3
$\begingroup$

If dark matter is made from fermions these should collide and cause dark matter to become denser in some places than others Bosons wont collide in this way so there should a different effect .So is there any observational evidence to suggest density changes of one kind or another?

$\endgroup$
3
  • 3
    $\begingroup$ "If dark matter is made from fermions these should collide", please outline why you think this is. Dark matter is generally assumed to have little/no self-interaction. $\endgroup$
    – ProfRob
    May 19, 2023 at 20:33
  • $\begingroup$ @ProfRob Fermions cannot occupy the same space at the same time but Bosons can $\endgroup$
    – user50918
    May 21, 2023 at 7:52
  • $\begingroup$ However, I now understand what you are asking about. $\endgroup$
    – ProfRob
    May 21, 2023 at 9:33

1 Answer 1

2
$\begingroup$

There is no law of physics that prevents fermions getting arbitrarily close to each other. The Pauli Exclusion Principle says that no two fermions can occupy the same quantum state. Basically, this means that two fermions can be in the same place if they have different spins (though this can only be up or down) or occupy different energy states.

What this means, is that closely packed fermions experience degeneracy pressure - the fermions have an average kinetic energy even at zero temperature. This degeneracy pressure would be there even if the fermions were non-interacting.

One way to work out whether this is an important effect is to compare the "phase space" occupied by each particle with the cube of Planck's constant. Essentially there will be a significant pressure if $$(\Delta x\ \Delta p)^3 \leq h^3\ , $$ where $\Delta x$ is a typical separation of the fermions and $\Delta p$ is a typical spread in their linear momenta, which depends on their mass and spread of speeds. Once the phase space occupied by each fermion gets down to this level then their wavefunctions overlap and fermion degeneracy becomes an issue.

We can play some back-of-the envelope games: Let's assume that the density of dark matter at the solar radius is $\sim 0.01 M_\odot$ per pc$^{3}$ and that the dark matter is WIMPS with a rest-mass energy of 100 GeV. These are reasonable numbers - although the density of dark matter in the halo of the Galaxy would be an order of magnitude or two lower than this. The velocity dispersion in the dark matter halo is let's say about 300 km/s.

Turning these numbers into self-consistent SI units, we have a WIMP mass $m_w \sim 1.8 \times 10^{-25}$ kg and a WIMP number density of $n_w \sim 4000$ m$^{-3}$. Thus $(\Delta x_w)^3 = n_w^{-1}$ and $\Delta p_w \sim 3\times 10^5 m_w$, which leads to $$ \left(\frac{\Delta x_w\ \Delta p_w}{h}\right)^3 \sim 10^{38}\ . $$

i.e. The phase space occupied by 1 WIMP is $10^{38}$ times bigger than a value that would generate significant degeneracy effects.

$\endgroup$
4
  • $\begingroup$ How is one wimp mass and velocity a spread of its momenta? I don't see how separation distance multiplied by momentum spread cubed is less than Planck constant cubed given your figures $\endgroup$
    – user50918
    May 21, 2023 at 11:06
  • $\begingroup$ Using "Viktor T. Toth - Hawking radiation calculator" vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator. wimps of 10^-25 kg would be the peak emission of a black hole with a starting temperature of about 3 K and a mass 10^12 kg. $\endgroup$
    – user50918
    May 21, 2023 at 11:36
  • 1
    $\begingroup$ It isn't, it's $10^{38}$ times bigger?? That's why the effect you are proposing doesn't happen. $\endgroup$
    – ProfRob
    May 21, 2023 at 12:07
  • $\begingroup$ Okay yes I got it the wrong way around. $\endgroup$
    – user50918
    May 21, 2023 at 12:49

You must log in to answer this question.