How warm and fuzzy dark matter erase small ripples

Read How a cusp or a core shapes a rotation curve first. You need waves and wavelength, powers and logarithms, and the idea of a mean square. The aims are to read a transfer function, convert a cutoff scale into a halo mass, and compute how much warm or fuzzy dark matter suppresses halo numbers at a chosen mass.

Ripples measured by wavenumber

Describe the matter by its density contrast $\delta=(\rho-\bar\rho)/\bar\rho$, the fractional excess over the cosmic mean $\bar\rho$. Any pattern of $\delta$ can be written as a sum of waves. A wave with wavelength $\lambda$ has wavenumber $k=2\pi/\lambda$, so large $k$ means small ripples. Cosmologists measure $k$ in comoving units, which divide out the overall expansion, usually in $h\,\mathrm{Mpc^{-1}}$. Here $h=0.6766$ is the Hubble constant in units of $100\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$, and one megaparsec is 1000 kiloparsecs.

The power spectrum $P(k)$ is the mean square amplitude of the waves with wavenumber near $k$. While ripples are small, each wave grows on its own, without mixing with the others. This is the linear regime, and it is where the differences between dark matter models are set.

The transfer function and the half-mode scale

Compare a model X with cold dark matter by the ratio of their linear power spectra:

$T^2=1$ means the model keeps the cold ripples; $T^2=0$ means it erases them. Nadler et al. 2021, equation 1 The half-mode wavenumber $k_{\mathrm{hm}}$ is where $T^2=1/4$, so the amplitude $T$ has fallen to one half. A sphere whose radius is half that wavelength, $\pi/k_{\mathrm{hm}}$, filled at the mean matter density, holds the half-mode mass

Below this mass, halos are significantly rarer than in cold dark matter. Nadler et al. 2021, equation 2 Remember that $k$ is in $h\,\mathrm{Mpc^{-1}}$: divide the radius by $h$ before using $\rho_{\mathrm{crit}}$ in solar masses per cubic kiloparsec.

Warm dark matter streams out of small ripples

Warm particles were still moving at a noticeable fraction of light speed when the first ripples began to grow. A particle that crosses a ripple before gravity can deepen it smooths that ripple away. The lighter the particle, the faster it moved, and the larger the ripples it erased. For a thermal relic, a particle that was once in thermal equilibrium, Viel and collaborators fitted Boltzmann-code results with

Viel et al. 2005, equations 6 and 7 Setting $T=1/2$ and solving gives the half-mode wavenumber in closed form:

Example 3.1. Take a 3 keV particle with $\Omega_{\mathrm{WDM}}=0.2607$, all the dark matter in our reference cosmology. Then $\alpha=0.01395\,h^{-1}\,\mathrm{Mpc}$. The bracket $(2^{0.224}-1)^{0.446}$ equals $0.451$, so $k_{\mathrm{hm}}\simeq32.3\,h\,\mathrm{Mpc^{-1}}$. The half wavelength is $\pi/k_{\mathrm{hm}}=0.0972\,h^{-1}\,\mathrm{Mpc}$, or $143.6\,\mathrm{kpc}$. With $\Omega_{\mathrm m}\rho_{\mathrm{crit}}=39.3\,\mathrm{M_\odot\,kpc^{-3}}$,

The fit is a fit. A full linear calculation with the Boltzmann code CLASS, saved for this course, puts the same half-mode at $34.6\,h\,\mathrm{Mpc^{-1}}$, so the fit sits about 7 per cent low. The method chapter shows how that comparison is made.

Schneider and collaborators measured how many halos form in warm simulations and fitted the ratio to cold dark matter as

Schneider et al. 2012, equation 28 For 3 keV this gives $0.63$ at $10^9\,\mathrm M_\odot$ and $0.13$ at $10^8\,\mathrm M_\odot$. The halos that do form are also less concentrated, by a factor $(1+15M_{\mathrm{hm}}/M)^{-0.3}$, about $0.53$ at $10^9\,\mathrm M_\odot$. Schneider et al. 2012, equation 39

Fuzzy dark matter resists being squeezed

If the dark matter is a boson with mass near $10^{-22}\,\mathrm{eV}$, its de Broglie wavelength $\lambda_{\mathrm{dB}}=h_{\mathrm P}/(mv)$ is astronomical. Here $h_{\mathrm P}$ is Planck's constant. At a speed of $100\,\mathrm{km\,s^{-1}}$ the wavelength is about $1.2\,\mathrm{kpc}$. A wave cannot be packed into a region smaller than its wavelength without raising its kinetic energy. That acts like an extra pressure, called quantum pressure, which pushes back against gravity on small scales.

In the linear regime each comoving wave of the fuzzy field obeys

where a dot is a time derivative, $H$ is the expansion rate, $a$ is the scale factor and $\hbar=h_{\mathrm P}/2\pi$. Gravity wins for small $k$ and the ripple grows. Quantum pressure wins above the Jeans wavenumber, where the bracket changes sign, and there the ripple oscillates instead of growing. Hu, Barkana and Gruzinov place this scale at matter-radiation equality at $k_{\mathrm{J,eq}}=9m_{22}^{1/2}\,\mathrm{Mpc^{-1}}$, with $m_{22}=m/(10^{-22}\,\mathrm{eV})$, and fit the resulting transfer function as

The power drops by a factor of two at $k_{1/2}=4.5\,m_{22}^{4/9}\,\mathrm{Mpc^{-1}}$. Hu, Barkana and Gruzinov 2000, equations 8 and 9 Beyond $x^3=\pi/2$ the cosine turns negative and the fit oscillates; it only describes a cutoff there, so the companion Lab sets $T=0$ beyond that point.

Example 3.2. For $m_{22}=1$, $k_{1/2}=4.5\,\mathrm{Mpc^{-1}}$, which is $6.65\,h\,\mathrm{Mpc^{-1}}$. A separate integration of the growth equation above, written for this course, finds the half-power point at $3.80\,\mathrm{Mpc^{-1}}$, 83 per cent of the fit. The two calculations treat radiation and baryons differently; the method chapter records the comparison.

Fuzzy halos are rarer still. Schive and collaborators started simulations from the fuzzy linear ripples, followed the matter as ordinary particles, and fitted the halo numbers at redshifts of four and above as

Schive et al. 2016, equation 7 For $m_{22}=1$ the ratio is $0.11$ at $10^{10}\,\mathrm M_\odot$ and $0.0011$ at $10^9\,\mathrm M_\odot$. Because quantum pressure enters only through the starting ripples, the suppression factor in their fit does not change with redshift, and the authors flag halos below about $10^9\,\mathrm M_\odot$ as uncertain. The companion Lab applies the fit today, which is an assumption.

Why the two cutoffs differ in shape

The warm transfer function falls as a power of $k$, while the fuzzy one falls off a cliff. At the same half-mode scale, fuzzy dark matter therefore removes small ripples more completely. Nadler and collaborators note this steeper suppression and fit fuzzy halo numbers with their own shape for that reason, instead of mapping fuzzy dark matter onto an equivalent warm mass. Nadler et al. 2021, fuzzy dark matter analysis

Neither cutoff touches the inside of an existing halo directly. A warm halo is less concentrated because it formed later. The flat core of a fuzzy halo comes from the same quantum pressure acting inside the halo, which is the subject of Three ways to make a dark matter core.

Lab tasks

1. Predict. In the transfer panel, where will the warm curve reach one quarter if you change 3 keV to 6.5 keV? Use the scaling of $\alpha$ before you move the slider.
2. Change one condition. Compare the density slices for warm dark matter at 1 keV and at 10 keV. Which features vanish first as the mass falls?
3. Compare. Set the fuzzy model so its half-power point roughly matches the 3 keV warm curve. Which curve is steeper?
4. Explain. The slices for every model use the same random phases. Why do the large features sit in the same places in each slice?

Exercises

1. Using $\alpha\propto m_{\mathrm{WDM}}^{-1.11}$, find $k_{\mathrm{hm}}$ and $M_{\mathrm{hm}}$ for 6.5 keV from Example 3.1.
2. A fuzzy particle thirty times heavier than $m_{22}=1$ is proposed. By what factor does $k_{1/2}$ grow, and by what factor does $M_0$ fall?
3. Why is the half-mode defined by $T^2=1/4$ and not by $T^2=1/2$? What does each describe?