Natural Science Forum / Physics / General Physics / July 2008

"Meanwhile, the hydroxide ions decompose to form oxygen. "

Electroneutrality requires charge balance for products against
reactants. This alleged 'decomposition' must therefore result in
negatively charged hydrogen ions in aqueous solution - never before
seen!!!

"Using an optical microscope, the Swedish team observed that the vortex
rings consist of water swirling around in very fine circles with
diameters ranging from 10 to 50 microns "

If they can optically observe vortices in 'pure' water, their optics
are nothing short of miraculous.

"The scientists say that the protons move along a spiral path in
solution, which corresponds to the hydrogen-bonding network between
water molecules, and that this leads the formation of the vortices
(figure 2)."

A charged particle moving in a spiral path must be subject to an
accelerating force normal to its path.  Maxwell's equations dictate
that such a force can only be magnetic, yet no magnetic fields are
reported.

I am dubious.

Tom Davidson
Richmond, VA

On 19 Jul, 22:40, Craig Markwardt

<craigm...@REMOVEcow.physics.wisc.edu> wrote:

Yes, it is virtually the same because I said *about* 10% (meaning of
the order of 10%). But even a 25% percent temperature drop still means
that cooling processes and the related loss of energy at the top of
the atmosphere are not overly important here, and that equipartition
of energy by means of elastic collisions dominates.

Such a scenario is debatable because it all depends on what
assumptions you make regarding the details of the cooling processes.
Without any inelastic collisions that can destroy the kinetic energy
of the atoms or molecules there simply won't be any cooling at all in
a bound system of particles.

But the comparison of the sun with the troposphere is misleading here
anyway: as indicated already earlier, the troposphere consists of
individual molecules that can absorb and re-emit infrared radiation.
No such coupling between matter and radiation could exist in the solar
interior, because a) no individual molecules and atoms can exist there
due to the high density (and temperature), and b) visible radiation
(which constitutes the bulk of the solar energy output) can not be
absorbed by atoms anyway.

Furthermore, the sun is, in contrast to the troposphere, a self-
gravitating system that can gain energy from gravitational
contraction:  if you assume a self-gravitating gas ball in hydrostatic
equilibrium, and somehow suddenly remove the kinetic energy of the
particles in some sub-volume, then this sub-volume will begin to
collapse at a free-fall rate until it has gained enough energy so that
a new equilibrium is established. So if you have a region where
cooling takes place (like in the solar photosphere), the information
about this cooling can never propagate to the sun's interior as
hydrostatic equilibrium is locally established much faster than the
heat can travel. So the effects of the cooling will only be restricted
to the photosphere itself, but not extend to regions below (where we
have the original 'virial' temperature).

What I said was that a collisionally dominated gas (in a state of
equilibrium) must be necessarily isothermal *in the absence of any
heat sources or sinks*. And in that case it will be *strictly*
isothermal. Of course this is an ideal condition which in practice
does not strictly hold, but according to my argument above, it should
hold quite closely for the interior of the sun, as the effect of
gravity will level out any temperature gradients potentially arising
from the cooling in the photosphere.

No, not really. In contrast to the troposphere, the sun doesn't need a
separate heating to be stable as it is a self-gravitating system and
thus produces the required energy itself.
Also, the fact that the sun obeys the same R~M^1/3 mass-radius
relationship as the giant planets in the solar system, indicates that
there is no difference in their density (and thus temperature-)
structure.

I quote from http://www.ap.smu.ca/~guenther/Level01/solar/what_is_ssm.html

"Nearly all stellar evolutionary calculations are calibrated with
respect to the standard solar model. The standard solar model is
derived from the conservation laws and energy transport equations of
physics, applied to a spherically symmetric gas (plasma) sphere and
constrained by the luminosity, radius, age and composition of the
Sun".

It should be obvious that if you apply physical parameters as
constraints to a model (like the luminosity (i.e. the surface
temperature)), then the model can not give an explanation for the
parameters, because they serve on the contrary as input values.
As I said already repeatedly, the temperature of the particles in the
sun can only be lowered below the 'virial' temperature if inelastic
collisions are present. The standard solar model can't address this
point in any way, and therefore it is not a valid physical model. On
the other hand, I only need the mass here to *derive* both the radius
and the surface temperature (the latter being 1/1800 of the 'virial
temperature').

You can only affect the kinetic energy of atoms/ions if you can
produce ionization (or dissociation or rotational/vibrational
excitation in case of molecules). Visible radiation can't produce
either of those.

Yes, that's what I said.

I said the *threshold density* is 3*10^23 cm^-3 , so the photospheric
density must be below that in order for atomic excitation processes to
be able to take place. This threshold density defines the 'edge' of
the sun. We may strictly speaking not be seeing it because there are a
few kilometers of less dense material sitting on top of it, but
nevertheless it defines the edge (and the mass-radius data for the sun
as well as the planets confirm this).

The electrostatic force for a point charge also diverges like 1/r^2 at
the origin, but nevertheless it is a valid and successful theoretical
model.
Of course I am not saying that the density is infinite at the center
of the sun. The assumption of an ideal gas in hydrostatic equilibrium
will surely break down if the density reaches the nuclear density for
instance, but this should only apply within 20m or so of the sun's
centre, so it will barely make any difference for the calculation.

If you want, you can do a series expansion in terms of power laws. The
argument would still then apply to each term. But there is hardly a
point to do this as the mass-radius data for the sun and the planets
are consistent with a 1/r^2 density decrease (at any rate an identical
structure)).

Thomas