Beyond Supergravity in AdS-CFT: An Application to Jet Quenching

Abstract

These notes are dedicated to the study of jet quenching in the strongly coupled limit using gauge/string duality. We are interested in corrections to the infinite coupling λ = ∞ result for the jet stopping, in powers of λ ^−1∕2. To estimate these corrections we need to go beyond supergravity in AdS-CFT, and include all higher-derivative corrections to the supergravity action which arise from the string α′ expansion. For the particular type of “jets” that we study, the expansion in λ ^−1∕2 is well behaved for jets whose stopping distance ℓ _stop is in the range $\lambda ^{-1/6}\ell_{\mathrm{max}} \ll \ell_{\mathrm{stop}} \lesssim \ell_{\mathrm{max}}$, but the expansion breaks down for jets created in such a way that $\ell_{\mathrm{stop}} \ll \lambda ^{-1/6}\ell_{\mathrm{max}}$. The reason for the breakdown of the λ ^−1∕2 expansion is caused by the excitation of massive string states. In particular, consider “jets” which are dual to high-momentum gravitons. In the black brane background the gravitons, which are closed string states, get stretched into relatively large classical strings by tidal forces. These stringy excitations of the graviton are not contained in the supergravity approximation, but the jet stopping problem can nonetheless still be solved by drawing on various string-theory methods (the eikonal approximation, the Penrose limit, string quantization in pp-wave backgrounds) to obtain a probability distribution for the late-time classical string loops.

Appendices

Appendix 1: What Happens for $\boldsymbol{z \gg z_{\star }}$?

By making the reasonable assumptions that we outlined in Sect. 14.2.3, we have managed to analyze the question of when corrections become important by focusing on particle trajectories at z ∼ z _⋆. As z increases beyond this scale, the forward progress of the trajectory slows to a stop.We previously asserted that at some scale z _bad ≫ z _⋆ the expansion in higher-derivative corrections would eventually break downeven if the expansion was well behaved at z _⋆.

Start by considering the cost of an α′D ² factor, which we analyzed in Sect. 14.4.3 for z ∼ z _⋆. At larger z, with $z_{\star } \lesssim z \ll z_{\mathrm{h}}$, (14.62) gives

$$\displaystyle{ q_{\mathsf{5}}\bigr |_{z\gtrsim z_{\star }} \sim \frac{z^{2}E} {z_{\mathrm{h}}^{2}}. }$$

(14.140)

Then the cost (14.61) of each α′D ² factor is

$$\displaystyle{ \alpha 'D^{2}\vert _{ z\gtrsim z_{\star }} \sim \frac{\alpha 'zq_{\mathsf{5}}} {\mathbf{R}^{2}} \sim \frac{z^{3}ET^{2}} {\lambda ^{1/2}} \,. }$$

(14.141)

This cost is unsuppressed for $z \gtrsim z_{\mathrm{bad}}$, where

$$\displaystyle{ z_{\mathrm{bad}} \sim \frac{\lambda ^{1/6}} {E^{1/3}T^{2/3}}\,. }$$

(14.142)

The same constraint arises from the other important corrections that we analyzed. For instance, the cost of adding an α′² D ² C factor was z ⁶ E ²∕λ z _h ⁴, which also becomes unsuppressed at the same $z \gtrsim z_{\mathrm{bad}}$.

Note that the requirement $\ell_{\mathrm{stop}} \gg \lambda ^{-1/6}\ell_{\mathrm{max}}$ for the expansion in corrections to be well-behaved in Fig. 14.3 is the same condition as requiring $z_{\mathrm{bad}} \gg z_{\star }$.

Appendix 2: Why (14.46) Cannot Precisely Determine $\boldsymbol{\varDelta \ell_{\mathrm{stop}}}$

Consider the safe region $\ell_{\mathrm{stop}} \gg \lambda ^{-1/6}\ell_{\mathrm{max}}$ of Fig. 14.3, where the effects of higher-dimensional supergravity interactions should be suppressed. The R ⁴ corrections then dominate the corrections at z ∼ z _⋆. We might then be tempted to use the explicit form (14.25) of the R ⁴ correction, combined with the particle-based formula (14.38) for the stopping distance, to explicitly calculate the first correction to the λ = ∞ result (14.18) for the stopping distance. In this appendix, we discuss why that does not work.

We start with (14.45),

$$\displaystyle{ \ell_{\mathrm{stop}} \simeq \int _{0}^{z_{\mathrm{h}} }\mathit{dz} \frac{\vert \boldsymbol{q}\vert \left [1 -\frac{2\varepsilon z^{10}} {z_{\mathrm{h}}^{8}} \vert \boldsymbol{q}\vert ^{2}\right ]} {\sqrt{-q^{2 } + \frac{z^{4 } } {z_{\mathrm{h}}^{4}} \vert \boldsymbol{q}\vert ^{2} + \frac{\varepsilon z^{10}} {z_{\mathrm{h}}^{8}} \vert \boldsymbol{q}\vert ^{4}f}}. }$$

(14.143)

First we explain why the potentially sign-changing behavior of the numerator correction in the R ⁴-corrected formula (14.143) for the stopping distance could be ignored. The disturbing features of this correction arise in the z range given by (14.48),

$$\displaystyle{ z \gg z_{\mathrm{disturbing}} \sim \left (\frac{\lambda ^{3/4}T} {E} \right )^{1/5}z_{\mathrm{ h}}. }$$

(14.144)

This difficulty only arises at all if z _disturbing < z _h, which requires

$$\displaystyle{ E \gg \lambda ^{3/4}T. }$$

(14.145)

Now compare (14.144) and (14.142):

$$\displaystyle{ z_{\mathrm{disturbing}} \sim \left ( \frac{E} {\lambda ^{1/8}T}\right )^{2/15}z_{\mathrm{ bad}}. }$$

(14.146)

The inequality (14.145) then gives

$$\displaystyle{ z_{\mathrm{disturbing}} \gg z_{\mathrm{bad}}, }$$

(14.147)

and so the numerator correction cannot be believed in the range of z for which it becomes disturbing.

Dropping the numerator correction from (14.143) leaves

$$\displaystyle{ \ell_{\mathrm{stop}} \simeq \int _{0}^{z_{\mathrm{h}} }\mathit{dz} \frac{\vert \boldsymbol{q}\vert } {\sqrt{-q^{2 } + \frac{z^{4 } } {z_{\mathrm{h}}^{4}} \vert \boldsymbol{q}\vert ^{2} + \frac{\varepsilon z^{10}} {z_{\mathrm{h}}^{8}} \vert \boldsymbol{q}\vert ^{4}f}}, }$$

(14.148)

For simplicity, in what follows we will just analyze the case E ≫ λ ^3∕4 T.

In the integrand, look at the expression under the square root in the denominator. The relative importance of the $\varepsilon z^{10}$ term grows with increasing z. For z ≫ z _⋆, the z ⁴ term under the square root dominates over the − q ² term, so we should compare the $\varepsilon z^{10}$ term to the z ⁴ term. These are the same size at a scale $z_{\star \star } \gg z_{\star }$ given by

$$\displaystyle{ z_{\star \star } \sim \frac{z_{\mathrm{h}}^{2/3}} {\varepsilon ^{1/6}\vert \boldsymbol{q}\vert ^{1/3}} \sim \frac{\lambda ^{1/4}} {E^{1/3}T^{2/3}} \sim \frac{\lambda ^{1/4}} {l_{\mathrm{max}}T^{2}}\,, }$$

(14.149)

assuming that z _⋆⋆ ≪ z _h so that f ≃ 1. But z _⋆⋆ ≪ z _h follows from (14.149) and our consideration of E ≫ λ ^3∕4 T.

Now calculate the correction Δ ℓ to the stopping distance by subtracting the λ = ∞ result (14.15) from (14.148),

$$\displaystyle{ \varDelta \ell\equiv \ell_{\mathrm{stop}} -\ell_{\mathrm{stop}}^{\lambda =\infty }\simeq \int _{ 0}^{z_{\mathrm{h}} }\mathit{dz}\left [ \frac{\vert \boldsymbol{q}\vert } {\sqrt{-q^{2 } + \frac{z^{4 } } {z_{\mathrm{h}}^{4}} \vert \boldsymbol{q}\vert ^{2} + \frac{\varepsilon z^{10}} {z_{\mathrm{h}}^{8}} \vert \boldsymbol{q}\vert ^{4}f}} - \frac{\vert \boldsymbol{q}\vert } {\sqrt{-q^{2 } + \frac{z^{4 } } {z_{\mathrm{h}}^{4}} \vert \boldsymbol{q}\vert ^{2}}}\right ]\!. }$$

(14.150)

This integral is dominated by z ∼ z _⋆⋆. So, to explicitly calculate Δ ℓ will require trusting the integrand at z ∼ z _⋆⋆ given by (14.149). Compare this to the z scale (14.142) where the expansion in supergravity corrections breaks down:

$$\displaystyle{ z_{\star \star } \sim \lambda ^{1/12}z_{\mathrm{ bad}} \gg z_{\mathrm{bad}}. }$$

(14.151)

So we cannot trust (14.150) in the range of z where we want to use it to get an explicit result for Δ ℓ.

Appendix 3: Other Higher-Derivative Terms

In Sect. 14.4.3, we addressed the α′Q ⁵ D ₅ piece of α′D ² when studying the importance of D ²ⁿ C ⁴. We also recycled our conclusion from that analysis when later considering applying extra powers of derivatives to D ^2k C ^4+k. The dominant terms involved α′Q ^I D _I where the D _I hits a background Weyl tensor. We motivated focusing on Q ⁵ D ₅ by noting that the background Weyl tensor depends only of the x ⁵ coordinate. If the D’s were ordinary derivatives instead of covariant derivatives, that would be the end of the story. However, the other components D _μ of the covariant derivative do not vanish when applied to the background Weyl tensor. In fact, they are parametrically of order 1∕z, just like D ₅ . As a result, for example,

$$\displaystyle{ \alpha 'Q^{\mathsf{3}}D_{\mathsf{ 3}} =\alpha 'Q_{\mathsf{3}}g^{\mathsf{3}\mathsf{3}}D_{\mathsf{ 3}} \sim \alpha '\times E \times \frac{z^{2}} {\mathbf{R}^{2}} \times \frac{1} {z} \sim \frac{Ez} {\lambda ^{1/2}} }$$

(14.152)

is actually parametrically larger than the derivative

$$\displaystyle{ \alpha 'Q^{\mathsf{5}}D_{\mathsf{ 5}} \sim \frac{q_{\mathsf{5}}z} {\lambda ^{1/2}} }$$

(14.153)

considered in the main text (14.61).

So why doesn’t this lead to much larger results for the importance of D ²ⁿ R ⁴ and other operators than shown in Fig. 14.3? Our answer requires thinking about how the indices of the background Weyl tensor C _JKLM hit by α′Q ^I D _I contract with everything else.

Because C _IJKL depends only on x ⁵, a non-zero value for Q ^μ D _μ C _IJKL arises only from the terms of D involving the Christoffel symbols:

$$\displaystyle{ Q^{\mu }D_{\mu }C_{\mathit{IJKL}} = -Q^{\mu }\varGamma ^{\bar{I}}_{ I\mu }C_{\bar{I}\mathit{JKL}} - Q^{\mu }\varGamma ^{\bar{J}}_{ J\mu }C_{I\bar{J}\mathit{KL}} -\cdots \,. }$$

(14.154)

Now write

$$\displaystyle{ \varGamma =\varGamma ^{\mathrm{(AdS)}}+\varDelta \varGamma, }$$

(14.155)

where Γ ^(AdS) is the zero-temperature, purely AdS expression for the connection Γ. The difference between AdS and AdS₅-Schwarzschild is the difference between taking f = 1 and f = 1 − (z∕z _h)⁴ in the metric (14.6). As a result, the Δ Γ piece of (14.155) is suppressed compared to the Γ ^(AdS) piece by order (z∕z _h)⁴. For studying the dominant corrections at z ∼ z _⋆ ≪ z _h, we should therefore focus on Γ ^(AdS). In particular,

$$\displaystyle{ \alpha 'Q^{\mu }\varDelta \varGamma ^{\bar{I}}_{ I\mu } \sim \alpha 'E \times \frac{z^{2}} {\mathbf{R}^{2}} \times \frac{1} {z}\left ( \frac{z} {z_{\mathrm{h}}}\right )^{4} \sim \frac{Ez^{5}} {\lambda ^{1/2}z_{\mathrm{h}}^{4}} }$$

(14.156)

is always less important at z ∼ z _⋆ than the α′Q ⁵ D ₅ term (14.153) that we considered in the main text.

So now focus on Γ ^AdS:

$$\displaystyle{ Q^{\mu }D_{\mu }C_{\mathit{IJKL}} \simeq -Q^{\mu }(\varGamma ^{\bar{I}}_{ I\mu })^{\mathrm{AdS}}C_{\bar{ I}\mathit{JKL}} - Q^{\mu }(\varGamma ^{\bar{J}}_{ J\mu })^{\mathrm{AdS}}C_{ I\bar{J}\mathit{KL}} -\cdots \,. }$$

(14.157)

Because AdS space has four-dimensional Lorentz invariance, the μ index on Q ^μ above must pass through to contract with something else. For example,

$$\displaystyle\begin{array}{rcl} Q^{\mu }(\varGamma ^{\bar{I}}_{ I\mu })^{\mathrm{AdS}}& & C_{\bar{ I}\mathit{JKL}} \times (\mbox{ other stuff})^{\mathit{IJKL}} \\ & & \simeq \frac{1} {z}\,Q^{\mu }C_{\mu \mathit{JKL}}\times (\mbox{ other stuff})^{\mathsf{5}\mathit{JKL}}-\frac{1} {z}\,C^{\mathsf{5}}_{ \mathit{JKL}}Q_{\mu }\times (\mbox{ other stuff})^{\mu \mathit{JKL}}{}\end{array}$$

(14.158)

and

$$\displaystyle\begin{array}{rcl} Q^{\mu }(\varGamma ^{\bar{J}}_{ J\mu })^{\mathrm{AdS}}& & C_{ I\bar{J}\mathit{KL}} \times (\mbox{ other stuff})^{\mathit{IJKL}} \\ & & \simeq \frac{1} {z}\,Q^{\mu }C_{I\mu \mathit{KL}}\times (\mbox{ other stuff})^{I\mathsf{5}\mathit{KL}} -\frac{1} {z}\,C_{I}^{\mathsf{5}}{}_{ \mathit{KL}}Q_{\mu }\times (\mbox{ other stuff})^{I\mu \mathit{KL}}{}\end{array}$$

(14.159)

But now recall that our dominant terms already had every C contracted with two Q’s. So the “other stuff” above had the form

$$\displaystyle{ (\mbox{ other stuff})^{\mathit{IJKL}} \sim Q^{I}Q^{K}(\mbox{ something})^{\mathit{JL}}, }$$

(14.160)

and these terms were dominant because both Q _I and Q _K were parametrically of order E when contracted with the Weyl tensor C _IJKL . We are currently worried about the possibility that the Q _μ factor above is also of order E. Now look at the first term in (14.158). The Q _μ Q _I Q _K  ∼ E ³ is contracted in such a way that it instead gives $Q_{\mu }Q_{\mathsf{5}}Q_{K} \sim q_{\mathsf{5}}E^{2} \ll E^{3}$, which is not problematical. The second term in (14.158) contracts two Q’s together to give a factor of Q _μ η ^{μ ν} Q _ν  ∼ q ² instead of an E ², and so it also is suppressed. Next look at the first term in (14.159). There we have Q ^μ Q ^I Q ^K C _{I μ KL} . Up to terms which are suppressed by q ₅  ≪ E, this is the same as Q ^J Q ^I Q ^K C _IJKL , which vanishes by the symmetry of the Weyl tensor. Finally, look at the second term of (14.159), which involves

$$\displaystyle{ Q_{\mu }(\mbox{ something})^{\mu L}. }$$

(14.161)

For the dominant terms analyzed in the main text of this paper, the “something” is made up of factors of Q and QQC. If (something)^{μ L} gives a factor of Q ^μ, then two of our Q’s that were supposed to be giving factors of E will instead give a factor of − q ² ≪ E. If (something)^{μ L} gives a factor of Q _N Q _P C ^{N μ P•}, then we’ll get a suppression as before because of the symmetry of C.

Appendix 4: Large $\boldsymbol{\xi }$ Behavior of $\boldsymbol{C(\xi )}$

For large ξ, the $\bar{z}^{6}$ term in the differential equation (14.104a) for ξ can be ignored until $\bar{z} \gg 1$. At that point, however, we may use the simple large-$\bar{z}$ result (14.110) for $\bar{z}$. Substituting this into (14.104a) gives

$$\displaystyle{ \frac{d^{2}\chi } {d\bar{\tau }^{2}} = -4\left [\xi ^{6} - \frac{1} {(-3\bar{\tau })^{2}}\right ]\chi, }$$

(14.162)

whose solution is

$$\displaystyle{ \chi = (-\pi \xi ^{3}\bar{\tau })^{1/2}H_{ 5/6}^{(2)}{\bigl ( - 2\xi ^{3}\bar{\tau }\bigr )}. }$$

(14.163)

The late-time behavior τ → 0 is

$$\displaystyle{ \vert \chi \vert \simeq \frac{\varGamma (\tfrac{5} {6})} {\pi ^{1/2}\xi (-\bar{\tau })^{1/3}}\,. }$$

(14.164)

Fig. 14.15

(a) Parallel geodesics in AdS₅ in the Lorentz frame where the excitation is at rest in 3-space. These geodesics maintain a constant proper separation as they fall into the bulk, and this separation should be thought of as of order the characteristic size ($\sim \sqrt{\alpha '}$) of the closed quantum string loop describing the graviton (or other massless string mode). The narrow red loops are meant to be suggestive of the closed string loop. (b) The same picture boosted to the original Lorentz frame. (c) A picture of how those geodesics evolve in AdS₅-Schwarzschild rather than AdS₅. The early-time behavior is the same as (b). [For classical oscillating string solutions, the strings depicted in (a) may be thought of as snapshots at moments when the string’s proper extent in x ³ is at, say, maximum (or half-maximum or whatever). Such solutions would similarly oscillate in the z ≪ z _⋆ part of (b) but not in the z ≫ z _⋆ part, where tidal forces dominate over tension]

Appendix 5: A Back-of-the-Envelope Estimate

In this appendix we give a parametric estimate of the amount of tidal stretching of the string compared to the size of the stopping distance ℓ _stop. Here the only thing we need to know is that the stopping distance given by following a null geodesic as in Fig. 14.5 is proportional to a power of the slope dx ³∕dx ⁵ of that geodesic where it starts, at the boundary. The more downward-directed one starts the trajectory in Fig. 14.5, the less distance it will travel in x ³ before reaching the horizon.

Now interpret the trajectory of Fig. 14.5 as a trajectory for the center of mass of a tiny, falling loop of string. Once the string gets far enough from the boundary that the tidal forces dominate over the string tension, then the string tension becomes ignorable, and different pieces of the string will fall independently along their own geodesics, the string stretching accordingly. Imagine plotting two such geodesics, for the two bits of the string loop that are most separated. The separation of those geodesics is a measure of the extent of the tidally-stretched loop of string as it falls towards the horizon. The proper size of the string should start out of order the quantum mechanical size Σ of the graviton, which is roughly set by dimensional analysis in terms of the string tension $\mathbb{T}$ as

$$\displaystyle{ \varSigma _{\mathrm{graviton}} \sim \mathbb{T}^{-1/2} \sim \sqrt{\alpha '}, }$$

(14.165)

where $\alpha ' = 1/2\pi \mathbb{T}$ is the string slope parameter.

Very close to the boundary, the tidal forces due to the black hole are negligible, and the closed loop of string is in its ground state. We can set up our two geodesics above so that, correspondingly, they maintain constant proper separation Σ _graviton near the boundary, where the AdS₅-Schwarzschild metric approaches a purely AdS₅ metric. To see how this works, imagine making a four-dimensional boost from (i) the plasma rest frame, in which we create an excitation with large 4-momentum q ^μ = (ω, 0, 0, q ₃ ) ≃ (E, 0, 0, E) and relatively small 4-virtuality − q ² ≪ E ², to (ii) the excitation’s initial rest frame, where the 4-momentum is instead $(\sqrt{-q^{2}},0,0,0)$. The Lorentz boost factor for this transformation is

$$\displaystyle{ \gamma = \sqrt{ \frac{\omega ^{2 } } {-q^{2}}} \simeq \sqrt{ \frac{E^{2 } } {-q^{2}}} \gg 1. }$$

(14.166)

In AdS₅, the trajectory in the new frame will drop straight down away from the boundary, as depicted by the dashed line in Fig. 14.15a.

Now consider the graviton as an extended object with proper size Σ. The two straight solid null lines in Fig. 14.15a depict the extent of the graviton in AdS₅ in the excitation’s rest frame at early times. In pure AdS₅ null geodesics are straight lines. We parametrize the two solid lines of Fig. 14.15a as

$$\displaystyle{ x^{I} ={\bigl (\gamma _{ +},0,0,\pm \beta _{+}\gamma _{+},1\bigr )}\,z }$$

(14.167)

with β ₊ ≪ 1 and γ ₊ ≡ (1 −β ₊ ²)^−1∕2 ≃ 1. Because of the warp factor in the metric, these two lines are parallel and maintain constant proper separation $\sqrt{\varDelta x^{\mathsf{3} } \,g_{\mathsf{3} \mathsf{3} } \,\varDelta x^{\mathsf{3}}} = 2\beta _{+}\gamma _{+}\mathbf{R} \simeq 2\beta _{+}\mathbf{R}$ as a function of the rest-frame time. Setting this proper separation to be of order the graviton size Σ given by (14.165) then gives

$$\displaystyle{ \beta _{+} \sim \frac{\varSigma _{\mathrm{graviton}}} {\mathbf{R}} \sim \frac{\sqrt{\alpha '}} {\mathbf{R}} \sim \lambda ^{-1/4}\,, }$$

(14.168)

and (14.167) gives

$$\displaystyle{ x^{I} \sim {\bigl ( 1,0,0,\pm \lambda ^{-1/4},1\bigr )}\,z. }$$

(14.169)

Now boost back to the original plasma frame using (14.166) to get the early-time trajectories depicted by solid lines in Fig. 14.15b: $x^{I} \sim {\Bigl (\gamma (1 \pm \lambda ^{-1/4}),0,0,\gamma (1 \pm \lambda ^{-1/4}),1\Bigr )}\,z$, where we have used γ ≫ 1 (14.166). Then

$$\displaystyle{ \frac{\varDelta (\mathit{dx}^{\mathsf{3}}/\mathit{dz})} {\mathit{dx}^{\mathsf{3}}/\mathit{dz}} \Bigg\vert _{\mathrm{initial}} \sim \lambda ^{-1/4}. }$$

(14.170)

As discussed before, the stopping distance (which requires a calculation in the full AdS₅-Schwarzschild metric) covered by a null geodesic is power-law related to this initial slope, and so the difference Δ ℓ _stop in how far the two bits of string travel also has the same small size (14.170) relative to ℓ _stop:

$$\displaystyle{ \frac{\varDelta \ell_{\mathrm{stop}}} {\ell_{\mathrm{stop}}} \sim \lambda ^{-1/4}. }$$

(14.171)

Appendix 6: Checking the Penrose Limit: Details

In order to check the validity of the Penrose limit, here we characterize the string by following null geodesics that roughly trace different bits of string and which deviate slightly from our reference geodesic. This approximation amounts to ignoring the tension in the string as in Appendix 5 (for an alternative check of the Penrose limit outside of this approximation see [37]).

From the null geodesic formula and the metric (14.6), the x ³ coordinate for such geodesics is given by

$$\displaystyle{ \frac{\mathit{dx}^{\mathsf{3}}} {\mathit{dz}} = \frac{\hat{q}_{\mathsf{3}}} {\sqrt{1 - f\hat{\boldsymbol{q}}^{2}}}, }$$

(14.172)

where

$$\displaystyle{ \hat{q}_{\mu } \equiv \frac{q_{\mu }} {\omega } = (-1,\hat{\boldsymbol{q}}). }$$

(14.173)

Remembering that $\varDelta x^{\mu } \equiv x^{\mu } -\bar{ x}^{\mu }(z)$ is the deviation relative to the reference geodesic, we have

$$\displaystyle{ \frac{d\varDelta x^{\mathsf{3}}} {\mathit{dz}} = \frac{\hat{q}_{\mathsf{3}}} {\sqrt{1 - f\hat{\boldsymbol{q}}^{2}}} - \frac{\bar{\hat{q}}_{\mathsf{3}}} {\sqrt{1 - f\bar{\hat{\boldsymbol{q}}}^{2}}}. }$$

(14.174)

Expand to first order in $\varDelta \hat{q}_{3} \equiv \hat{ q}_{3} -\bar{\hat{ q}}_{3}$:

$$\displaystyle{ \frac{d\varDelta x^{\mathsf{3}}} {\mathit{dz}} \simeq \frac{\varDelta \hat{q}_{\mathsf{3}}} {(1 - f\bar{\hat{\boldsymbol{q}}}^{2})^{3/2}}. }$$

(14.175)

Then using (14.85) (and defining u with respect to the reference geodesic $\bar{x}$),

$$\displaystyle{ \frac{f\mathbf{R}^{2}} {z^{2}} \,\frac{d\varDelta x^{\mathsf{3}}} {\mathit{du}} \simeq \frac{f\,\varDelta \hat{q}_{\mathsf{3}}} {1 - f\hat{\boldsymbol{q}}^{2}}. }$$

(14.176)

Since $1 - f\hat{\boldsymbol{q}}^{2} \simeq (z_{\star }^{4} + z^{4})/z_{\mathrm{h}}^{4}$, the combination (14.176) is largest for $z \lesssim z_{\star }$, and the d Δ x ³∕du condition in (14.128) requires

$$\displaystyle{ \varDelta \hat{q}_{\mathsf{3}} \ll \frac{z_{\star }^{4}} {z_{\mathrm{h}}^{4}} }$$

(14.177)

for the Penrose limit. Use (14.18) to relate this to the stopping distance:

$$\displaystyle{ \ell_{\mathrm{stop}} \sim \frac{z_{\mathrm{h}}^{2}} {z_{\star }} \sim z_{\mathrm{h}}\left ( \frac{E^{2}} {-q^{2}}\right )^{1/4} \sim \frac{z_{\mathrm{h}}} {(1 -\hat{ q}_{\mathsf{3}})^{1/4}}, }$$

(14.178)

so that

$$\displaystyle{ \varDelta \ell_{\mathrm{stop}} \sim \frac{z_{\mathrm{h}}\,\varDelta \hat{q}_{\mathsf{3}}} {(1 -\hat{ q}_{\mathsf{3}})^{5/4}} \sim \frac{\varDelta \hat{q}_{\mathsf{3}}\,\ell_{\mathrm{stop}}} {1 -\hat{ q}_{\mathsf{3}}} \sim \varDelta \hat{ q}_{\mathsf{3}}\,\ell_{\mathrm{stop}}\,\frac{z_{\mathrm{h}}^{4}} {z_{\star }^{4}}. }$$

(14.179)

Combining (14.177) and (14.179) gives the condition

$$\displaystyle{ \varDelta \ell_{\mathrm{stop}} \ll \ell_{\mathrm{stop}} }$$

(14.180)

quoted in (14.130).

Now turn to the condition on dv∕du in (14.128). The definition (14.77) of v gives

$$\displaystyle{ \mathit{dv} =\bar{\hat{ q}}_{\mu }\,d(\varDelta x^{\mu }) = -d(\varDelta x^{\mathsf{0}}) +\bar{\hat{ q}}_{\mathsf{ 3}}\,d(\varDelta x^{\mathsf{3}}), }$$

(14.181)

and so we need a formula for d(Δ x ⁰). The analog of (14.172) is

$$\displaystyle{ \frac{\mathit{dx}^{\mathsf{0}}} {\mathit{dz}} = \frac{-f^{-1}\hat{q}_{\mathsf{0}}} {\sqrt{1 - f\hat{\boldsymbol{q}}^{2}}}, }$$

(14.182)

with expansion

$$\displaystyle{ \frac{d\varDelta x^{\mathsf{0}}} {\mathit{dz}} \simeq \frac{\hat{q}_{\mathsf{3}}\,\varDelta \hat{q}_{\mathsf{3}}} {(1 - f\bar{\hat{q}}^{2})^{3/2}}. }$$

(14.183)

Combining (14.175), (14.181), and (14.183), gives dv∕du ≃ 0. We therefore have to go back and make our expansions to second-order in $\varDelta \hat{\boldsymbol{q}}$. The result is

$$\displaystyle{ \frac{\mathit{dv}} {\mathit{dz}} \simeq - \frac{(\varDelta \hat{\boldsymbol{q}}_{\perp })^{2}} {2(1 - f\bar{\hat{q}}_{3}^{2})^{1/2}} - \frac{(\varDelta \hat{q}_{\mathsf{3}})^{2}} {2(1 - f\bar{\hat{q}}_{3}^{2})^{3/2}}\,, }$$

(14.184)

and so

$$\displaystyle{ \frac{f\mathbf{R}^{2}} {z^{2}} \left \vert \frac{\mathit{dv}} {\mathit{du}}\right \vert \simeq \frac{f(\varDelta \hat{\boldsymbol{q}}_{\perp })^{2}} {2} + \frac{f(\varDelta \hat{q}_{\mathsf{3}})^{2}} {2(1 - f\bar{\hat{q}}_{3}^{2})}\,. }$$

(14.185)

The corresponding condition on dv∕du in (14.128) is then

$$\displaystyle{ f(\varDelta \hat{\boldsymbol{q}}_{\perp })^{2}\quad \mbox{ and}\quad \frac{f(\varDelta \hat{q}_{\mathsf{3}})^{2}} {(1 - f\bar{\hat{q}}_{3}^{2})}\quad \ll \quad 1. }$$

(14.186)

The first condition is strongest for z ≪ z _h and the second for $z \lesssim z_{\star }$, giving

$$\displaystyle{ \vert \varDelta \hat{\boldsymbol{q}}_{\perp }\vert \quad \mbox{ and}\quad \vert \varDelta \hat{q}_{\mathsf{3}}\vert \,\frac{z_{\mathrm{h}}^{2}} {z_{\star }^{2}} \quad \ll \quad 1. }$$

(14.187)

Using (14.179), the condition involving $\varDelta \hat{q}_{\mathsf{3}}$ becomes

$$\displaystyle{ \varDelta \ell_{\mathrm{stop}} \ll \ell_{\mathrm{stop}}\frac{z_{\mathrm{h}}^{2}} {z_{\star }^{2}} \,. }$$

(14.188)

Since z _⋆ ≪ z _h, this is weaker than the previous condition (14.180).

Lastly, consider the other condition, $\vert \varDelta \boldsymbol{q}_{\perp }\vert \ll 1$ in (14.187). To estimate $\vert \varDelta \boldsymbol{q}_{\perp }\vert$, return to the arguments of Appendix 5, but now, in the rest frame, include an initial proper displacement of the two geodesics in $\boldsymbol{x}^{\perp }$ of the same parametric size as the initial proper displacement in x ³. Following through the argument, one finds $x^{I} \simeq {\Bigl (\gamma (1 +\beta \beta _{+}),\boldsymbol{\beta }_{\perp },\gamma (\beta +\beta _{+}),1\Bigr )}\,z$ with β _⊥ ∼ β ₊. Then

$$\displaystyle{ \varDelta \hat{q}_{\mathsf{3}} =\varDelta \frac{q_{\mathsf{3}}} {q_{\mathsf{0}}} =\varDelta \frac{\mathit{dx}^{\mathsf{3}}/\mathit{dz}} {\mathit{dx}^{\mathsf{0}}/\mathit{dz}} \simeq \varDelta \frac{\gamma (\beta +\beta _{+})} {\gamma (1 +\beta \beta _{+})} \simeq \frac{\beta _{+}} {\gamma ^{2}} }$$

(14.189)

and

$$\displaystyle{ \varDelta \hat{\boldsymbol{q}}_{\perp } =\varDelta \frac{\boldsymbol{q}_{\perp }} {q_{\mathsf{0}}} =\varDelta \frac{d\boldsymbol{x}^{\perp }/\mathit{dz}} {\mathit{dx}^{\mathsf{0}}/\mathit{dz}} \simeq \varDelta \frac{\boldsymbol{\beta }_{\perp }} {\gamma (1 +\beta \beta _{+})} \simeq \frac{\boldsymbol{\beta }_{\perp }} {\gamma }, }$$

(14.190)

so that

$$\displaystyle{ \frac{\vert \varDelta \hat{\boldsymbol{q}}_{\perp }\vert } {\vert \varDelta \hat{q}_{\mathsf{3}}\vert } \sim \gamma \sim \sqrt{ \frac{E^{2 } } {-q^{2}}} \sim \frac{z_{\mathrm{h}}^{2}} {z_{\star }^{2}} \,. }$$

(14.191)

So, using (14.179),

$$\displaystyle{ \vert \varDelta \hat{\boldsymbol{q}}_{\perp }\vert \sim \vert \varDelta \hat{q}_{\mathsf{3}}\vert \,\frac{z^{2}} {z_{\star }^{2}} \sim \frac{\varDelta \ell_{\mathrm{stop}}} {\ell_{\mathrm{stop}}}\, \frac{z_{\star }^{2}} {z_{\mathrm{h}}^{2}}\,. }$$

(14.192)

The condition $\vert \varDelta \hat{\boldsymbol{q}}_{\perp }\vert \ll 1$ is therefore the same as the previous condition (14.188) and so is also weaker than (14.180).