Chapter 10 · Section 5 of 7
Estimate the WIMPs' flow onto the Earth's surface.
Using \( \rho _0 \approx 0.3\mbox{GeV/cm}^3 \) under assumption $m_\chi\approx 100\mbox{GeV}$, one obtains for the WIMP number density the following expression \( n = \rho _0 /m_\chi \approx 3 \times 10^{ - 3} \mbox{cm}^{ - 3} \). For \( \left\langle v \right\rangle \simeq 250\mbox{km/s} \) the flow equals \(J = n\left\langle v \right\rangle \sim 10^5 \mbox{cm}^{ - 2} \mbox{s}^{ - 1}\)
What are the main processes due to which the WIMPs can be detected?
WIMP can dissipate their energy via the following three main processes:
1) creation of ionized system which then can emit light.
2) Motion of the recoil nucleus in the lattice can induce creation of the vibrational phonons.
3) The energy converted into the phonons can eventually thermalize and lead to minor increase in the detector temperature.
Therefore the recoil nucleus can induce three kinds of signals:
1) ionization (charge);
2) scintillation (light);
3) heat (phonons).
Show that WIMPs of mass $\sim100GeV$ being elastically scattered on Xenon nuclei with mass $\sim130GeV$ lead to energy recoil $\le40keV$.
As WIMP are gravitationally coupled to the Galaxy (they move in the common gravitational well), it is naturally to assume that the average velocity of WIMP is approximately the same as that of stars in vicinity of the Sun: \(\left\langle v \right\rangle \simeq 270\mbox{ km/s}.\) Then one obtains for mean kinetic energy of WIMP prior to the collision \( \left\langle {E_\chi } \right\rangle \approx 40\mbox{ keV}.\) The recoil energy of the target nucleus acquired in elastic collision equals to \[ Q = \left[ {\frac{4m_\chi m_N }{\left( {m_\chi + m_N }\right)^2 }\cos \theta _{Lab} } \right]E_\chi, \] where $\theta _{Lab} $ is the scattering angle in the laboratory reference frame. Then for $m_\chi\sim 100\mbox{GeV}$ and $m_N \sim 130\mbox{GeV}$ one finds $Q_{\max } \simeq 40\mbox{ keV}$.
Show that the WIMPs' elastic scattering experiments will be most efficient if the target nuclei's mass is comparable with the WIMPs' mass.
Consider the recoil energy as a function of the nucleus mass at given WIMP mass (see problem ), and show that it has maximum at $m_N = m_\chi.$
Determine the minimum velocity of WIMPs that can transmit energy $Q$ to a nucleus with mass $m_N$.
The transmitted momentum $q$ is related to the scattering angle in the center-of-mass reference frame $\theta _{CM} $ by the following \[ q = \frac{2m_N m_\chi v}{m_\chi + m_N } \sin \frac{\theta _{CM} }{2} = 2m_r v\sqrt {\frac{{1 - \cos \theta _{CM} }}{2}} \] As $0 \le 1 - \cos \theta _{CM} \le 2$ then for the given value $Q$ transmitted to the detector one gets $\left( {q = \sqrt {2m_N Q} } \right)$ and thus \[ v_{\min } = \frac{\sqrt {2m_N Q}}{2m_r}. \]
Estimate the count rate for the detector registering elastic events of WIMPs.
The counting rate per unit mass of the detector is proportional to the incident flow of
WIMP and cross-section of the WIMP-nucleus elastic scattering $\sigma_{\chi N} \simeq \alpha ^2 /m_\chi ^2 $ , where $\alpha
\simeq {\rm O}\left( {10^{ - 2} } \right)$ is the weak interaction constant. Therefore $\sigma \approx 10^{ - 9} \mbox{GeV}^{ -
2} \approx 1\mbox{pb}$. (Recall that $1\mbox{GeV}^{ - 1}
\approx 2 \times 10^{ - 14} \mbox{cm}$.)
Note that the experimenters prefer deal with the
WIMP-nucleon cross-section rather than WIMP-nucleus one. The former is of order of
$\sigma_{\chi n} \approx 10^{ - 8} \mbox{pb}.$ The very rough estimate of the detector counting rate gives $R \sim J\sigma /m_N \approx 10 \mbox{events}\,\mbox{kg}^{ - 1} \mbox{year}^{ - 1} $
(for a target with nuclei containing about $100$ nucleons each, i.e. $m_N
\sim 100\mbox{GeV} = 1.77 \times 10^{ - 27} \mbox{kg}$ ). It means that every day several WIMP (precise number depends on the detector material) will hit the atomic nuclei which the detector is composed of. Of course the above given speculations are nothing ore than an estimate. Accurate calculations should take into account the interaction of WIMP with quarks and gluons, then its consequent translation into the interaction with nucleons and at last to the interaction with the nuclei.
Obtain the expression for the count rate of the detector registering elastic events of WIMPs.
Differential counting rate for the detector equals
\[
\frac{dR}{dQ} = \frac{1}{m_N }n\left\langle v \right\rangle
\frac{d\sigma }{dQ},\] where $Q$
is the energy transmitted to the detector nucleus in elastic collision with the
WIMP.
The differential cross-section depends on the elastic nuclear form-factor
$F(q)$
\[
d\sigma = \frac{1}{v^2}\frac{\sigma _0}{4m_r^2}F^2
(q)dq^2,
\]
where $\sigma _0 $ is the total cross-section, \( m_r = m_\chi m_N /(m_\chi+ m_N ) \) is the reduced mass, $q = \sqrt {2m_N Q}$ is the transmitted momentum. Thus the differential counting rate per unit mass of the detector can be presented in the following form:
$$
\frac{dR}{dQ} = \frac{\rho _0 }{m_\chi m_N }\int vf_1 (v)d\sigma dv
= \frac{\rho _0 \sigma _0 }{2m_\chi m_r^2 }F^2 (Q)\int \frac{f_1
(v)}{v} dv,
$$
where $f_1 (v)$ is one-dimensional velocity distribution of WIMP, colliding the detector,
$v$ is absolute value of the WIMP velocity in the reference frame of the Earth. For the simplest case of spherical isothermal halo with Maxwellian velocity distribution of WIMP one obtains
\[
f(v) = \frac{1}{\pi ^{3/2} v_0^3 } exp\left( { - \frac{v^2}{v_0^2}} \right),
\]
where $v_0 $ is orbital velocity of the Sun in the glactocentric reference frame. As $d^3 v = v^2 dvd\Omega $ , then for arbitrary function $F(v)$
\[
\int {F(v)f(v)d^3 v = \int {F(v)} } f_1 (v)dv,
\]
where
\[
f_1 (v) = \frac{4}{\sqrt \pi }\frac{v^2 }{v_0^3 } \exp
\left( { - \frac{v^2 }{v_0^2 }} \right).
\]
it is easy to obtain that
\[
\begin{array}{l}
\left\langle v \right\rangle _{(1)} = \int_0^\infty {vf_1 (v)dv = \frac{\sqrt 2}{\pi } } v_0 ; \\
\left\langle {v^2 } \right\rangle _{(1)}^{1/2} = \left( {\int_0^\infty {v^2 f_1 (v)dv} } \right)^{1/2} = \sqrt {\frac{3}{2}}
v_0.
\end{array}
\]
One should integrate the expression $dR/dQ$ over all possible velocities. In the problem of the present chapter we have shown that the minimum velocity of WIMP, that can transmit energy $Q$ to a nucleus with mass $m_N $, equals \[ v_{\min } = \frac{\sqrt{2m_N Q}}{2m_r}.
\] Therefore the differential counting rate can be presented in the form
\[
\frac{dR}{dQ} = AF^2 (Q)\int_{v_{\min } }^\infty { {\frac{f_1(v)}{v}} dv},
\]
where the constant coefficient is
\[
A \equiv \frac{\rho _0 \sigma _0 }{2m_\chi m_r^2 }.
\]
Finally one obtains for total events number of the elastic scattering registered by the detector unit mass per unit time the following
\[
R = \int_{Q_{th} }^\infty {\left( \frac{dR}{dQ} \right)dQ},
\]
where $Q_{th} $ is the threshold value of the detector energy.
The result is obtained under the assumption that the target is homogeneous with respect to its isotopic composition. if the target contains different isotopes
(for example $NaI$ crystals), then the ultimate expression should be replaced by the sum of terms that account for contribution of separate isotopes.
Reconstruct the one-dimensional WIMPs' velocities distribution $f_1(v)$ (see the previous problem) basing on the given count rate of the elastic events.
Construct a model-independent scheme for WIMPs' mass determination using the results of WIMPs' elastic scattering events detection for two or more sets of experimental data with detectors of different composition.
The main idea of the scheme is to use the fact that the velocity distributions of the incident WIMP (in particular $f_1 (v)$ ), obtained from the measurements of the recoil spectrum of the detector nuclei, in the low energy range must be the same for detectors made of different materials.
Show that if a WIMP has the mass of the order $100GeV$ and velocity of the order of $300km/s$, then it coherently interacts with nucleons of a detector nuclei.
It is sufficient to show that the WIMP wave-length is of order of or greater than the size of the nuceus. The latter is \[ R = 1.25\,A^{1/3} 10^{ - 13} \mbox{cm}, \] where $A$ is number of nucleons in the nucleus. For $A = 100$ \[ R \simeq 0.58 \times 10^{ - 12} \mbox{cm} \] The wavelength of the WIMP is \[ \lambda = \frac{h}{p} = 2\pi \frac{\hbar }{m_p c}\frac{m_p}{m_\chi}\frac{c}{v} \simeq 2\pi \cdot 2 \cdot 10^{ - 14} \mbox{cm} \cdot 10^{ - 2} \cdot 10^3 \simeq 1.3 \cdot 10^{ -12} \mbox{ cm} \] It is easy to see that the coherence conditions are fulfilled.
Find the total WIMP-nucleus cross-section determining the elastic events' count rate.
The WIMP-nucleus elastic scattering process in the non-relativistic case can be naturally divided to spin-dependent (SD) and spin-independent (SI) part. The latter coherent scattering is amplified by the factor $A^2$, which is especially manifested in targets of heavy nuclei. This amplification is due to the fact that the WIMP wavelength is of order of the size of nucleus (see problem). Therefore the scattering amplitudes on individual nucleons add coherently,
\[\sigma _0^{(SI)} = \frac{4m_r^2}{\pi }\left[{Zf_p + (A - Z)f_n}\right]^2,\]
where $f_{p,n} $ is the amplitude of the interaction of WIMP with proton and neutron. Under assumption $f_p \simeq f_n $, one can see that $ \sigma _0^{(SI)} \propto A^2,$ i.e. due to the coherence effect the spin-independent cross-section depends quadratically on the target nucleus mass. As the consequence the targets of heavier nuclei are more preferable to investigate the spin-independent contribution into the total cross-section of the elastic scattering.
The spin-dependent contribution into the total cross-section due the spin-spin interaction equals
\[
\sigma _0^{(SD)} = \frac{32m_r^2}{\pi }{\Lambda ^2 J(J + 1)},\]
where the constant $\Lambda $ and the total angular momentum $J$ of the nucleus are determined by the axial coupling constant of WIMPs with quarks.
The total cross-section of elastic scattering of WIMP on nuclei equals to sum of spin-dependent and spin-independent parts. In the latter case a universal parametrization for nuclear form-factor can be found, while for the spin-dependent case an additional difficulty arises due to the fact that the form-factor changes irregularly from one nucleus to another. For nuclei with $A \ge 30$ the spin-independent interaction dominates due to the coherence effect $\left( { \propto A^2 } \right)$. However it is the utilization of detectors with non-zero spin nuclei that will possibly help to solve the dark matter detection problem. in particular, the specific dependence of the nucleus form-factor on the transmitted momentum in the case of spin-dependent interaction will allow to extract the useful signal. From the microscopic point of view the amplitudes $f_{p,n} $ for interaction of WIMP with proton and neutron can be presented in the form:
\[
f_{p,n} = \sum\limits_{q = u,d,s} {f_{T_q }^{(p,n)} a_q
\frac{m_{p,n}}{m_q} + \frac{2}{27}f_{TG}^{(p,n)}
\sum\limits_{q = c,b,t} {a_q \frac{m_{p,n}}{m_q}} },
\]
where $a_q $ are the WIMP-quark coupling constants and the quantity $f_{T_q
}^{(p,n)}$ is determined by the quark structure of the nucleon. The first term corresponds to interaction of WIMP with quarks in the target nucleus and the second - to that with gluons.
How will the number of counts for the elastic events detector be affected by transition to heavier target nuclei at fixed detector mass?
For fixed detector mass, transition to heavier target nuclei effectively reduces number of objects interacting with the incident WIMP and thus the number of registered events reduces too.
Show that if the WIMPs' mass is of the order of $100GeV$ then the elastic spin-independent cross-section for WIMPs on a nucleus with $A\sim100$ is eight orders of magnitude larger then the corresponding cross-section on a nucleon
As $\sigma \propto m^2_r A^2 $ , then amplification in four orders of magnitude is due to the transition $1 \to A$ and else four orders - to the transition $m_r^{nucleon} \to m_r^{nuclei}.$
Estimate the annual modulations' amplitude of the WIMPs elastic scattering cross-section.
Estimate the diurnal modulations' amplitude of the WIMPs elastic scattering cross-section.