Is dark matter real?

Mike Hewson
Mike Hewson
Moderator
Joined: 1 Dec 05
Posts: 6,585
Credit: 310,873,724
RAC: 97,016

RE: I've read elsewhere

Message 38342 in response to message 38341

Quote:
I've read elsewhere that helium (in the 'helium II' state, cooled below ~2.1 K) conducts heat ........... it looks like there's plenty of hydrogen and helium out there. But it's hard to tell it's there when it isn't ionized.


Helium II has to be real pure to have those properties, and undisturbed to boot. The Helium atoms will stay in the relevant ( quantum mechanical ) phase if they have only a few 'options'. It's properties derive from that. In any case the cosmic microwave background photons ( at least ) would keep it above 2.1K. You don't have to have an electron orbital transition for a photon to interact - it can simply punt off of it like a billiard ball ( that's the reference to Compton scattering ). Even if it is a purely elastic process ( kinetic energies unchanged ) there can be polarization changes. Some signature of the interaction would exist. There are plenty of photons of all energies flying around the universe, and there has been plenty of time for them to interact with those substances that do. The 'dark' in dark matter means not only a dearth of emission for us to see, but a paucity of electromagnetic interaction overall ......
Cheers, Mike.

( edit ) Chipper, there's a good discussion of the awesome properties of Helium, in various forms, as used for the Large Hadron Collider here. Go to the talk by L.Evans "The LHC" - it has a nifty overview of their magnetic arrangements too. I was stunned to discover the stored energy in the beam, when fully operational, will be of the order of 80 kilos of TNT! The Helium, and other systems have to withstand any accidental sudden quenching of that... phew!

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Joachim Schmidt
Joachim Schmidt
Joined: 19 Feb 05
Posts: 35
Credit: 391,050
RAC: 0

RE: If dark matter is

Message 38343 in response to message 38341

Quote:
If dark matter is 'weakly interacting', here's a quote from an abstract concerning the interaction between helium and x-rays:
Quote:
X rays in the keV energy regime interact only very weakly with helium atoms. Starting at about 6 keV x-ray energy, the absorption of x rays becomes less important in the ionization of helium than Compton scattering. Whereas for photoabsorption the high-energy asymptotic value for the fraction of helium double ionization is practically reached for keV x rays, the fraction of helium double ionization in Compton scattering is expected to be dependent on the x-ray energy up to much higher energies.

I've read elsewhere that helium (in the 'helium II' state, cooled below ~2.1 K) conducts heat better than any other substance known, as its thermal conductivity is several hundred times that of copper. Maybe a wrong conclusion on my part, but to me this suggests that no scattering is taking place (for helium II).

Seems like it would take the energy of a nova's shockwave to stir it up enough for it to be 'visible'. The reason is that anything less is so weakly interacting and easily conducted. Helium (in the Helium II state) has no boiling point, and the dark matter region is one of tremendous volume and almost no pressure.

I understand what you're saying about emission and absorption, Solomon (and thanks for your patience). From this work on primordial abundance using light from quasars, it looks like there's plenty of hydrogen and helium out there. But it's hard to tell it's there when it isn't ionized.

Well, with "dark matter is only weakly interacting with other matter" it is ment that it only interacts with other particles over the gravitational force and the weak nuclear force.
As illustration:
The weak nuclear force is 10^-11 times weaker than the electromagnetic force,
the gravitational forc is 10^-36 times weaker than the electromagnetic force.

The weakly interacting of your X-rays with Helium atoms is a complete different thing.
X-rays interact with helium atoms over the electromagnetic force.

Just a few examples to show you the difference between weak nuclear force an electromagnet force:
Neutrinos also only interact over the weak nuclear and gravitational force.
They are produced in the middle of the sun and can escape from there without interacting with the sun. So they only need 8 minutes from the middle of the sund to the earth. The normal energy, which interacts over the electromagnetic force, needs a few 100000 years from the middle to the surface of the sun, because it can interact with it.

Also here at earth you have a solar neutrino flux of 7*10^10 per cm² per s and also with huge detectors (tanks with about 50000 tons of water) you only see a few (perhaps 1-10) neutrinos per day.

In contrary to that, you could measure every single helium atom that flys through your detector.

And yes you are wrong with your conclusion :) there is scattering with helium atoms irrelevant in which state they are.
Furthermore you don't need much energy to see Helium. In your abstract they speak about some keV. A supernova emitts in a few secons over 10^46 Joule, this is over 10^64 eV ...

Furthermore it's very easy to detect hydrogen and helium which isn't ionized because you can see absorption lines of light which had to go through the gas clouds on it's way to earth.

greets

tullio
tullio
Joined: 22 Jan 05
Posts: 2,118
Credit: 61,407,735
RAC: 0

RE: Axions are far too

Message 38344 in response to message 38340

Quote:

Axions are far too light to explain the gravity of dark matter


What about Higgs bosons? They should be heavy enough!
Tullio

Chipper Q
Chipper Q
Joined: 20 Feb 05
Posts: 1,540
Credit: 708,571
RAC: 0

Hmm. The purity's not a

Hmm. The purity's not a problem since He-3 and He-4 are immiscible, albeit at temperatures below 0.8 K. As for the CMB with regard to temperature: it couldn't be as cold as it is now (~2.7K?) without there being something colder than that to begin with, and there must still be regions below the CMB temperature since the background is anisotropic, right? This is to say that, for every region I can see that is above the CMB temperature (which includes all the stars in all the galaxies), I can then say there must be corresponding regions which must be below the CMB temperature, to balance it out.

As for the EM interactions (scattering, spectral emission/absorption, and polarization): wouldn't the interaction between photons (bosons) and helium-II (bosons) be quantum mechanically null? From the article (link in previous post) about imparting rotational energy to a molecule suspended in the helium-II (by punting photons off it :) ), it's believed that some of the helium atoms rotate with the molecule, but no mention is made of any of these photons scattering off any helium atoms. Wouldn't all photons be mediated through the helium via Cooper pairs?

It was a shot in the dark, -er at the dark, anyway :) Thanks for the link to the LHC lectures, Mike; I'm very much looking forward to watching them!

Joachim Schmidt
Joachim Schmidt
Joined: 19 Feb 05
Posts: 35
Credit: 391,050
RAC: 0

RE: RE: Axions are far

Message 38346 in response to message 38344

Quote:
Quote:

Axions are far too light to explain the gravity of dark matter


What about Higgs bosons? They should be heavy enough!
Tullio

No way tullio, sorry :) the expected mass of the higgs boson is between 100 - 200 GeV (it's not yet discoverd). But Higgs Bosons are not stable, they decay very fast. So they aren't candidates for dark matter

greets

Solomon
Solomon
Joined: 6 Nov 05
Posts: 28
Credit: 19,976
RAC: 0

RE: Hmm. The purity's not

Message 38347 in response to message 38345

Quote:
Hmm. The purity's not a problem since He-3 and He-4 are immiscible, albeit at temperatures below 0.8 K. As for the CMB with regard to temperature: it couldn't be as cold as it is now (~2.7K?) without there being something colder than that to begin with, and there must still be regions below the CMB temperature since the background is anisotropic, right? This is to say that, for every region I can see that is above the CMB temperature (which includes all the stars in all the galaxies), I can then say there must be corresponding regions which must be below the CMB temperature, to balance it out.

This would be a good argument if the "cooling" of the CMB were a thermal process to begin with. It isn't. When we talk about the temperature of the CMB, what we really mean is the temperature of a blackbody that would emit radiation with the CMB spectrum. The cooling we talk about is actually the light itself getting stretched out by the expansion of the universe. Since hotter objects tend to give off more energetic, and thus higher wavelength, light, as the wavelength of the CMB light stretches, the spectrum matches that of a blackbody of progressively cooler temperature.

The important point here is that the cooling of the CMB does not require any cold reservoir to transfer heat to because it is a process which does not require the transfer of heat at all.

Quote:
As for the EM interactions (scattering, spectral emission/absorption, and polarization): wouldn't the interaction between photons (bosons) and helium-II (bosons) be quantum mechanically null? From the article (link in previous post) about imparting rotational energy to a molecule suspended in the helium-II (by punting photons off it :) ), it's believed that some of the helium atoms rotate with the molecule, but no mention is made of any of these photons scattering off any helium atoms. Wouldn't all photons be mediated through the helium via Cooper pairs?

When a photon interacts with an atom, what is really happening is that the photon directly interacts with one of the charged particles in the atom. Any effects felt by the whole atom are mediated by the forces between the particles that make up the atom.

Joachim Schmidt
Joachim Schmidt
Joined: 19 Feb 05
Posts: 35
Credit: 391,050
RAC: 0

RE: Hmm. The purity's not

Message 38348 in response to message 38345

Quote:

Hmm. The purity's not a problem since He-3 and He-4 are immiscible, albeit at temperatures below 0.8 K. As for the CMB with regard to temperature: it couldn't be as cold as it is now (~2.7K?) without there being something colder than that to begin with, and there must still be regions below the CMB temperature since the background is anisotropic, right? This is to say that, for every region I can see that is above the CMB temperature (which includes all the stars in all the galaxies), I can then say there must be corresponding regions which must be below the CMB temperature, to balance it out.

As for the EM interactions (scattering, spectral emission/absorption, and polarization): wouldn't the interaction between photons (bosons) and helium-II (bosons) be quantum mechanically null? From the article (link in previous post) about imparting rotational energy to a molecule suspended in the helium-II (by punting photons off it :) ), it's believed that some of the helium atoms rotate with the molecule, but no mention is made of any of these photons scattering off any helium atoms. Wouldn't all photons be mediated through the helium via Cooper pairs?

It was a shot in the dark, -er at the dark, anyway :) Thanks for the link to the LHC lectures, Mike; I'm very much looking forward to watching them!

Well, unfortunately i guess i have to dissapoint you :) The CMB is the afterglow of the big bang, so it was unbeliveably hot at the beginning, and decreased with time because of the inflation of the universe. So it wasn't colder any time before. You are right that there is an anisotropic of the CMB but it's not very high, dT/T = 10^-5 or in other words, it's up to 50 micro Kelvins. Followly the coldest place still has 2,725 K.

Helium II is not a boson it's a fermion. It also has an electromagnetic charge, so it interacts with photons.

greets

Solomon
Solomon
Joined: 6 Nov 05
Posts: 28
Credit: 19,976
RAC: 0

RE: Helium II is not a

Message 38349 in response to message 38348

Quote:

Helium II is not a boson it's a fermion. It also has an electromagnetic charge, so it interacts with photons.

HeII consists of 6 fermions (2 protons, 2 neutrons, and 2 electrons), so its total spin must be an integer. Hence, HeII, when considered as a whole, is a boson. This accounts for its superfluid phase.

Also, it would probably be better to say that, while the atoms have no net charge, they do have electrical structure due to separation of charge.

Chipper Q
Chipper Q
Joined: 20 Feb 05
Posts: 1,540
Credit: 708,571
RAC: 0

Is it correct to say that

Is it correct to say that dark matter is not only weakly interacting with normal matter, but that it is also weakly interacting with itself, since it apparently doesn't undergo gravitational collapse?

Solomon
Solomon
Joined: 6 Nov 05
Posts: 28
Credit: 19,976
RAC: 0

RE: Is it correct to say

Message 38351 in response to message 38350

Quote:
Is it correct to say that dark matter is not only weakly interacting with normal matter, but that it is also weakly interacting with itself, since it apparently doesn't undergo gravitational collapse?

From the perspective of quantum field theory, one should really only talk about the forces by which a particle interacts, since interactions (as written in the field theory) are never directly with the other particles, but are mediated by the force-carrying particles. So, to say that something is weakly interacting means that, at most, it can only interact by means of the weak and/or gravitational forces. This means that it will interact with whatever other particles are present that can feel one of these forces.

Comment viewing options

Select your preferred way to display the comments and click "Save settings" to activate your changes.