Re: Pair production of neutrinos

Vladimir Dergachev <volodya@xxxxxxxxxxxxxx> wrote in
news:uvydnbQngrBsHpXanZ2dnUVZ_g2dnZ2d@xxxxxxxxxxx:

John Schutkeker wrote:

Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> wrote in
news:200709111721.l8BHLt4i015722@xxxxxxxxxx:

John S wrote:

PD <TheDraperFamily@xxxxxxxxx> wrote in
news:1189452119.175254.111980@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

Feynman diagrams are mnemonics for the terms in a path integral.
Once you draw the diagrams, you should know how to write down the
formula to calculate the scattering matrix element and hence the
differential cross-section. If you don't know how that's done,
then you should start from scratch and learn some skills about
* calculating cross-sections from scattering matrix elements.
* calculating scattering matrix elements from Feynman diagrams.
Since the calculations generally involve integrals of complex
functions, you may want to learn something about functional
integrals and complex variables as well.

Sorry, no shortcuts.

I haven't taken that class. My quantum is very weak.

"No shortcuts" is the salient point. This beginning QFT is what you
need to get the answer you asked for. I last did this in student
days, in the early 1980s. I don't remember the details but the
ideas are what PD said. We had to do these to use information on
the neutron decay to deduce limits on the Weinberg angle and/or the
W and/or Z masses. I only remember doing it, not the detailed steps
and not how much info we needed to get the numbers.

All the basic Feynmann diagrams I know are shaped roughly like the
letter 'H.' I posted this question over a year ago, and actually got
an intelligible answer, but unfortunately, I can't find those posts
with google. I was informed that the pair production reaction could
only occur in the vicinity of a third particle, and I accepted that
answer, although I don't know how to verify it.

That third particle screws up my attempt to draw the Feynmann
diagram, because now it is no longer in the shape of an 'H'. On the
left side, I would have a photon and a quark, but on the right side,
I would have a quark and two neutrinos. That's what throws the
wrench into my analysis.

Also, I have no idea how to draw a Feynman diagram with ascii
characters, so I could post it here and ask someone knowledgeable to
look at it. ?:p

I'm not actually interested in pair production in the vicinity of a
quark or electron, but all pair production that's not virtual.
Perhaps the Feynmann diagram wouldn't involve a quark or electron,
but a second photon. In that case, there would be two photons on the
bottom of the 'H' and two neutrinos on the top, and I'd finally have
my nice 'H' shaped Feynmann diagram.

Assuming that's the approach that I should take, what's the equation
that results from this diagram?

nu anti-nu
^ ^
| |
| WFB |
---->
| |
| |
^ ^
ph ph

Here ph=photon, WFB=weak force boson, and nu=neutrino, of course. I
still don't know which boson I'm dealing with yet, but I've got a
reference coming that I hope will answer that.


There is one more component to it - the edges correspond to
propagators and the vertices correspond to interaction terms in the
Lagrangian.

If the vertex term is 0 (not present in the Lagrangian) then the
entire diagram evaluates to 0.

How do we know which terms are present in the Lagrangian ? Well, one
can write out (or look up) the Lagrangian for the standard model, but,
an easier way is to use conservation laws - for example charge is
conserved, and as both nu and gamma are neutral then WFB must be
neutral too, i.e. Z.

However, gamma - photon - is the mediator of electromagnetic
interaction and thus only has terms that involve charged particles, or
itself. Since neither nu nor Z are charged this contribution is 0 too.

Here is a diagram which is likely non-zero (but very, very small - and
likely compensated with a conjugate one):
anti-nu
/
gamma--- e- --- W-
\/ || <- virtual pair
/\ ||
gamma--- e+ --- W+
\
nu

This diagram is small because for this event to happen we must create
an e-/e+ pair *and* a W-/W+ pair - which are very heavy, 160 GeV
total.

Also, I would like to point out that people made measurements of
neutrino flux from (large) commercial nuclear reactors and were able
to account for neutrinos based solely on fission byproducts (they were
trying to create an independent way to estimate reactor power).

So any neutrino creation by processes different from n -> p + e +
anti-nu is much smaller than the usual reaction - which is quite
common in astrophysics as well.

best

Vladimir Dergachev

Thank you very much Vladimir. :) This is exactly the sort of answer I
was looking for. You've saved me a lot of work, trying to chase down a
reaction pathway that doesn't happen. :) Of course, now I have no
excuse to teach myself QFT, so that'll have to await a new idea, but
nonetheless I learned a lot in the process.

Best Wishes To You,
John Schutkeker
.

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