{Im}G_R=N^2s^2/(32\pi)>0$, reproducing the state sum at $\mathbf q=0$. Changing the subtraction scale $\mu$ adds a local polynomial and leaves the finite-frequency absorption unchanged. This also explains the power law: a dimension-four operator gives $\rho\sim\omega^4$, and the incident-flux factor converts it to $\sigma\sim\omega^3$.

What is protected

Conformal symmetry fixes the power of frequency once the operator dimension is known. It does not, by itself, fix the coefficient of a two-point function. In this example the dilaton operator is a descendant in the protected $\mathcal N=4$ stress-tensor multiplet. With its normalization fixed by the bulk source, supersymmetric nonrenormalization relates its two-point coefficient to the protected stress-tensor normalization. This is the additional reason the leading weak-coupling coefficient agrees with the supergravity value. The relation between absorption and protected correlators is discussed in Gubser and Klebanov and in Klebanov’s review, sections 1–2.

The weak-coupling calculation above includes the $U(N)$ adjoint, of dimension $N^2$. The interacting $SU(N)$ sector has $N^2-1$ generators; the additional free $U(1)$ is the center-of-mass multiplet. Classical throat supergravity resolves the leading large-$N$ answer. Inferring an exact finite-$N$ geometric result would require the corresponding quantum corrections and careful treatment of the decoupled sector.

Protection of this vacuum two-point coefficient does not make generic correlators, finite-temperature observables, or higher-frequency corrections equal to their free-field values. In particular, the logarithmic corrections to low-energy absorption probe information beyond the leading conformal result.

From the throat to a boundary source in AdS

The absorption calculation uses the entire asymptotically flat geometry. Its inner region already contains the spacetime that appears in the AdS/CFT correspondence. At $r\ll R$, drop the $1$ in $H$ and introduce a length-valued coordinate $u=R^2/r$. Then

$$ds^2=\frac{R^2}{u^2} \left(\eta_{ab}dx^a dx^b+du^2\right) +R^2d\Omega_5^2.$$

This is $AdS_5\times S^5$ with common radius $R$. The earlier Bessel coordinate is $z=\omega u$. The Poincaré horizon lies at $u\to\infty$. In the isolated near-horizon geometry, the conformal boundary is $u\to0$; obtaining that complete geometry requires the decoupling limit developed on the next page.

For an $S^5$-independent dilaton fluctuation, integrate its ten-dimensional kinetic action over the unit sphere. The determinant and inverse metric supply $R^{10}u^{-5}$ and $u^2/R^2$, respectively, while $\Omega_5=\pi^3$. Hence

$$\begin{aligned} S_\phi^{(2)} &=-\frac{\pi^3R^8}{4\kappa_{10}^2} \int d^4x\int_0^\infty\frac{du}{u^3} \left[(\partial_u\phi)^2+\eta^{ab}\partial_a\phi\partial_b\phi\right]\\ &=-\frac{N^2}{16\pi^2} \int d^4x\int_0^\infty\frac{du}{u^3} \left[(\partial_u\phi)^2+\eta^{ab}\partial_a\phi\partial_b\phi\right]. \end{aligned}$$

The overall minus sign is the Lorentzian mostly-plus scalar-action sign. Wick rotation produces the positive Euclidean action. The coefficient proportional to $N^2$ is the same normalization already detected by absorption.

The equation for a mode with boundary momentum $(\omega,\mathbf q)$ is

$$\partial_u^2\phi-\frac3u\partial_u\phi+s\phi=0.$$

At zero momentum its independent behaviors are $1$ and $u^4$. The leading boundary value supplies a dimensionless source, and the response corresponds to an operator of dimension four. For Euclidean momentum magnitude $k>0$, the solution regular at the Poincaré interior is

$$\phi(u,k)=\phi_0(k)\,\frac{k^2u^2}{2}K_2(ku),$$

normalized so that $\phi(u,k)\to\phi_0(k)$ as $u\to0$. Inserting it into the action reduces the on-shell integral to a boundary term. Regulating at $u=\delta$, subtracting the divergent local terms, and differentiating with respect to $\phi_0$ yields the nonlocal $k^4\log k^2$ dependence. In Lorentzian signature, the retarded response instead uses the ingoing condition in the interior—the same condition that selected $H_2^{(1)}$ in the absorption calculation. The precise boundary prescription and its renormalization are developed on page 32; its original formulation is given by Gubser, Klebanov, and Polyakov.

Exercises

1. Angular momentum and operator dimensions

Replace the s-wave by $e^{-i\omega t+i\mathbf q\cdot\mathbf x}\phi_\ell(r)Y_\ell(\Omega)$, where $\nabla_{S^5}^2Y_\ell=-\ell(\ell+4)Y_\ell$. Derive the radial equation in the full D3 geometry. Reduce it in the throat and identify the five-dimensional mass. Use $m^2R^2=\Delta(\Delta-4)$ to determine the physical dimension $\Delta$. Explain why the s-wave cross-section cannot simply be reused for $\ell>0$.

Solution

The worldvolume derivatives give $s=\omega^2-\mathbf q^2$, and the sphere Laplacian contributes $-\ell(\ell+4)/r^2$ after multiplying the equation by $H^{1/2}$. Therefore

$$\frac1{r^5}\partial_r(r^5\partial_r\phi_\ell) +\left[sH(r)-\frac{\ell(\ell+4)}{r^2}\right]\phi_\ell=0.$$

In the throat, put $H=R^4/r^4$ and $u=R^2/r$. Using $\partial_r=-(u^2/R^2)\partial_u$ and dividing out the common factor $u^4/R^4$ gives

$$\partial_u^2\phi_\ell-\frac3u\partial_u\phi_\ell +\left[s-\frac{\ell(\ell+4)}{u^2}\right]\phi_\ell=0.$$

This is the equation of an $AdS_5$ scalar with $m_\ell^2R^2=\ell(\ell+4)$. Substituting $\phi_\ell\sim u^\beta$ near the boundary gives $\beta(\beta-4)=\ell(\ell+4)$, whose roots are $-\ell$ and $\ell+4$. The standard source and response behaviors are thus $u^{4-\Delta}$ and $u^\Delta$, with $\Delta=\ell+4$.

The full-geometry radial barrier now contains an additional positive angular-momentum term, so its transmission probability changes. In addition, a plane wave has angular degeneracies beyond the s-wave. Microscopically, the incident angular dependence probes transverse derivatives of the bulk field and different worldvolume operators. Both ingredients must be included to obtain a higher-partial-wave cross-section.

2. Gauge invariance and a small final-state mass

Verify the Ward identity of the two-gauge-boson amplitude without choosing a polarization basis. Separately, repeat the phase-space integral for two identical particles of mass $m$, keeping the squared matrix element symbolic. Determine the threshold and the factor that replaces $1/(8\pi)$. Which part of the massless absorption argument then ceases to follow from scale invariance?

Solution

Replacing $e_1$ by $p_1$ turns the bracket in the amplitude into

$$(p_1\cdot p_2)(p_1\cdot e_2) -(p_1\cdot e_2)(p_2\cdot p_1)=0.$$

The cancellation for $e_2\to p_2$ is identical. This checks the vertex before any sum over physical polarizations.

For the separate massive phase-space problem, let $E_k=\sqrt{k^2+m^2}$. After using the spatial delta function,

$$\int d\Pi_2 =\frac1{4(2\pi)^2}\int d\Omega\int_0^\infty dk\, \frac{k^2}{E_k^2}\delta(\omega-2E_k).$$

There is support only for $\omega\geq2m$. At the root, $E_*=\omega/2$, $k_*=\sqrt{\omega^2/4-m^2}$, and the derivative of $2E_k$ is $2k_*/E_*$. Hence

$$\int d\Pi_2 =\frac{k_*}{8\pi E_*} =\frac1{8\pi}\sqrt{1-\frac{4m^2}{\omega^2}}\, \theta(\omega-2m).$$

For an angle-independent squared matrix element, including any internal-state sum, the identical-particle cross-section is

$$\sigma=\frac{\overline{|\mathcal M|^2}}{32\pi\omega} \sqrt{1-\frac{4m^2}{\omega^2}}\, \theta(\omega-2m).$$

Here the overbar denotes the angular average, which can also be used when the matrix element depends on the emission direction. This exercise changes only the kinematics; a theory of massive vector particles would also require a new interaction and polarization analysis. The mass introduces the dimensionless ratio $m/\omega$ and a threshold, so scale invariance no longer determines a single frequency power law.

3. The special role of a D3-brane

Apply the same string-to-Einstein-frame transformation to a D$p$-brane DBI action. Show that the zero-field determinant carries an overall factor $e^{(p-3)\phi/4}$, while the quadratic gauge term carries $e^{(p-7)\phi/4}$. Explain why the direct dilaton coupling to the transverse-scalar kinetic term cancels specifically at $p=3$.

Solution

There are $p+1$ rows in the worldvolume determinant. Factoring out $e^{\phi/2}$ therefore supplies $e^{(p+1)\phi/4}$ after the square root. Combining it with $e^{-\Phi}=g_s^{-1}e^{-\phi}$ gives

$$S_p=-T_p\int d^{p+1}x\,e^{(p-3)\phi/4} \sqrt{-\det\bigl(P[g]+2\pi\alpha'e^{-\phi/2}F\bigr)}.$$

The terms quadratic in a transverse displacement and in the gauge field are

$$\begin{aligned} \mathcal L_Y&=-\frac{T_p}{2}e^{(p-3)\phi/4} \partial_aY^i\partial^aY^i,\\ \mathcal L_F&=-\frac{T_p(2\pi\alpha')^2}{4} e^{(p-7)\phi/4}F_{ab}F^{ab}. \end{aligned}$$

The extra $e^{-\phi}$ in the second line comes from the two factors of $F$ in the determinant expansion. At $p=3$, the scalar exponent vanishes, whereas the gauge exponent is $-1$. At $p=7$ it is instead the direct quadratic gauge coupling that has zero exponent. These statements concern the specified Einstein-frame action at fixed fields; they do not remove other interactions with bulk fluctuations.

4. Spectral scaling and contact terms

For a scalar operator of dimension $\Delta$ in a four-dimensional conformal theory, determine the timelike spectral scaling and the resulting frequency dependence of absorption for a coupling $g\varphi\mathcal O$ with fixed $g$. Identify the logarithm required when $\Delta-2$ is a nonnegative integer. Show explicitly why adding a real polynomial in $p^2$ to the retarded correlator does not change absorption at positive timelike frequency.

Solution

The separated-point correlator scales as $(x^2)^{-\Delta}$. A four-dimensional Fourier transform consequently has mass dimension $2\Delta-4$. For a noninteger $\Delta$, its nonlocal part is proportional to $(p^2)^{\Delta-2}$, with the retarded branch specified by $\mathbf q^2-(\omega+i0)^2$.

For a scalar with a nonzero continuum spectral density, positivity and scale invariance then give, at positive frequency,

$$\rho_{\mathcal O}(\omega,\mathbf q) =C_\rho\,\theta(s)s^{\Delta-2}, \qquad C_\rho>0.$$

Thus at $\mathbf q=0$,

$$\sigma(\omega)=\frac{g^2C_\rho}{2}\,\omega^{2\Delta-5}.$$

If $n=\Delta-2$ is a nonnegative integer, a pure $(p^2)^n$ would be local and have no cut. Renormalization instead leaves a nonlocal term proportional to $(p^2)^n\log(p^2/\mu^2)$, whose discontinuity has the required $s^n$ dependence. The dimension-one free scalar is an exception to the continuum formula: its spectral weight is concentrated on the light cone, rather than at positive $s$.

A real local counterterm shifts $G_R$ by a polynomial $P(p^2)$ with real coefficients. At real, strictly timelike frequency, $\operatorname{Im}P(p^2)=0$. It therefore changes contact terms but not $-2\operatorname{Im}G_R$ or absorption. The overall $C_\rho$ remains a dynamical normalization; scale invariance alone does not fix it.

5. A controlled numerical matching check

Integrate the full radial equation for several decreasing values of $\epsilon$, imposing ingoing data at a large value of $z$. At large $\rho$, fit independent incoming and outgoing Hankel waves. Compute absorption both from the throat-to-incident flux ratio and from $1-|C_{\rm out}/C_{\rm in}|^2$. Test convergence to $\mathcal P/\epsilon^8=\pi^2/256$, varying both integration endpoints. Explain why subtracting nearly equal reflected and incident probabilities loses accuracy when $\epsilon$ is small.

Solution

A logarithmic integration coordinate $x=\log\rho$ resolves the widely separated inner and outer scales. With dots denoting $x$ derivatives, the full equation and its conserved flux become

$$\ddot\phi+4\dot\phi+ \left(e^{2x}+\epsilon^4e^{-2x}\right)\phi=0, \qquad \mathcal F=\rho^4\operatorname{Im}(\phi^*\dot\phi).$$

Choose an inner endpoint using $\rho_{\rm start}=\epsilon^2/z_{\rm start}$, where $z_{\rm start}\gg1$. The initial data from the ingoing approximate solution are

$$\begin{aligned} \phi_{\rm start}&=iz_{\rm start}^2H_2^{(1)}(z_{\rm start}),\\ \dot\phi_{\rm start} &=-i\left[2z_{\rm start}^2H_2^{(1)}(z_{\rm start}) +z_{\rm start}^3H_2^{(1)\prime}(z_{\rm start})\right]. \end{aligned}$$

At an outer endpoint $\rho_{\rm end}\gg1$, fit both $\phi$ and $\dot\phi$ to

$$\phi=C_{\rm out}\rho^{-2}H_2^{(1)}(\rho) +C_{\rm in}\rho^{-2}H_2^{(2)}(\rho).$$

The following Python implementation uses NumPy and SciPy. It returns the flux result, the reflection-subtraction result, and their underlying flux-conservation check.

import numpy as np
from scipy.integrate import solve_ivp
from scipy.special import hankel1, hankel2, h1vp, h2vp

def absorption(eps, z0=60.0, r1=40.0, tol=5e-12):
    r0 = eps**2 / z0                    # r0 and r1 denote rho
    y0 = np.array([
        1j*z0**2*hankel1(2, z0),
        -1j*(2*z0**2*hankel1(2, z0) + z0**3*h1vp(2, z0))
    ])
    def rhs(x, y):
        return [y[1], -4*y[1]
                - (np.exp(2*x) + eps**4*np.exp(-2*x))*y[0]]
    sol = solve_ivp(rhs, [np.log(r0), np.log(r1)], y0,
                    method="DOP853", rtol=tol, atol=tol)
    if not sol.success:
        raise RuntimeError(sol.message)
    phi, dotphi = sol.y[:, -1]
    waves = np.array([hankel1(2, r1), hankel2(2, r1)]) / r1**2
    dots = -2*waves + np.array([h1vp(2, r1), h2vp(2, r1)]) / r1
    cout, cin = np.linalg.solve(np.array([waves, dots]),
                                np.array([phi, dotphi]))
    f0 = r0**4*np.imag(np.conj(y0[0])*y0[1])
    f1 = r1**4*np.imag(np.conj(phi)*dotphi)
    pflux = abs(f0) / (2/np.pi * abs(cin)**2)
    preflection = 1 - abs(cout/cin)**2
    return pflux, preflection, abs(f1/f0 - 1)

for eps in [0.3, 0.2, 0.1]:
    p, preflection, error = absorption(eps)
    leading = np.pi**2 * eps**8 / 256
    print(eps, p/leading, preflection, error)

A converged calculation gives approximately

$\epsilon$	$\mathcal P/(\pi^2\epsilon^8/256)$
$0.3$	$1.0025743$
$0.2$	$1.0006160$
$0.1$	$1.0000500$

The ratios approach one without being exactly one at finite frequency. Repeat the calculation, for example, with $(z_{\rm start},\rho_{\rm end})=(40,30)$ and $(60,40)$ and tighter integration tolerances. The endpoint and tolerance changes should affect the answer much less than the finite-$\epsilon$ deviation from one. Also monitor flux conservation: reducing the integration error cannot by itself remove an inaccurate asymptotic boundary condition.

The flux relation implies

$$\mathcal P=1-\left|\frac{C_{\rm out}}{C_{\rm in}}\right|^2.$$

But $\mathcal P$ is of order $\epsilon^8$, so the second method subtracts numbers increasingly close to one. At $\epsilon=0.1$, the probability is only about $3.86\times10^{-10}$. The direct throat-to-incident flux ratio avoids that subtraction and is the more stable way to test the leading coefficient.

6. The boundary logarithm

Expand $(k^2u^2/2)K_2(ku)$ through its first logarithmic term. Insert the result into the regulated on-shell action, identify the terms that are local in boundary momentum, and isolate the nonlocal logarithm. Continue it to the retarded frequency prescription used above, checking the sign of the spectral density.

Solution

Write $F(u,k)=(k^2u^2/2)K_2(ku)$. Its expansion is

$$\begin{aligned} F(u,k)={}&1-\frac{k^2u^2}{4} -\frac{k^4u^4}{16}\log\frac{ku}{2}\\ &+\frac{3-4\gamma_{\rm E}}{64}k^4u^4 +O(k^6u^6\log ku). \end{aligned}$$

For a boundary condition imposed at $u=\delta$, use $\phi(u,k)=\phi_\delta(k)F(u,k)/F(\delta,k)$. This holds the cutoff value fixed while the bulk field is varied. Define $C=N^2/(16\pi^2)$ and $\int_k=\int d^4k/(2\pi)^4$. Integrating the Euclidean action by parts and using the equation of motion gives its lower-boundary term:

$$S_{\rm E,on\text{-}shell} =-C\int_k\phi_\delta(k)\phi_\delta(-k) \left.\frac{F'(u,k)}{u^3F(u,k)}\right|_{u=\delta}.$$

The minus sign is the orientation of the boundary at the lower endpoint of the $u$ integral. The derivative ratio is

$$\left.\frac{F'}{u^3F}\right|_{u=\delta} =-\frac{k^2}{2\delta^2} -\frac{k^4}{4}\log\frac{k\delta}{2} +c\,k^4+\cdots,$$

where the numerical constant $c$ affects only a local term. The $k^2/\delta^2$ and $k^4\log\delta$ divergences can be subtracted by local boundary counterterms. Finite polynomials in $k^2$ are also scheme dependent. The remaining nonlocal term is

$$S_{\rm E,ren} =\frac{N^2}{128\pi^2}\int_k \phi_0(k)\phi_0(-k)\,k^4\log\frac{k^2}{\mu^2} +\text{local terms}.$$

With $Z_{\rm CFT}[\phi_0]=\exp[-S_{\rm E,ren}[\phi_0]]$, two source derivatives of $\log Z$ supply both a minus sign and a factor of two from the quadratic functional:

$$G_{\rm E}(k) =-\frac{N^2}{64\pi^2}k^4\log\frac{k^2}{\mu^2} +\text{local terms}.$$

Using the shared retarded convention, the continuation is $G_R=-G_{\rm E}$ evaluated at $k^2=\mathbf q^2-(\omega+i0)^2$. Thus the retarded logarithm has positive coefficient. At $\omega>0$ and $s=\omega^2-\mathbf q^2>0$,

$$\operatorname{Im}G_R=-\frac{N^2}{64\pi}s^2, \qquad \rho_{\mathcal O_\phi}=-2\operatorname{Im}G_R =\frac{N^2}{32\pi}s^2.$$

The positive spectral density agrees with the inclusive worldvolume calculation. The sign of the nonlocal Euclidean momentum expression by itself is not a negative spectral density: analytic continuation and the retarded convention are essential.