Disclaimer: Conservative motion
& non-spinning (for most of the talk)
Boundary Data Meets Bound States
Work with Rafael Porto in
[1910.03008]&
[1911.09130]
QCD meets Gravity 2019
UCLA
meets
[see also Bjerrum-Bohr, Cristofoli, Damgaard 19]
Impetus formula
$$\mathbf{p}^2(r,E) = p_\infty^2(E) + \widetilde{\cM}(r,E)$$
$${\widetilde{\cM}}(r,E) \equiv \frac{1}{2E}\int \frac{\dd^3\bq}{(2\pi)^3}\, {\cal M}(\bq,\bp^2=p_\infty^2(E)) e^{-i\bq\cdot \br}$$ IR-finite classical part of the scattering amplitude
Sketch of proof: Match to effective potential scattering $$H_{\rm eff}|\psi_\bp(p_\infty)\rangle = \left(\bp^2 + V_{\rm eff}\right) |\psi_\bp(p_\infty)\rangle= p_\infty^2(E) |\psi_\bp(p_\infty)\rangle\,$$ with $$V_{\rm eff} = -\sum_i P_i(E) \frac{G^i}{r^i}\quad \textrm{(PM expansion)}$$
Iterate Lippman-Schwinger equation:
$$\begin{align*}
\frac{4\pi}{\rm Vol} \tilde f(\bp^2,\bq) &=- \langle \bp+\bq| V_{\rm eff} |\bp\rangle + \dots\\
&= \frac{1}{\rm Vol} \sum_i \int \dd^3\br \left( P_i(E)\frac{G^i}{r^i}\right) e^{i\bq\cdot \br}+\dots
\end{align*}$$
Higher iterations can be shown to be IR-divergent (superclassical). Match cross-section
$$\frac{\dd \sigma}{\dd\Omega}=|\tilde{f}|^2=|\cM|^2$$
\(\Rightarrow\) Born approximation is exact in the classical limit.
Use iterations to match/subtract IR-divergent terms from the amplitude.
More physically: Follow [Kosower, Maybee, O'Connell 19] to split the classical, localized and instantaneous scattered momentum into: $$\bp_{\rm sc}^2(r,p_\infty^2) \sim \psi^\dagger _\bp(\br,p_\infty)\, (-\nabla^2 - p_\infty^2) \psi_\bp(\br,p_\infty) = I_{(1)}(r,E) + I_{(2)}(r,E)$$ \(I_1\): linear in amplitude, conservative potential \(I_2\): square in amplitude, radiation reaction (+conservative)
In PM expanded form: $$\widetilde{\cal M}_n(E) = P_n(E)$$ using $$\widetilde{\cal M}(r,E) = \sum_{n=1}^\infty \widetilde{\cal M}_n(E) \left(\frac{G}{r}\right)^n,\quad\bp^2(r,E) = p_\infty^2(E) +\sum_{n=1}^\infty P_n(E) \left(\frac{G}{r}\right)^n$$
$$\chi(b,E) +\pi %= 2b \int_{r_{\rm min}}^\infty \frac{\dd r}{r\sqrt{r^2\bar\bp^2(r,E)-b^2}}\\
= 2b \int_{r_{\rm min}}^\infty \frac{\dd r}{r\sqrt{r^2 (1+\cM(r,E)/p_\infty^2)-b^2}}$$
\(r_\textrm{min}\) is the positive real root of \(p_r\): \(p_r^2(r,E)=\bp^2(r,E)-J^2/r^2\) \(\,\,(J=p_\infty b)\)
Or PM expanded (using \(f_n (E) = \frac{\cM_n(E)}{p_\infty^2(E)M^n}\) ): $$\chi_b^{(n)} = \frac{\sqrt{\pi}}{2} \Gamma\left(\frac{n+1}{2}\right)\sum_{\sigma\in\mathcal{P}(n)}\frac{1}{\Gamma\left(1+\frac{n}{2} -\Sigma^\ell\right)}\prod_{\ell} \frac{f_{\sigma_{\ell}}^{\sigma^{\ell}}}{\sigma^{\ell}!}$$ with $$\frac{1}{2} \chi(b,E) = \sum_n \chi^{(n)}_b (E) \left(\frac{GM}{b}\right)^n =\sum_n \chi^{(n)}_{j} (E)\frac{1}{j^n}$$
[Firsov 53]
Firsov's inversion formula
$$\bar\bp^2(r,E) = \exp\left[ \frac{2}{\pi} \int_{r|\bar{\bp}(r,E)|}^\infty \frac{\chi(\tilde b,E)\dd\tilde b}{\sqrt{\tilde b^2-r^2\bar\bp^2(r,E)}}\right]$$
$$\frac{1}{2E\,p_\infty^2} \int \frac{\dd^3\bq}{(2\pi)^3} {\cal M}(\bq,\bp)e^{-i\bq\cdot\br_{\rm min}(b,E)} = \exp\left[ \frac{2}{\pi} \int_{b}^\infty \frac{\chi(\tilde b,E)\dd\tilde b}{\sqrt{\tilde b^2-b^2}}\right]-1$$
$$f_n = \sum_{\sigma\in\mathcal{P}(n)}\frac{2(2-n)^{\Sigma^{\ell} - 1}}{\prod_{\ell} (2\sigma^{\ell})!!} \prod_{\ell} \left(\frac{2}{\sqrt{\pi}}\frac{\Gamma(\frac{\sigma_\ell}{2})}{\Gamma(\frac{\sigma_\ell+1}{2})}\chi^{(\sigma_\ell)}_b\right)^{\sigma^{\ell}}$$
meets
Radial action via analytic continuation \({\cE} = \frac{E-M}{\mu} < 0\)
[GK, Porto 19] $$\begin{align*} {\cal S}_r(J,{\cal E}) &\equiv \frac{1}{\pi} \int_{r_-}^{r_+} p_r \dd r = \frac{1}{\pi} \int_{r_-}^{r_+} \sqrt{\bp^2(r,{\cal E})-J^2/r^2} \, \dd r\\ &= \frac{1}{\pi} \int_{r_-}^{r_+} \sqrt {p_\infty^2({\cal E}) + \widetilde{\cal M}(r,{\cal E})-J^2/r^2} \end{align*}$$Orbital Elements
From Firsovs formula find root(s) of \(p_r\): $$r_{\textrm{min}} = \tilde r_- = b \exp\left[ -\frac{1}{\pi} \int_{b}^\infty \frac{\chi(\tilde b,E)\dd\tilde b}{\sqrt{\tilde b^2-b^2}}\right] =b \prod_{n=1}^\infty e^{-\frac{(GM)^n\chi_b^{(n)}(\beta)\Gamma\left(\frac{n}{2}\right)}{b^n\sqrt{\pi}\Gamma\left(\frac{n+1}{2}\right)}}$$
Do an analytic continuation to \(J=p_\infty b\) with \(b\in i\mathbb{R}\) (\(p_\infty^2\leq0\)): $$\begin{align*} r_-(J) &= r_{\textrm{min}}(b)\\ r_+(J) &= r_-(-J) = r_{\textrm{min}}(-b) \end{align*}$$ These are the two real positive roots of \(p_r\) the radial momentum.
PM expand: $${\cS}_r(J,{\cE}) = \frac{1}{\pi} \int_{r_-}^{r_+} \dd r \sqrt{ Q(J,{\cE},r) + \lambda \sum_{\ell=1}^\infty \frac{D_\ell({\cE})}{r^{\ell+2}}}$$ $$\begin{align*} Q(J,{\cal E},r) &\equiv A ({\cal E}) + \frac{2B({\cal E})}{r} + \frac{C(J,{\cal E})}{r^2}\\ A({\cal E}) &\equiv p_\infty^2({\cal E}) & 2B({\cal E}) &\equiv \widetilde{M}_1({\cal E}) G \\ C(J,{\cal E}) &\equiv \widetilde{M}_2({\cal E}) G^2 -J^2 & D_n({\cal E}) &\equiv \widetilde{M}_{n+2} ({\cal E}) G^{n+2} \end{align*}$$
Master integrals (Hypergeometric function!): $$\begin{align} &\cS_{\{m,q\}} = \frac{1}{2\pi} \oint_C \frac{\dd r}{r^m}Q^{\tfrac{1}{2}-q}\\ &= \cS_{\{m,q\}}^\infty \!+\!\sum_{k} \frac{(-1)^q i^{k+m+1}2^k\Gamma\left(\frac{m+k-1}{2}\right)}{\Gamma(k+1)\Gamma\left(\frac{2+m-k-2q}{2}\right)\Gamma\left(q-\frac{1}{2}\right)}\frac{A(\mathcal E)^{\frac{m-k-2q}{2}}B(\mathcal E)^k}{C(J,\mathcal E)^{\frac{m+k-1}{2}}} \end{align}$$ Leads to the PM expanded radial action: $$\cS_r =-\sum_{n=0}^\infty \sum_{\sigma\in\cP(n)} \frac{(-1)^{ \Sigma^{\ell}}\Gamma\left(\Sigma^{\ell} - \frac{1}{2}\right)}{2\sqrt{\pi}}\cS_{\left\{n+2\Sigma^{\ell},\Sigma^{\ell}\right\}}(J,\cE)\prod_{\ell} \frac{D_{\sigma_{\ell}}^{\sigma^{\ell}}(\cE)}{\sigma^{\ell}!}$$
meets
Compare the scattering case \(\cE \geq 0\): $$\frac{\chi+\pi}{2\pi} = -{\partial \cS_r(J,\cE) \over \partial J} = \frac{1}{\pi} \int_{{\tilde r}_-(J,\cE)}^\infty \,\, \frac{J}{r^2\sqrt{\bp^2(\cE,r)-J^2/r^2}}\dd r$$ to closed orbits \(\cE \leq 0\): $$\frac{\Delta \Phi+2\pi}{2\pi} = - {\partial \cS_r(J,\cE) \over \partial J} = \frac{1}{\pi} \int_{r_-(J,\cE)}^{r_+(J,\cE)} \frac{J}{r^2\sqrt{\bp^2(\cE,r)-J^2/r^2}}\dd r$$
Periastron advance from Scattering Angle
$$\Delta \Phi(J,\cE<0) = \chi(J,\cE<0) + \chi(-J,\cE<0)$$Works nicely perturbatively (PM), e.g.: $$\Delta\Phi = \pi \frac{\widetilde\cM_2}{\mu^2M^2 j^2} +\frac{3\pi}{4}\frac{1}{M^4 \mu^4 j^4}\big(\widetilde\cM_2^2+2\widetilde\cM_1\widetilde\cM_3+2p_\infty^2\widetilde\cM_4\big)+\cdots$$ \(j = J/(GM \mu)\)
meets
Having obtained the periastron advance from the scattering angle we can integrate in \(J\) to get \(\cS_r\) (+boundary term at \(\infty\)):
$$\frac{{\cal S}_r}{GM\mu} = {\rm sg}(\hat p_\infty )\chi^{(1)}_j(\cE) - j \left(1 + \frac{2}{\pi} \sum_{n=1} \frac{\chi^{(2n)}_j({\cE})}{(1-2n)j^{2n}}\right)$$ \(j = J/(GM \mu)\)
Agrees with previously obtained expression in the PM regime! $$\frac{{\cal S}_r}{GM\mu} = -j + \frac{{\widetilde \cM}_1}{2 \sqrt{ -\hat p^2_\infty}M \mu^2} + \frac{{\widetilde \cM}_2}{2 jM^2\mu^2} + \frac{{\widetilde \cM}_2^2 +2 {\widetilde \cM}_1{\widetilde \cM}_3+2 p_\infty^2 {\widetilde \cM}_4}{8j^3M^4\mu^4}+\cdots$$
meets
- Periastron advance: \(\frac{\Delta\Phi}{2\pi} = -\frac{\partial}{\partial j} \cS_r(j,\cE)-1\)
- Periastron-to-periastron period: \(\frac{T_p}{2\pi} = \frac{1}{\mu}\frac{\partial}{\partial \cE} \cS_r(j,\cE)\)
- Radial frequency: \(\Omega_r (j,\cE) = \frac{2\pi}{T_p}\)
- Periastron frequency: \(\Omega_p (j,\cE) = \frac{\Delta\Phi}{T_p}\) [LeTiec 15]
- Averaged redshift: \(\langle z_a\rangle = 1+\frac{\partial \mu}{\partial m_a} \cE -\Omega_r \frac{\partial}{\partial m_a}\cS_r(j,\cE,m_a) \)
$$\small\begin{aligned}
\frac{\Delta \Phi_{(L=2)}}{2\pi}&= \frac{3}{4j^2} \frac{5\gamma^2-1}{\Gamma} + \frac{27}{32j^4}\frac{(5\gamma^2-1)^2}{\Gamma^2} -\frac{1}{4 j^4} \frac{2\gamma^2-1}{\Gamma^2} \Bigg[ 3-54\gamma^2 \\ &+ \nu \left(-6+206 \gamma + 108 \gamma^2+4\gamma^3-\frac{18\Gamma(1-2\gamma^2)(1-5\gamma^2)}{(1+\Gamma)(1+\gamma)} \right)\\
&\left.- 48 \nu(3+12\gamma^2-4\gamma^4)\frac{\arcsin\sqrt{\frac{1-\gamma}{2}}}{\sqrt{1-\gamma^2}}\right]
\end{aligned}$$
$$\small\begin{aligned}
\frac{GM \Omega^{(L=2)}_r}{\epsilon^{\frac{3}{2}}} &= \textcolor{blue}{1-
\frac{(15-\nu)}{8}\epsilon+\frac{555+30\nu+11\nu^2}{128 } \epsilon^2} \\
&+ \left(\textcolor{blue}{\frac{3(2\nu-5)}{2 j}}\textcolor{red}{-\frac{194-184\nu+23\nu^2}{4 j^3}}\right)\epsilon^{\frac{3}{2}}\\
&+\left(\textcolor{blue}{\frac{15(17-9\nu+2\nu^2)}{8 j}}\textcolor{green}{+\frac{21620-28592\nu+8765\nu^2-865\nu^3}{80 j^3}}\right)\epsilon^{\frac{5}{2}}
+ \cdots \\
\frac{G M \Omega^{(L=2)}_\phi}{\epsilon^{\frac{3}{2}}}
&=\textcolor{blue}{1+\frac{3}{j^2}-\frac{15(2\nu-7)}{4j^4}}+\left(\textcolor{blue}{\frac{1}{8}(\nu-15)+\frac{15(\nu-5)}{8j^2}} \textcolor{red}{-\frac{3(1301-921\nu+102\nu^2)}{32j^4}}\right)\epsilon\\
&+\left(\textcolor{blue}{\frac{3(2\nu-5)}{2j}}\textcolor{red}{+\frac{-284+220\nu-23\nu^2}{4j^3}}\textcolor{green}{+\frac{3(913-728\nu+106\nu^2)}{j^5}}\right)\epsilon^{\frac{3}{2}}
\\
% &\left.\textcolor{blue}{+\frac{15(2\nu-7)(194-184\nu+23\nu^2)}{16j^7}}\right)\epsilon^{\frac{3}{2}}
&+\left(\textcolor{blue}{\frac{1}{128}(555+30\nu+11\nu^2)+\frac{3(895-150\nu+51\nu^2)}{128j^2}}\right.\\
&- \left. \textcolor{green}{\frac{3(-270085+251236\nu-70545\nu^2+7470\nu^3)}{2560j^4}}\right)\epsilon^2\\
&+\left(\textcolor{blue}{\frac{15(17-9\nu+2\nu^2)}{8j}}
\textcolor{green}{+\frac{31520-34442\nu+10025\nu^2-865\nu^3}{80j^3}}\right)\epsilon^{\frac{5}{2}}\,.\label{eq:omegas}
% &\quad +\left.\textcolor{blue}{\frac{3(126665-145879\nu+48850\nu^2-5070\nu^3)}{160j^5}} % &\quad +\left. \textcolor{blue}{\frac{3(226612-386168\nu+204099\nu^2-40513\nu^3+2788\nu^4)}{64j^7}} \right)\epsilon^{\frac{5}{2}}+\dots
\end{aligned}$$
Circular Orbit
Solve \(r_+(J) = r_-(J)\) $$\Leftrightarrow -2 \sum_{n=0}^\infty \left(\frac{1}{\sqrt{\pi}}\left(\frac{GM}{b}\right)^{2n+1}\frac{\Gamma\left(\frac{2n+1}{2}\right)}{\Gamma(n+1)}\chi_b^{(2n+1)}\right) = i \pi + 2\pi i \mathbb{N}$$ to find \(j(\cE) \) and compute the radial frequency: $$GM\Omega_{\rm circ} = \left(\frac{d\, j({\cal E})}{d\, {\cal E}}\right)^{-1}$$ Need to resum for a truncated theory \(f_i=0\) for \(i\geq n\)! (More later)
We can invert to write the binding energy \(\epsilon\equiv -2 \cE\): $$\begin{aligned} \epsilon =&x \left[1 - \frac{x}{12}(9+\nu) - \frac{x^2}{8}\left(27 -19\nu + \frac{\nu^2}{3}\right)\right. \\ &+\frac{x^3}{32}\left(\frac{535}{6}-\frac{5585\nu}{6}+135\nu^2-\frac{35\nu^3}{162}\right) \\ &+ \left.\frac{x^4}{384}\left(-10171+\frac{559993}{15}\nu-\frac{34027\nu^2}{3}+\frac{11354\nu^3}{9}+\frac{77\nu^4}{81}\right) + {\cal O}(x^5)\right] \end{aligned}$$ using the standard PN parameter \(x \equiv (GM \Omega_{\rm circ})^{2/3}\).
Aligned spins
Idea: extend our map to the aligned spins for binary BH problem.
Motion is still in a plane!
$$\frac{\chi(J,\cE)+\chi(-J,\cE)}{2\pi} = \frac{\Delta\Phi(J,\cE)}{2\pi}$$
where \(J\) is now the total the total angular momentum, i.e. orbital angular momentum + spins.
- Explicit checks for known PN and PM results work neatly!
- Relies on the invariance of the (canonical) radial momentum \(p_r\) under \(J\rightarrow -J\), which is true for a quasi-isotropic gauge (given to us automatically by the amplitudes construction).
- We propose a version of the impetus formula to also hold for the aligned spin case.
Resummation
Let us truncate our theory at given order \(n\), i.e. \(\cM_m=f_m=0\) for \(m \geq n\).
We can try to resum contributions to all orders in \(G\), e.g. for the scattering angle:
$$\begin{align*}
\frac{\chi[f_1]}{2} &= \Arctan(y/2)\\
\frac{\chi[f_{1,2}]+\pi}{2} &= \frac{1}{\sqrt{1-{\cF}_2 y^2}}\left(\frac{\pi}{2} + \Arctan\left(\frac{y}{2\sqrt{1-{\cF}_2 y^2}}\right)\right)
\end{align*}$$
with \(y \equiv G M f_1/b\) and \(\cF_2 \equiv f_2/f_1^2\)
- Resummation of \(\Delta\phi\) works similar to \(\chi\).
- We can resum parts of \(\cS_r\).
- We can resum \(f_{1,2}\) contributions for \(r_\textrm{min}\) and \(r_\pm\). ("closed form" for real positive roots of arbitrary polynomial?)
- It becomes hard. More work needs to be done!
Conclusions and Outlook
- Mapping scattering data to orbital observables does not require a Hamiltonian. Analytic continuation does the job.
- The (classical) amplitude, the scattering angle, and the momentum contain the same information.
- Non-perturbative map invites for numerical studies.
- We can resum. But more work needs to be done!
- Generalization works for aligned spins.
- Continue to arbitrary spin? Radiation? Radiation-reaction?
How would the proposed correction in [Damour 19] change the results?
Discrepancy at 6PN! Momentum/amplitude coefficients: $$\delta f_3 = f_3^\textrm{(BCRSSZ)} - f_3^\textrm{(Damour)} = -\frac{1024\nu}{315}\cE^3 + \cO(\cE^4)$$ Periastron advance: $$\delta (\Delta\Phi) = \Delta\Phi^\textrm{(BCRSSZ)} - \Delta\Phi^\textrm{(Damour)} = -\frac{1024\nu}{105j^4}\cE^4 + \cO(\cE^5)$$ Binding energy for the circular orbits: $$\delta\epsilon= \epsilon^{\textrm{(BCRSSZ)}}-\epsilon^{\textrm{(Damour)}} = \frac{1408 x^7 \nu}{945} +\cO(x^8)$$