Axil--
Note the following conclusion of the Wiki item on neutron magnetic moment:
"While the results of this calculation are encouraging, the masses of
the up or down quarks were assumed to be 1/3 the mass of a nucleon,[31] whereas
the masses of these quarks are only about 1% that of a nucleon.[32] The
discrepancy stems from the complexity of the Standard Model for nucleons, where
most of their mass originates in the gluon fields and virtual particles that
are essential aspects of the strong force.[32] Further, the complex system of
quarks and gluons that constitute a neutron requires a relativistic treatment.
A calculation of nucleon magnetic moments from first principles is not yet
available."
How do you know about the magnetic properties of isospin particles and their
reaction to a magnetic field?
Bob
----- Original Message -----
From: Axil Axil
To: vortex-l
Sent: Tuesday, April 07, 2015 10:47 PM
Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea
Rossi
isospin is a product of the strong force and of the quarks inside the protons
and neutrons. It is fixed no matter how the atoms spins. An atom might be
induced to spin using EMF but usually that spin cannot effect the isospin of
the nucleus. with zero isospin. Only non zero isospins are effected by RF.
On Wed, Apr 8, 2015 at 1:37 AM, Bob Cook <frobertc...@hotmail.com> wrote:
Two particles spinning anti-parallel equal 0 spin if they each have an
equal spin energy. Angular momentum is a vector quantity, not a scalar one.
Bob
----- Original Message -----
From: Axil Axil
To: vortex-l
Sent: Tuesday, April 07, 2015 10:16 PM
Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author
Andrea Rossi
I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin.
How can a nuclear reaction involving them have huge annular momentum?
On Wed, Apr 8, 2015 at 12:28 AM, Eric Walker <eric.wal...@gmail.com>
wrote:
Hi Bob,
The possibility you've been drawing attention to, that the result of
the decay of the [8Be]* compound nucleus into two 4He nuclei with little linear
momentum and a great deal of angular momentum makes for an interesting thought
experiment. Out of curiosity, I calculated the energy that would be needed to
break up an alpha particle into either tritium and a proton or 3He and a
neutron, which would be the reverse of these two reactions:
3He + n → 4He + Q (19.3 MeV)
t + p → 4He + Q (20.5 MeV)
As I understand it, this implies that angular momentum sufficient to
produce ~ 19 MeV of centripetal force would be needed to break apart a 4He into
either 3He and a neutron or tritium and a proton. This suggests that a 4He can
carry a large amount of angular momentum before it is likely to break apart. (I
assume the process is probabilistic and that the force needed lies along a
distribution.)
Further comments inline.
Eric
On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook <frobertc...@hotmail.com>
wrote:
However, I know of know reason why the light nuclei cannot have any
spin quantum number--high or low. Any spin quantum is available.
Further to the thought experiment, I think we should make a clear
distinction between two types of "spin" -- there's the actual spinning motion
of a nucleus (e.g., 4He), and there is the spin state of the nucleus. At
higher rates of rotation, a heavy nucleus such as an isotope of nickel will
reconfigure into a higher spin state, presumably through deformation. In such
a state a photon may be emitted, with the nucleus relaxing into a lower spin
state. Here my mental model is of neodymium magnets spinning around in a
clump. When they snap together into a lower-energy configuration, a photon is
emitted through the movement of the magnets as they snap together. The photon
is emitted in a direction and carries away energy in such a way as to slow the
angular movement of the spinning nucleus a little (by the amount of energy
carried away by the photon). The participants involved in such a transition
are the nucleons, and the energy of the photon that is emitted will
correspondingly be in the keV or MeV range, which is in the nuclear range.
A light nucleus, such as 4He, does not have a bound excited state. My
understanding is that it cannot deform under high angular momentum into a
higher energy state which will emit a photon when it relaxes. The 4He will
either break apart into lighter constituents under centrifugal forces or it
will not. But I'm guessing that the actual moment-to-moment velocity of the
4He about its axis of motion is in principle a continuous quantity. If this is
true, perhaps the energy could be released to the environment in small amounts.
Where the thought experiment gets interesting is in the supposition
that you and others have already offered in this thread, that charged body such
as a 4He nucleus that is spinning at an incredible rate will set up a magnetic
field. This magnetic field could disturb nearby electrons, causing them to
emit lower energy photons in the process.
Although I do not see anything special in the 7Li+p to 8Be transition
that has been proposed (and note Jones's point about the gamma that would be
omitted in the process), I think the more general notion of the energy of a
nuclear transition somehow being deposited in angular momentum and then
released in small amounts is a very interesting one.
