Re: Is electromagnetic field theory unified?
From: Franz Heymann (notfranz.heymann_at_btopenworld.com)
Date: 03/04/05
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Date: Fri, 4 Mar 2005 22:20:45 +0000 (UTC)
"Creighton Hogg" <wchogg@hep.wisc.edu> wrote in message
news:Pine.LNX.4.44.0503031701280.10753-100000@dill.hep.wisc.edu...
>
>
> On 2 Mar 2005, Ken S. Tucker wrote:
>
> > Creighton Hogg wrote:
> > > On 1 Mar 2005, Ken S. Tucker wrote:
> > >
> > > > Creighton Hogg wrote:
> > > > > On 1 Mar 2005, Ken S. Tucker wrote:
> > > > >
> > > > > > Creighton Hogg wrote:
> > > > > > ...
> > > > > >
> > > > > > >You
> > > > > > > realize you are talking to someone who insists that the
> > Minkowski
> > > > > > >metric
> > > > > > > is not flat?
> > > > > >
> > > > > > Of course it's not "flat", nor is it curved,
> > > > > > it is however nonorthogonal, and I'd be careful
> > > > > > about describing 4D nonorthogonal spacetime as
> > > > > > flat without a definition.
> > > > > > Of course Riemann's B^a_bcd=0 is true, but
> > > > > > that detail escapes most.
> > > > >
> > > > > Flat is defined as the Riemann curvature being uniformly
zero.
> > > > Everyone
> > > > > understands that.
> > > >
> > > > Are *flat* metrics transformable to a
> > > > cartesian metric? Is that always true?
> > >
> > > I believe that if the riemannian curvature is uniformly zero,
then
> > you can
> > > have riemann normal coordinates covering the entire manifold.
I'd
> > have to
> > > check my copy of Jost's book on differential geometry though.
> >
> > In a orthogonal CS's, the covariant and contravariant
> > basis vectors are equal, but not in a nonorthogonal CS.
> >
> > Therefore
> >
> > g^uv = g_uv (is orthogonal)
> >
> > g^uv =/= g_uv (in nonorthogonal).
> >
> > In both cases Riemann Curvature tensor can be
> > B^a_bcd=0, however I would hesitate to assume
> > a nonorthogonal CS can be transformed to a
> > cartesian CS, and so the term "flat", for
> > nonorthogonal spacetime CS's can be a bit
> > misleading.
> >
> > > > >If you're describing a vacuum solution to the Einstein
> > > > > field equations, you'll have a spacetime that is Ricci flat
but
> > not
> > > > > (Riemann) flat, like the Swarzchild solution.
> > > > >
> > > > > > >Do you remember the time he thought gluon exchange
diagrams
> > > > > > > were springs?
> > > > > >
> > > > > > Recently, a bright gamma ray burst hit our solar
> > > > > > system (Dec 27). It is theorized to result from
> > > > > > a change in a neutron star called a magnetar.
> > > > > >
> > > > > > You can look up the figures yourself, and if
> > > > > > you can't I'll get some refs if requested.
> > > > > > Hogg, do you know what the effect of a 10^15 Tesla
> > > > > > magnetic field is on a nucleus, apparently not.
> > > > >
> > > > > You tend to assume strange things.
> > > > >
> > > > > > The gluon exchange becomes elastic, the nucleus
> > > > > > is streched to an ellipsoid, and the nucleus is
> > > > > > eventually shredded under higher power, the
> > > > > > Strong Force mediating gluon itself becomes a
> > > > > > mere elastic band stretched to the breaking
> > > > > > point.
> > > > >
> > > > > Nope, not elastic at all. The potential rise is linear at
long
> > > > distances,
> > > > > with a spring it's quadratic. Gluon screening leads to an
> > > > essentially
> > > > > constant force at long distances. Not exactly Hooke's law,
is
> > it?
> > > > In
> > > > > addition, at short distances there is no repulsive force, as
> > you'd
> > > > expect
> > > > > from a compressed spring. This description fails pretty
badly.
> > > >
> > > > For a good read, check out this site...
> > > >
> > > > http://solomon.as.utexas.edu/~duncan/magnetar.html
> > >
> > > Yes, but what I was talking about was pretty much unrelated to
that
> > site.
> > > I was saying that your spring analogy of fragmentation and
> > hadronization
> > > doesn't really work that well. Read about the Lund string
model, the
> > most
> > > commonly used implementation of fragmentation of partons.
> >
> > Well the point is a field is quite able
> > to rip apart an Fe nucleus, ie. N-stars,
> > so now we're discussing brittleness and
> > elasticity of that nucleus.
>
> Okay, but pulling apart a nucleus doesn't directly have anything to
do
> with gluon exchange. The force *between* nucleons is given by the
> exchange of colorless combinations of quarks and gluons, for example
> pions. This was how the pion was first predicted I believe, before
strong
> interactions were understood as well as they are today. That being
said,
> there is still alot of work being done today to connect QCD with
observed
> nuclear physics. It's a non-trivial task. However, my point here
is that
> the interactions between *nucleons* are not going to exhibit
> hadronization, fragmentation, and confinement because all objects
involved
> are colorless combinations of other particles. So let's go a step
lower
> in scale.
> There's something called deep inelastic scattering that has been
studied
> at HERA, an electron/positron proton collider where Zeus is an
experiment.
> Now, what happens is that you have an incoming electron exchanging a
> virtual photon with one of the quarks in the proton. This photon is
> exchanging a very large momentum between the two, so it is very
short
> wavelength. In essence, it doesn't see the proton as a solid
object, but
> as a bag of valence quarks, gluons, and quark anti-quark pairs.
Now, when
> this photon is absorbed by one of the quarks, it kicks it out hard
and
> fast from the proton. Now, this is where confinement and all that
stuff
> like the string model come in. The quark that was kicked out is
still
> interacting via gluon exchange with the rest of the proton, and you
can
> model this by a string of *constant* tension connecting the quark
with one
> of the other particles in the proton. As this string gets longer,
the
> total energy contained in the string grows, since the energy
contained is
> essentially the tension times the length of the string. Eventually,
if
> the quark was kicked out hard enough, it will become energetically
> favorable for the string to "snap" and create a particle
anti-particle
> pair. So what you get now is two shorter strings with a quark on
each end
> of them. This is the basic picture of fragmentation. Let this
process
> keep repeating until eventually the strings are short enough and the
> momenta similar enough that all the particles connected by strings
are
> going to recombine in a process called hadronization. The end
result of
> this process is something called a hadron jet. Jets are very
important to
> measurements in modern particle physics.
> Now, what was the point of this entire digression? Well, my point
is that
> the idea of "elasticity" may work on the level of a whole nucleus,
I'm
> not sure really, but on the level of the individual protons and
neutrons
> it is going to fail in high energy interactions. Also, gluon
behavior is
> not elastic. Elastic implies a force growing over distance, which
isn't
> right. What you're seeing in the production of jets is the effect
of
> gluon screening, essentially it's because gluons themselves carry
color
> charge and interact with themselves.
>
> (Franz and Bjoern, if you have anything to add or correct please do
so
> with my blessing.)
That was a nice exposition, fit for the more intelligent non-experts
who may br lurkoing.
-- Franz "A first-rate laboratory is one in which mediocre scientists can produce outstanding work" P.M.S. Blackett
- Next message: Rand Simberg: "Re: Runaway Global Warming Possible!"
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