## May 27, 2006

Previously, I wrote about AdS/CFT computations of the jet-quenching parameter. A related quantity is the energy loss of a heavy quark moving through the quark-gluon plasma. Herzog, Karch, Kovtun, Kozcaz and Yaffe have a beautiful recent paper treating this latter problem.

First, there’s the matter of introducing a heavy quark into the $𝒩=4$ system. This is accomplished by introducing a probe D7-brane which which fills the ${\text{AdS}}_{5}$ from $u={u}_{m}$ to $u=\infty$ and wraps a $u$-dependent ${S}^{3}\subset {S}^{5}$.

You might worry that a space-filling D7-brane violates some charge-conservation condition on the ${S}^{5}$, and you might wonder how the D7-brane can “end” at $u={u}_{m}$. The answer to both these questions is most easily understood at zero temperature. The ${\text{AdS}}_{5}×{S}^{5}$ metric is

(1)
${\mathrm{ds}}^{2}={L}^{2}\left(\frac{{\mathrm{du}}^{2}}{{u}^{2}}+{u}^{2}{\mathrm{dx}}_{\mu }{\mathrm{dx}}^{\mu }\right)+{L}^{2}{d\Omega }_{5}^{2}$

where $L={\lambda }^{1/4}{l}_{s}$ and the ${S}^{5}$ metric is ${d\Omega }_{5}^{2}={d\psi }^{2}+{\mathrm{cos}}^{2}\psi {\theta }^{2}+{\mathrm{sin}}^{2}\psi {d\Omega }_{3}^{2}$ The D7-brane is located at $\theta =0$, $\psi ={\mathrm{cos}}^{-1}\left({u}_{m}/u\right)$. This corresponds simply to a D7-brane, parallel to the stack of D3-branes, displaced a distance $u\mathrm{cos}\left(\psi \right)={u}_{m}$ in the transverse direction. The mass of the quark is

(2)
$m=\frac{\sqrt{\lambda }}{2\pi }{u}_{m}$

At finite temperature, $T$, we replace the ${\text{AdS}}_{5}$ geometry by AdS-Schwarzschild,

(3)
${\mathrm{ds}}^{2}={L}^{2}\left(\frac{{\mathrm{du}}^{2}}{h\left(u\right)}+h\left(u\right){\mathrm{dx}}_{\mu }{\mathrm{dx}}^{\mu }\right)+{L}^{2}{d\Omega }_{5}^{2},\phantom{\rule{1em}{0ex}}h\left(u\right)={u}^{2}\left[1-\left({u}_{h}/u{\right)}^{4}\right]$

A static heavy quark in this setup is a string which stretches from ${u}_{m}$ down to the blackhole horizon, ${u}_{h}=\pi T$. The relationship between its Lagrangian mass and ${u}_{m}$ is modified from the zero temperature expression (2) to

(4)
$\frac{\sqrt{\lambda }{u}_{m}}{2\pi m}=1+g\left(\frac{\sqrt{\lambda }{u}_{h}}{2\pi m}\right)$

where $g\left(x\right)=\frac{1}{8}{x}^{4}-\frac{5}{128}{x}^{8}+O\left({x}^{12}\right)$ Its rest-energy is ${M}_{\text{rest}}\left(T\right)=E=\frac{1}{2\pi {l}_{s}^{2}}{\int }_{{u}_{h}}^{{u}_{m}}{L}^{2}\mathrm{du}=\frac{1}{2\pi {l}_{s}^{2}}{L}^{2}\left({u}_{m}-{u}_{h}\right)=m-\Delta m\left(T\right)+mg\left(\frac{\Delta m\left(T\right)}{m}\right)$ The thermal contribution to the mass is $\Delta m\left(T\right)=\frac{\sqrt{\lambda }{u}_{h}}{2\pi m}=\frac{1}{2}\sqrt{\lambda }T$

They then go on to study moving strings, using the Nambu-Goto action in static gauge ($\sigma =u$), with an ansatz of the form $x\left(u,t\right)=x\left(u\right)+vt$ The Nambu-Goto equations of motion reduce to $\frac{\partial }{\partial u}\left(h\left(u\right){u}^{2}\frac{x\prime }{\sqrt{-\gamma }}\right)$ where $\sqrt{-\gamma }={L}^{2}\left[1-{h}^{-1}{u}^{2}{v}^{2}+h{u}^{2}{x}^{\prime 2}{\right]}^{1/2}$ The solution is $x\left(u,t\right)={x}_{0}±vF\left(u\right)+vt$ where $F\left(u\right)=\frac{1}{2{u}_{h}}\left[\pi /2-{\mathrm{tan}}^{-1}\left(u/{u}_{h}\right)-{\mathrm{tanh}}^{-1}\left(u/{u}_{h}\right)\right]$ This solution does not satisfy the standard Neumann boundary condition at the D7-brane ($u={u}_{m}$). There is energy- and momentum-flux down the string, supplied by a constant electric field on the D7-brane. At the horizon, the energy and momentum flux along the string are, respectively $ℰ{\mid }_{u={u}_{h}}=\frac{\pi }{2}\sqrt{\lambda }{T}^{2}\frac{{v}^{2}}{\sqrt{1-{v}^{2}}}$ and

(5)
$𝒫{\mid }_{u={u}_{h}}=-\frac{\pi }{2}\sqrt{\lambda }{T}^{2}\frac{v}{\sqrt{1-{v}^{2}}}$

Treating the heavy quark as having effective mass, ${M}_{\text{kin}}$, and momentum $p=\frac{{M}_{\text{kin}}v}{\sqrt{1-{v}^{2}}}$, the steady-state motion, under the influence of viscous drag and an external driving force, $f$, $0=\frac{dp}{dt}=f-\mu p$ Comparing with (5), we learn

(6)
$\mu {M}_{\text{kin}}=\frac{\pi }{2}\sqrt{\lambda }{T}^{2}$

independent of the Lagrangian mass, $m$. The “flavour diffusion constant” is $D=\frac{T}{\mu {M}_{\text{kin}}}=\frac{2}{\pi \sqrt{\lambda }T}$

To extract the viscous damping coefficient, $\mu$, itself, one needs to go on and study the damping of small fluctuations about the long, straight static string. The linearized Nambu-Goto equation of motion is $\frac{\partial }{\partial u}\left(h{u}^{2}x\prime \right)=\frac{{u}^{2}}{h}\stackrel{¨}{x}$ Assuming a time-dependence of the form ${e}^{-\mu t}$ reduces this to an ODE. Demanding a purely outgoing solution at the horizon, $u={u}_{h}$, and Neumann boundary conditions at the D7-brane, $u={u}_{m}$, yields a discrete eigenvalue problem for the quasinormal mode frequencies. In the limit of large quark mass, one finds $\mu =\frac{\pi \sqrt{\lambda }{T}^{2}}{2m}$ which, unsurprisingly, agrees with the previous result (6), since ${M}_{\text{kin}}\to m$ in this limit. In the limit of low quark mass (below ${u}_{m}\sim 1.02{u}_{h}$, the D7-brane becomes unstable), $\mu \to 2\pi T$

This, apparently, is the upper limit on the viscous damping parameter, and may be universal. But it’s a little unclear that one should trust the quasinormal mode analysis in this limit, as quantum fluctuations of the short string are important, in contrast to the long-string (heavy-quark) limit.

Chris Herzog has a very nice little followup paper in which he, among other things, studies the same problem in the background of a certain non-extremal ${\text{AdS}}_{5}$ blackhole background.

Posted by distler at May 27, 2006 12:41 AM

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Thank you for nice words! Many months and a lot of hard work went into this project.

I wanted to point out that there is an apparent discrepancy between the results of the MIT group and our result for the friction coefficient. Using the Langevin equation, our result for the friction coefficient can be converted into a result for the jet-quenching parameter. While both our and the MIT group’s results have the same temperature and ‘t Hooft coupling dependence, the numerical coefficient is different. The coefficient for the MIT group depends on Gamma functions while we find only a factor of 2 pi. (These Gamma functions are the same type of Gamma functions that showed up in the quark-antiquark potential in early papers on Wilson loops and AdS/CFT.)

One curious issue which we here at Washington have not completely got our heads around yet is the fact that the MIT group essentially works in Euclidean signature. Their Nambu-Goto action has no minus one in the square root of the determinant, and their path integral depends on an exponential of minus the action rather than i times the action. The string configurations they consider would have imaginary action in Minkowski signature.

A hopeful resolution of the discrepancy might be that we are considering different limits – the MIT group’s results are relevant for relativistic, nearby quark-antiquark pairs while our results are valid for slower, well separated quarks.

While I’m on my soap box, I thought I would point out another interesting entry in the quark energy loss debate. The calculation involved is similar to our quasinormal mode calculation, but the interpretation involves the Schwinger-Keldysh formalism.

Posted by: Christopher Herzog on May 28, 2006 2:17 PM | Permalink | Reply to this

### Various calculations

A hopeful resolution of the discrepancy might be that we are considering different limits – the MIT group’s results are relevant for relativistic, nearby quark-antiquark pairs while our results are valid for slower, well separated quarks.

That was my understanding: their calculation was relevant to relativistic (light) quarks, whereas yours are relevant for (nonrelativistic) heavy quarks. It’s interesting that the temperature and 't Hooft coupling dependence is the same, though. Perhaps the calculations are more closely-related than I thought!

While I’m on my soap box, I thought I would point out another interesting entry in the quark energy loss debate.

I would love to spark a discussion and comparison of these various different calculations. Unfortunately, I’m not enough of an expert. But now that I have your attention …

Posted by: Jacques Distler on May 28, 2006 2:37 PM | Permalink | PGP Sig | Reply to this