Duda Jarek Posted February 20, 2021 Posted February 20, 2021 (edited) Quote Can infinities exist in nature? At least some of physicists believe in energy conservation, requiring source e.g. for hypothetical infinite energy. For example there is pair creation: of electron + positron from 2x511 keV only, so energy of electric field of such electron shouldn't exceed 511keV. But electric field of point charge is E~1/r^2, energy density is ~|E|^2, what integrates to infinite energy. So is there infinite energy in electric field of electron? QFT "repairs" this problem by just subtracting this infinity in renormalization procedure - successfully swept under the rug. But maybe it contains some regularization - deformation of electric field to indeed integrate to at most 511keV energy? Edited February 20, 2021 by Duda Jarek
swansont Posted February 20, 2021 Posted February 20, 2021 This assumes the energy of the electric field can somehow be separated from the mass energy of the electron.
Duda Jarek Posted February 20, 2021 Author Posted February 20, 2021 Energy conservation says that energy before and after should be the same. For 2gamma -> electron + positron pair creation: - energy before is 2x511keV in photons, which are EM waves - so this energy is initially in EM field, - energy after is, among others, energy of electric field of these two charges: integrating over energy density of E~1/r^2, we get infinite energy after this way. There is a problem, and just subtracting infinity in renormalization is only a mathematical trick without really understanding. We need to regularize this singularity: deform electric field to prevent E->infinity in the center. Running coupling ( https://en.wikipedia.org/wiki/Coupling_constant#Running_coupling ) can be seen as a consequence of such deformation: alpha increases e.g. to 1/127 in 90GeV - perfect point charges wouldn't need such deformation. Some discussion: https://physics.stackexchange.com/questions/386760/the-problem-of-infinite-energy-of-electron-as-point-charge 2 hours ago, swansont said: This assumes the energy of the electric field can somehow be separated from the mass energy of the electron. Indeed the question of total energy of electric field of electron is extremely interesting - we only know the upper bound: 511keV. There are also other fields like magnetic of magnetic dipole: integrating energy density of all the fields, should we get 511keV or smaller? In other words: are particles more then just their fields? What more? Nice quote from 1961 Infeld "Evolution of Physics: The Growth of Ideas from Early Concepts to Relativity and Quanta.":
swansont Posted February 20, 2021 Posted February 20, 2021 27 minutes ago, Duda Jarek said: Energy conservation says that energy before and after should be the same. For 2gamma -> electron + positron pair creation: One gamma, typically. With a massive particle nearby so that momentum will be conserved. Quote - energy before is 2x511keV in photons, which are EM waves - so this energy is initially in EM field, You will never actually get pair production at this energy. It’s the theoretical asymptotic limit (i.e. you know you can’t get PP below that threshold) Quote - energy after is, among others, energy of electric field of these two charges: integrating over energy density of E~1/r^2, we get infinite energy after this way. You only get infinity if you naively apply classical equations, which is one reason we know classical physics fails at small distances.
Duda Jarek Posted February 20, 2021 Author Posted February 20, 2021 2 minutes ago, swansont said: One gamma, typically. With a massive particle nearby so that momentum will be conserved. Ok, but in annihilation there are usually two (for para-positronium, +1 for ortho to reach zero angular momentum) ... e.g. used in PET imaging. 6 minutes ago, swansont said: You will never actually get pair production at this energy. It’s the theoretical asymptotic limit Sure, in practice we need a bit larger energy to get pair in a distance ... ... but, this final energy also includes Coulomb potential V(r) ~ -1/r, again infinity in the beginning r=0 ... requiring some regularization. 11 minutes ago, swansont said: You only get infinity if you naively apply classical equations, which is one reason we know classical physics fails at small distances. From one side, "subtracting infinity by hand" in renormalization is only mathematical trick - there still remains the question "what is mean energy density of electric field in distance r from electron?" Asymptotically ~|E|^2 = 1/r^4, but it leads to infinity - requires some deformation in the center. From the other, there are classical regularization methods. For example use Higgs potential V(u) = (|u|^2 -1)^2. It has |u|=1 vacuum (energy minimum), but e.g. in the center of topological singularity like hedgehog u(x) = x/|x|, to prevent discontinuity (infinite energy) it allows to get to v=0 by activating potential:
MigL Posted February 20, 2021 Posted February 20, 2021 The problem, as I see it, is that you cannot measure the field of a single charged particle. You need to 'probe' it with a second charged particle. This then gives rise to more and more Feinman paths ( virtual ) as they approach, which when summed, lead to infinities. ( and must be removed by renormalization ) This is only a problem with the model, and how we handle it. I don't expect fields to have any infinities anywhere; just a density distribution of energy, which, if above a threshold makes that virtual interaction, a real particle.
Duda Jarek Posted February 20, 2021 Author Posted February 20, 2021 (edited) Regarding probing with other charged particle, there appears Coulomb potential V(r) ~ -1/r ... again having this infinity problem. There is scattering - extrapolating to energy of resting particles to remove Lorentz contraction contribution, we get ~100mb corresponding to size of electron in femtometers - we have discussed it here: But cannot we directly ask about field? Its mean energy density in r distance? Edited February 20, 2021 by Duda Jarek
studiot Posted February 20, 2021 Posted February 20, 2021 (edited) 33 minutes ago, MigL said: The problem, as I see it, is that you cannot measure the field of a single charged particle. You need to 'probe' it with a second charged particle. This then gives rise to more and more Feinman paths ( virtual ) as they approach, which when summed, lead to infinities. ( and must be removed by renormalization ) This is only a problem with the model, and how we handle it. I don't expect fields to have any infinities anywhere; just a density distribution of energy, which, if above a threshold makes that virtual interaction, a real particle. 20 minutes ago, Duda Jarek said: Regarding probing with other charged particle, there appears Coulomb potential V(r) ~ -1/r ... again having this infinity problem. There is scattering - extrapolating to energy of resting particles to remove Lorentz contraction contribution, we get ~100mb corresponding to size of electron in femtometers - we have discussed it here: But cannot we directly ask about field? Its mean energy density in r distance? You asked specifically about electric field energy. Not any other form of energy. A single charged particle has no electric field energy, nada, zilch, nothing. Electric field energy is a potential energy of a system with more than one charged particle (which is due to the system configuration). So no other particle no energy. Edited February 20, 2021 by studiot
Duda Jarek Posted February 20, 2021 Author Posted February 20, 2021 (edited) https://en.wikipedia.org/wiki/Electric_field#Energy_in_the_electric_field Or this F_munu F^munu in QED https://en.wikipedia.org/wiki/Quantum_electrodynamics#Mathematical_formulation Edited February 20, 2021 by Duda Jarek
swansont Posted February 20, 2021 Posted February 20, 2021 1 hour ago, Duda Jarek said: Ok, but in annihilation there are usually two (for para-positronium, +1 for ortho to reach zero angular momentum) ... e.g. used in PET imaging. Sure, in practice we need a bit larger energy to get pair in a distance ... ... but, this final energy also includes Coulomb potential V(r) ~ -1/r, again infinity in the beginning r=0 ... requiring some regularization. At r=0? Has the Heisenberg uncertainty principle been revoked? (i.e. you continue to ignore QM and then for some reason be surprised that classical physics fails at small scales)
Duda Jarek Posted February 20, 2021 Author Posted February 20, 2021 (edited) Regarding Heisenberg principle, notice that I am using "mean": what is mean energy density of electric field in distance r from electron? If one really needs a probe particle, e.g.: what is mean energy of electron - positron pair in distance r? Edited February 20, 2021 by Duda Jarek
swansont Posted February 20, 2021 Posted February 20, 2021 1 minute ago, Duda Jarek said: Regarding Heisenberg principle, notice that I am using "mean": what is mean energy density of electric field in distance r from electron? If one really needs a probe particle, e.g.: what is mean energy of electron - positron pair in distance r? r=0 is inconsistent with this
Duda Jarek Posted February 20, 2021 Author Posted February 20, 2021 Ok, you can skip r=0 case ... but such energy density should be continuous also there.
studiot Posted February 20, 2021 Posted February 20, 2021 32 minutes ago, Duda Jarek said: https://en.wikipedia.org/wiki/Electric_field#Energy_in_the_electric_field Or this F_munu F^munu in QED https://en.wikipedia.org/wiki/Quantum_electrodynamics#Mathematical_formulation Thank you for your quotations. Do you understand them ? The first refers to Griffiths in the Wiki article you quoted. Quote Introduction to Electrodynamics (3rd Edition), D.J. Griffiths, Pearson Education, Dorling Kindersley, 2007, ISBN 81-7758-293-3 But it does not tell you where. If you read Griffiths properly (pages 60 - 61) you would see that this 'total energy' is the energy you would have to input or withdraw to move a second charge in the field of of a first charge. I agree that Griffiths, like many before him, call it a test charge. But at least two charges are necessary for the equation to have validity. Griffiths deals with the general case of many charges in the section quoted. 1
Duda Jarek Posted February 20, 2021 Author Posted February 20, 2021 (edited) Photons are EM waves, antenna produces EM waves, wireless charger ... which carry energy proportional to |E|^2 without need for charged particles on the go. Edited February 20, 2021 by Duda Jarek
studiot Posted February 20, 2021 Posted February 20, 2021 25 minutes ago, Duda Jarek said: Photons are EM waves, antenna produces EM waves, wireless charger ... which carry energy proportional to |E|^2 without need for charged particles on the go. So why is this thread entitled ? Quote Electric field energy of the electron EM radiation, and E fields are entirely different things. 44 minutes ago, Duda Jarek said: Photons are EM waves, antenna produces EM waves, wireless charger ... which carry energy proportional to |E|^2 without need for charged particles on the go. For explanation, (radio) photons are produced when a free electron is accelerated. The energy carried by these photons comes from whatever is accelerating the free electrons in the antenna metal. They are being accelerated because they are classically performing simple harmonic motion (SHM) in the metal. Do you understand the mechanics of SHM ?
swansont Posted February 20, 2021 Posted February 20, 2021 1 hour ago, studiot said: So why is this thread entitled ? Because that seemed a reasonable summary of the post when I split this off.
Duda Jarek Posted February 21, 2021 Author Posted February 21, 2021 studiot, I haven't met before with someone questioning energy of EM field alone - this is the energy in optical photon, power of radio transmitter, wireless charger, F_munu F^munu e.g. of QED, in derivation of Casimir effect, in 2.7K temperature of cosmic microwave background radiation, etc. But if you really deny energy of EM field alone, you can also use probe particle: Coulomb potential for electron + positron has V(r) ~ -1/r asymptotic behavior, what happens with it for r->0? Here is a diagram from classical Faber's model ( https://iopscience.iop.org/article/10.1088/1742-6596/361/1/012022/pdf ) : we model electric charge as topological charge of e.g. field of unitary vectors with Higgs potential V(u) = (|u|^2-1)^2 for regularization. Defining curvature of this field as electric field and using this standard F_munu F^munu Lagrangian (containing energy density), Gauss law becomes Gauss-Bonnet: counts winding number/topological charge - leading to Maxwell equations with built in charge quantization. So we take field configurations e.g. with two opposite charges in various distances, and calculate energy of this configuration: asymptotically he gets Coulomb, but it is deformed for very small distances (as in observed running coupling effect):
Markus Hanke Posted February 21, 2021 Posted February 21, 2021 14 hours ago, Duda Jarek said: Regarding probing with other charged particle, there appears Coulomb potential V(r) ~ -1/r ... again having this infinity problem. That’s because once you fix r to any exact value, the associated momentum of the particle in question becomes infinite, because this isn’t a classical system. So attempting to define the field energy in this way is meaningless, which is why it’s not done that way in QFT. You can’t calculate the vacuum energy of a quantum field simply by integrating over a volume, as you would in classical field theory. It’s very much more complicated than that, I’m afraid.
studiot Posted February 21, 2021 Posted February 21, 2021 (edited) 6 hours ago, Duda Jarek said: studiot, I haven't met before with someone questioning energy of EM field alone - this is the energy in optical photon, power of radio transmitter, wireless charger, F_munu F^munu e.g. of QED, in derivation of Casimir effect, in 2.7K temperature of cosmic microwave background radiation, etc. But if you really deny energy of EM field alone, you can also use probe particle: Coulomb potential for electron + positron has V(r) ~ -1/r asymptotic behavior, what happens with it for r->0? I have not 'denied that the EM field alone' , whatever that means. I have questioned you understanding of the energy of the electric field of an electron. On 2/20/2021 at 8:34 AM, Duda Jarek said: energy of electric field of such electron Why have you moved the goalposts ? I really think it goes right back to your understanding of the question what is energy ? And to what 12/13 year old schoolboys learn to regurgitate to gain a mark in an examination. "Energy is the capacity to do work" I certainly didn't appreciate what that means until I was much older or how deep and insightful that definition really is. I would also appreciate you addressing my other points, which are all designed to be helpful to you and your understanding, rather than posting fancy diagram by others that add nothing to the discussion. Edited February 21, 2021 by studiot
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