Enthalpy

Amazing! Mild steel containing very little carbon is known to resist corrosion better (but a month in brine?) - though, could your production process make low-carbon steel? One other direction would be an alloying element that precipitates carbon, but I've read it for small amounts of carbon. Silicon would have better reasons to be in your steel and is known to help against corrosion. Easy to test: from 1.8%Si up, the steel is impossible to forge.

Plastic pipes and pressure? Beware! Here a corrosion table for sulfur dioxide and metals: http://www.google.de/search?q=%22Sulfur+dioxide%22+Copper+compatibility&ie=utf-8&oe=utf-8&rls=org.mozilla:fr:official&client=firefox-a&gws_rd=cr&ei=NRO7UpzZOMebtAbd9IH4AQ "stainless steel" isn't precise enough.

If I interpret properly, these got dates around 1000BP using 14C: http://www.bcin.ca/Interface/openbcin.cgi?submit=submit&Chinkey=177355 http://www.jstor.org/discover/10.2307/4602441?uid=3737864&uid=2129&uid=2&uid=70&uid=4&sid=21103172325027 Here a list of methods: http://quizlet.com/28825730/antropology-dating-midterm-2-flash-cards/ it doesn't look professional but the method names can be search words.

Changed my mind about a single wavefunction describing an (or several) emitted photon. If using the electric field as a wavefunction for the photon, sometimes one function doesn't suffice. If using the ubiquitous Psi function, it does. More thoughts there http://www.scienceforums.net/topic/80780-spin-2-what-wavefunction/#entry783582 after the second ----------. I would use E2 = m02c4 + p2c2 only for one particle, or for a group with identical speed vectors. Not for the summed vector p of several particles. Then each photon can keep its zero mass. ----------------------------------- There is more in the original question. If we take a more intuitive system of several objects, say, a lead atom, or a satellite with a spinning momentum wheel, instead of two photons: The electrons it the atom, or the momentum wheel in the satellite, have kinetic energy. If we decide to accelerate the atom or satellite, this kinetic energy adds inertia to what the rest mass of the electrons or wheel would bring. It also creates more gravity than the rest mass alone. As seen globally, we can ignore (for not knowing the details, or not wanting to know them) the kinetic energy of the electrons or the momentum wheel, and just take an atom or satellite mass bigger than the constituents' rest mass. This bigger mass serves both to accelerate the global object and to compute the gravity it creates. An interesting case are the particles of unclear composition, as are still the proton and the neutron: no consensus about how much kinetic energy, rest mass of constituents (if this makes sense) and so on. But we know a rest mass for the global particle, perfectly useable. Just compute with its center of inertia and observed mass. When computing just the movements of the composite object's center of inertia, what is kinetic energy and interaction energy of its constituents becomes rest mass of the composite. This means also that we can't tell by that way if a particle is fundamental. Now, for your two photons: If someone decided to consider the system of two photons, with their vector sum p=0, then he would attribute a rest mass to the system. But not to each photon: the rest mass of the system results from the kinetic energy p*c of the constituents instead. This would be useful if one can act on a system of two photons: "accelerate" both, measure the gravity field they create... Nothing easy, and computing with separate photons is more natural since the system is free, not bound. In that sense, E2 = m02c4 + p2c2 still applies to several objects with different speed vectors, but then you must take a system rest mass m0 which is not the sum of the constituents' rest mass. m0 must include the kinetic energies of the constituents and their interaction energy.

Yes, speed=c removes the possibility that the electric field is parallel to the propagation. Massive particles with spin=1 could be polarized horizontally, vertically or along speed. ---------- We use to write the wavefunction of photons as Electricfield (position, time) because it preexisted QM, is more explicit, and is more convenient to deduce the properties. Though, QM has a uniform formulation, applicable to photons as well: Psi (polarization, position, time) and then Psi is just the complex function used for other particles. Instead of telling "the electric field is that vector" it answers "your guess is probable". The guess to detect the photon there, then, with that polarization - with the chances to detect a photon related with the electric field intensity. ---------- The formulation as Psi (polarization etc) is often less convenient but it fits entangled photons more naturally. Remember the two spin-entangled photons which can be linearly or circularly polarized: If one is detected horizontal, the other as well (with probabilities). If vertical, the other as well. From the same source, if one is detected right, the other as well. If left, the other as well. If writing each photon wavefunction as Electricfield (position, time) one has to choose between Linear electric fields. A phased sum synthesizes a circular field where needed, but this sum must be decided upon detection. Circular fields. The phased sum if detecting linearly must be decided upon detection. I'm not so pleased with a weighted sum of wavefunctions whose coefficients are random. That is, a unique Electricfield (position, time) wavefunction does not describe the particle and can be written only upon detection. As opposed, the wavefunction for both photons A and B Psi (polarization A, position A, polarization B, position B, time) can be fully written before knowing the random outcome. ---------- So, the wavefunction for a particle with spin=2 must just be Psi (PolarizationMeaningfulWithSpinTwo, position, time) and my initial quest would rather be re-written like: "can we use some mathematical object, as we use a vector for the electric field of a single photon, to represent a single particle with spin=2, more conveniently than the wavefunction does?"

"With equations", asked the first message... The one I know is quite disappointing. It consists in writing one single wavefunction for both particles A and B: psi (position A, position B, state A, state B, time) where state includes the angular momentum. Psi can then give a probability density, serve to compute time evolution and eigenstates... The difference with PsiA (position A, state A, time) * PsiB (position B, state B, time) is the entanglement. That is, the chances to observe the particle A in state A or position A depend on particle B being in state B or at position B - plus, if particles A and B are of the same kind the chances to observe BA instead of AB, computed properly depnding on bosons or fermions. This single wavefunction can formulate nearly anything, hence I wrote "disappointing": as is, it gives very little hints about what is independent or linked in the particles' behaviour. It's more a general frame where on can model, by writing details of Psi, how he believes the particles are linked. What is happening: the link between the particle results from the process that emitted them. For instance the intermediate state of an electron that defines the polarisations and directions of two photons. Though, what makes entanglement special is that particles can "decide" their state late, upon detection for instance. Ancient comprehension would have wanted the electron's intermediate state, or whatever decides the states of the emitted particles, to be certain - this is disproven. One interpretation of QM wanted the states be already decided upon particle emission, but not observed by us - and this as well is disproven by EPR-type experiments. That is, the entangled particles are in a superposition of states until their state is measured. Disproven "by experiments" means also "nobody knows why it is happening", yes. No justification by theory, but "superposition until measured" is a formulation that fits experiments.

Diesel engines avoid the ignition hardware but need high-pressure injection pumps (plus separate valves now) that are expensive. The higher chamber pressure also demands a stronger construction. The net result is an engine a bit more expensive at identical power. Power per volume unit is now the same, per mass unit nearly the same. And because Diesel uses its fuel better, thus saves refueling stops, Diesel cars win endurance races presently. Diesel engines are but absent from individual cars in teh US, but make half of this market in Germany as a gut feeling, and are dominant in France. This results much of which technology national car makers have developed and how governments favoured them. The vehicle fleet must also fit the available fuels. Oil distillers can adapt their production much, but within limilts. The trend is to exploit heavier oil (Alberta, Orinoco...) that gives heavy fuels more easily, favouring Diesel in the future.

It's something all piano tuners claim. They allege that they have to stretch out the intervals at the higher notes of a piano so they sound in tune to human ear. I have big doubts about that, but did not experiment it in detail. Anyway, the highest octave of a piano makes such short sounds that one doesn't perceive properly the pitch. These notes are just too high for strings, because efficient radiation dampens them too quickly; a glockenspiel or a celesta play them properly.

Electrons continue to accelerate all the way to the screen in a cathode ray tube: the electron gun uses just a few 100V over 20kV. The electrons must emit a bit on the way, but this is a limited voltage over a long time. Significant emission is rather with 100keV+ energy and deviation over 1pm, and even better if the electron moves toward the observer at nearly c. Also, as the electron throughput is random, their emission is uncorrelated and adds poorly. As opposed, a free electron laser has its electrons organized in bunches that combine well with the successive magnets.

Screens with cold spikes would already be there, if a getter did suffice... Pumps achieve a much harder vacuum, but not sufficient for long operation of cold spikes. They get destroyed by a single molecule stuck at the tip. Researchers make the best possible vacuum, bake the spike for long in the vacuum, carry their experiment in a quarter hour and publish. Little to do with a screen whose million pixels must work for 10 years.

Things like cyclohexadiene (and polycycles) are a techological hydrogen carrier, as the compound turns easily into benzene. No idea whether it's any useable, what kind of catalyst it takes, nor if sensible chemists would consider it instead of hydrogen... Lithium (?) helps move double bonds within a molecule and make metathesis. Would it move hydrogen from cyclohexadiene to propene? Please take with much skepticism, until chemists give an opinion.

Nice of you to take care of presbyopic readers, but I wear my glasses, thanks. You answered it: quick electrons radiate in the electrostatic field of a nucleus.

Electric energy in vacuum: yes. When a distant star radiates light, it loses energy immediately, while the telescope gets (a part of) the energy years later. Meanwhile, the energy was in vacuum, needing no matter for it. Atom filled with dielectric: some situations are modelled that way. For instance a phosphorus atom in a silicon crystal. Phosphorus brings one electron more, one proton more, and only four electrons participate in the crystal bond. If cold enough like 4K, the extra electron stays attracted by the extra proton, but at a "big" distance due to the "permittivity" of silicon. "Big" means here a few atomic radii, so using the macroscopic permittivity is a bit cheeky, but the computed energy to separate the electron isn't bad. An air capacitor has some 1+10-4 times the capacitance of a vacuum one, so the air brings only worries here. Vacuum capacitors do exist commercially. Because discharges do exist in vacuum as well - the processes aren't well understood - the energy density isn't significantly better than ceramic ones, but losses are smaller. Such capacitors are used at the final stages of RF transmitters (SW, maybe MW and LW) of highest power and antenna matching, to ease cooling.

The beat persists at integer ratio relationship when the ratio isn't very simple, say 9/8. Though, it does get unpleasant when the equal-tempered interval differs from an integer ratio, like the small third: (1.0595)3 = 1.1892 <> 6/5 = 1.2000 Violonists train to follow the equal-tempered scale in order to play with other instruments, yes. But not only. The equal-tempered scale is also necessary for a lone violonist. If he plays successively several intervals according to simple ratios, he quickly gets badly out of tune. Imagine four small thirds played as simple ratios: (6/5)4 = 2.0736 <> 2.000 that's 4% away from the expected octave, more than a quarter tone. While no violonist would be off after four notes, some pieces are feared for long series of such intervals without a reference like an empty string, and surprises do happen - before training that piece.

Hello you all! For an electromagnetic wave (spin=1) we write the wavefunction as a vector (or little more), depending on position and time. For an electron (spin=1/2), as a scalar (though the angular momentum is treated separately then). What mathematical object shall represent a particle with spin=2? Thank you!

Musical synthesizers presently sample a real instrument and reproduce the record when the keyboard demands a note. This works well for a piano, less so for a violin, whose expression capabilities exceed a keyboard, and because two violin notes reproduced from samples sound like made from two instruments, due to the lack of beat. Sound synthesis could imitate a violin's beat better than samples do. If using a jittered signal, say the sawtooth plus some frequency filters, the synthesis could just displace a bit the transition of one sound waveform according to the time interval to the transition of the other sound. Marc Schaefer, aka Enthalpy I like this idea. It will distinguish the two first couplings I propose from the third. Two bows on one violin will be difficult for lack of room, but violonists are always available for such stupid challenges.

With the growing success of polypropylene, oil cracking doesn't suffice to provide propene, which is additionally produced from other short olefins - so economics can't be a good reason.

Exact, electron pairs don't condense in an atom, so pairs being bosons does not suffice in an atom, and may not suffice in a superconductor. Pairing in an atom just permits two electrons, the maximum number, to occupy the most favourable available orbital, in response to the nucleus' confinement potential. The evidence in favour of pairing in BCS, notably the heat capacity curve versus the temperature, speaks against a Bose-Einstein condensate. I still don't know if the BEC, which I doubt takes place, was a part of BCS, which I have nothing against. I know that science relies on models working better or worse - and this is your third option in addition to believing -, not on the reputation of people (or rather: it shouldn't...), and it is my opinion that a science forum can as well.

Hi dear music lovers! The violin and all bowed instruments has an important non-linear behaviour. It is noticed when playing two notes simultaneously on one instrument: the nonlinearity produces an additional beat, absent if two instruments produce the notes separately. When the frequencies are far from a ratio of integers, for instance in an equal-tempered small third interval, the beat is unpleasantly fast and seems out of tune, so violonists must train to play equal-tempered. The equal-tempered octave, fourth and fifth in contrast are nearly ratios of integers, the beat can be unnoticeable for being slow; violonists use it to tune the instrument. No source of non-linearity is expected in a violin, because all vibration strains and stresses are small. I claim (others may have done it) that the bow-string contact is the source. In favour of this: we don't hear the beat if playing pizzicato, that is, by plucking the strings. Also: the beat is still audible if playing piano, while a non-linearity in the violin would decrease the relative strength of the beat when the notes weaken. The bow-string contact operates in a well-understood fashion. Rosin makes the bow sticky; the moving bow pulls the string to the side and gives it energy until the string's stiffness wins, then the sticky contact breaks, the string relaxes back, until one-half to one period later, the speeds again similar permit the string to stick again to the bow. By the way, this stick-slip process is somewhat random as sticking uses to be. As a consequence, sticking ceases within a period at a moment that fluctuates, and the sound isn't really periodic. It is known meanwhile (but not enough) that for a listener to recognize a violin, the sound must be non-periodic. Attempts to synthesize it from a harmonic spectrum, hence producing a periodic sound, consistently failed over decades - but a primitive synthesis with a simple sawtooth signal shape, whose transition had intentional jitter, was immediately "violin-like". This experiment (in a French university I believe) was among the first proofs that musical sounds are, and must be, non-periodic. Relying on a threshold, the stick-slip is as nonlinear as possible. Several processes can couple the strings so that the instant phase at one string influences the moment when the other string tears off the bow: - The strings' movement is knowingly elliptic, not just parallel to the bow's speed. As one string moves perpendicularly to the bow, it changes the bow's pressure on the other, hence the maximum sticking force, and influences the tear-off instant. - The horsehairs at the bow have some elasticity. When a first string tears off the bow, the horsehairs reduce their strain, pulling suddenly stronger the other string, which tears off if it was about to do so. - The violin's bridge is compliant. Less so than a string, to make there a displacement node, but enough to move the sounding boards. The movement of one string displaces the other a bit, which again influences the instant when the other string tears off. These processes keep a beat amplitude essentially constant relative to both notes, which a nonlinearity downstream the oscillator can't. Marc Schaefer, aka Enthalpy

I do argue that a single wave (including a weighed sum of base waves) does not suffice to describe some emitted photons. With two polarization-entangled photons, if you write the wave as linear polarized, you can't explain to correlation between circularly polarized detectors. If you write the wave as circularly polarized, you can't explain the correlation between linear detectors. You have to say "it can be this wave, it can be an other". Then, accepting that a single wave does not suffice, you can write a spherical sum of waves that do have a polarization - just the usual EM fields for photons.

Swansont, I appreciate you. On most occurences, I agree with you, even if I don't feel the need to tell it everytime. Sometimes I disagree, and that's normal life: no reason to loose one's good mood. Electrons making pairs doesn't suffice to let them condensate to some ground state. Atoms prove it. Electrons are paired in atoms, making bosons, but these pairs do not condensate to the 1s orbital. On the other hand, supercooled atoms form Bose-Einstein condensates. The difference must reside in the density of the condensation candidate. Realized BEC use gas, where the atoms have much space between them. At enough distance, the bosonic atom behaves like one single boson, because the effects of its constituent fermions decrease quickly. So bosonic atoms can condensate, provided they have enough empty space between them - and provided one doesn't check for instance if two atoms are close enough to overlap their shells, which is a departure from the simple boson condensate picture. As opposed, electrons in an atom are about as wide as the atom, and their mean distance as well, so each pair cannot react with an other pair as global bosons would. Pairs act as two fermions, and don't condensate to the same state which is as small as the pairs themselves. Now, if I consider a metal or superconductor: The electrons near the Fermi level are very delocalized, much broader than their mean distance (I don't consider the electrons clearly deeper than the Fermi level, since for them, "localized" and "distance" has no interesting meaning, because we can't attribute a weighed sum of states to one particular electron; the best to say is "all states are occupied"). Their interaction distance as well exceeds their mean geometric distance. Because these electrons near the Fermi level - and pairs if applicable - are so much bigger than the mean distance, they must be in the atom-like situation, where no condensation can take place. The heat capacity at the superconducting transition is a strong argument against a Bose-Einstein condensate. This heat capacity confirms the law of a single energy gap. This is consistent with electrons pairing and nothing more, with approximately the same energy for each pair. It is not consistent with pairs condensing further once formed, which would leave more heat capacity than observed at temperatures below the superconducting transition. The Bose-Einstein condensate in superconductivity was part of the London theory, superseded by BCS. The BEC can well have been added later in the presentations of BCS.

What experiments say is that - There is a condensation energy - The quantum flux shows TWO electrons together, not 1023 They don't say that ALL electrons are in the same state. Maybe they don't even tell that the paired electrons are of opposite spin, nor even in the same state except for spin: some reports tell that the paired electrons have opposite momentums, which then doesn't need paired spins. But ONE ground state does mean no movement. Movement and acceleration need a change of state. And energy does not mean quantity of movement. Cool down! BCS is known not to work for higher temperature superconductors. And what I see everywhere, be it on Wiki; Hyperphysics, or the ideas here gained there, is not BCS, with a very high probability. So defending things like "bosons all in the same state" is not necessarily defending BCS.

I've just read about 14C giving a date around 1000 BP, so it might be useable after all. The article did not tell the accuracy, though.

You can't put several electron pairs into a single state, because electrons are fermions. However they're "paired". Anyway, electrons, or pairs, or many pairs, in the ground state, would not move, hence no current. Maybe I should read the BCS theory some day. It must differ seriously from the explanations given usually, includung at Hyperphysics and Wiki. I know this kind of interpretations, and believe the interpretations are wrong.

Imagine, instead of one photon exceeding 1022keV, two photons slightly over 511keV each, that are to make an electron-positron pair. Wiki describes it as "first photon fluctuates into a virtual pair, second photon interacts with the pair". The virtual pair would have -250keV per electron. - This gives a life expectancy of 10-21s, during which light propagates by 0.4pm, so the second photon has little time to be there. - The electrons can "move" from their virtual appearance point, or rather delocalize as an evanescent wave does, by 0.4pm. Over such a distance, the second photon must bring the missing 250keV per electron, so the second photon must concentrate itself a lot, making the event improbable. (Both 0.4pm coincide by chance, just because the virtual electrons miss half their mass of energy) Imagine a second 500keV photon in a collider, spread over (1mm)2: it must concentrate to (0.4pm)2 to provide the missing energy, which is about 10-19 probable.