# The Pirate Solution

## WIMP Dark Matter Direct Detection

All direct WIMP dark matter detectors aim to observe nuclear recoils produced by WIMP-nucleon scattering. WIMPs interact with all target materials in exactly the same way as neutrons and generally scatter off nuclei rather than electrons due to their relatively large mass. There are a few different methods that can be used to detect nuclear recoils, including collecting ionization, scintillation or thermal energy deposition data. The energy spectra of nuclear recoils for different target materials and WIMP masses can be simulated and calculated, then used to compare with real data from detectors to aid in the identification, or in placing an upper limit on characteristics (usually mass and WIMP-nucleon scattering cross-section) of WIMPs.

### Rates for Direct Detection of Dark Matter

#### Direct Detection Formalism

In this exercise we want to compute the cross section relevant for “direct” detection of Dark Matter WIMPs ${\chi},$ which is based on the elastic scattering of WIMPs off nuclei in a detector. In the case of spin-independent interactions, the procedure is fairly straightforward. The nucleus can be approximated as a collection of nucleons with overlapping wavefunctions, so an effective WIMP-nucleus interaction may be written:

${{\scriptstyle\mathcal{L}}_{{\chi}{(n,p)}} = f_{(n,p)} {\overline {\chi}}{\chi}{\sum_{k=1}^{A}{({\overline n}, {\overline p})}_k}(n,p)_k}.$

$f_{(n,p)}$ are the coupling strengths of the dark matter particle to neutrons and protons respectively. $f_{(n,p)}$ are calculated from a coherent sum over the couplings to the quark model constituents of the nucleon. $A$ is the mass number of the nucleus. Integrating this interaction-Lagrangian density over space to get the matrix element for a transition from a nucleus at rest to a nucleus with momentum $\vec {q}$ gives

$M(\vec{q}) = [{Z}{f_p} + {(A - Z)}{f_n}]\int {\rho}(\vec{r}){e^{i\vec{q}\cdot\vec{r}}}\,d^{3}\vec{r}$
$\equiv [{Z}{f_p} + {(A - Z)}{f_n}]{F(\vec{q})},$

${\rho}(\vec {r})$ is the probability of a given nucleon being at $\vec {r},$ averaged over all nucleons. $F(\vec {q}) = F(q)$ is the Fourier transform of ${\rho}(\vec {r})$ which is called the “nuclear form factor” because it expresses the dependence of the interaction on the shape of the nucleus, given by ${\rho}(\vec {r}).$ $Z$ is the proton number of the nucleus. Under the assumption that electric charge is distributed in the same way as the nucleons (i.e., there is no segregation of neutrons and protons), low-energy lepton-nucleus elastic scattering can be used to determine $F(\vec {q})$ and ${\rho}(\vec {r}).$ It is observed that low-energy scattering is isotropic, so $F(\vec {q}) = F(q)$ and ${\rho}(\vec {r}) = {\rho}(r).$ Calculation of the differential cross section from the above matrix element for elastic ${\chi}{N}$ → ${\chi}{N}$ scattering and Fermi's Golden Rule gives

$\frac{d{\sigma}}{dq^2} = \frac{1}{{\pi}{v^2}}|M (\vec {q})|^2 = \frac{1}{{\pi}{v^2}} {[{Z}{f_p} + {(A - Z)}{f_n}]^2}{F^2(q^2)}.$

The factor $\frac{1}{{\pi}{v^2}}$ arises from final-state density of states and the standard Golden Rule ${2}{\pi}/{\hbar}$ factor. This may be written in the form

$\frac{d{\sigma}}{dq^2} = {\frac{{\sigma}_0}{{4}{{\mu}^2}{v^2}}}{F^2(q^2)}$

where ${\mu} = {m_N}{m_{\chi}}/{m_N + m_{\chi}}$ is the reduced mass of the nucleus and WIMP and $v$ is the relative WIMP-nucleus velocity. ${\sigma}_0$ gives the "pointlike" total cross section if there were no form-factor suppression which corrects for the above formula being strictly correct only as the momentum transfer $q^2 = 2{m_N}{E_R}$ → $0:$

${\sigma}_0 = \frac{4}{\pi}{{{\mu}_N}^2}{[{Z}{f_p} + {(A - Z)}{f_n}]^2}.$

Direct DM search collaborations quote constraints on WIMP-nuclei cross sections normalized to the WIMP-nucleon cross section. The total WIMP-nucleon cross section is given by

${\sigma}_p = \frac{4}{\pi}{{{\mu}_p}^2}{f_p}^2.$

The tiny mass difference between a proton and a neutron has been neglected. The above definitions yield :${\sigma}_0 \approx {(\frac{{\mu}_N}{{\mu}_p})^2}{{\sigma}_p}{\frac{[{Z}{f_p} + {(A - Z)}{f_n}]^2}{{f_p}^2}}.$ The WIMP-nucleus cross section is usually written differentially in the recoil energy ${E_R}$, i.e., ${E_R}$ is the kinetic energy of the struck nucleus in the lab frame:

$\frac{d{\sigma}}{dE_R} = {\frac{m_N}{{2}{v^2}}}{\frac{{\sigma}_p}{{\mu}_n^2}{\frac{[{Z}{f_p} + {(A - Z)}{f_n}]^2}{f_n^2}}{F^2(E_R)}}$          $(1)$

if $f_n = f_p$ (which is true in most WIMP models, up to small corrections, e.g., for the lightest supersymmetric neutralino, and for all WIMPs which interact primarily through Higgs exchange) and ${\mu}_n^2 = {\mu}_p^2$ are assumed. The differential WIMP-nucleus scattering cross section depends on astrophysical, particle physics, and nuclear physics inputs. We calculated the differential cross section for the scattering of a WIMP with given energy on a nucleus. Here we want to calculate the differential scattering rate, for given cross section and given WIMP distribution function. Recall that the scattering rate is the product of cross section and incident flux. In general, the rate of interactions per unit mass of target material of a particle with interaction cross section ${\sigma}$ (particle physics input) is given by

$dR = {N_T}{\sigma}{v}{dn}$

where $N_T$ is the number of target nuclei per unit mass (detector physics input), $v$ is the dark matter particle velocity in the lab frame (astrophysics input), and $dn$ is the differential particle density. Including the normalization factor $k$ gives

$dn = {\frac{n_0}{k}}{f(\vec {v}, \vec {v}_{\oplus})}d^3v$

with

$n_0 \equiv \int_{0}^{v_{esc}} dn$

and

$k = {\int_{0}^{2 {\pi}} d{\phi}}{\int_{-1}^{+1} d(cos {\theta})}{\int_{0}^{v_{esc}} d^3{v}{v^2}{f(\vec {v}, \vec {v}_{\oplus})}}.$ Here $n_0$ is the mean dark matter particle number density ($= {{\rho}_{DM}}/{m_{DM}}$ for dark matter particle mass $m_{DM}$ [particle physics input], local dark matter halo density ${\rho}_{DM} = .3 GeV/{cm}^3$ [astrophysics input]), $v$ is velocity onto the (Earth-borne) target, $v_{\oplus}$ is Earth (target) velocity relative to the dark matter distribution, and $v_{esc}$ is the local Galactic escape velocity (astrophysics inputs); $dn$ is the particle density of dark matter particles with relative velocities within velocity space volume element $d^3{v}$ about $\vec {v}.$ Then, the total event rate is
$R = {N_T}{\sigma}{\int vdn} \equiv {N_T}{\frac{{\rho}_{DM}}{m_{DM}}}{\sigma}{\langle v \rangle},$

where $\langle v \rangle$ is the average dark matter particle velocity in the lab frame. The differential rate per unit detector mass in nuclear recoil energy is given by

$\frac {dR}{dE_R} = {N_T} \frac {{\rho}_{DM}}{m_{DM}}{{k^{-1}}(v_{esc})} \int_{\vert \vec {v} \vert > {v_{min}}} d^3{v}{v}f(\vec {v}, \vec {v}_{\oplus}){\frac {d{\sigma}}{dE_R}}.$

The quantity $k (v_{esc})$ is the normalization $k$ with the dependence on $v_{esc}$ shown explicitly. The WIMP velocity distribution is given by the WIMP phase-space distribution. The full phase-space distribution function for an isothermal halo composed of particles of mass $m_{DM}$ is

$f(\vec {r}, \vec {v}){d^3r}{d^3v} \propto exp \Bigg(-\frac{{m_{DM}}{v^2}/{2} + {m_{DM}}{{\phi} (\vec r)}}{{k_B}{T}}\Bigg)$

which is a Boltzmann distribution with the gravitational potential ${\phi}(\vec {r})$ included. At a given point, the position-dependent factor is fixed and can be included in the normalization, leaving a simple Maxwellian velocity distribution: :$f(\vec {v}, \vec {v}_{\oplus}) = \frac {1}{({\pi}{v_0}^2)^{3/2}}e^{-{(\vec {v} + \vec {v}_{\oplus})^2}/{v_0}^2}.$ From the observational side, the most relevant piece of information coming from astrophysics is related to the rotational velocity of objects bounded to the Galaxy: ${m}{v_{\oplus}}^2/2 = {m_{DM}}{\phi}.$ The quantity ${v_0}^2$ is characteristic of the WIMP kinetic energy $m{v_0}^2/2 = {k_B}T.$ It corresponds to the most probable velocity and is known as the velocity dispersion. For $v_{esc} = \infty, k = k_0 = ({\pi}{v_0^2})^{3/2}.$ The velocity distribution entering the differential rate is evaluated in the frame of the detector, therefore we need to transform the galactic velocity distribution to the Earth frame via a Galilean transformation. "Galactic" $(-{\hat {r}}, \hat {\phi}, \hat {z})$ coordinates are usually used. $f(\vec {v}, \vec {v}_{\oplus})$ and $f(\vec {v})$ are simply related through: :$f(\vec {v}, \vec {v}_{\oplus}) = f(\vec {v} + \vec {v}_{\oplus})$ where $\vec {v}_{\oplus} = \vec {v}_{\odot} + \vec {v}_{{\oplus}orb}$ ($*$). $\vec {v}_{\oplus}$ and $\vec {v}_{\odot}$ denote the velocities of the Earth and the Sun in the Galactic rest frame and $\vec {v}_{{\oplus}orb}$ is the Earth’s orbital velocity around the Sun. $\vec {v}_{\odot} = \vec {v}_{{\odot}circ} + \vec {v}_{{\odot}pec}$ is the sum of the Sun's circular velocity $\vec {v}_{{\odot}circ}$ relative to the galactic center, and peculiar velocity $\vec {v}_{{\odot}pec}$ relative to Sun's circular velocity. Projecting $Eq. (*)$ in the galactic plane, one gets the Earth’s speed relative to the galactic halo: :$\vert {\vec {v}_{\oplus}} \vert = \vert {\vec {v}_{\odot}} \vert + {\vert {\vec {v}_{{\oplus}orb}} \vert}{{cos}{\gamma}{cos}[{\omega}{(t - t_0)}]}$ where $\gamma$ is the inclination of the plane of rotation with respect to the galactic one and $t_0$ corresponds to the day when the Earth’s velocity is at its maximum. $v_{\odot} = v_0 + 12 km/s,$ $v_{{\oplus}orb} = 30km/s,$ ${cos}{\gamma} = 0.51,$ $t_0 = June \ 2nd,$ and $\omega = 2{\pi}/year$. We set the most probable dark matter speed in the galactic frame to $v_0 = 220 km/s$. The Maxwellian distribution is cut off at $\vert \vec {v} + \vec {v}_{\oplus} \vert = v_{esc}$ by the halo escape velocity. Note that the cutoff is isotropic in the galactocentric WIMP velocity $v + v_{\oplus}$ not in the Earth-centric $v.$ Heuristically, we may write :$\# \ recoil \ events = {\frac {rate}{nucleus}} \cdot {time}.$ The total number of nuclear recoil events in a recoil energy range between $E_{r}^{1}$ and $E_{r}^{2}$ is

$N = \sum_{i} \int_{E_{r}^{1}}^{E_{r}^2} {\frac {{d}{R}_{i}}{{d}{E}_{r}}}{\frac {{\varepsilon}_{i}}{{N_T}{m_i}} {d}{E_r}}$          $(2)$

where the sum is over each nuclear species $i$ in the detector. ${\varepsilon}_i$ is the effective exposure of species $i$ expressed in kg-days. We can parametrize ${\varepsilon}_i = {m_i}{t}{\epsilon},$ where $t$ is the time of exposure, $m_i$ is the target mass of species $i,$ and $\epsilon$ is a detection efficiency.

Sheldon was calculating the rate for the wrong target material, xenon not sodium. Some experiments competing to find dark matter by observing directly the extremely small amount of energy a dark matter particle might deposit in a detector as they pass through Earth use Sodium (which has an atomic mass 23) and others use Xenon (with atomic mass 131). This is why Raj erases 131 and changes it to 23.

In The Pirate Solution, British astronomer Professor Laughlin's first name is never stated. However, the plaque on his door reads: PROF. ___ LAUGHLIN, PhD. (first line) ASTRONOMY (second line). From this Blu Ray HD image, it appears that the obscured name on the plaque is IAN. If this is the case, then he has the same name as a Marvel Comics colorist.