sci.physics.research Google Group

Re: Why does the renormalization group work?

arnold.neuma...@univie.ac.at (Arnold Neumaier) — Fri, 11 Nov 2009 21:23:50 UT

The renormalization group is simply the consequence of having in a
theory a parameter E whose value does not affect the ideal theory itself
but only the approximation method used. In each such case (whether it
arises from quantum field theory or from a simple algebraic problem)
one has an associated transfromation group that transforms the results

Re: Your thoughts on idea that may kill string theory.

john_m_stan...@yahoo.co.uk (JohnMS) — Fri, 11 Nov 2009 21:23:25 UT

On Nov 26, 9:08 pm, "Jonathan Thornburg [remove -animal to reply]"
I see that Schiller writes about neutrino mixing on pages 280 to 281
of [link] .
He predicts neutrino mixing from the model,
and he states that knot calculations can determine
the numbers. Hm...

Re: relation between quantum mechanics and classical mechanics

xphysic...@gmail.com (X-Phy) — Fri, 11 Nov 2009 21:22:43 UT

On 26 nov, 19:32, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:
I was too radical effectively. But it remains that incoherent
superposition occurs in classical mechanics, and is formalized by
plain classical probabilities.
That is a contingency I spoke about. We were discussing the theories

Re: Charged or Neutral Strong Nuclear Force Exchanges?

lester.we...@gmail.com (Lester Welch) — Fri, 11 Nov 2009 21:22:06 UT

No. The strong nuclear force causes an attraction between any
combination of p's & n's.
The strength depends of the spin. Some particle/spin combinations are
not allowed because of
Fermi statistics.

Re: Quantum Field Theory: The Big, Simple Picture?

arnold.neuma...@univie.ac.at (Arnold Neumaier) — Fri, 11 Nov 2009 21:20:46 UT

As long as it has not been shown to be well-defined, it _is_
ill-defined.
No. it means that, at present, it is ill-defined, and that therefore
the question of its existence is, strictly speaking, meaningless.
What I meant in precise terms is, that nobody has been able to figure
out a definition of the path integral that makes logically sense.

Re: Who said Work = F d and Kinetic energy KE = (1/2)mv^2?

rds...@att.net (Richard D. Saam) — Fri, 11 Nov 2009 21:20:14 UT

Perhaps
Emilie du Ch'telet (December 17, 1706 -- September 10, 1749)
She did some basic work indicating Energy related to v^2 and not v
by dropping ball masses from different heights into clay
and checking deformation.
Richard D. Saam

Virtual Photons and the Infinite Range of Electromagnetic Interaction

palit.anami...@gmail.com (Anamitra Palit) — Fri, 11 Nov 2009 21:19:55 UT

Virtual Photons are responsible for electromagnetic interactions which
extend for an infinitely large distance.When these virtual photons are
shot out of charged particles the process conforms to the principle of
momentum conservation but energy conervation is violated. They move
with "borrowed energy" which is accommodated with the help of the

Re: Quantum Field Theory: The Big, Simple Picture?[repost]

jyab...@nycap.rr.com (Jay R. Yablon) — Fri, 11 Nov 2009 21:18:40 UT

Yes, see, e.g., the first page of the file linked below.
[link]
Jay

Re: The Mathematical Substructure Underlying Quantum Field Theory, and Why I Like Anthony Zee's Book so Much

jyab...@nycap.rr.com (Jay R. Yablon) — Fri, 11 Nov 2009 21:18:29 UT

For those who like conciseness, the core of what I am driving at in
[link]
can be put into a one-equation, one page synopsis, linked below:
[link]

Re: Quantum Field Theory: The Big, Simple Picture?

igor...@gmail.com (Igor Khavkine) — Fri, 11 Nov 2009 21:17:31 UT

Jay, as I pointed out in another post. This result is wrong. No matter
what steps you believe lead you to this result, it is incorrect. This
equality can only hold if S_EH were linear in the integration
variable, be it g_uv, sqrt(-g)g_uv, or just sqrt(-g). And it should be
quite clear that S_EH is not linear in any of these variables.

Re: relation between quantum mechanics and classical mechanics

xphysic...@gmail.com (X-Phy) — Fri, 11 Nov 2009 21:17:10 UT

On 25 nov, 07:52, Bob_for_short <vladimir.kalitvian...@wanadoo .fr>
wrote:
You seem to forget the uncertainty relation. For a single particle,
the position can be measured with any precision, even if its quantum
state is dispersed. Yet, the quantum state is afterward changed,
projected onto an eigenfunction, and thus it is impossible the measure

Re: Your thoughts on idea that may kill string theory.

hel...@astro.multiclothesvax.de (Phillip Helbig---undress to reply) — Fri, 11 Nov 2009 21:16:40 UT

In article
<3e988b94-e2be-4207-9d55-d1070 1453...@p19g2000vbq.googlegrou ps.com>,
Keep in mind that the neutrino was first detected a couple of decades
after it was first theoretically postulated. Isaac Newton mentioned the
bending of light by gravitation, and it took over 200 years before that
was observed.

Why does the renormalization group work?

fizix2...@gmail.com (Fizix 2009) — Fri, 11 Nov 2009 10:16:15 UT

Why does the renormalization group work so well? In fact, why does it
even work at all? Before you think this is a dumb question on my part,
please consider this for a moment. The renormalization group involves
"decimating", i.e. integrating away the ultraviolet degrees of
freedom. The basic idea of the renormalization group is that as long

Re: Your thoughts on idea that may kill string theory.

jth...@astro.indiana-zebra.edu (Jonathan Thornburg [remove -animal to reply]) — Thu, 11 Nov 2009 20:08:22 UT

In a moderator's note earlier in this thread, I wrote
Gordon Stangler <gordon.stang...@gmail.com> asked
Yes. "K2K" was a brain glitch on my part; the new experiment is T2K.
There's a bit more information about T2K at
[link]
and at
[link]

Re: Charged or Neutral Strong Nuclear Force Exchanges?

chalkys...@bleachboys.co.uk (Chalky) — Thu, 11 Nov 2009 18:32:03 UT

This would seem to imply that protons can only bind to neutrons and
vice versa in heavy nuclei.
Is that correct?
If so, I guess my original question would resolve into:
How often is this interaction neutral, and how often charged (ie
proton becomes neutron and vice versa)