Article View: sci.physics.fusion
Article #1316Quick replies
From: jonesse
Date: Wed, 13 Jan 1993 00:26
Date: Wed, 13 Jan 1993 00:26
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7048 bytes
Dear Colleagues,
With the start of the new term at BYU -- and teaching a class I've not
previously taught -- I find considerable difficulty in keeping up with
questions and comments apropos (did I misspell, Mitchell?) to the net.
I regret this since I enjoy the ongoing give-and-take, but will try a
few quick responses here.
1. Natural fusion in the earth
Paul J. Karol asks whether we considered a tritium-producing branch in
uranium and thorium fission -- as a prosaic explanation for tritium now
being detected in volcanic emissions (see Jones' earlier postings).
We looked at this some time ago (ca 1989) and found the 3H branch to be
far too small. I don't find quantitative notes. Note that some
(e.g. Harmon Craig) suggested
that the tritium which we noted was detected in Kilaeua serendipitously
in 1972 (see Jones et al. J. Fusion Energy Dec. 1990 paper) was due to
release of radioactive gasses from a nuclear sub stationed off the coast
of Hawaii (big island). This colorful explanation failed for several
reasons -- including the large distance from the shore to the monitoring
station near the top of Mauna Loa, and a sharp denial by the U.S. navy--
one of which was the small branching ratio for fission into 3H.
However, since one should be quantitative, would someone do some homework
on this? (For extra credit, of course.)
2. Fracto- vs piezo-nuclear fusion (or, Hot vs Cold fusion)
Dieter asks whether fracto-fusion is related to piezonuclear fusion (so called
in a joint paper I wrote in 1985 -yes, 1985 - with Clint Van Siclen, J. Physics
G12:213-221, published March 1986). No: fracto-fusion is a speculative
sub-order of hot fusion. It involves acceleration
of deuterons due to electric fields produced during fracture of a deuterided
solid, so as to induce microscopic hot fusion in the crack. (BTW, I think
Dick Blue is correct about electrons being accelerated preferentially to d's,
so as to wipe out any accelerating E field. But electrons moving in a metal,
e.g. Pd, should neutralize fields in a conductor when a crack forms, precluding
formation of a large E field to begin with.)
On the other hand, piezonuclear fusion, if it exists pour de vrai, would
involve a cold-fusion process, of which
the prime example is muon-catalyzed fusion.
What is the difference between hot and cold fusion? Glad you asked.
Essentially, the difference is that in hot fusion the nuclei are unbound and
mostly undergo elastic collisions, whereas in cold fusion [which we earlier
called "piezonuclear" fusion, using Greek term for squeezing], the nuclei are
bound by some agent, such as muons (or perhaps a lattice). Therefore, in
cold fusion, the nuclei continually bounce against the Coulomb barrier so that
barrier penetration ("tunnelling") has a higher probability to occur than would
otherwise be the case. E.g., in muon catalysis, deuterons are bound by a muon
and separated by about 0.004 angstroms (1/200th the separation in D2 bound by
electrons). The equivalent temperature is about 250 eV, of the deuterons so
bound. Fusion occurs in about a nanosecond -- while the deuterium target
itself may very well be near room temperature. Hence the term 'cold fusion'
has been applied to muon-induced fusion for decades. (See article in Sci.
American, July 1987 by Johann Rafelski and myself: "Cold Nuclear Fusion.")
Someone calculate the fusion rate for
these conditions assuming hot fusion -- another exercise.
3. Someone at Johns Hopkins (Nod Sivad?) objects to a statement of John
Huizenga, which I quoted:
"Room temperature nuclear fusion without commensurate amounts of fusion
products is a delusion and qualifies as pathological science."
This is indeed a strong statement, but I think it hits the mark pretty closely.
Sivad objects: "Of course, if one defines fusion in terms of current theory
and calls anything else a non-fusion nuclear process, then the statement
stands. If there is anything to CF, it may very be some special case
undreamed of by our Horatio imaginations. I wouldn't be surprised if it
turned out to be some "intermediate" nuclear process residing somewhere
between chemistry and fusion. -- me" (8 Jan. 1993 posting,
ded@aplcen.apl.jhu.edu)
>>> Please give one example of an ' "intermediate" nuclear process residing
somewhere between chemistry and fusion'. Thanks.
I don't see why current theory is needed here, beyond E=Mc2 (which is strongly
supported by a large experimental data base). That is, take ANY nuclear
reaction, for discussion, say
A + B ---> C + D.
In a nuclear reaction, nucleons are rearranged, the binding (nuclear) energy
changes, and energy(E) may be released. If so, mass of (C+D) is less than that
of (A+B), by an exact amount E/c2. But this also means that (C+D) are
different species than (A+B) -- unless we are talking about de-excitation.
Even then, the de-excited species (hydrinos, anyone?) must be produced in
quantities commensurate with E/c2.
Ash or products must correspond in quantity to the heat released, says
E=Mc2. Huizenga does not need to presuppose any particular
reaction. E.g., d+d --> helium4 + lattice heating is allowed in
his statement (and it can be broadened beyond "fusion" of hydrogen isotopes
while retaining full validity) -- but even in this reaction which certainly
violates current paradigms, there must be helium4 produced commensurate with
the heat released. (A quick calculation for this particular reaction is
instructive since it shows that a great deal of 4He is made, 2.6 X 10E11 atoms
4He per watt of "excess heat". Quantities of 4He associated with megajoules
of xs heat now claimed are EASILY measurable.) Where's the helium4???? This
is an example of what Huizenga is demanding -- and rightly so.
If one is
claiming nuclear reactions as the basis for xs heat, then nuclear ash
(products) must be shown in quantities commensurate with the heat.
This is what Huizenga is demanding, and I agree.
Note that for ANY nuclear reaction, products are produced at a rate
6.24 X 10E12/Q per second, for each watt of xs power,
where Q is the energy released in the reaction in MeV.
Since nuclear reactions release of order MeV, we find that a LARGE quantity
of products -- helium isotopes, or Be, or transmuted isotopes -- whatever--
MUST be produced.
These products of nuclear reactions are nowhere found in
quantities commensurate with xs heat claimed. That's the bottom line.
--SteveJones
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Path: arc1-ca.pugleaf.net!sparky!uunet!haven.umd.edu!darwin.sura.net!paladin.american.edu!howland.reston.ans.net!usc!elroy.jpl.nasa.gov!swrinde!emory!sol.ctr.columbia.edu!hamblin.math.byu.edu!yvax.byu.edu!physc1.byu.edu!jonesse Newsgroups: sci.physics.fusion Subject: Quick replies Message-ID: <1993Jan12.172603.331@physc1.byu.edu> From: jonesse@physc1.byu.edu Date: 12 Jan 93 17:26:02 -0700 Distribution: world Organization: Brigham Young University Lines: 112