Widdekind Posted July 22, 2013 Posted July 22, 2013 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 ??
swansont Posted July 22, 2013 Posted July 22, 2013 in the fusing centers of stars, plasma particles are attracted to each other (at close range)... VdW is a dipole interaction term applied to atoms and molecules. Where are these dipoles?
Enthalpy Posted July 22, 2013 Posted July 22, 2013 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. ----- 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.
Widdekind Posted July 22, 2013 Author Posted July 22, 2013 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. ----- 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
swansont Posted July 23, 2013 Posted July 23, 2013 Yes, you model it that way, if you have molecular dipoles. Where are they? i.e. where in a star do you have molecules?
Enthalpy Posted July 23, 2013 Posted July 23, 2013 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 ? 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. 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 ?? 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.
Widdekind Posted July 24, 2013 Author Posted July 24, 2013 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.
swansont Posted July 24, 2013 Posted July 24, 2013 i understand, that the SF keeps them together, whilst the EM interaction supplies the energy, required for pair production No, that's not correct.
Widdekind Posted July 25, 2013 Author Posted July 25, 2013 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 ?
swansont Posted July 25, 2013 Posted July 25, 2013 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 ? 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.
Widdekind Posted July 25, 2013 Author Posted July 25, 2013 (edited) 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 .................... + {ve} 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 Edited July 25, 2013 by Widdekind
swansont Posted July 26, 2013 Posted July 26, 2013 "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 .................... + {ve} Beta decay doesn't require an electron or positron be present. Are "vampirically sucks the charge from" and "blood-charge" quotes from the book? 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). And if you want to have energy released from a system, you need an attractive interaction, so the final system can have less energy. This occurs because of the loss of the EM repulsion. IOW, you have a sign error in your interpretation. Wikipedia provides a page: http://en.wikipedia.org/wiki/Pair_production#Photon.E2.80.93nucleus_interaction That's an interaction of a photon, not a proton.
Enthalpy Posted July 27, 2013 Posted July 27, 2013 (edited) How would you mathematically model the proton-proton interaction ? As far as I ignore (which is a lot) no justifiable model exists at all for the proton-proton force. Even the forces within a proton are badly modelled, and not by algebraic solutions. Usual models tell that the force between nucleons works only at contact: they vanish very quickly over distance, and get repulsive quickly if nearer. This explains why all nuclei have nearly the same density. It's a strong contrast with electrostatic repulsion. With such a model, you can forget any solution in terms of orbitals. "At contact" does not allow for a solution as a wave function. Only the force versus distance could, but this dependency is not known. You suggested "slightly attractive", Wiki puts "unstable" and the protons separate quickly in experiments that seem to have observed it http://en.wikipedia.org/wiki/Diproton#Helium-2_.28diproton.29 one might argue that the experiments produced an excited state. Maybe the diproton has a probability to stay stuck that decreases exponentially over time depending on how much energy it lacks (like any tunnel effect), and the improbable transformation into a deuteron depends on the transformation into a neutron within this short lapse. ----- Shall a positron-negaton pair be created to help the fusion of two protons? - This would need additional 1.22MeV if the pair must stay for some time, or until the positron is expelled - There are already negative electrons in a star. About as many as protons. That is, about as dense as in a solid on Earth, and in 7Be http://www.webelements.com/beryllium/isotopes.html the density of the 2s (valence) electrons, or the electron density in a solid, contributes measurably to the frequency of electron capture. - The standard theory seems to be: emission of a positron by one proton, nothing less direct. ----- Van der Waals' equation for dense stars: I suppose others have already answered this, like Chandrasekhar. The pressure of the electron gas must be stronger. http://en.wikipedia.org/wiki/Degenerate_gases#Electron_degeneracy Also, when a star is near to collapsing, its core is no more hydrogen, but heavier elements. Please take with care, I'm uneasy with these questions. Edited July 27, 2013 by Enthalpy
Widdekind Posted August 1, 2013 Author Posted August 1, 2013 (edited) .... 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 Edited August 1, 2013 by Widdekind
swansont Posted August 1, 2013 Posted August 1, 2013 dE for p ----> n = 1.29 MeV dE for e+ + e- = 1.22 MeV What are these numbers, and how did you calculate them? dE, I assume, is a difference in energy. Of what?
Widdekind Posted August 2, 2013 Author Posted August 2, 2013 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 + ve + e+ + 0.42 MeV = (p:p + 1.29 MeV) + ve + 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
swansont Posted August 2, 2013 Posted August 2, 2013 the mass-energy difference, between protons to neutrons, is ~1.29 MeV; and the (minimum) energy required for pair-production is ~1.22 MeV The mass energy difference of p and n is about 0.78 MeV, and the minimum energy for PP is 1.02 MeV. http://en.wikipedia.org/wiki/Neutron#Free_neutron_decay http://en.wikipedia.org/wiki/Pair_production#Energy
Enthalpy Posted August 2, 2013 Posted August 2, 2013 The distance at which strong interaction is positive is known - but not for proton and proton I guess. Have a look at Yukawa, and at the tunnel theory of alpha radioactivity. And because fusion involves tunnelling, this distance must be shorter than what the temperature provides; on the other hand, this scarce interaction can and does relies on kinetic energies that are several times bigger than the mean value: again a statistical tail. Pair production prior to emission of a positron? I don't feel it necessary. With gamma rays, pair production is rather efficient when the photon energy suffices, but it is very sensitive to the atomic number: efficient with lead, very poor with hydrogen and helium, whose near electric field is weaker. Electron capture: my suggestion is only that electron capture does happen (in the beryllium-7 example) with an electron density (from the 2s shell) comparable to the electron density of the plasma in a star core, so a temporary di-proton, or an almost di-proton, could use this process just as 7Be does. This could help fusion just as muons are known to do. Would this process be a significant competitor to the complete formation of a short-lived di-proton which then transforms quickly enough into a deuteron? No idea. The answer must be already known.
Widdekind Posted August 2, 2013 Author Posted August 2, 2013 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
swansont Posted August 3, 2013 Posted August 3, 2013 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... (?) 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)
Widdekind Posted August 5, 2013 Author Posted August 5, 2013 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)
swansont Posted August 5, 2013 Posted August 5, 2013 the mass... of a neutron... exceeds that... of a proton... by 1.29 MeV... yes ? Yes. It was not (and is not) clear from your post that you were doing the energy balance properly, because you stated this as an energy of a reaction. And this posting style is very annoying
Widdekind Posted August 8, 2013 Author Posted August 8, 2013 the mass excess, of 2p > p:n = D = H2, is ~0.9 MeV. But the EM force => p:p is unstable... logically, ergo, the EM positive potential energy > 0.9 MeV i.e. EM energy of interaction ~MeV or greater MeV EM interaction could create electron-positron pairs
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