dalemiller Posted October 15, 2009 Posted October 15, 2009 (edited) A previous posting here, “Exothermic migration of charged particles” has had time to draw any discouraging words but stands so far with no sustained contradictions. A mind experiment ventured to show that charged particles can join a Faraday cage without starting at an outside surface and so demonstrate that a traction exists to reposition them so as to be found according to the words of Michael Faraday about his ice pail. (Excess electrons arrived at the interior before release to the common premises of the Faraday cage and were saddled with a forceful migration process in order to fit his stipulation.) Such outward travel of electrons is demonstrated by the meteorologists’ Fair Weather Current that is underestimated as some two picoamperes per square meter of Earth’s surface. Hence Earth bears a negative charge. Notice that the ionosphere is deflected by the sun. Hence, Sol bears a negative charge. Hence Sol too should have a core of protons with hardly an electron. Hence the story about positrons annihilating against electrons where helium is formed would be an old wives’ tail. Hence we should know why Sol would maintain her level of negative charge and show an overflow of electrons. Edited October 15, 2009 by dalemiller Correcting scope of conclusion etc.
swansont Posted October 15, 2009 Posted October 15, 2009 What does fair-weather current have to do with a Faraday cage?
dalemiller Posted October 15, 2009 Author Posted October 15, 2009 The Fair Weather Current is a manifestation of electrons rising to the electrosphere/ionosphere. Such a phenomenon suggests that Earth functions as a Faraday cage bearing a negative charge. The rising electrons are taking their place amongst the majority charge particles. Meteorologists still wonder about what causes FWC but it is quite clear to an electronic technician.
swansont Posted October 15, 2009 Posted October 15, 2009 If the atmosphere is a Faraday cage, how do we communicate with satellites, and how do we do radio astronomy? The sky would look opaque.
dalemiller Posted October 17, 2009 Author Posted October 17, 2009 The entire earth is Faraday cage. It even has a positive core that easily escapes our notice. The electromagnetic shielding implied by existing nomenclature for a body with surrounding electrical discontinuity, “Faraday cage”, is mere semantic mischief. Shielding per se assures neither total blockage nor full effect across the spectrum. The ionosphere can reflect enough short wave radio wave energy to accommodate long-range communications but nevertheless passes lots of sunshine on into our world. Explaining lightning formation, it helps to identify raindrops as little Faraday cages simply to emphasize restriction of negative ions to their outer surfaces. The issues there and those concerning our planet are about concentric ionic arrays rather than any preoccupation with shielding effects.
swansont Posted October 17, 2009 Posted October 17, 2009 (edited) Such outward travel of electrons is demonstrated by the meteorologists’ Fair Weather Current that is underestimated as some two picoamperes per square meter of Earth’s surface. Hence Earth bears a negative charge. If the electron travel is outwards, why is the net charge negative? Are you talking about the surface, or the whole system? Merged post follows: Consecutive posts mergedThe entire earth is Faraday cage. It even has a positive core that easily escapes our notice. What positive core? The direction of the electric field of the atmosphere is inward. Edited October 17, 2009 by swansont Consecutive posts merged.
dalemiller Posted October 17, 2009 Author Posted October 17, 2009 (edited) In saying that the net charge of a body is negative, I mean to be saying that there are more electrons than protons contained by that body. That would be the whole system as you spoke of it. The surface of the body would have an electron to match every local proton where they engage within neutral atomic structure, an aditional electron for every ionic proton located within the core or “heading” that way, plus the electrons making up the charge which are the only electrons without proton counterparts somewhere in that body. Not to say that we can tell the difference between them on sight.. Hence the isolated array of protons caught in the center are complimented by an equal count of isolated electrons arrayed in the outer shell augmenting the count of electrons there making up the overall charge on the hosting body. The best way to check the charge on a Faraday cage from an inside view is to see what polarity of charged particles is trying to get out of there. That would be the majority charge. Too many electrons sends other electrons climbing the walls. A particle of the opposite charge however, would head in for the middle or propagate its charge in that direction. If we switch to two dimensions by considering a ring or a disk, we can say that a positive charge inside responds to a greater arc of attraction by heading for that and leaving a lesser pull behind it until it gets to the center. That is how electron guns in old TV sets work if we may switch example to that of a negative minority charge: The focusing anode is a ring or sleeve connected to positive voltage (dead giveaway – anode). Electrons called through that shape maintain equidistance from all points around the circle on the way through and far beyond. Speaking of this, ask about our solution to the polar jet mystery sometime. As to the “what positive core” question: propagation of positive charge to the center of ball or disk is natural for a negative majority charge. In the case of a spherical host, it would be tricky for any protons to escape without a proxy. (Thermal activity could promote just such an option.) As time goes on a substantial core would build up as long as the hosting body maintains its negative charge. Within a stellar hosting body however there would be another way out, but that is another story. Edited October 17, 2009 by dalemiller Punctuated
swansont Posted October 17, 2009 Posted October 17, 2009 The best way to check the charge on a Faraday cage from an inside view is to see what polarity of charged particles is trying to get out of there. That would be the majority charge. Too many electrons sends other electrons climbing the walls. A particle of the opposite charge however, would head in for the middle or propagate its charge in that direction. No. There is no way you can tell what the charge on the exterior of a Faraday cage is if you are on the interior. The charges on the exterior contribute nothing to the field inside.
dalemiller Posted October 19, 2009 Author Posted October 19, 2009 All of my information is first-hand. The word “field”, as a second-hand term, does not serve my brass tacks little mind. It is as though a cause-and-effect muse must deal with esoteric doctrine. The charges on the exterior are held there by mutual repulsion between like-charged particles all lying on what we call the exterior of the hosting body. The shortest path for such influence would seem to be, for almost any charged particle involved, right straight through the interior field and any other entity within, in order to exert appropriate push upon any other particles of the same electrical persuasion. It is difficult to begin an attempt to understand how a sole electon within a negatively charged Faraday cage could roam at random within such a chamber.
CaptainPanic Posted October 19, 2009 Posted October 19, 2009 A previous posting here, “Exothermic migration of charged particles” has had time to draw any discouraging words but stands so far with no sustained contradictions. A mind experiment ventured to show that charged particles can join a Faraday cage without starting at an outside surface and so demonstrate that a traction exists to reposition them so as to be found according to the words of Michael Faraday about his ice pail. (Excess electrons arrived at the interior before release to the common premises of the Faraday cage and were saddled with a forceful migration process in order to fit his stipulation.) Such outward travel of electrons is demonstrated by the meteorologists’ Fair Weather Current that is underestimated as some two picoamperes per square meter of Earth’s surface. Hence Earth bears a negative charge. Notice that the ionosphere is deflected by the sun. Hence, Sol bears a negative charge. Hence Sol too should have a core of protons with hardly an electron. Hence the story about positrons annihilating against electrons where helium is formed would be an old wives’ tail. Hence we should know why Sol would maintain her level of negative charge and show an overflow of electrons. You start with a false claim: because your previous thread did not draw any discouraging words does not mean it's all true. It's like my thermodynamics professor who would always end his lectures with the question: "Any questions"? (silence) He then followed with the claim: "Ok, you have no questions, so you all understood". That last claim couldn't have been further from the truth. We were all so confused that we couldn't formulate a proper question. This is a very common mistake. Meanwhile, I don't understand which claim is made here in this thead. But I took time to reply to the previous thread, which contains some mistakes.
swansont Posted October 19, 2009 Posted October 19, 2009 All of my information is first-hand. The word “field”, as a second-hand term, does not serve my brass tacks little mind. It is as though a cause-and-effect muse must deal with esoteric doctrine. The charges on the exterior are held there by mutual repulsion between like-charged particles all lying on what we call the exterior of the hosting body. The shortest path for such influence would seem to be, for almost any charged particle involved, right straight through the interior field and any other entity within, in order to exert appropriate push upon any other particles of the same electrical persuasion. It is difficult to begin an attempt to understand how a sole electon within a negatively charged Faraday cage could roam at random within such a chamber. Being difficult to understand is not necessarily correlated with it being true. The field inside is zero because the individual contributions to the field go in both directions and, being a vector, they cancel out.
dalemiller Posted October 19, 2009 Author Posted October 19, 2009 Then the fields do not matter. Since electrons introduced into an isolated bucket as part of charged raindrops become added to the charge of said bucket, and all of the charge becomes presented to the outer surface according to widely accepted scientific lore, then an exothermic migration must then have prevailed. Of course there is no free lunch. The endothermic event supporting such traction is the disruption of an electron from its electrical bond to a positive neucleus.
dalemiller Posted July 15, 2010 Author Posted July 15, 2010 No. There is no way you can tell what the charge on the exterior of a Faraday cage is if you are on the interior. The charges on the exterior contribute nothing to the field inside. Upon ever beginning to host a negative bias, an isolated body would undergo an outward migration of its surplus of electrons. They would be travelling away from each other due to mutual repulsion. Until they had all reached the outer surface, the hosting body would not fit all of the special cases claimed for a Faraday cage: all of the charge would not yet be on the outside, there would be electric fields not balanced by equal and opposite vectors, each travelling electron would be departing from the greater number of electrons below it and toward the lesser number of electrons above it. Until each migrating electron completed its journey we might be wrong to call the hosting body a Faraday cage. Such a transient status endures for significant duration if the dimensions of the hosting body are very large, or if the dynamics of continual disturbances to electrically charged particles prevails.
swansont Posted July 15, 2010 Posted July 15, 2010 Upon ever beginning to host a negative bias, an isolated body would undergo an outward migration of its surplus of electrons. They would be travelling away from each other due to mutual repulsion. Until they had all reached the outer surface, the hosting body would not fit all of the special cases claimed for a Faraday cage: all of the charge would not yet be on the outside, there would be electric fields not balanced by equal and opposite vectors, each travelling electron would be departing from the greater number of electrons below it and toward the lesser number of electrons above it. Until each migrating electron completed its journey we might be wrong to call the hosting body a Faraday cage. Such a transient status endures for significant duration if the dimensions of the hosting body are very large, or if the dynamics of continual disturbances to electrically charged particles prevails. Since the application in this case is electrostatics, the transient behavior is not applicable. It is assumed that you are in steady state.
dalemiller Posted July 15, 2010 Author Posted July 15, 2010 Since the application in this case is electrostatics, the transient behavior is not applicable. It is assumed that you are in steady state. Who assumes me to be in a steady state and why the assumption?
swansont Posted July 15, 2010 Posted July 15, 2010 The static electric field inside an ideal conductor is zero. It's an application of electrostatics. Steady state. However, if you perturb it, the transients will be very short-lived.
dalemiller Posted July 15, 2010 Author Posted July 15, 2010 The static electric field inside an ideal conductor is zero. It's an application of electrostatics. Steady state. However, if you perturb it, the transients will be very short-lived. My thread has not taken up any presumption of an ideal conductor. It deals with information we can interpret from Earth's Fair Weather Current. The fact that electrostatic force is involved does not restrict discussion from consideration of dynamic disturbances. A negative electric bias upon the earth would present excess electrons aloft that would continually be subject to being driven back down to the surface with rain. The atmosphere is not a perfect conductor but nothing but a perfect insulator would bar these contentions. No matter how short lived transient disturbances might be, the restoration of electrons into the atmosphere would have been accomplished as a result of termination of just such transients. Some one hundred or so of lightning bolts occur every second, and many more negatively charged are always falling somewhere. We need not refer to the negatively charged earth as a Faraday cage if semantics is all that results from such a reference.
swansont Posted July 15, 2010 Posted July 15, 2010 Well, you keep talking about the earth being a Faraday cage, and I'm trying to point out to you that it isn't.
dalemiller Posted July 15, 2010 Author Posted July 15, 2010 Well, you keep talking about the earth being a Faraday cage, and I'm trying to point out to you that it isn't. Let us have it your way. That is what my last sentence proposed in the post you just answered. What counts is that we can know that Earth and Sol do both carry a negative electrical bias. That bias does cause surplus electrons to rise to their surfaces because of repulsion. A planet or star that repulses electrons to its surface would attract a positive charge toward its center. Likewise, I invite you to please overlook my reference to so many other things such as raindrops and galaxies as Faraday cages and simply indulge such language as signifying bodies of some isolation that carry any charge upon their outer surfaces. I am convinced that I have blundered into some insight that has been overlooked but that does not deviate from mainstream science. My last seven years have been consumed in seeking to project my findings, and other things remain for me to put in order. You might suppose that a page of the best work of an electronic technician with 60 years' experience could interest certain thinking people, but might escape their attention if every posting is met with immediate denigration. Were you to reserve judgement instead of killing the ball as soon as it comes over the net, the intellectual climate here might be even more refreshing. If you let other people judge before reading your standard grim responses, one or more of them might come forward to craft the explanations to you more favorably.
dalemiller Posted July 17, 2010 Author Posted July 17, 2010 The static electric field inside an ideal conductor is zero. It's an application of electrostatics. Steady state. However, if you perturb it, the transients will be very short-lived. Although I am ignorant of need to limit our scope to ideal conductors, that shouldn't interfere with our logic. If a single perturbation were to disrupt the steady state only for the short-lived duration of that perturbation, then a purist might conclude that electrostatic status has been momentarily suspended. However, were that perturbation to have been constituted merely by my insertion, somehow, of an offending excess internal electron, and if I were able to detect the direction taken by that electron during the disturbance, then when the transient became history, I would already have my information about the electrostatic state surrounding my chamber.
dalemiller Posted March 5, 2011 Author Posted March 5, 2011 No. There is no way you can tell what the charge on the exterior of a Faraday cage is if you are on the interior. The charges on the exterior contribute nothing to the field inside. Experiments by Michael Faraday upon his ice pail should suggest that whereby for any negative charge retained within an ice pail, an equivalent number of electrons are presented onto the external surface. Let us assume such a pail to be encountered to contain equal counts of protons and electrons (hence electrically perturbed only by virtue of it surrounding an immobile negative electrical charge within). The immobile negative charge would reposition some electrons originally contained by metallic molecules microscopically of neutral electrical charge. For every such repositioning of an electron away from the inner surface, an electron will have been added to molecules upon the outer surface. The pail would continue to be macroscopically of neutral electric charge: Every additional electron appearing upon the outer surface would be balanced by an equivalent microscopically positive charge upon the inner surface. (That positive microscopic charge can likewise be accounted for consequential macroscopic neutrality for internal electrical status embracing the internal premises of the pail as the inner surface and the inner contents.) From the perspective of the immobile negative electric charge within the pail's contents, we can honor the contention that external electrical influence denied to the interior applies to the outer coat of electrons as well as charges further removed. The positive charge upon the inner surface is thus not distracted from its natural attraction for the negative charge within. Hence, an electron released from the inner contents would travel outward in a manner resembling electron travel detected with our Fair Weather Current of meteorological significance. Furthermore, if two ice pails, both of negative charge were approaching each other, their outer microscopic charges would repel each other, slewing electron population densities away from facing hemispheres. It should follow that the inner surface microscopic charges of positive polarity would assume comparable displacement. As a result, any released electron in either pail could be influenced by the coming collision. Nevertheless, none of this should traumatize persons entering one of those safety cages within high voltage electrodes. We could look the other way when their councilors advise of absolutely no internal effects when man-made lightning goes boom. It is just a little white lie.
swansont Posted March 5, 2011 Posted March 5, 2011 Experiments by Michael Faraday upon his ice pail should suggest that whereby for any negative charge retained within an ice pail, an equivalent number of electrons are presented onto the external surface. Let us assume such a pail to be encountered to contain equal counts of protons and electrons (hence electrically perturbed only by virtue of it surrounding an immobile negative electrical charge within). The immobile negative charge would reposition some electrons originally contained by metallic molecules microscopically of neutral electrical charge. For every such repositioning of an electron away from the inner surface, an electron will have been added to molecules upon the outer surface. The pail would continue to be macroscopically of neutral electric charge: Every additional electron appearing upon the outer surface would be balanced by an equivalent microscopically positive charge upon the inner surface. (That positive microscopic charge can likewise be accounted for consequential macroscopic neutrality for internal electrical status embracing the internal premises of the pail as the inner surface and the inner contents.) From the perspective of the immobile negative electric charge within the pail's contents, we can honor the contention that external electrical influence denied to the interior applies to the outer coat of electrons as well as charges further removed. The positive charge upon the inner surface is thus not distracted from its natural attraction for the negative charge within. Hence, an electron released from the inner contents would travel outward in a manner resembling electron travel detected with our Fair Weather Current of meteorological significance. Furthermore, if two ice pails, both of negative charge were approaching each other, their outer microscopic charges would repel each other, slewing electron population densities away from facing hemispheres. It should follow that the inner surface microscopic charges of positive polarity would assume comparable displacement. As a result, any released electron in either pail could be influenced by the coming collision. Nevertheless, none of this should traumatize persons entering one of those safety cages within high voltage electrodes. We could look the other way when their councilors advise of absolutely no internal effects when man-made lightning goes boom. It is just a little white lie. I don't see how this rebuts or contradicts what I've said. If you have immobile negative charges, you don't have a conductor. You will have a field due to charges inside the volume, but you don't know that the conductor is electrically neutral; the field they produce is not dependent on the charge on the conductor.
dalemiller Posted March 6, 2011 Author Posted March 6, 2011 (edited) I don't see how this rebuts or contradicts what I've said. If you have immobile negative charges, you don't have a conductor. You will have a field due to charges inside the volume, but you don't know that the conductor is electrically neutral; the field they produce is not dependent on the charge on the conductor. Would it help if I changed my description of the internal charge as that of a stationary charge? The concern is not that such a charge is eternally prevented from launching charged particles toward the outer shell, but only that for some limited time the charged particles remain, perhaps toward the center of the pail. I do not see why you are concerned about whether or not the charge resides in a conductor. I do know that the pail is macroscopically neutral because that was my premise. We are dealing with a pail with no excess electrical particles of either polarity. The microscopic negative charge upon the exterior surface: electrons positioned like goosebumps; represents electrons displaced from the interior surface of the pail causing that surface to bear a microscopically positive charge. As soon as any electron(s) can make the trip, the pail does begin to take on a macroscopic charge. It is toward some part of that positive microscopic charge toward which an electron released from the interior would travel. The overall result of these relationships between electrical particles is that in any isolated body of any conductivity at all, the electrical particles that outnumber those of opposite polarity present themselves at least by proxy to the outermost surfaces of the body. Such action is continuously operable. In insurgent electron will repel an inner surface electron outwardly to present an attraction for that electron upon that inner surface. Edit: I have removed the presumptive incidental paragraph that is the sole material that I suspect caused my arguments to be ejected from classic physics. All that remains is consistent with findings of Michael Faraday. It would seem fair play for me to be appraised of what technical disagreement Swansont takes against my remaining rationale above. Edited March 7, 2011 by dalemiller
dalemiller Posted March 18, 2011 Author Posted March 18, 2011 I don't see how this rebuts or contradicts what I've said. If you have immobile negative charges, you don't have a conductor. You will have a field due to charges inside the volume, but you don't know that the conductor is electrically neutral; the field they produce is not dependent on the charge on the conductor. My reply to your response above received the mute critique of relocation. It is assumed here that you take exception to something that I had posted. Did the paragraph reading ("I do know that the pail is macroscopically neutral because that was my premise. We are dealing with a pail with no excess electrical particles of either polarity. The microscopic negative charge upon the exterior surface: electrons positioned like goosebumps; represents electrons displaced from the interior surface of the pail causing that surface to bear a microscopically positive charge. As soon as any electron(s) can make the trip, the pail does begin to take on a macroscopic charge. It is toward some part of that positive microscopic charge toward which an electron released from the interior would travel." ) offend you or cross some forbidden boundary?
swansont Posted March 18, 2011 Posted March 18, 2011 Did the paragraph reading … offend you or cross some forbidden boundary? It crossed the boundary of being supported by physics. It's one thing to ask questions about physics, but if you are going to insist that physics itself is wrong, you get moved to speculations.
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