Trouble For Dark Energy Hypothesis?

[Jonathan Thornburg [remove -animal to reply]]

[[
Meta-comment: This discussion started in sci.physics.research, but
its "natural home" is in sci.astro.research. I've cross-posted this
article to both newsgroups, and set the Followup-To: header so further
discussion should be in s.a.r.
]]

Gary Harnagel <***@yahoo.com> wrote:
> I'm having trouble picturing why we should see the CMBR at all. Since it's
> traveling at the speed of light but we're moving somewhat slower, shouldn't
> it have passed us long ago? I know, the FLWR metric must have something to
> do with it, but ...

To try to answer Gary Harnagel's question:

<<<begin analogy>>>

Imagine an infinite static Euclidean universe (i.e., flat spacetime,
no gravity involved) filled with (stationary) fog which both emits and
scatters (visible) light, and consider a (stationary) observer in that
fog. Now suppose that at a time which we will label t=0, two things
happen:
* all the fog suddenly condenses into larger water droplets, and
* those water droplets no longer emit light.
Since the scattering cross-section of large water droplets is vastly
smaller than that of fog, the result is that at t>0, the sea-of-droplets
is mostly transparent to light (certainly much more transparent than the
original fog was). In other words, at times t>0 light basically travels
in straight lines, with little scattering, emission, or absorption.

What will our observer see at t=1 year?

Since at t>0 there is minimal scattering, emission, or absorption, we
see that at t=1 year our observer will see (receive) those photons, and
only those photons, which were
(a) exactly 1 light-year away from her at t=0, and
(b) travelling directly towards her at t=0.
This holds in any direction our observer looks. In other words, at
t=1 year our observer will see a uniform glow on her "sky".

At t=2 years our observer will will see those photons, and only those
photons, which were
(a) exactly 2 light-years away from her at t=0, and
(b) travelling directly towards her at t=0.
This holds in any direction our observer looks. In other words, at
t=2 year our observer will see a uniform glow on her "sky". But that
glow is comprised of a *different set of photons, emitted at a different
set of events* than was the glow she saw at t=1 year.

Etc etc for any other time t>0.

<<<end analogy>>>

As you can see, this analogy reproduces many of the features of the
CMBR. It doesn't reproduce the CMBR's temperature -- for that you need
a cosmological redshift between the last-scattering time (t=0 in the
analogy, approximately 0.5 million years after the big bang in standard
cosmology) and today. But the analogy does produce an all-sky uniform
glow seen by all observers, even at far-future times.

I hope this makes things a bit clearer (no pun intended).

--
-- "Jonathan Thornburg [remove -animal to reply]" <***@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
currently visiting Max-Plack-Institute fuer Gravitationsphysik
(Albert-Einstein-Institut), Potsdam-Golm, Germany
"There was of course no way of knowing whether you were being watched
at any given moment. How often, or on what system, the Thought Police
plugged in on any individual wire was guesswork. It was even conceivable
that they watched everybody all the time." -- George Orwell, "1984"

[Gary Harnagel]

On Sunday, January 14, 2018 at 9:42:10 AM UTC-7, Jonathan Thornburg [remove -animal to reply] wrote:
> [[
> Meta-comment: This discussion started in sci.physics.research, but
> its "natural home" is in sci.astro.research. I've cross-posted this
> article to both newsgroups, and set the Followup-To: header so further
> discussion should be in s.a.r.
> ]]
>
> Gary Harnagel <***@yahoo.com> wrote:
> > I'm having trouble picturing why we should see the CMBR at all. Since it's
> > traveling at the speed of light but we're moving somewhat slower, shouldn't
> > it have passed us long ago? I know, the FLWR metric must have something to
> > do with it, but ...
>
> To try to answer Gary Harnagel's question:
>
> <<<begin analogy>>>
>
> Imagine an infinite static Euclidean universe (i.e., flat spacetime,
> no gravity involved) filled with (stationary) fog which both emits and
> scatters (visible) light, and consider a (stationary) observer in that
> fog. Now suppose that at a time which we will label t=0, two things
> happen:
> * all the fog suddenly condenses into larger water droplets, and
> * those water droplets no longer emit light.
> Since the scattering cross-section of large water droplets is vastly
> smaller than that of fog, the result is that at t>0, the sea-of-droplets
> is mostly transparent to light (certainly much more transparent than the
> original fog was). In other words, at times t>0 light basically travels
> in straight lines, with little scattering, emission, or absorption.
>
> What will our observer see at t=1 year?
>
> Since at t>0 there is minimal scattering, emission, or absorption, we
> see that at t=1 year our observer will see (receive) those photons, and
> only those photons, which were
> (a) exactly 1 light-year away from her at t=0, and
> (b) travelling directly towards her at t=0.
> This holds in any direction our observer looks. In other words, at
> t=1 year our observer will see a uniform glow on her "sky".
>
> At t=2 years our observer will will see those photons, and only those
> photons, which were
> (a) exactly 2 light-years away from her at t=0, and
> (b) travelling directly towards her at t=0.
> This holds in any direction our observer looks. In other words, at
> t=2 year our observer will see a uniform glow on her "sky". But that
> glow is comprised of a *different set of photons, emitted at a different
> set of events* than was the glow she saw at t=1 year.
>
> Etc etc for any other time t>0.
>
> <<<end analogy>>>
>
> As you can see, this analogy reproduces many of the features of the
> CMBR. It doesn't reproduce the CMBR's temperature -- for that you need
> a cosmological redshift between the last-scattering time (t=0 in the
> analogy, approximately 0.5 million years after the big bang in standard
> cosmology) and today. But the analogy does produce an all-sky uniform
> glow seen by all observers, even at far-future times.
>
> I hope this makes things a bit clearer (no pun intended).
>
> --
> -- "Jonathan Thornburg [remove -animal to reply]" <***@astro.indiana-zebra.edu>
> Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
> currently visiting Max-Plack-Institute fuer Gravitationsphysik
> (Albert-Einstein-Institut), Potsdam-Golm, Germany

Thanks, JT.

I like your fog analogy; however, let's consider the case where the fog
consists of photons which begin in some finite volume of space. They
would be moving in random directions at c and, presumably, would
interact in some process to create particles with mass, conserving
energy and momentum. But pair production can't satisfy both energy and
momentum conservation unless there is some other mass that can absorb
the excess of one or the other, yes? Of course, the big bang has the
same problem, as well as the problem of having equal parts matter and
antimatter.

Anyway, the created particles will still have kinetic energy and will
disperse, but at lower speeds than the unconverted photons. So my
original question is still unanswered by the fog analogy.

[Tom Roberts]

On 1/15/18 1/15/18 3:30 PM, Gary Harnagel wrote:
> On Sunday, January 14, 2018 at 9:42:10 AM UTC-7, Jonathan Thornburg [remove
> -animal to reply] wrote:
>> <<<begin analogy>>>
>>
>> Imagine an infinite static Euclidean universe (i.e., flat spacetime, no
>> gravity involved) filled with (stationary) fog which both emits and
>> scatters (visible) light, and consider a (stationary) observer in that
>> fog. Now suppose that at a time which we will label t=0, two things happen:
>> * all the fog suddenly condenses into larger water droplets, and * those
>> water droplets no longer emit light. Since the scattering cross-section of
>> large water droplets is vastly smaller than that of fog, the result is that
>> at t>0, the sea-of-droplets is mostly transparent to light (certainly much
>> more transparent than the original fog was). In other words, at times t>0
>> light basically travels in straight lines, with little scattering,
>> emission, or absorption.
>>
>> What will our observer see at t=1 year?
>>
>> Since at t>0 there is minimal scattering, emission, or absorption, we see
>> that at t=1 year our observer will see (receive) those photons, and only
>> those photons, which were (a) exactly 1 light-year away from her at t=0,
>> and (b) travelling directly towards her at t=0. This holds in any
>> direction our observer looks. In other words, at t=1 year our observer
>> will see a uniform glow on her "sky".
>>
>> At t=2 years our observer will will see those photons, and only those
>> photons, which were (a) exactly 2 light-years away from her at t=0, and
>> (b) travelling directly towards her at t=0. This holds in any direction
>> our observer looks. In other words, at t=2 year our observer will see a
>> uniform glow on her "sky". But that glow is comprised of a *different set
>> of photons, emitted at a different set of events* than was the glow she saw
>> at t=1 year.
>>
>> Etc etc for any other time t>0.
>>
>> <<<end analogy>>>
>>
>> As you can see, this analogy reproduces many of the features of the CMBR.
>> It doesn't reproduce the CMBR's temperature -- for that you need a
>> cosmological redshift between the last-scattering time (t=0 in the
>> analogy, approximately 0.5 million years after the big bang in standard
>> cosmology) and today. But the analogy does produce an all-sky uniform glow
>> seen by all observers, even at far-future times.
>
> I like your fog analogy; however, let's consider the case where the fog
> consists of photons [...]

That is not the analogy. As the first sentence says, the fog "both
emits and scatters (visible) light". The correspondence to the
cosmological model and CMBR:
fog <=> primordial particles
water droplets <=> atoms
(visible) light <=> CMBR

At "recombination", z ~ 1100, the cosmological expansion had reduced
the average energies of primordial particles (protons, deuterons,
alphas, electrons, ...) so much that they could form stable atoms
(mostly hydrogen and helium). Suddenly the universe became nearly
transparent to electromagnetic radiation. Today we see CMBR photons
whose time of last scatter was ~ 13 gigayears ago.

At and after recombination, the energy of the CMBR photons is far
too low for pair production (threshold = 1.022E6 eV). During
recombination the CMBR photon energy peak was a few eV; today it
is 6.624E-4 eV.

BTW there should also be a cosmic neutrino background, which decoupled
much earlier than photons. It would be very interesting if this
could be detected.

Tom Roberts

[Gary Harnagel]

On Tuesday, January 16, 2018 at 9:57:35 AM UTC-7, Tom Roberts wrote:
>
> On 1/15/18 1/15/18 3:30 PM, Gary Harnagel wrote:
> >
> > ....
>
> BTW there should also be a cosmic neutrino background, which decoupled
> much earlier than photons. It would be very interesting if this
> could be detected.
>
> Tom Roberts

The neutrino flux would be red-shifted by z ~ 1100 also, but today's neutrino
detectors aren't sensitive to energies below a few 100 KeV, yes? That means
primordial neutrinos would have to have had energies above 100 MeV.

BTW, did you get the email I sent you last November about my equipment?

Gary

[[Mod. note --
1. I am not an expert on this, but I think you're right: if you want
any direction-of-travel angular resolution (necessary to distinguish
other sources from the Sun's neutrino flux), today's neutrino
detectors have essentially no sensitivity to neutrinos with energies
below a few MeV.
2. Whoever the last line is directed to, replies should be private
(e.g., email); the newsgroup is for *public* conversations.
-- jt]]

[Steve Willner]

In article <9a148735-0045-426e-8b2f-***@googlegroups.com>,
Gary Harnagel <***@yahoo.com> writes:
> The neutrino flux would be red-shifted by z ~ 1100 also,

This ignores the part about the neutrinos having decoupled long
before the photons. One source, which seems to be a textbook by
Daniel Baumann at Cambridge:
http://www.damtp.cam.ac.uk/user/db275/Cosmology/Chapter3.pdf
gives a neutrino decoupling redshift of 6E9. That corresponds to an
energy of about 1 Mev and a time about 1 s after the Big Bang.
Cosmological neutrinos should therefore have a kinetic energy today
of about 1/6 meV (i.e., milli-, not mega-). As the OP wrote, that's
very far from detectable.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone [phone] ***@cfa.harvard.edu
Cambridge, MA 02138 USA

[Martin Brown]

On 15/01/2018 21:30, Gary Harnagel wrote:
> On Sunday, January 14, 2018 at 9:42:10 AM UTC-7, Jonathan Thornburg [remove -animal to reply] wrote:
>> [[
>> Meta-comment: This discussion started in sci.physics.research, but
>> its "natural home" is in sci.astro.research. I've cross-posted this
>> article to both newsgroups, and set the Followup-To: header so further
>> discussion should be in s.a.r.
>> ]]
>>
>> Gary Harnagel <***@yahoo.com> wrote:
>>> I'm having trouble picturing why we should see the CMBR at all. Since it's
>>> traveling at the speed of light but we're moving somewhat slower, shouldn't
>>> it have passed us long ago? I know, the FLWR metric must have something to
>>> do with it, but ...
>>
>> To try to answer Gary Harnagel's question:
>>
>> <<<begin analogy>>>
>>
>> Imagine an infinite static Euclidean universe (i.e., flat spacetime,
>> no gravity involved) filled with (stationary) fog which both emits and
>> scatters (visible) light, and consider a (stationary) observer in that
>> fog. Now suppose that at a time which we will label t=0, two things
>> happen:
>> * all the fog suddenly condenses into larger water droplets, and
>> * those water droplets no longer emit light.
>> Since the scattering cross-section of large water droplets is vastly
>> smaller than that of fog, the result is that at t>0, the sea-of-droplets
>> is mostly transparent to light (certainly much more transparent than the
>> original fog was). In other words, at times t>0 light basically travels
>> in straight lines, with little scattering, emission, or absorption.
>>
>> What will our observer see at t=1 year?
>>
>> Since at t>0 there is minimal scattering, emission, or absorption, we
>> see that at t=1 year our observer will see (receive) those photons, and
>> only those photons, which were
>> (a) exactly 1 light-year away from her at t=0, and
>> (b) travelling directly towards her at t=0.
>> This holds in any direction our observer looks. In other words, at
>> t=1 year our observer will see a uniform glow on her "sky".
>>
>> At t=2 years our observer will will see those photons, and only those
>> photons, which were
>> (a) exactly 2 light-years away from her at t=0, and
>> (b) travelling directly towards her at t=0.
>> This holds in any direction our observer looks. In other words, at
>> t=2 year our observer will see a uniform glow on her "sky". But that
>> glow is comprised of a *different set of photons, emitted at a different
>> set of events* than was the glow she saw at t=1 year.
>>
>> Etc etc for any other time t>0.
>>
>> <<<end analogy>>>
>>
>> As you can see, this analogy reproduces many of the features of the
>> CMBR. It doesn't reproduce the CMBR's temperature -- for that you need
>> a cosmological redshift between the last-scattering time (t=0 in the
>> analogy, approximately 0.5 million years after the big bang in standard
>> cosmology) and today. But the analogy does produce an all-sky uniform
>> glow seen by all observers, even at far-future times.
>>
>> I hope this makes things a bit clearer (no pun intended).
>>
>> --
>> -- "Jonathan Thornburg [remove -animal to reply]" <***@astro.indiana-zebra.edu>
>> Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
>> currently visiting Max-Plack-Institute fuer Gravitationsphysik
>> (Albert-Einstein-Institut), Potsdam-Golm, Germany
>
> Thanks, JT.
>
> I like your fog analogy; however, let's consider the case where the fog
> consists of photons which begin in some finite volume of space. They

In an infinite universe (or for that matter even a finite one that is
much larger in size than c*age_of_universe) there is always a surface of
last scattering where photons can escape from the previously opaque
plasma and travel more or less unhindered until they reach us.

The fog analogy is about as good as it gets to explain this.

> would be moving in random directions at c and, presumably, would
> interact in some process to create particles with mass, conserving
> energy and momentum. But pair production can't satisfy both energy and
> momentum conservation unless there is some other mass that can absorb
> the excess of one or the other, yes? Of course, the big bang has the
> same problem, as well as the problem of having equal parts matter and
> antimatter.

Observationally near to us it appears that there was an excess of what
we call matter - otherwise there would be a lot more matter-antimatter
annihilation events seen. Other alternatives would require some sorting
of matter and antimatter on a galactic cluster scale. I don't think you
can observationally determine if this is the case.

Is there any way even in principle to determine observationally if all
clusters are made of matter as opposed to some being of antimatter?

> Anyway, the created particles will still have kinetic energy and will
> disperse, but at lower speeds than the unconverted photons. So my
> original question is still unanswered by the fog analogy.

The photons from the cosmic microwave background have been going past
the our position ever since the universe first became transparent. Their
energy gradually decreasing as the universe expands and cools and also
by redshifting of the receding surface of emitters. There is always a
layer at just the right distance for its photons to be reaching us now.

I suppose if the universe was of finite extent the CMBR could go out
suddenly tomorrow. It does beg an interesting question in the very long
term future - will the CMBR gradually fade out to T=0 and time dilate as
the surface of last scattering approaches the particle horizon or will
the dark energy acceleration make it switch off sharply?

--
Regards,
Martin Brown

[Steve Willner]

In article <p3kmkr$58q$***@gioia.aioe.org>,
Martin Brown <'''newspam'''@nezumi.demon.co.uk> writes:
> Is there any way even in principle to determine observationally if all
> clusters are made of matter as opposed to some being of antimatter?

I don't think there's any way to distinguish for any one
cluster. However, if a matter galaxy falls into an anti-matter
cluster or vice versa, there should be lots of gamma rays, which
should (I expect) heat the X-ray-emitting cluster gas in an obvious
pattern. Even more spectacularly, entire galaxy clusters sometimes
collide. If a collision between two "opposite types" took place, I
expect that would be obvious.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone [phone] ***@cfa.harvard.edu
Cambridge, MA 02138 USA

[Jonathan Thornburg [remove -animal to reply]]

Gary Harnagel <***@yahoo.com> wrote:
> I'm having trouble picturing why we should see the CMBR at all. Since it's
> traveling at the speed of light but we're moving somewhat slower, shouldn't
> it have passed us long ago? I know, the FLWR metric must have something to
> do with it, but ...

I replied with an analogy:
# Imagine an infinite static Euclidean universe (i.e., flat spacetime,
# no gravity involved) filled with (stationary) fog which both emits and
# scatters (visible) light, and consider a (stationary) observer in that
# fog. [[...]]

Gary Harnagel replied:
> I like your fog analogy; however, let's consider the case where the fog
> consists of photons which begin in some finite volume of space.

Stop here.

In my analogy the fog consisted of matter (water) + a bunch of photons,
not just photons. And the matter+photons are everywhere in space, not
restricted to "some finite volume of space". The case that you're now
describing (photons only, restricted to a finite spatial volume) is not
a good analogy to the hot-big-bang-model CMBR formation.

In the hot-big-bang model, the "fog" consists of matter (mostly a plasma
of protons, alpha particles, and electrons) in radiative equilibrium
with am ambient field of (approximately black-body) photons. These
matter and photons are everywhere in space, not restricted to a finite
spatial volume.

In the hot-big-bang model the occurence analogous to my analogy's
"fog condensing" is (was) the temperature dropping low enough so that
the ambient matter could "condense" and form (mostly) hydrogen & helium
atoms. This happened roughly 0.5e6 years after the big bang.

ciao,
--
-- "Jonathan Thornburg [remove -animal to reply]" <***@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
currently visiting Max-Plack-Institute fuer Gravitationsphysik
(Albert-Einstein-Institut), Potsdam-Golm, Germany
"There was of course no way of knowing whether you were being watched
at any given moment. How often, or on what system, the Thought Police
plugged in on any individual wire was guesswork. It was even conceivable
that they watched everybody all the time." -- George Orwell, "1984"

----- End forwarded message -----

[Phillip Helbig (undress to reply)]

In article <p3kmkr$58q$***@gioia.aioe.org>, Martin Brown
<'''newspam'''@nezumi.demon.co.uk> writes:

> I suppose if the universe was of finite extent the CMBR could go out
> suddenly tomorrow. It does beg an interesting question in the very long
> term future

I have to point out that it merely asks the question. Begging the
question means essentially answering a question with a statement which
repeats the question, e.g. "Why did you come here?" "Because I had a
desire to do so."

> - will the CMBR gradually fade out to T=0 and time dilate as
> the surface of last scattering approaches the particle horizon or will
> the dark energy acceleration make it switch off sharply?

The particle horizon always increases in comoving coordinates and, in an
expanding universe, in proper distance as well. It is always behind the
surface of last scattering. The redshifts of both increase with time,
but it doesn't make sense to say that the latter approaches the particle
horizon.

If the cosmological constant comes to dominate the expansion, which
appears will happen in our universe, then the universe will
asymptotically approach the de Sitter model. This model has an event
horizon at a fixed proper distance. As the universe expands, objects
approach it, becoming infinitely redshifted when reaching it. There is
no sharp switch-off.

[jacobnavia]

Le 23/01/2018 22:33, Steve Willner a écrit :
> In article <9a148735-0045-426e-8b2f-***@googlegroups.com>,
> Gary Harnagel <***@yahoo.com> writes:
>> The neutrino flux would be red-shifted by z ~ 1100 also,
>
> This ignores the part about the neutrinos having decoupled long
> before the photons. One source, which seems to be a textbook by
> Daniel Baumann at Cambridge:
> http://www.damtp.cam.ac.uk/user/db275/Cosmology/Chapter3.pdf
> gives a neutrino decoupling redshift of 6E9. That corresponds to an
> energy of about 1 Mev and a time about 1 s after the Big Bang.
> Cosmological neutrinos should therefore have a kinetic energy today
> of about 1/6 meV (i.e., milli-, not mega-). As the OP wrote, that's
> very far from detectable.
>

It depends on your antena's neutrino sensitivity.

Why do neutrinos react with some Chlorate compounds?

Isn't it a consequence of the geometry of the collision?

What about putting a ring of neutrino sensitive atoms orientable with an
outer magnetic field and just trying to point to the sun?

We could turn around the chlorine with its ring until we see what
direction and position should the chlorine have to intercept at best
neutrinos coming from a specific direction.

Why does underground chlorine detectors work?

Bceause among the millions of atoms, one has the right orientation to
exactly trap a neutrino.

Having an array of neutrino detectors at a molecular level would
increase sensitivity and positioning.

Or not?

[Martin Brown]

On 24/01/2018 22:55, jacobnavia wrote:
> Le 23/01/2018 22:33, Steve Willner a =C3=A9crit :
>> In article <9a148735-0045-426e-8b2f-***@googlegroups.com>,
>> Gary Harnagel <***@yahoo.com> writes:
>>> The neutrino flux would be red-shifted by z ~ 1100 also,
>>
>> This ignores the part about the neutrinos having decoupled long
>> before the photons. One source, which seems to be a textbook by
>> Daniel Baumann at Cambridge:
>> http://www.damtp.cam.ac.uk/user/db275/Cosmology/Chapter3.pdf
>> gives a neutrino decoupling redshift of 6E9. That corresponds to an
>> energy of about 1 Mev and a time about 1 s after the Big Bang.
>> Cosmological neutrinos should therefore have a kinetic energy today
>> of about 1/6 meV (i.e., milli-, not mega-). As the OP wrote, that's
>> very far from detectable.
>
> It depends on your antena's neutrino sensitivity.

Which is so poor that they evaded detection for quite a while. It took
bulk tanks and exquisite experimental technique to see anything at all.
When they did it was about a third of what was expected.

> Why do neutrinos react with some Chlorate compounds?

Neutrinos and anti-neutrinos barely interact with anything but when they
do if they hit the right thing with enough energy they drive the proton
to neutron conversion that underpins fusion backwards.

Proton proton fusion is : p + p + e- = pn + e-neutrino + energy

Neutron to proton is: n + e-neutrino + energy = p + e-

The significance of chlorine and gallium is mainly that they are easily
purified liquids and the product of a neutrino interaction can be
isolated and identified from the bulk unchanged material (as argon gas).

I think there may be a little bit of resonance enhancement in some
nuclei too and in a tight spot every little helps.

The crucial point about the whole exercise is that you have to be able
to detect the absolutely miniscule amount of product which limits your
choices to something where the resulting species is radioactive with a
suitable half life and separable from the bulk material. If they are
lucky they get a dozen or so atoms converted in many tonnes of material
in each experimental run. See:

http://www.slac.stanford.edu/econf/C0805263/ProcContrib/hahn_r.pdf

> Isn't it a consequence of the geometry of the collision?

It takes a direct hit and a lot of luck for anything to happen. It also
depends critically on the energy of the incident neutrino - higher
energy ones giving you more options for detection a la super Kamiokande.
>
> What about putting a ring of neutrino sensitive atoms orientable with an
> outer magnetic field and just trying to point to the sun?

Neutrinos mostly pass right through the earth without hindrance. You
can't point a detector at anything. The best you could hope for would be
to arrange a series of big detectors so that they partially self shadow
and use timing details to do indirect imaging. Supernova neutrino
detection where there is a good pulse at high energy is now realistic:

https://arxiv.org/pdf/1205.6003.pdf

And with a bit of cunning they can get some directionality out of it.

> We could turn around the chlorine with its ring until we see what
> direction and position should the chlorine have to intercept at best
> neutrinos coming from a specific direction.
>
> Why does underground chlorine detectors work?

Enough deep shielding to prevent confusing side reactions of protons and
muons from masquerading as neutrino events. Together with the right
combination of properties in the Cl37 atom and Ar37 product to allow
detection of a reaction which is happening to neutrons in everything.

If the product of a neutrino reaction is stable, too radioactive or not
radioactive enough you have no way of distinguishing it.

> Bceause among the millions of atoms, one has the right orientation to
> exactly trap a neutrino.

Because the size nucleus is tiny compared to the volume of an atom and
the probability of a neutrino reaction occurring is even smaller.
Neutrinos mostly go straight through the Earth as if it wasn't there.

> Having an array of neutrino detectors at a molecular level would
> increase sensitivity and positioning.
>
> Or not?

Not. There might be something in forcing their *nuclear* spins to align
with a strong magnetic field but I doubt if it is realistically possible
or would give any worthwhile advantage in sensitivity.

--
Regards,
Martin Brown

[Steve Willner]

In article <p48fec$4m6$***@dont-email.me>,
jacobnavia <***@jacob.remcomp.fr> writes:
> Why do neutrinos react with some Chlorate compounds?
>
> Isn't it a consequence of the geometry of the collision?

Not that I'm aware of.

The chlorine-37 experiment has an energy threshold of 814 keV.
Neutrinos with lower energy cannot be detected by that type of
detector.

Other detectors have lower thresholds but still in the many-keV
range. I can't imagine any hope of detecting milli-eV neutrinos,
though perhaps some very advanced civilization might find a way.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone [phone] ***@cfa.harvard.edu
Cambridge, MA 02138 USA

[Phillip Helbig (undress to reply)]

In article <p48fec$4m6$***@dont-email.me>, jacobnavia
<***@jacob.remcomp.fr> writes:

> Le 23/01/2018 22:33, Steve Willner a écrit :
> > In article <9a148735-0045-426e-8b2f-***@googlegroups.com>,
> > Gary Harnagel <***@yahoo.com> writes:
> >> The neutrino flux would be red-shifted by z ~ 1100 also,
> >
> > This ignores the part about the neutrinos having decoupled long
> > before the photons. One source, which seems to be a textbook by
> > Daniel Baumann at Cambridge:
> > http://www.damtp.cam.ac.uk/user/db275/Cosmology/Chapter3.pdf
> > gives a neutrino decoupling redshift of 6E9. That corresponds to an
> > energy of about 1 Mev and a time about 1 s after the Big Bang.
> > Cosmological neutrinos should therefore have a kinetic energy today
> > of about 1/6 meV (i.e., milli-, not mega-). As the OP wrote, that's
> > very far from detectable.
> >
>
> It depends on your antena's neutrino sensitivity.

I'm no expert here, but the energy of neutrinos in the cosmic neutrino
background have energies orders of magnitude below that needed for the
reactions used in conventional detectors.

> Why do neutrinos react with some Chlorate compounds?
>
> Isn't it a consequence of the geometry of the collision?

I'm not sure what you mean here, but even if it were, the energies are
too low.

I'm sure some expert will weigh in on the "direction" stuff.