Widdekind

http://en.wikipedia.org/wiki/Work_hardening

[math]\sigma = \sigma_y + G \alpha b \rho_{\perp}^{1/2}[/math]

[math]\sigma \approx \sigma_y + K \epsilon_{p}^{0.2 - 0.5}[/math]

So, simplistically stated, the perpendicular / transverse dislocation density (# / m²) is proportional to the plastic strain ?

[math]\rho_{\perp} \approx \epsilon_p[/math]

Now, [math]\rho_{\perp}^{1/2} = 1/<d>[/math] where <d> = average distance between dislocations. So, seemingly,

[math]\sigma \approx \sigma_y + \frac{\left( \alpha G b \right)}{<d>}[/math]

And, a 1/d dependency resembles viscous fluid flow, [math]F = \mu A v / \Delta h[/math].

So, the material under strain is "trying" to flow away from the middle of the material body, "left & right" towards the two vice grips grabbing the specimen studied...

but, dislocations impede the plastic fluid-like flow, somewhat similar to viscous drag friction forces ? i.e. the material is forced to flow "up over (down under) and around" the dislocations (hence the presence of the shear modulus G) ?

Also, the "necking down" of the strained specimen seemingly concentrates dislocations, increasing the perpendicular dislocation density ? So, would Poisson's ratio, commonly close to a half, have some sort of conceivable connection, to this particular process ? For [math]\nu = 1/2[/math], if the strain doubled, the perpendicular area would have to halve, and so for a constant number of dislocations, their perpendicular density would double, [math]\rho_{\perp} \approx \epsilon[/math]. And, if [math]\nu \ll 1/2[/math], then nearly no "necking down" occurs, and so a constant number of dislocations would be concentrated allot less, seemingly suggesting, that the strain's dependency on stress, would decrease, as seems said in the second equation considered.

( producing a plot, i notice no clear correlation, between n <----> v )

why are you so sure that electrons are moving w/ linear translation at 0K ?

at 0K, "building" bonding orbitals, from individual atomic orbitals, could construct co-orbitals, involving cycloidal motions, "do-se-do dancing" in "Sufi circles" (for want of worthier words) "loop-de-looping" through the lattice

all of the probability currents would be closed, loop-de-looping through the lattice, back to some beginning

for example, you could conceivably construct a big bonding orbital, wherein one electron was (as a super-position of individual orbitals) rotating around the whole entire lattice, in (say) the counter-clockwise direction...

"you stare at a gold ingot bar, and one electron is in a wave-function state encompassing the brick, and rotating around right-handedly"...

why is such a picture impossible ? why could not all linear Maxwellian-like motions be attributed to thermal energy, over and above the baseline orbitals at 0K ?

the mass...

of a neutron...

exceeds that...

of a proton...

by 1.29 MeV...

yes ?

so, in n ----> p + e + v

that 1.29 MeV gets distributed, into 0.511 MeV for the mc2 of the electron, and 0.78 MeV of KE shared between the electron & neutrino, yes ?

in deuterium formation,

pp ----> pn + positron + neutrino + 0.42 MeV,

yes ?

so, w.h.t.:

pp ----> (pn = pp + 1.29 MeV) + (positron = 0.511+ MeV) + neutrino + 0.42 MeV = pp + 2.22 MeV

so, i guess the depth of the SF potential = 2.22 MeV = 1/4th maximum binding energy per nucleon

of that 2.22 MeV, 1.29 ups the proton to neutron, and 0.511 forms the positron

No.

If you want to change a neutron into a proton, you must also create an electron and the antineutrino. You ignored the mass energy of the electron. The Q of the reaction is 0.78 MeV, also mentioned in that link. That's the total energy released. It's not the energy the electron receives. The energy the electron gets is shown in a curve at that link and is a continuous function from 0 to 0.78 MeV, as it has to be, because it's a three-particle sharing of momentum. (which is one of the reasons the antineutrino was postulated in the fist place)

the formula for thermal conductivity is proportional, both to the temperature, and then also to the temperature gradient...

no heat flows, at absolute zero K... or with no temperature difference...

in post #10 i tried to show, that the formula for thermal conductivity decomposes, into pieces, which suggest that electrons possess no translational motion, at absolute 0K...

if so, then 0K bonding orbitals involve orbital-like motions, of electrons, circling around ions... but all "Bloch wave" motion e^ikx is due to thermal heat

i don't know those words to be true, yet that is what the equations seemingly suggest

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/proton.html

mass difference = 1.29 MeV

http://wiki.answers.com/Q/What_is_the_mass_of_the_Proton_Neutron_and_Electrons

dm c2 = 1.28 MeV

evidently, in neutron decay, the exiting electron receives 0.78 MeV of that energy... (?)

also, electron mass = 0.511 MeV

x2 = 1.022 MeV

so oops on the multiply...

yet that would only make PP more energetically likely & feasible (?)

EM interactions at 1fm <----> 1MeV ~= PP threshold

relying on electron capture turns the He-2 --> H-2 process into a less-likely three-body problem, yes ? a suggestion involving only the original reactants, relying on 1fm <----> 1MeV <----> PP, seems sound

simplistically re-stated, if the macroscopic bonding-like orbitals, of electrons, in metals, are "built" out of the valence orbitals, of all of the individual atoms...

which individual orbitals possess no linear motion e^ikx...

but do possess "orbiting" revolving motions...

then why would the resulting "Bloch waves" more resemble the former, than the latter ?

perhaps the motion, of electrons, through metals (at 0K), more resembles a planet, wandering through a large star cluster, from star to star, swinging around one lattice ion, over to another, revolving around it a couple of times, hooking one way to another ion, and so forth... more of a "wandering while spinning dance" do-se-do-ing thru the ions in the lattice ?

the mass-energy difference, between protons to neutrons, is ~1.29 MeV; and the (minimum) energy required for pair-production is ~1.22 MeV

in the production of deuterium:

products ----> reactants

p+p ----> p:n + v_e + e⁺ + 0.42 MeV = (p:p + 1.29 MeV) + v_e + e⁺ + 0.42 MeV

evidently, ~1.71 MeV would be released, if He-2 = p:p was a stable isotope...

but, since He-2 is not stable, ~1.29 MeV, of that amount, is consumed, in converting one proton into a neutron... => net energy released = 0.42 MeV

EM interactions that occur over distances of ~1 fm <----> ~1 MeV

thus, all of the EM interactions inside nuclei are MeV intense, sufficient for pair-production to occur...

in Beta(plus) decay, matter (neutrino) + antimatter (positron) is emitted, suggesting some sort of prior pair production of matter/antimatter...

intense EM interactions are known to produce pairs of positrons / nega-elec-trons...

such seems to suggest, that nuclei are quite commonly generating positron/electron pairs, the electrons of which (or their charge) can be absorbed into protons, via Weak Force interaction, generating neutrons in beta(plus) decay

otherwise, the Strong Force interaction's potential appears to be about ~1.7 MeV per nucleon-neighbor deep, and about ~1 fm wide... an attractive square-well-like potential, nestled into the center, of the EM 1/r potential...

in stars, protons tunnel through ~1000 fm of the EM potential, to (occasionally) fuse via the SF, inside the central 1 fm-wide SF potential well

....

 

In the formation of a neutron star in contrast, (negative) electrons are absorbed to make neutrons, while in the proton-proton cycle according to Wiki, a positron is emitted; the mass balance differs a lot. ...

dE for p ----> n = 1.29 MeV

dE for e+ + e- = 1.22 MeV

so, the turn-around distance, where KE = V, at typical temperatures of tens of MK, would be about 1pm=1000fm ? so, in stars, all "He-2" fusion would be quite cold... and could occur, only w/ tunneling... only the "tails" of the protons' wave-functions could conceivably overlap enough, to (partially) perceive any Strong Force attraction... the "bodies" of the wave-functions would be repelled, whilst the "tail tips" thereof would touch and become attracted...

if an additional electron is not required to collide into the "complex" of proton-proton wave-functions... then the comparability, of the energies, of pair-production, and proton-neutron conversion, seems suspiciously auspicious

evidently, p+p ----> p:n + 0.42MeV

if the n soaks up 1.29 MeV... then the actual p+p attractive Strong Force potential could conceivably be about 1.7 MeV of which most is consumed in converting one proton into a more massive neutron...

the p-p potential would have a width of ~1fm, and a depth of ~1.7MeV ?

(if the max binding energy per nucleon is ~9MeV, which is ~= 5-6 x 1.7MeV... then perhaps in more massive nuclei, individual nucleons typically possess 5-6 partner neighbors, with which to engage in attractive SF interactions ?)

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/proton.html

http://en.wikipedia.org/wiki/Proton–proton_chain_reaction

a metal = "mega molecule"... qualitatively similar to a smaller molecule, w/ only a few atoms...

the bonding orbitals, in all such "molecules", do stretch around and span all the nuclei there-in...

e.g. in H2, the bonding orbitals stretch around both protons... in a macroscopically massive amount of metal, the orbitals stretch through the whole block of material...

at zero K, the thermal conductivity = 0... there is no more paradox, in such a statement, than claiming a heat gradient exists... at a single temperature (?!)... yes, technically, if the two ends of some metal are maintained at different temperatures, then what is the one temperature of the material?? But, at zero K, if one end were heated slightly, no heat would flow thru (the regions at 0K)...

you say the electrons are moving... but not storing thermal energy... at the minimum, would not extra thermal energy, affect their motions ?

if the bonding orbital, in H2 does NOT possess any e^(ikx) term (at 0K)...

then are you completely certain & sure, that the bonding orbitals in macroscopic metal "mega molecules" DO possess e^(iix) terms (at 0K) ?? what would account, for that qualitative difference (no motion ----> motion) ?

...

when you "build" a bonding orbital, for H2, from the two 1S orbitals, of the separate protons... your "ingredients" are orbitals having no organized, coherent motion... so the resulting composite combined hybrid orbital, embodies no more motion, than its constituents...

but, if you built an "excited bonding orbital", for H2*, from (say) the two 2P₊ orbitals, of the two protons, which orbitals look like "donuts" w/ phase circling about the z-axis in a right-handed sense...

then the resulting hybrid orbital, would (logically) be a double-donut, within which the electron would basically be circling about both protons, a little like a planet orbiting two stars...

and, in macroscopic metal blocks, bonding orbitals are built, out of such high-n, high-l, high-m orbitals...

so, i guess you could imagine macroscopic metal bonding orbitals, within which electrons circled all the way up some line of atoms (say), hooked around the last one, and then propagated all the way back down the other side of the ions, until hooking back around at the opposite edge...

in the hydrogen wave functions, the only e^(i ...) terms, are [math]e^{\imath m \phi}[/math], representing circulation, about the nucleus... if you built bonding orbitals, up out of such individual orbitals, then why would you start with orbital circulation motion of bound electrons... and end w/ linear free-particle-like motion ?

You were lastly speaking about water molecules, in which context I replied. A metal is different.

 

"Conduct heat at zero kelvin" is a formal paradox. A question of logic, not physics.

 

At arbitrary low temperature, electrons conduct electricity, and very well. Again your old misconception. Why stick to it?

And again, heat conductivity is low because these very mobile eelctrons store little heat, proportional to kT.

 

exp(ikx) exists at any temperature, even cold, and many such states are occupied. The unique state with k=0 is negligibly scarce. The states in a metal are usually not called "orbitals", and besides naming, their states are delocalized to the full metal part. Your old misconception, again.

 

Well, I stop for this time. A metal is not a small molecule, and you have to understand that electronic states are not bound to one atom, but spread over all atoms. These states are the lowest accessible ones to electrons and are occupied at any temperature.

Oxygen is electronegative, and "steals" (up to) two electrons, to effect an electron configuration similar to neon. In such a state, the full valence shell is spherically symmetric... and spherically symmetric charge distributions, perceived from afar, are those of point charges, of the same total charge, located at the centers of those charge distributions. So, the potential, of O^-- would be

[math]\approx \frac{-2 q_e}{4 \pi \epsilon_0 r}[/math]

But then you must re-remove two electrons, from two 2P orbitals... given that Oxygen forms bonds about its equator, into P_x + P_y (hybrid) orbitals, the equator seems the appropriate place to re-remove two units of charge from... the P₊ + P_- orbitals' charge distributions, are "donuts" centered on the Oxygen nucleus, evidently w/ radii of roughly 100 pm, which is the size of the oxygen atom, namely the level 2 orbitals of the oxygen.

Now, at least in the equatorial plane, the potential from a ring of charge, of radius R, out at a distance D > R from the origin, is:

[math]= \int \frac{dq}{4 \pi \epsilon_0 r} = \int_{\phi=0}^{2 pi} \frac{\mu R d\phi}{4 \pi \epsilon_0 r(\phi)}[/math]

[math]= \frac{\mu R}{4 \pi \epsilon_0} \int d\phi \frac{1}{D}\sum_{l=0}^{\infty} \left( \frac{R}{D} \right)^l P_l(cos \theta)[/math]

The odd-numbered Legendre Polynomials all integrate to zero. But, the first few even LP's give rise to:

[math]= \frac{\mu R}{4 \pi \epsilon_0 D} \int d\phi \left( 1 + \left(\frac{R}{D}\right)^2\frac{1}{2}\left( 3 cos^2\phi - 1\right) + ...\right)[/math]

[math]= \frac{\mu R}{4 \pi \epsilon_0} \left( 2 \pi + \frac{1}{2} \left( 3 \frac{1}{2} 2 \pi - 2 \pi \right)\left(\frac{R}{D}\right)^2 + ...\right)[/math]

[math]= \frac{2 \pi \mu R}{4 \pi \epsilon_0D} \left(1 + \frac{1}{4}\left(\frac{R}{D}\right)^2 + ...\right)[/math]

[math]\approx \frac{Q}{4 \pi \epsilon_0 D} \left( 1 + \frac{1}{4}\left(\frac{R}{D}\right)^2 \right)[/math]

Now, if the ring of charge, mathematically modeling the absence (presence) of two electrons (holes) from the Oxygen's other-wise-spherically-symmetric charge distribution, contains two units worth of charge (the missing two electrons), then the total potential, from a normal neutral Oxygen atom, would be:

[math]V_O(equator) \approx \frac{-2 q_e}{4 \pi \epsilon_0 D} + \frac{+2 q_e}{4\pi\epsilon_0D}\left(1 + \frac{1}{4}\left(\frac{R}{D}\right)^2\right)[/math]

[math]= \frac{\frac{1}{2} q_e}{4 \pi \epsilon_0 D} \left(\frac{R}{D}\right)^2[/math]

That equation implies, that (near the equatorial plane) the missing electrons, generate a positive potential, which falls off as 1/r³... such a potential could account for the electro-negativity ("electro-positivity"?) of oxygen, which "wants" to grab (up to) two other electrons, into equatorial P_x,y orbitals...

does that equation seem potentially qualitatively accurate (if not quantitatively precise) ?

i didn't suggest that electrons are actually in fully excited states...

but bonding orbitals, at absolute zero T, are static / stationary... the electrons in metals conduct no heat, at absolute zero T...

but at finite T, electrons must begin moving, acquiring some amount of thermal motion, explaining why they can then begin to conduct heat... and also their "Bloch" waves...

at absolute zero T, there are no Bloch states, all the electron orbitals are static => no thermal heat conduction, no moving electrons

at >0K, electrons' wave-functions acquire [math]\times e^{\imath k x}[/math] => begin moving, begin conducting heat

if something similar occurred w/in water molecules, then the electrons' wave-functions would be "perturbed" above the ground state... w/o being fully excited... "perturbation theory" could become applicable, and could conceivably account, for thermal conduction, Bloch-wave-motion, and the depolarization of water molecules...

the electrons are not fully in the 0K ground state... they acquire a "small perturbation" worth of energy, which "ruffles" or "fluffs up" the wave-functions (for want of worthier words) somewhat slightly

the electrons then begin to move thru their molecular lattice, a little like a satellite orbiting earth (say) acquiring some amount of thrust

are you saying, that electrons in metals are moving, at absolute 0K ? if so, then why is thermal conductivity [math]\propto k_B T[/math] and so zero at zero K ?

if you build the bonding orbital, in H2, from two 1S states, localized to each separate proton...

then would you build the bonding orbital, between O:H w/in water, from a 2P state localized to the oxygen + 1S state localized to the hydrogen ?

inexpertly, the O in water is basically an O2- ion... w/ two protons attached to two 2P orbitals, w/in the level-2 spherical shell of negative charge excess

that O2- ion would be spherically symmetric (closed shells), and so would generate basically a -2 charged point source' field...

but, take away the extra two electrons, borrowed from the 2 hydrogens, into the two 2P orbitals...

and the resulting charge distribution would be "spherically symmetric -2" plus "a positive charge worth of charge in 2P-x" plus "a positive charge worth of charge in 2P-y"...

which would produce a negative charge shell, surrounding the oxygen nucleus, having "holes" w/in it, along the x-y axes around the equator...

thru which positive field lines would protrude...

which would account for oxygen's electronegativity... around its equator, the "fog" of the electrons' charge distribution is "lifted", and so you could electro-statically "see" straight down towards the +8 nucleus, at least down two orthogonal axes

someone who could do the math would probably find, that building a bonding orbital, out of the oxygen's 2P + hydrogen's 1S, and then differentiating the answer w.r.t. bonding distance / bond length, could closely account for the 5eV bond energy, and the 100pm bond length... essentially similar to the calculation for H2

how would you account, for all of the other electrons? the bonding orbital for H2+ is actually simple... but O:H has numerous other electrons partially screening out, but partially not, the nuclei

(over above that, the 0.05eV of the hydrogen bonds / critical temperature could conceivably be treated as a small perturbation; if you did the calculation correctly, you would evidently find, that the perturbed ground bond state gains 1% in energy, and is no longer localized preferentially to the oxygen... evidently the perturbed T>0K ground state is "more of the H's 1S, less of the O's 2P" ?)

if you estimate the quantum Hooke's constant "k", for many metals, from their Young's moduli & sound-speed, then you find that the quantum oscillator frequency ~THz, and energy 0.001-0.01 eV...

at STP = 300K = 1/30th eV, most lattice ions are probably in the lowest few oscillator states, i.e. n = 1,2,3... "a few"...

to melt the metals = 1eV = thousands of K = oscillator energy level in the dozens, i.e. n = 30 => metal is melting

i think the calculation of the "k" constant is straightforward & simple, but am not completely confident

if i calculated correctly, then the 0.05eV energy of water's H-bonding + surface tension is of the appropriate order of magnitude... at 374C = critical temperature, the O's + H's are jiggling around in n~= 5 (roughly) states... and the electrons, perhaps entrained w/ their ions' motions, no longer generate polarized charge distributions => surface tension ----> 0

perhaps perturbation theory could add such harmonic oscillator potentials, to the 0K bonding orbital ground states, to account for such effects ??

At moderate temperature, electrons are not excited.

[by the way, hydrogen has no 1s and 2s in water - only molecular orbitals]

 

Heat is stored in translation, rotation, in PV (for the Cp heat), little in vibrations at room temperature. In the liquid, breaking more hydrogen bonds at a higher temperature makes a significant heat capacity, which you can estimate by comparing the liquid's capacity with the vapour's Cv (or Cp - P*V).

 

Heat is very unlikely to excite electrons, even in molecules bigger than water. Dyes for instance are specially made: long and with electrons delocalized over dozens of atoms; and then the excitation energy is in the visible spectrum, which corresponds to thousands of kelvins (Sun's chromosphere: 6000K).

jet-fighter exhaust is still "too cold" to be highly ionized, and so still presents little opacity:

http://community.warplanes.com/wp-content/uploads/2010/06/Lockheed-F-35-Lightning-IIside.jpg

i think the same is sayable, for space shuttle thrusters, through which one can see the other in-side surface of the nozzle

what would be least unlike stars, would be ion thrusters... perhaps people have ascertained the opacity of ion-thruster-exhaust streams ?

looking online:

for water molecules:

http://en.wikipedia.org/wiki/Bond-dissociation_energy

k_BT_C ~= 0.05 eV

Hydrogen-bond energy ~= 0.05 eV

H:O bond energy ~= 5 eV

prima facie, the critical temperature, for water molecules, reflects the energies of the hydrogen-bonds that cause the surface tension...

the seeming "depolarization" of water molecules, hotter than the critical temperature, seems to reflect something similar to "excitation" of electrons in atoms, from (say) 1S orbitals to 2S orbitals...

when thermal energy supplies ~0.05 eV to the electrons, they have entered into an "excited" state w/in the water molecule, within which state their wave-functions expand -- somewhat similar to 1S ----> 2S excitations -- to more fully encompass all the nuclei, so that sufficiently hot water molecules are electro-statically non-polar

~100x more energy is required, to "ionize" electrons out of the water-molecule bonding orbitals, to break said bonds

such seems reasonable, is it also correct ?

i think this discussion pertains to surface tension:

surface tension [math]\left( \gamma \right)[/math] = force per length = energy per area

surface tension = inter-molecular forces from polarization of electron-cloud-charge distribution

order-of-magnitude, w/in water, the lattice length (L) ~= molecular bond lengths (B)

order-of-magnitude, charge excess on oxygen ~= charge deficit on hydrogens

order-of-magnitude, per molecule of water:

[math]F = \gamma L \approx \frac{\delta q^2}{4 \pi \epsilon_0 L^2}[/math]

[math]E = \gamma L^2 \approx \frac{\delta q^2}{4 \pi \epsilon_0 L}[/math]

from Poisson's equation:

[math]\nabla^2 \phi = \frac{\rho_q}{ \epsilon_0} \longrightarrow \frac{\phi}{L^2} \approx \frac{\delta q}{\epsilon_0 L^3}[/math] and [math]E \approx \delta q \phi[/math]

Now, [math]\gamma L^2 \approx k_B \left( T_C - T\right)[/math]

but, from above:

[math]\gamma \approx \frac{\delta q^2}{4 \pi \epsilon_0 L^3}[/math]

[math]\therefore \frac{k_B \left( T_C - T\right)}{L^2} \approx \frac{\delta q^2}{4 \pi \epsilon_0 L^3}[/math]

[math] k_B \Delta T \approx \frac{\delta q^2}{4 \pi \epsilon_0 L}[/math]

when [math]T \ge T_C \implies \gamma = 0 \implies \delta q = 0[/math] since the molecules within the water remain at finite distance from each other...

ipso facto i conclude, that above the critical temperature, water molecules become completely de-polarized, having no more charge excess on the oxygens, and no more charge deficits on the hydrogens (?)

in PPs, Enthalpy seemed adamant, that electrons in macroscopically massive "mega-molecules" of metallic materials can "roam around" and "wander through" the ionic lattice, in semi-Maxwellian "Bloch wave" states...

i conclude, that at absolute zero temperature, electrons in multi-nucleus "molecules", remain fixed & frozen in (semi-)bonding orbitals, which are static & stationary... but at hotter temperatures, electrons acquire thermal motions, within-and-through their "molecules"...

i conclude, that this picture can account for surface tension... because at [math]T \ge T_C[/math], the electrons within water molecules have acquired sufficient thermal energy, to "overcome the electronegativity of the oxygen nuclei" (for want of worthier words)...

in space-satellite analogy, whereas within water when cold, the electrons are tightly bound in "low earth orbit" near to "earth = oxygen", at higher hotter temperatures, the electrons have "rocket thrusted out to higher orbits" and so spend more time "circling the moon = hydrogen"... then above the critical temperature, the electrons "escape the electro-negativity gravity-well of earth = oxygen", and zip between the "moon = hydrogen & earth = oxygen" with equal frequency...

so that the overall charge distribution of the electrons becomes depolarized... above T_C, water molecules are electrically non-polar... hence unable to exert (VdW or) surface-tension forces, from the same said hydrogen-bonding-polarity

is that true ?

------------------------------------

[math]\gamma \approx \frac{\delta q^2}{4 \pi \epsilon_0 L^3}[/math]

[math]\delta q \approx \sqrt{\gamma \epsilon_0 L^3}[/math]

[math]\frac{\delta q}{q_e}\approx 0.1[/math]

and

[math]\frac{\phi} \approx \frac{\delta q}{\epsilon_0 L} \approx 0.05 V[/math]

are those calculations correct ?

http://www.elmhurst.edu/~chm/vchembook/images/210dipole.gif

[math]\frac{ \delta q}{q_e} = 0.179[/math]

--------------------------------

hypothetically, could you create a kind of metallic armor plate, through which a current was driven by applied voltage...

such that the photons w/in any incident laser beam, would be continually encountering "new electrons", which electrons would be continually drifting thru the ionic lattice...

such that the incident energy would be continually dispersed, borne away by the electrical current ?

some material having high conductivity, high electron density, and w/ high voltage, could conceivably constantly put new electrons in the way of the photons, dispersing the incident energy of the beam

http://www.physicsforums.com/showthread.php?t=203289&page=3

fire = "semi-plasma", ionization % = PPB

if fire was 100% ionized, then the equations would imply perceivable opacity... but the attenuation distance [math]\equiv 1/\rho \kappa \propto 1/\rho^2[/math]

so PPB => km's to attenuate

i tried flashlight + match => no shadow from flame, no obscuration thru the flame from flashlight

is fire opaque, to EM radiation ?

can you project a laser-pointer, through the flame, of (say) some cigarette lighter, or a candle ?

http://www.astro.princeton.edu/~burrows/classes/514/opac.2.ps

As a thought experiment, if the cores of big bright blue super-massive stars are ~1 g/cm3, ~100MK, then the Kramers' Opacity implies a mean-free-path of ~10-100m...

so if you could somehow bury yourself inside the center of some super-massive star, then w/ some sort of super-dark star-glasses you could see through the star, like fog on a football field, from post to post

are those numbers accurate, for said supermassive stars ?

 

This is a different argument than you gave above; you claimed that the EM interaction was the source of the energy, and I don't see how you can make this assignment. Also, the reaction for p:p fusion does not produce electron/positron pairs.

"emit" [math]\neq[/math] "produce"

p + p --------> p:p --------> p: {e+/e-}: p

i.e. the intense EM interaction, between two protons partially bound into an "He-2" diproton, produces pairs of positrons/electrons, as happens between electrons and large positively charged nuclei, in atoms, according to QM for Dummies

then, the same intense EM interaction propels the positively charged positron "odd man out" out of the diproton-complex:

p :{e-}: p .................... + {e+}

then, the electron "torn" between two protons, has four up-quarks with which to "roll the quantum dice" and possibly interact Weakly...

if the Weak interaction occurs, then one of the up-quarks, in one of the protons, "vampirically sucks the charge from" the electron, which "drained of its blood-charge" becomes an electron neutrino, which, w/o further interactions, "drifts away listlessly"

p:n .................... + {v_e}

the energy of the deuterium p:n is less than the energy of the diproton p:p, because of the intense EM interaction...

otherwise neutrons are higher energy than protons...

so, somehow, seemingly, the EM interaction "energetically finances" the Weak interaction, and transmutation, of a proton (up-quark) into a neutron (down-quark).

Wikipedia provides a page:

http://en.wikipedia.org/wiki/Pair_production#Photon.E2.80.93nucleus_interaction

 

No, that's not correct.

how would you mathematically model the proton-proton interaction ?

unlike the Schrodinger solutions for hydrogen, the potential term would involve two pieces, one for EM, one for SF, yes ?

and, the two particles (protons) are of equal mass, so the CM frame would more resemble positrons, wherewithin the e+ / e- have the same mass

so, you have a deep attractive potential (SF, linear? square-well? ) plus an nearly-as-deep repulsive potential (EM, 1/r)...

evidently, no truly stable states for such exist... square-well + 1/r has a "spike" in the center, perhaps the wave-function would be a spherically-symmetric-but-hollow shell ?

where would you put the WF into the equation ?

in intense EM interactions, virtual photons can "split" into (virtual?) electron-positron pairs, yes ?

so the intense EM interaction would seem to be the "source", for positrons emitted in p:p ----> p:n, i.e. "diproton" to "deuteron"... where else would the positrons "appear" from ?

p + p ----> p:p

"diproton"

Strong Force attracts them, EM repels them

in such an intense EM interaction, could not virtual photons, mediating the intense interaction, pair produce e+/e- ? Doesn't pair production occur, when electrons interact intensely, w/ large highly positive nuclei ?

i understand, that the SF keeps them together, whilst the EM interaction supplies the energy, required for pair production

Yes, in the proton-proton cycle, strong interaction attracts two protons together; most often they separate again (which is in favour of an "equilibrium" and an equation of state), sometimes the "di-proton" becomes a deuteron by weak interaction, that is beta plus emission.

 

That's at least what Wiki tells. A di-proton has never been observed, as far as I know. Maybe one proton must undergo the beta decay at the same time as it fuses with the other proton. Beta before should be excluded since it would require several billion K.

 

In the formation of a neutron star in contrast, (negative) electrons are absorbed to make neutrons, while in the proton-proton cycle according to Wiki, a positron is emitted; the mass balance differs a lot. There, the strong force can't be the main cause, since atoms heavier than lead are radioactive even with the best possible proportion of neutrons; gravity provides the energy.

No, the strong force alone doesn't produce nuclei past iron approximately. Gravity provides the energy, but to make a neutron star, it needs the weak interaction to produce neutrons. If I get it properly (take with care), light electrons take more room than heavy neutrons.

 

As nearly all nuclear reactions that provide net energy are exhausted, the star's (largely radiative) equilibrium is lost and the star collapses. The but-last reactions would still have fuel but they don't burn stably.

VdW is a dipole interaction term applied to atoms and molecules. Where are these dipoles?

that's what i was told...

but according to Wikipedia, the attractive potential is modeled as [math]-\epsilon \left( \frac{r}{R} \right)^6[/math]

[math]n \delta V \equiv \eta[/math]

[math]\frac{k_B T}{\epsilon} \equiv \tau[/math]

[math]P = \frac{\epsilon}{\delta V} \left( \frac{\eta\tau}{1-\eta}-\eta^2\right)[/math]

[math]\approx \frac{\epsilon}{\delta V} \left( \eta \tau \left( 1 + \eta + \eta^2 \right) - \eta^2 \right)[/math]

[math]\frac{\partial P}{\partial \eta} \approx \frac{\epsilon}{\delta V} \left( \tau \left( 1 + 2 \eta + 3 \eta^2 \right) - 2 \eta \right) \longrightarrow 0[/math]

[math]\left( 3 \tau \right) \eta^2 + \left( 2 \tau - 2 \right) \eta + \left( \tau \right) = 0[/math]

[math] \eta = \frac{ 2 \left( 1 - \tau \right) \pm \sqrt{4 \left( 1 - \tau \right)^2 - 12 \tau^2} }{6 \tau }[/math]

[math]= \frac{ \left( 1 - \tau \right) - \sqrt{ \left( 1 - \tau \right)^2 - 3 \tau^2} }{3 \tau }[/math]

[math] = \frac{ \left( 1 - \tau \right) - \sqrt{ 1 - 2\tau - 2 \tau^2} }{3 \tau }[/math]

from the Wikipedia page proffered per PP, we want the lower root, for the lower "condensation density", at which the pressure tops out, and begins to decrease, for increasing density (the squared term dominates)...

the higher root, represents the "contact density", at which the gas particles begin to basically be in complete contact, and pressure (in the mathematical model) goes to infinity, as the gas transitions to liquid / solid phase, and pressure ----> Bulk Modulus thereof (effective P ~ 100GPa characteristically) (the 1/(1-x) term dominates)...

from the above equation, and per the proffered page, there exists a temperature, above which there is no initial "condensation" density... perhaps that represents the "critical point", in PT diagrams, above which there is no distinction, between gas vs. liquid ??

if [math]\epsilon \approx 1-10 MeV \longleftrightarrow 10-100GK[/math], then [math]\tau \approx 0.001 - 0.1[/math] if i understand star core temperatures accurately...

so, treating [math]\tau \ll 1[/math] as a small parameter...

[math]\eta = \frac{\tau^2}{3 \tau} = \frac{\tau }{3} [/math]

so, since [math]\eta = 1[/math] is basically nuclear density, then at inter-particle spacings of ~10x nuclear density, the mathematical model makes the gas "condense", in the sense, that increasing density is no longer compensated by increasing pressure... squeeze on the plasma particles, and they begin to "deflate" and "condense" into nuclear density "neutronium" (for want of worthier words)...

so, including a VDW like term could help mathematically model star-core-collapse as a "condensation" process -- the plasma particles get too close, and "condense" out into a non-gas-like phase...

looking for the zero-crossing, and looking for the minima, both imply densities, w/ inter-particle spacings of ~10x the "contact" spacing size... that would be 10fm for star cores, so 'twould only logically be likely to apply, practicably, to core-collapse sorts of scenarios...

for normal terrestrial gases, all the above math still seemingly applies, so perhaps VDW explains the "condensation" of gases into liquids ?? When the particles are compressed close enough, they "deflate" and "condense" out of gas phase ?

Again the same old story... Widdekind, you need to understand that the weak interaction transforms protons and electrons to neutrons. The strong force doesn't.

 

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An attractive strong interaction could maybe have been treated to the plasma's equation of state as a corrective Van der Waal's force is in gas... but: why?

 

- Our Sun takes 10 billion years to fuse hydrogen to helium, and only through indirect processes. Not by shocks between protons. Now consider how many shocks per second for a proton in the Sun: what can be the significance of this correction?

 

- Fusion is generally not reversible. It creates new species like ³He or n. This is a strong contrast with molecular shocks. One couldn't use such a fusion as a corrective pressure for ²H.

 

- More generally, an evolutive interaction has not its place in an equilibrium equation. The star is not at nuclear equilibrium, which would be a chunk of iron. Fusion is essentially a dissipation mechanism; helium won't give hydrogen back if you lower the pressure, even at Sun's temperature. Only non-nuclear interaction define a plasma equation of state.

when two protons collide, and fuse, the STRONG interaction is what attracts them together ??

that attractive interaction is what "supplies" the energy, for the Weak interaction, to generate required leptons ?

if you included angle terms, appropriate to a dipole field, into the derivation on Wiki, then they'd probably integrate out, and only make minor modifications, to the overall coefficient, ignored on Wiki, for sake of simplicity

[math]\phi_Q = \kappa \nabla T \propto \left( k_B T \right) \frac{n_e}{m_e} \times \tau \times \nabla \left( k_B T \right) [/math]

[math]\approx n_e \frac{k_B T}{m_e} \tau \nabla \left( k_B T\right) [/math]

[math]\approx n_e C_S^2 \tau \nabla \left( k_B T\right) [/math]

[math] \approx \left( n_e C_S \right) \left( C_S \tau \right) \nabla \left( k_B T \right) [/math]

the first term = # per area per time = electron flow flux...

the second term = distance per electron in between scatterings...

second term x third term = extra heat energy acquired by electron, one scattering distance up-thermal-gradient, transported down-thermal gradient...

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apparently, this picture explains the physical phenomena...

apparently, at absolute zero Kelvin, electrons in metallic "mega molecule" crystal lattices, would actually be completely static & stationary & still, w/o motion or movement, analogous to the static bonding orbitals of H2...

but, at non-zero temperature, electrons "warm up" and acquire a Maxwellian-like motion, thru the metallic "mega molecule", behaving somewhat similar, to a Maxwellian-like gas, of "free" electrons...

that the electrons move constrained to the contours of their crystal lattice, conceivably explains, why the exact numeric coefficients, are often somewhat different, from the Maxwellian "ideal"

http://en.wikipedia.org/wiki/Van_der_Waals_equation

the VDW modifications, to the ideal-gas-law equation-of-state, imply that pressure is reduced, by attractive particle-particle interactions...

in the fusing centers of stars, plasma particles are attracted to each other (at close range)...

so, according to the VDW equation-of-state, at some density, those attractive nuclear forces could conceivably "deflate" the plasma pressure...

could such actually occur, in the cores of stars ??

the VDW e-o-s:

[math]P = n \times \left( \frac{k_B T}{1 - n \delta V}-n \delta V \epsilon \right)[/math]

where [math]\delta V[/math] is the effective particle volume ( of order the cube of the radius of the interaction ), and [math]\epsilon[/math] is the depth of the potential energy well describing said interaction...

inexpertly, for fusing plasma particles, [math]\delta V \approx [/math] 1 fm³, and [math]\epsilon \approx[/math] 1-10 MeV...

for P=0 w.h.t.

[math] k_B T = n \delta V \epsilon \left( 1 - n \delta V \right) [/math]

[math]\epsilon \left( n \delta V \right)^2 - \epsilon \left( n \delta V \right) + k_B T = 0[/math]

[math]n \delta V = \frac{\epsilon \pm \sqrt{\epsilon^2 - 4 \epsilon k_B T} }{2 \epsilon}[/math]

[math] = \frac{1}{2} \times \left( 1 - \sqrt{1 - \frac{4 k_B T}{\epsilon}} \right) [/math]

where we take the negative root, so that T=0 => n dV = 0 for P=0...

so, as T ----> 1-10 MeV = 10-100 GK...

P = 0 if n dV = 1/2...

i.e. n = 1/2dV...

which is basically near-nuclear density...

so, is the VDW attraction term, what "deflates" and causes the collapse, of cores, of super-massive stars, whose central temperatures >1GK, and whose densities are enormous ??

that's precisely the picture i presently cannot comprehend...

in an H2 molecule, the electrons occupy bonding orbitals, "built" from the individual separate atomic orbitals...

the latter are stationary static orbitals, w/in which the electrons have zero overall average momentum / speed...

and so the latter are also stationary static "co-" orbitals...

nobody ever talks about electrons moving, or drifting, around, in H2, yes ?

so, by (egregious) extrapolation, if you "build" up "mega-molecular" mega-orbitals, for macroscopically massive metallic lattices, in essentially the same / similar way...

then why would "mega-molecular" mega-orbitals "suddenly" acquire overall average drifting velocities / momenta ?

the simplistic straightforward extrapolation, from H2 bonding orbitals, to bonding-like mega-orbitals in macroscopic metal lattices... would be stationary static orbitals, "stretching" around and throughout the lattice, but not possessing any overall average drifting speed / momentum.

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re: times...

oops - WebElements was giving me electrical resistivities... what i wanted was conductivities...

that's x10¹⁶...

so, yes, all the calculated drifting collision times are of order 10^-15 s

w/ Fermi speeds of order 1000s km/s, those drift-times => drift-distances of dozens of lattice lengths

(except for manganese, whose resistivity is very high, and whose lattice length is very long)

if you take the metallic Young's moduli...

and estimate therefrom the Hooke's constant (k), then the quantum oscillator frequency ([math]m \omega^2 = k[/math]) is of order THz, 1000x too slow for the electrical scattering time-scale...

BUT the electrical collision time-scale is closely comparable to, and statistically highly correlated with (R2 ~ 0.7), the "Heisenberg time", derived from the "Heisenberg energy" (derived from the valence electron density => "Heisenberg distance" => "Heisenberg momentum")...

Q: so, in thermo-electric conduction, the electrons, in the valence "conduction" band, are NOT scattering off of slowly sluggishly sloshing nuclei (THz)...

but are scattering, or "quantum decohering", over "Heisenberg" state-decoherence time-scales (1000 THz) ?

i personally predicted, from that presumption, that low-valence => low-valence/conduction-electron-density => low Heisenberg-energy => high Heisenberg-time-scale metals (i.e. column 1 alkali's) would have high therm-electric drift time scales...

which turned out to be true (according to data from Web-Elements + Wikipedia)

so, if the number of massive particles is conserved, e.g. D+T ----> n + He, then no other particles (photons, neutrinos) need be emitted ?

[math]\sigma = \frac{J}{E}[/math]

[math] = \frac{q_e^2 n_e v_e}{q_e E}[/math]

[math] = q_e^2 \times \frac{n_e v_e}{\frac{m_e v_e}{\tau}}[/math]

[math] = q_e^2 \times \frac{n_e}{m_e} \times \tau[/math]

[math]\tau = \sigma \times \frac{m_e}{n_e q_e^2} \approx 10^{-30} s[/math]

what would such a femto-femto-scopic time-scale suggest ?? In a femto-femto-second, electrons can barely budge by billionths of a fm = width of a lattice nucleus

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by (egregious) extrapolation, to the "mega molecule" of a metallic crystal lattice, from a humble hydrogen diatomic molecule...

the molecular orbitals, of H₂, are constructed from the individual separate atomic orbitals, of the two H's...

the "bonding" orbital = [math]\Psi_1 + \Psi_2[/math], and the "anti-bonding" orbital = [math]\Psi_1 - \Psi_2[/math], which are orthogonal...

by (egregious) extrapolation, to the "mega molecule" of a metallic crystal lattice, the "mega molecular orbitals" ought to be constructed, from the (enormous numbers of) individual atomic orbitals...

into "mega bonding" and "mega anti bonding" orbitals...

roughly resembling [math]\Psi_1 \pm \Psi_2 \pm \Psi_3 \pm[/math]...

which would presumably be partly "bonding", partly "anti bonding", and so represent orbitals which "wrapped around" and "enveloped":

• all the metallic ion nuclei (pure bonding), orbital ~= whole bulk block of metallic material
  
• 
  
• some of the metallic ion nuclei (semi-bonding), orbital ~= planes of nuclei (horizontal, vertical, diagonal), or ~= lines of nuclei (H,V,D)
  
• 
  
• none of the nuclei (pure anti-bonding), orbital ~= localized "clumps" of wave-function centered on each nucleus, w/ no wave-function in between them, bonding them together
  
• 
  

or some such. In the "anti-bonding" orbital in H₂, the electron wave-function occupies space near and around each nucleus, but not between them. By egregious extrapolation, a pure-anti-bonding orbital, in a metallic "mega molecule", would occupy space near & around each lattice nucleus, but not between them. Likewise, semi-bonding orbitals would occupy space near, around, and between some lattice nuclei (lines (1D) or planes (2D)), but not between those blocks of "orbitally connected" nuclei (i.e. not between the lines or planes of atoms). Only pure bonding orbitals would occupy space near, around, and between all lattice nuclei (3D).

Q: is that an accurate (if imprecise) extrapolation ?

Aluminum is an obvious metal to consider. Z=13, so the 1S electrons have Hydrogen-like wave-functions, with bonding energies ~Z² x 13.6 eV ~= 2000 eV = 2KeV. If Aluminum's trio of level-3 (n=3) valence electrons experienced the full Z=13 charge of the nucleus, then they would have energies ~= 2KeV / 3² ~= 200 eV. But, the Fermi energy of Aluminum is only 12 eV, implying that the "effective charge" "felt" by the valence electrons is only Z ~= 3, i.e. the 10 inner closed-shell electrons screen out 10 protons' worth of charge, from the nucleus -- the three valence electrons only "see" (approx.) three units of charge, from out of the "fog" of inner closed-shell electrons.

According to the Virial Theorem, |U| = 2 |K|, so that U+K = -|K| = Fermi energy, i.e. the valence "conduction band" electrons in Aluminum, whose Fermi energy is 12 eV, presumably possess about 12 eV worth of quantum KE. Yet, those same electrons have wave-functions which overlap & interfere, causing a "Heisenberg energy", according to [math]\Delta p \approx \frac{\hbar}{2 \Delta x} = \frac{\hbar}{2} \left( \frac{n_e}{2} \right)^{1/3}[/math], [math]\frac{\Delta p^2}{2m_e} \approx 0.1 eV[/math].

Q: that "Heisenberg energy" represents the energy-width, of the conduction band, yes ? (for each electron,

[math]\frac{p^2}{2m} = \frac{\left( p + \Delta p \right)^2}{2m} \approx \frac{ \left( p^2 + \Delta p^2 \right) }{2m} [/math]

by ignoring the cross term, which is statistically on average zero, since the uncertainty in electron momentum is statistically uncorrelated w/ average momentum)

If that "Heisenberg energy" is of order 0.1eV, then the "Heisenberg time" is of order a fs, according to the Heisenberg energy-time uncertainty principle.

Q: So, the "conduction" band electrons, in Aluminum, exist in a given sub-orbital, w/in the band, for of order a fs, before they decohere, into some other sub-orbital, w/in the band, yes ?