graybear13 Posted July 4 Posted July 4 22 hours ago, MigL said: You are welcome to ask as many questions as you wish. Thank you MigL, Is it possible that dark matter, dark energy and zero point energy are all the same thing, just neutrinos arranged in different ways and moving in different ways?
Mordred Posted July 4 Posted July 4 Dark matter no plausible on DE and zero point energy though the zero point energy led to the vacuum catastrophe where it's calculated energy was 120 orders of magnitude too high. It's still not discounted however as it's still a viable possibility
graybear13 Posted July 4 Author Posted July 4 We know that every time atomic nuclei come together(like in the sun) or break apart(like in a fission reactor or particle accelerator), they produce neutrinos. So we know that as radioactive decay happens neutrinos are released. Doesn't this suggest that the entire atom is made of neutrinos arranged and held together in different ways? Can neutrinos be degraded by any heat we know of?
exchemist Posted July 4 Posted July 4 35 minutes ago, graybear13 said: We know that every time atomic nuclei come together(like in the sun) or break apart(like in a fission reactor or particle accelerator), they produce neutrinos. So we know that as radioactive decay happens neutrinos are released. Doesn't this suggest that the entire atom is made of neutrinos arranged and held together in different ways? Can neutrinos be degraded by any heat we know of? No.
MigL Posted July 4 Posted July 4 (edited) No, fundamental particles like neutrinos electrons, and quarks have no internal structure and are accurately defined as points with 'diffuse edges', for all intents and purposes. No structure means they cannot be subdivided into smaller parts. Released particles are not necessarily constituent particles of the incident composite particles that take part in a reaction. Quite often they are a result of the energy of the reaction and the need to balance it, and momentum, before and after said reaction. However these sort of questions are now off topic in this thread. Start another if you wish more details, or have further questions. All this stuff is covered in any introductory Physics textbook; or even Wiki. Edited July 4 by MigL 1
swansont Posted July 4 Posted July 4 1 hour ago, graybear13 said: We know that every time atomic nuclei come together(like in the sun) or break apart(like in a fission reactor or particle accelerator), they produce neutrinos. We don’t, in fact, “know” this. Neutrinos would only be present if the decay involved the weak interaction. 1 hour ago, graybear13 said: So we know that as radioactive decay happens neutrinos are released. Doesn't this suggest that the entire atom is made of neutrinos arranged and held together in different ways? Nope. This is fiction, based on flawed understanding.
joigus Posted July 4 Posted July 4 1 hour ago, graybear13 said: We know that every time atomic nuclei come together(like in the sun) or break apart(like in a fission reactor or particle accelerator), they produce neutrinos. So we know that as radioactive decay happens neutrinos are released. Doesn't this suggest that the entire atom is made of neutrinos arranged and held together in different ways? Can neutrinos be degraded by any heat we know of? Wrong. Beta decay must be involved, as said. And, 49 minutes ago, MigL said: Released particles are not necessarily constituent particles of the incident composite particles that take part in a reaction. There's a cute explanation in Surely you're joking Mr Feynman, about his father asking him, "when an electron falls from a higher to a lower orbit, where was the photon before it happened?"
Mordred Posted July 4 Posted July 4 It helps to understand that particles are not little bullets with corpuscular (matter like) constituents. It's best to think of them as field excitations. This is the QFT view but when you get into the Feymann path integrals you quickly learn this is a very accurate description. Though keep in mind those integrals also incorporate probability functions as well as the particle state.
MigL Posted July 4 Posted July 4 I think you lost Graybear, Mordred. I like to use the KISS principle until he follows up with another question.
Mordred Posted July 4 Posted July 4 Possibly but we will see the main thing is to throw away the billiard ball or bullet image of a particle.
Sensei Posted July 4 Posted July 4 2 hours ago, graybear13 said: We know that every time atomic nuclei come together(like in the sun) or break apart(like in a fission reactor or particle accelerator), they produce neutrinos. So we know that as radioactive decay happens neutrinos are released. p+ + p+ -> D+ + e+ + ve + 0.42 MeV (average half of it goes to positron and half with neutrino) i.e. 0.42 MeV = 420 keV from initial 938.272 MeV/c2 x 2 = 1876.544 MeV/c2 (~ 1.9 GeV/c2) 1876.544 MeV / 0.42 MeV = 4468 x as much... @Mordred Neutrinos with higher (kinetic) energies are emitted in rarer situations (compare the amount of hydrogen and helium with all other elements).
graybear13 Posted July 4 Author Posted July 4 1 hour ago, Mordred said: Though keep in mind those integrals also incorporate probability functions as well as the particle state. Are there clusters of neutrinos: maybe 100 neutrinos clustered together to create an electron?
joigus Posted July 4 Posted July 4 1 hour ago, graybear13 said: Are there clusters of neutrinos: maybe 100 neutrinos clustered together to create an electron? Neutrinos cannot cluster. They hardly interact at all.
swansont Posted July 4 Posted July 4 1 hour ago, graybear13 said: Are there clusters of neutrinos: maybe 100 neutrinos clustered together to create an electron? No. There are multiple problems with that scenario. Spin, Pauli exclusion, a mechanism to form a bound state, lack of evidence of an electron being a composite particle even if you could. Probably more. Attacking this from a position of ignorance about physics isn’t going to go anywhere useful.
Mordred Posted July 4 Posted July 4 No that would violate numerous conservation laws of particle physics. For example conservation of isospin, charge, lepton number, energy/mass momentum, color, flavor. There are others but the incoming particles and outgoing particles must obey those laws.
joigus Posted July 4 Posted July 4 Well, they can cluster gravitationally, therefore very weakly. At galactic scale. x-posted with mordred. Plus all the other reasons they're telling you...
swansont Posted July 4 Posted July 4 1 minute ago, joigus said: Well, they can cluster gravitationally, therefore very weakly. At galactic scale. Cold neutrinos, perhaps. But where are you going to find a space with only neutrinos in it?
joigus Posted July 4 Posted July 4 43 minutes ago, swansont said: Cold neutrinos, perhaps. But where are you going to find a space with only neutrinos in it? Agreed. I think the reasons having to do with PEP and conservation laws were very robust already. I sometimes succumb to the drive to try to help the idea as much as I can before it dies. I suppose it's because after a while I get too tired of always being quickly dismissive. I sometimes want to make a point that it's not that you have an extreme dislike for the idea. Gravitation is indeed the only interaction that could have neutrinos clustering at some level in some scenario. If you had a universe with no dark matter, no scatterers of any kind, nothing, huge distances before you could find the next galactic halo, I guess at some point neutrinos would start to cluster in huge mega-halos gravitationally... Anyway. As you said, it would fail on so many levels.
Mordred Posted July 4 Posted July 4 (edited) Well assuming DM is sterile neutrinos then due to the Right hand singlet state and applying Majoronna mass with the Higgs metastability then the sterile neutrinos are more massive than matter neutrinos. So this would be the only viable option for cold neutrinos though we have yet to detect an antineutrino (right hand/sterile). Though I know the OP won't understand the mathematics (few will ) the related math is here (Just an FYI It takes considerable familiarity with particle physics to understand below) \[m\overline{\Psi}\Psi=(m\overline{\Psi_l}\Psi_r+\overline{\Psi_r}\Psi)\] \[\mathcal{L}=(D_\mu\Phi^\dagger)(D_\mu\Phi)-V(\Phi^\dagger\Phi)\] 4 effective degrees of freedom doublet complex scalar field. with \[D_\mu\Phi=(\partial_\mu+igW_\mu-\frac{i}{2}\acute{g}B_\mu)\Phi\]\ \[V(\Phi^\dagger\Phi)=-\mu^2\Phi^\dagger\Phi+\frac{1}{2}\lambda(\Phi^\dagger\Phi)^2,\mu^2>0\] in Unitary gauge \[\mathcal{L}=\frac{\lambda}{4}v^4\] \[+\frac{1}{2}\partial_\mu H \partial^\mu H-\lambda v^2H^2+\frac{\lambda}{\sqrt{2}}vH^3+\frac{\lambda}{8}H^4\] \[+\frac{1}{4}(v+(\frac{1}{2}H)^2(W_mu^1W_\mu^2W_\mu^3B_\mu)\begin{pmatrix}g^2&0&0&0\\0&g^2&0&0\\0&0&g^2&g\acute{g}\\0&0&\acute{g}g&\acute{g}^2 \end{pmatrix}\begin{pmatrix}W^{1\mu}\\W^{2\mu}\\W^{3\mu}\\B^\mu\end{pmatrix}\] Right hand neutrino singlet needs charge conjugate for Majorana mass term (singlet requirement) \[\Psi^c=C\overline{\Psi}^T\] charge conjugate spinor \[C=i\gamma^2\gamma^0\] Chirality \[P_L\Psi_R^C=\Psi_R\] mass term requires \[\overline\Psi^C\Psi\] grants gauge invariance for singlets only. \[\mathcal{L}_{v.mass}=hv_{ij}\overline{I}_{Li}V_{Rj}\Phi+\frac{1}{2}M_{ij}\overline{V_{ri}}V_{rj}+h.c\] Higgs expectation value turns the Higgs coupling matrix into the Dirac mass matrix. Majorana mass matrix eugenvalues can be much higher than the Dirac mass. diagonal of \[\Psi^L,\Psi_R\] leads to three light modes v_i with mass matrix \[m_v=-MD^{-1}M_D^T\] MajorN mass in typical GUT \[M\propto10^{15},,GeV\] further details on Majorana mass matrix https://arxiv.org/pdf/1307.0988.pdf https://arxiv.org/pdf/hep-ph/9702253.pdf Edited July 4 by Mordred
MigL Posted July 5 Posted July 5 Oh, now you've really lost him with your sterile neutrino analysis, Mordred. Maybe now he'll ask if electrons are actually bound neutrinos ...
Mordred Posted July 5 Posted July 5 Lol should be clear that post was more for @Joigus. In terms of DM possible solutions it does have viability.
graybear13 Posted July 5 Author Posted July 5 23 hours ago, joigus said: There's a cute explanation in Surely you're joking Mr Feynman, about his father asking him, "when an electron falls from a higher to a lower orbit, where was the photon before it happened?" Could it be that when electrons jump to a different orbit a quanta of energy is given off? Perhaps a photon is created. Does gravity create mass? And does mass create gravity? It seems to me that, at first cause, gravity must create mass. Then at some point the mass creates it's own gravity. What does that transition look like?
swansont Posted July 5 Posted July 5 8 minutes ago, graybear13 said: Could it be that when electrons jump to a different orbit a quanta of energy is given off? Perhaps a photon is created. Yes, we’ve known this for quite some time.
Mordred Posted July 5 Posted July 5 39 minutes ago, graybear13 said: Does gravity create mass? And does mass create gravity? It seems to me that, at first cause, gravity must create mass. Then at some point the mass creates it's own gravity. What does that transition look like? Swansont answered the top part. I wouldn't think of it as strictly mass causes spacetime curvature (gravity). It's more accurate to use the stress energy momentum tensor. Massless particles can cause curvature as well hence the above statement.
MigL Posted July 5 Posted July 5 44 minutes ago, graybear13 said: Does gravity create mass? No. Gravity 'creates' weight, not mass. Mass is an intrinsic property of the system; equivalent to its 'internal' energy. This is not the only energy a system can possess; there are also 'external' sources, such as momentum and stresses/strains, which, all taken together, modify space-time to produce the geodesic paths we attribute to gravity. 1
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