15
$\begingroup$

According to Wikipedia, neutrinos separated from other matter seconds after the Big Bang and formed a separate background radiation field which now fills space at a temperature ~2 K.

Supposing neutrinos have a rest mass of 2 eV/c2, they should be traveling at an average speed

$$v_\nu = c \sqrt{2 \cdot k \cdot \frac{2\mathrm{K}}{2 \mathrm{eV}}} \approx 4 \cdot 10^6 \mathrm{m/s}$$

Seems to me that as neutrinos slow down due to the metric expansion of space, they should be clustering into orbits around galaxies. Distant galaxies should show the effect less because the neutrinos were going faster at the time.

Is this correct?

According to the same Wikipedia article, the background neutrinos originally had an energy of 2.5 MeV and were in thermal equilibrium with electrons, positrons, and photons. So whereas they once accounted for a quarter of the mass-energy of the Universe (or half, because there are three flavors?), their current, essentially rest mass is less than a millionth of it.

That's too little to put a dent in the dark matter mystery, or even observe the time-dependent effect I just mentioned, right?

Also, galactic orbits aren't metrically expanding so this essentially traps the neutrinos around any galaxy into a spherical shell with the orbital radius and speed dependent on the galaxy mass. A neutrino which is going too fast will escape and pass a galaxy, but by the time it reaches the next galaxy, it will be further redshifted, and might be gravitationally trapped. Then it won't slow down any further because the neighborhood of a galaxy is geometrically stable.

If there were something larger for neutrinos to orbit, they could stabilize at a higher speed, but it still couldn't be relativistic, right?

Since nothing (besides lots of empty space) can slow down a neutrino, this sets a lower speed limit on all neutrinos in a galaxy or galaxy cluster, right?

|cite|improve this question
$\endgroup$
  • 5
    $\begingroup$ How timely. This paper, Neutrino clustering around spherical dark matter halos, was put on the Arxiv just a week ago. I think the short answer is "there is a small but nonvanishing amount of neutrinos." Perhaps someone can use this or some other paper as a launching-off point for a complete answer. $\endgroup$ – user10851 Oct 31 '13 at 10:43
  • $\begingroup$ @ChrisWhite Thanks, looks like it hits the spot! I'm just an armchair physicist so I'm not really asking that an answer be "launched," although more can be better :) $\endgroup$ – Blackbody Blacklight Oct 31 '13 at 11:03
  • $\begingroup$ This is in line with how people have been thinking about cold neutrinos, but I believe that current thinking favors much lower masses for the three known mass states. See for instance arxiv.org/abs/1306.5544. $\endgroup$ – dmckee♦ Oct 31 '13 at 15:42
  • $\begingroup$ @dmckee According to the paper linked by Chris, thermal distribution provides ample halo neutrinos at masses down to 0.05 eV. If I understand Figure 4 correctly, there should be a million to a billion suns' worth of neutrino mass around the Milky Way. I don't really understand when that becomes true, but it seems to be roughly the present time. $\endgroup$ – Blackbody Blacklight Nov 1 '13 at 1:01
3
$\begingroup$

There are several question marks, here I try to analyze one by one.

  1. 3rd paragraph: yes and no. The clustering depends on the following two energies -- (absolute value of) gravitational potential of the galaxy $\phi$, and the kinetical energy of neutrinos $E_k$. The average energy of relic neutrinos $\bar{E}_k$ decreases as the universe expands. $\phi$ would grow if there are only normal matter and dark matter. Combining these two, the clustering of neutrinos does become significant in recent universe -- giving the 'yes' part. But, if our standard $\Lambda$CDM model proves correct, the growth of structures is slowing down on large scales, and will slow down on galactic scales too, and finally $\phi$ will decrease. It then needs some calculation to see whether the clustering of neutrinos will slow down or speed up sometime in the future -- hence the 'no' part. But again, if $\Lambda$CDM model is correct, the final state of the universe is uniform and without any structure -- i.e., there must be a slowing-down period for neutrino clustering.

  2. 4th paragraph: The average energy of neutrinos increase with the temperature of the universe, so it can be well above 2.5 MeV -- this value is of some significance perhaps because around this value the neutrinos become free from collisions with electrons etc. At such high temperature, according to the equipartition theorem we calculate the energy fraction through counting degrees of freedom (dof) of all particles including electrons (dof=2), positrons (dof=2), neutrinos (three generations, each with dof 1 giving total dof=3), antiparticles(dof=3), as well as photons (dof=2). -- Now I see the logic of your result. If there is no this factor of 7/8 attached to the electrons/neutrinos, the fraction of energy in neutrinos will indeed be 1/2 at that time. The 7/8 factor comes from different statistics satisfied by neutrinos/electrons (Fermi-Dirac) and photons (Bose-Einstein); because of this factor, the fraction becomes 21/43.

  3. 5th paragraph: the paper mentioned by @Chris White calculated the time evolution of neutrino clustering. The conclusion however seemed to be about the current density profile; but even the neutrino clustering today is difficult to observe -- the neutrinos are still too few, not to mention the time evolution.

  4. two questions in the last two paragraphs: the temperature, 1.9K, indicates an average kinetic energy. There can be arbitrarily higher- and arbitrarily lower- energy relic neutrinos than this value indicates; this is a Fermi-Dirac distribution problem.

[edit note]: Low velocity of the neutrinos and the strong gravitational field of the galaxy make the capture easier but are insufficient for the capture to happen. Imagine a neutrino coming from a zero-$\phi$ place; its energy will always be positive. It will fall in the galaxy and will fall out eventually, if there are no collisions happened to it. To capture a neutrino for a galaxy, is like to capture a satellite for a planet, which is a classic problem and involves 3-body interactions.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks for bearing with me… it's so easy to ask too many questions :) The idea behind (4) is that the cold neutrinos will leave the thermodynamic distribution as capture selectively halts the cooling of neutrinos. As long as galactic neighborhoods are dense compared to the distance scale required for appreciable redshifting, and fairly homogeneous, every relic neutrino is captured into the observer's neighborhood soon after its energy reaches the critical value. $\endgroup$ – Blackbody Blacklight Dec 11 '13 at 14:01
  • $\begingroup$ … it's rather speculative, though. If ever there was an academic question :P $\endgroup$ – Blackbody Blacklight Dec 11 '13 at 14:09
  • $\begingroup$ @Potatoswatter: I agree with the no-redshift-after-capture argument. Neutrinos in free space follow geodesics, when outside the galaxy the metric is the (ever-expanding) homogeneous, Robertson-Walker metric; and once inside the galaxy, the local metric is more spherical, Schwarzschild metric, which does not expand. And IMO your questions are very academic. $\endgroup$ – Hao Wang Dec 12 '13 at 6:56
  • $\begingroup$ Your edit not is interesting; I suppose another interpretation be that either 1. the last bit of energy must be sapped by redshift immediately before capture (causing apparent negative initial potential energy), or 2. that the neutrino was slowed by a slingshot effect from a passing galaxy, which subsequently recedes from the scene and leaves the neutrino trapped by its neighbor. $\endgroup$ – Blackbody Blacklight Dec 12 '13 at 7:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.