understanding thermal conductivity ?

[Widdekind]

for electrically conductive metals, the thermal conductivity ( Watts per area per temperature gradient ) seems to be two orders of magnitude more, than one would expect, from heat capacity, propagating at the sound speed:

 

[math] thermal conductivity \gg C_m \rho L^3 L L^{-2} \frac{C_S}{L}[/math]

 

where L = lattice length, typically several hundred picometers. So, i conclude, that heat flows far faster, than the nuclei are jostling back and forth, fro' and to.

 

Q: do conduction-band electrons play an important part in thermal conductivity (in electrically conductive metals) ? do nuclei, in the hotter zones, impart energy to conduction electrons, which then zip thru the lattice, until they collide with another nucleus, far far away, in the colder zones ? i.e. is thermal conductivity, the transfer of thermal heat energy, from hotter nuclei, to colder nuclei, via mediating conduction electrons:

 

hot nucleus ----> electron ----> cold nucleus

 

?

 

If so, then are conduction electrons propagating hundreds of times more quickly, than the sound speed, of most metals, i.e. hundreds of km / sec <-----> eV scale energies ? eV scale energies seems plausible, for conduction electrons, near the Fermi energy, of their "host" lattice material.

[EdEarl]

I do not know the answer to your question, but pure diamonds are electrical insulators and have high thermal conductivity.

[Enthalpy]

Metals conduct electricity by their electrons, under normal conditions.

 

So much that there is a fixed relationship between thermal and electric conductivity

http://en.wikipedia.org/wiki/Wiedemann%E2%80%93Franz_law

because electrons conduct and kT at the same time.

[Widdekind]

[math] \kappa \propto k_B T \sigma = k_B T \frac{J}{E}[/math]

 

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

 

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

 

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

 

relating the mean-free-collision-time [math]\left( \tau \right)[/math] to the mean-free-collision-distance [math]\left( \delta \right)[/math]. Now, the actual measured heat flow flux (Watts per square meter) is the product of the thermal conductivity, multiplied by the actual temperature gradient:

[math] \vec{\phi_Q} = \kappa \nabla T \propto q_e^2 \times k_B T \frac{n_e v_e}{m_e v_e^2} \times \delta \nabla \left( k_B T \right)[/math]

[math] = q_e^2 \frac{k_B T}{m_e v_e^2} n_e v_e \Delta E[/math]

where [math]\Delta E \equiv \delta \nabla \left( k_B T \right)[/math] is the extra energy (Joules) electrons acquire, from one mean-free-collision-distance up the temperature gradient, i.e. from the hotter region, which energy the electrons then transport, one mean-free-collision-distance down the temperature gradient, i.e. to the colder region. Then, [math]n_e v_e \Delta E[/math] is a heat flux (Watts per square meter), transported by the conduction electrons.

But, that electron velocity [math]\left( v_e \right)[/math] is the drift velocity, of electrons, through the lattice of nuclei, under the influence, of an applied electric field; the former is proportional to the latter; w/ no applied electric field, no drift velocity occurs. Yet, if [math]v_e \longrightarrow 0[/math] in the above equation, then the equation breaks down.

What weirdness has happened here ? i have made no mathematical mistakes, doing the derivation, yes ?

[studiot]

Read here about phonons.

 

http://www-sp.phy.cam.ac.uk/~wa14/camonly/statistical/Lecture13.pdf

[Widdekind]

[math] \kappa \propto k_B T \sigma = k_B T \frac{J}{E}[/math]

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

[math] = q_e^2 \times k_B T \frac{n_e v_e}{\left( \frac{m_e v_e}{\tau} \right)}[/math]

[math] = q_e^2 \times \left( k_B T \right) \frac{n_e}{m_e} \times \tau[/math]

so, according to the Wikipedia page provided per PP, the mean-free-collision-time is invariant, for a given metallic material... as more electric field is applied, the electrons are accelerated to higher drift velocities... but they collide, with lattice ions (?), at the same (average) rate, of (on average) once per [math]\tau[/math] seconds...

so, i guess, that when you consider converting the calculation, over to thermal energy conduction...

then you have to re-insert an electron velocity, which would be the "Fermi velocity", of conduction electrons, at the "top" of the "Fermi sea":

[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] = \left( k_B T \right) \frac{n_e v_F}{m_e v_F^2} \times \delta \times \nabla \left( k_B T \right) [/math]

[math] \approx \frac{k_B T}{E_F} \times n_e v_F \times \Delta E [/math]

which states, that the heat flux (Watts per square meter) is proportional to the flux of conduction electrons (# per area per time), multiplied by the extra thermal energy those electrons acquire from one mean-free-collision-distance up the thermal gradient, multiplied by a "Fermi fraction" [math]\left( \frac{k_B T}{E_F} \right) [/math], which represents the fraction of electrons, residing in the thermal "froth" on the "top" of the "Fermi sea", according to Fermi-Dirac statistics, which demand, that the width, in energy, of the transition region, from fully occupied states near zero momentum, to fully unoccupied states at infinite momentum, has an energy width proportional to the thermal energy. Low energy electrons are "locked in" to their states, by the Pauli exclusion principle, and the completely-filled nature of nearby momentum states, into which they would "want" to scatter...

only electrons with energy near the Fermi energy, in the "froth" on the "Fermi sea", have available states (of comparable energy) to scatter into...

only those electrons in the "Fermi froth" can scatter, off from higher energy ions in the hot zone, to lower energy ions in the cold zone...

only they can contribute to the (electronic) thermal conductivity...

so, of all available conduction electrons per unit volume [math]\left( n_e \right)[/math], only a fraction can contribute [math]\left( n_e \times \frac{k_B T}{E_F} \right)[/math]...

is that the correct interpretation, of the Wikipedia page provided per PP ?

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi.html

Edited July 13, 2013 by Widdekind

[Widdekind]

[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

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

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 ?

Edited July 17, 2013 by Widdekind

[Enthalpy]

10^-30s is no known time for electrical and thermal conductivity. But 1fs is encountered.

 

The wavefunctions accessible to electrons in crystals are computed by summing individual atomic wavefunctions...

- With a constant phase hop between adjacent atoms, corresponding to k in the "metallic" electron's propagation, not just with +- signs for each atom

- Textbooks do it with s functions and linear or cubic crystals. Serious work take the real atoms' wavefunctions (several ones, not just the lowest available) and the real atoms' positions, and they do obtain band diagrams that match measurements more or less.

- The true wavefunction is necessary to explai why diamon, silicon and germanium have very different band structures.

- Aluminium, as a metal, has no simple conduction band. It has many bands that overlap hence are partly filled.

 

Electrons, or available states, exist forever and can be occupied in theory for ever. Deep states are indeed occupied for very long. Relaxation times tell how long an electron near the Fermi level stays in a state before something (for instance a phonon) sends it to an other state.

people avoid this subject and introduce a time ex nothing.

 

There is more. It seems that electrons interact very often with an other, but because this interaction conserves the momentum sum and the current sum is proportional to the momentum sum as soon as the electron mass doesn't vary (which would happenin a solid far before relativistic speed), models neglect completely the interactions between electrons and care only about the crystal. One exception are "hot electrons" whose mass is uncommon.

 

With the band structure, you get only state densities and electron masses. To evaluate electric and thermal conductivities, you also need the mean free path or the time of free flight for electrons, which depends on temperature, alloying elements, metallurgical state. Reasonable

 

Some people claim to model resistivity against temperature... but superconductivity isn't well explained, so resistivity model can't be complete.

 

Why choose aluminium? It's a complicated case. Silver is easier.

[Widdekind]

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.

 

 

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

 

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)

[Widdekind]

[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...

 

 

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

 

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"

[Widdekind]

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

[Widdekind]

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 ?

[Enthalpy]

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).

[Widdekind]

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).

[Enthalpy]

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.

[Widdekind]

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.

[Widdekind]

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 ?

[studiot]

 

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

 

Classical thermodynamics requires a temperature difference for heat to flow.

 

 

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

 

but do possess "orbiting" revolving motions...

 

 

All curved motion can be constructed from two linear components in 2D or three in 3D.

Edited August 3, 2013 by studiot

[Widdekind]

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

[Enthalpy]

are you completely certain & sure, that the bonding orbitals in macroscopic metal "mega molecules" DO possess e^(iix) terms (at 0K) ?

 

Yes, I am.

2P₊ orbitals which look like "donuts" w/ phase circling about the z-axis in a right-handed sense...

 

The doughnut is only one possible shape of a 2p orbital. The peacock, with one side positive and the opposite negative, is an other. Chemists use to take the peacock shape to think at molecular orbitals.

 

A peacock is a sum of two doughnuts, a doughnut is a sum of two peacocks.

 

Just like exp(ikx) is a sum of cos(kx) and sin(kx) - from where the oriented impulse is absent, be happy - and cos(kx) is a sum of exp(ikx) and of exp(-ikx).

[Widdekind]

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 ?

[Enthalpy]

This state or a similar one must be possible, as are around 6*10²³ more. Since eddy currents exist in metals, some states resembling this one must exist. As does the one that rotates in the opposite direction. And because both have the same energy, both are occupied or not without external influence on the metal.

 

Though I doubt one gets the eddy current through a sum of such 2p orbitals. A phased sum of any orbital is at least as good.