Advanced Topics in Cosmology

A galaxy cluster at $z = 0.5$ has a gravitational potential well with depth $\delta\Phi/c^2 = -10^{-4}$.

(a) Sachs—Wolfe contribution at last scattering (from primordial potential):

$\frac{\Delta T}{T} \approx \frac{\delta\Phi}{3c^2} = \frac{-10^{-4}}{3} \approx -3.3 \times 10^{-5}$

This is of the same order as the observed CMB anisotropy ($\Delta T/T \sim 10^{-5}$).

(b) The Sunyaev—Zel”dovich (SZ) effect from hot electrons in the cluster:

$\frac{\Delta T}{T}\bigg|_{\text{SZ} = -2\int \frac{k_B T_e}{m_e c^2}\, n_e \sigma_T\, dl}$

For a typical cluster with $k_BT_e \sim 5$ keV, $n_e \sim 10^3$ m$^{-3}$, $l \sim 1$ Mpc:

$y = \frac{k_BT_e}{m_e c^2} n_e \sigma_T l = \frac{5 \times 10^3 \times 1.602 \times 10^{-16}}{9.109 \times 10^{-31} \times 9 \times 10^{16}} \times 10^3 \times 6.65 \times 10^{-29} \times 3.086 \times 10^{22}$

$= 8.8 \times 10^{-2} \times 10^3 \times 6.65 \times 10^{-29} \times 3.086 \times 10^{22} \approx 1.8 \times 10^{-4}$

The SZ effect ($\Delta T/T \approx -2y$ at low frequency) gives a temperature decrement of $\sim 3.6 \times 10^{-4}$Significantly larger than the primary CMB anisotropy.

The type-I seesaw mechanism adds three right-handed neutrinos $N_R$ with Majorana mass $M$. The light neutrino mass matrix:

$m_\nu \approx -m_D M^{-1} m_D^T$

For a single generation with $m_D \sim 173$ GeV (top Yukawa-scale Dirac mass) and $M = 10^{15}$ GeV:

$m_\nu \sim \frac{(173\ \text{GeV})^2}{10^{15}\ \text{GeV} = \frac{2.99 \times 10^4}{10^{15}}\ \text{GeV} = 3.0 \times 10^{-11}\ \text{GeV} = 0.030\ \text{eV}}$

This is in the right ballpark for atmospheric neutrino oscillations ($\Delta m_{32}^2 \approx 2.5 \times 10^{-3}$ eV$^2$Giving $m_3 \sim 0.05$ eV).

For three degenerate right-handed neutrinos with $M \sim 10^{14}$ GeV:

$m_\nu \sim 0.3\ \text{eV}$

This is at the upper edge of the cosmological bound $\sum m_\nu < 0.12$ eV, showing that the seesaw with $M \sim 10^{14}$—$10^{15}$ GeV explains the tiny neutrino masses.

The quark distribution function $q(x, Q^2)$ evolves according to the DGLAP equation. At leading order:

$\frac{\partial q(x, Q^2)}{\partial \ln Q^2} = \frac{\alpha_s(Q^2)}{2\pi}\int_x^1 \frac{dz}{z}\, P_{qq}(z)\, q\!\left(\frac{x}{z}, Q^2\right)$

(a) Show that the number sum rule $\int_0^1 q(x, Q^2)\,dx$ is independent of $Q^2$.

(b) Using $P_{qq}(z) = C_F\left[\frac{1+z^2}{(1-z)_+}\right]$ where $(1-z)_+$ is the plus prescription, and the leading-order running $\alpha_s(Q^2) = \alpha_s(Q_0^2)/(1 + b_0\alpha_s(Q_0^2)\ln(Q^2/Q_0^2)/2\pi)$Show that the average momentum fraction $\langle x \rangle_q$ decreases with $Q^2$.

Solution:

(a) Integrating both sides over $x$:

$\int_0^1 dx\,\frac{\partial q}{\partial \ln Q^2} = \frac{\alpha_s}{2\pi}\int_0^1 dx\int_x^1 \frac{dz}{z}\, P_{qq}(z)\, q\!\left(\frac{x}{z}\right)$

Change variables $y = x/z$, $x = yz$, $dx = z\,dy$:

$= \frac{\alpha_s}{2\pi}\int_0^1 dz\, P_{qq}(z)\int_0^1 dy\, q(y) = \frac{\alpha_s}{2\pi}\left[\int_0^1 P_{qq}(z)\,dz\right]\left[\int_0^1 q(y)\,dy\right]$

The integral of the splitting function $\int_0^1 P_{qq}(z)\,dz = 0$ (the plus prescription ensures this), so the RHS vanishes. Therefore $\partial/\partial(\ln Q^2) \int_0^1 q(x)\,dx = 0$.

(b) Multiply the DGLAP equation by $x$ and integrate:

$\frac{\partial}{\partial\ln Q^2}\int_0^1 xq(x)\,dx = \frac{\alpha_s}{2\pi}\int_0^1 dz\, P_{qq}(z)\int_0^1 yz\, q(y)\,z\,dy$

Wait, more carefully. With $x = yz$:

$= \frac{\alpha_s}{2\pi}\int_0^1 dz\, z\, P_{qq}(z)\int_0^1 y\, q(y)\, dy$

The first moment of $P_{qq}$ is $\int_0^1 z\,P_{qq}(z)\,dz = -A_q < 0$ (the quark loses momentum to gluons). Since $\alpha_s > 0$:

$\frac{\partial\langle x\rangle_q}{\partial\ln Q^2} = \frac{\alpha_s}{2\pi}(-A_q)\langle x\rangle_q < 0$

The quark momentum fraction decreases with $Q^2$ because quarks radiate gluons.

(a) Estimate the initial temperature of the QGP produced in Pb—Pb collisions at the LHC, given that $\sim 1600$ charged particles per unit rapidity are produced and the Bjorken energy density estimate gives $\epsilon = \frac{1}{\tau_0 A_T}\frac{dE_T}{dy}$ with $\tau_0 = 1$ fm/c and $A_T = \pi(7.1\ \text{fm})^2$.

(b) Calculate the Debye screening mass $m_D$ at this temperature and estimate the screening length.

Solution:

(a) The transverse energy per unit rapidity is roughly $dE_T/dy \sim 1600 \times 0.5\ \text{GeV} \sim 800$ GeV (each charged particle carries $\sim 0.5$ GeV of $E_T$ on average).

\epsilon = \frac{800\ \text{GeV}{(1\ \text{fm}/c) \times \pi \times (7.1\ \text{fm})^2} = \frac{800 \times 1.602 \times 10^{-10}\ \text{J}{10^{-15}\,\text{s} \times \pi \times 5.04 \times 10^{-30}\,\text{m}^2}}}

$= \frac{1.28 \times 10^{-7}}{1.58 \times 10^{-44}} = 8.1 \times 10^{36}\ \text{J}/m^3$

Using the Stefan—Boltzmann relation for an ideal QGP with $N_f = 2.5$ effective flavours (including gluons):

$\epsilon = \frac{\pi^2}{30}\left(2 \times 8 + \frac{7}{4} \times 2 \times 2.5 \times 2\right)T^4 = \frac{\pi^2}{30}(16 + 35)T^4 = \frac{\pi^2}{30} \times 51\, T^4$

$T = \left(\frac{30\epsilon}{51\pi^2 k_B^4}\right)^{1/4}$

In natural units ($\epsilon$ in GeV/fm$^3$):

$\epsilon \approx \frac{800}{\pi \times 50.4}\ \text{GeV}/fm^3 \approx 5.0\ \text{GeV}/fm^3$

$T = \left(\frac{5.0 \times 30}{51\pi^2}\right)^{1/4}\ \text{GeV} \approx (0.94)^{1/4}\ \text{GeV} \approx 0.985\ \text{GeV} \times 197\ \text{MeV}/fm^{1/4}$

Actually, using $\epsilon = a_{\text{QGP} T^4}$ with $a_{\text{QGP} \approx 47.5\pi^2/30}$:

$T \approx \left(\frac{5.0}{15.6}\right)^{1/4} \approx 0.79^{1/4} \approx 0.94\ \text{GeV}^{1/4}$

This gives $T \approx 300$—$400$ MeV, consistent with LHC measurements.

(b) The Debye screening mass in QCD:

$m_D^2 = g_s^2 T^2\left(\frac{N_c}{3} + \frac{N_f}{6}\right) = g_s^2 T^2\left(1 + \frac{N_f}{6}\right)$

With $N_c = 3$, $N_f = 3$, $g_s^2/(4\pi) = \alpha_s \approx 0.3$ at $T \sim 300$ MeV:

$m_D^2 = 4\pi \times 0.3 \times (0.3\ \text{GeV})^2 \times (1 + 0.5) = 3.77 \times 0.09 \times 1.5 = 0.509\ \text{GeV}^2$

$m_D \approx 0.71\ \text{GeV}, \quad \lambda_D = 1/m_D \approx 0.28\ \text{fm}$

This screening length ($\sim 0.3$ fm) is much shorter than the typical hadron size ($\sim 1$ fm), confirming that colour forces are screened in the QGP and quarkonium states are dissociated.