Coulomb's law and a voltage frame of reference - Page 1
Forum section: electronics / general

author: John Heath
author_rank: Frequent Contributor
timestamp: April 24, 2016, 12:49:50 pm
content: In special relativity the laws of physics are the same for all frames of reference. Can one assume from this that laws of physics are the same for all voltage frames of reference? I will charge a Faraday cage to 10 K volts positive then enter it. From within the 10 K volt cage there is no experiment or measurement that can be made to indicate the voltage frame of reference is 10 K volts above earth. Can not reach outside the Faraday cage with a voltage probe as that would cheating. Must be from within the Faraday cage. Is there any way to tell?Hair should start to stand out at 10 K volts as like charges repel. However the entire Faraday cage is 10 K volts so it cancels out. Yes and no. Yes it cancels out but Coulomb's law of like charges repelling is Q x Q / R^2 = force where Q is charge and R is distance. R^2 is the problem. The hairs on my head are 1/10 of an inch apart but the Faraday wall is 5 feet away. According to Coulomb's law my hair should be standing out. However if my hair were standing out I would be in violation of the said postulate that the laws of physics are the same for all voltage frames of reference as I would know that my Faraday cage is 10 K volts above earth ground. On the other hand if my hair does not stand out then we are in violation of Coulomb's law Q x Q / R^2 = force. Therein is the rub.I can venture a guess that my hair would not stand out in this Faraday experiment as an airplane at 32 thousand feet has a sky charge of 50 to 100 K volts. If hair were standing out on airplanes it would be known. This leaves Coulomb's law of Q X Q / R^2 = force in question with focus on R^2 distance . Then again Coulomb's law is known to be true.If it helps charge a classic gold leaf to 10 K volts then seal it electrically in a glass jar. It should be standing out from a Coulomb force of 10 K volts. Then place the sealed gold leaf inside the 10 K volt charge Faraday cage. It should still be standing out with Q X Q / R^2 = force but if it did the one inside the Faraday cage would be in violation of the laws of physics being then same in all voltage frames of reference as he would know he is in a different voltage frame of reference? Any thoughts on this would be appreciated.

author: ruffy91
author_rank: Regular Contributor
timestamp: April 24, 2016, 01:25:44 pm
content: If you bring the jar inside the cage you change the frame of reference. It's still charged 10kV above reference (So cage is at 10kV and gold at 20kV -> still same force as potential difference is still 10kV)

author: Andy Watson
author_rank: Super Contributor
timestamp: April 24, 2016, 01:42:43 pm
content: Quote from: John Heath on April 24, 2016, 12:49:50 pmNo way to tell for a complete Faraday cage. Experimentally, the absence of a detectable E field within an enclosed conducting surface is used to prove the inverse square law. Rather than say that like charges repel, if you consider the action of a single charge to be the result of interacting with the E field, once you make the E field zero there is no force on the charge.Bleaney and Bleaney give description of your Faraday cage problem under "Experimental proof of the inverse square law" - chapter 1.https://archive.org/details/ElectricityMagnetism2ndEd

author: Marco
author_rank: Super Contributor
timestamp: April 24, 2016, 01:47:58 pm
content: Mass of the cage would change slightly.

author: orolo
author_rank: Frequent Contributor
timestamp: April 24, 2016, 02:00:24 pm
content: A Faraday cage is charged. Inside, owing to Gauss' Law, there is no E field. You are inside and start rotating around yourself in place. That is equivalent to the charges in the cage rotating around you in the opposite direction: moving charges cause a magnetic field. So you should be able to measure a magnetic field while spinning. In other words, a rotating charged sphere should have a measurable magnetic moment.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 24, 2016, 02:49:42 pm
content: Quote from: Marco on April 24, 2016, 01:47:58 pmHmmm, due to the difference in electrons? Possibly.There will also be a change in mass due to the energy stored in the electric field, but this is a property of the space itself where those fields exist, not just of the cage. Indeed, there's no field and thus no mass change from within the cage.I suppose one could determine the mass change "from the inside" by shaking the cage -- assuming it's on a sufficiently compliant suspension, or say, floating in space. But this wouldn't uniquely identify the mass change; it could be due to other fields or effects instead.Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: April 24, 2016, 03:41:43 pm
content: Quote from: Andy Watson on April 24, 2016, 01:42:43 pmLet us set some limits. The stray capacity when charging the gold leaf was 10 p Farads and the current to charge the gold leaf to 10 K volts was 1 p amp. 1 amp is 6.241×10^18 electrons so 1 p amp is 6.24 X 10^6 electrons. The gold leaf is missing 6.24 X 10^6 electrons and electrically sealed to prevent electrons from returning. Let us also also set the stray capacity of the gold leaf in the Faraday cage at 10 p Farad for simplicity. With all this in place I would have to agree with ruffy91. All we did was change the voltage reference from 0 to 10 K volts therefore the gold leaf is now 20 K volts. Your detectable E field , voltage gradient , is between the Faraday wall and the gold leaf that is electrically sealed and still missing 6.24 X 10^6 electrons when I charged it outside the Faraday cage. Did not mean to complicate the issue by counting missing electrons , 6.24 X 10^6 , but Coulomb's laws is all about charges not voltage where Q in Q X Q / R^2 is number of charges , number of missing electrons , not the voltage gradient. Can I say that +10 K volts and + 10 K volts relative to ones voltage frame of reference always equals a given repulsive force at distance X or do we have to stick to Coulomb's law that charge parity , accounting of number of extra electrons or missing electrons equals force with Q X Q / r^2= force ?? I am not sure. If I had a infinitely high impedance meter to measure the voltage of 1 electron then yes maybe. What is the voltage of one electron relative to an infinite number of match electrons to protons such as earth?There is a larger issue here that I did not want to bring up as it just complicates it too much. I am going to bring it up anyways, apologies in advance. An electron and positron have the exact same mass. If I drag both Mr positron and electron into a 10 K volt + charged Faraday cage the electron will go in willingly but the positron will resist as positive charge do not like a 10 K volt Faraday cage. This means the positron must be forced into the positive Faraday cage but the electron not. For energy conservation laws we have to reduce the mass of the electron and increase the mass of the positron to justify this. Now energy is conserved. The chances of earth being a perfect 0 voltage frame of reference are extremely unlikely so how could we have a electron and positron of precisely the same mass ?? No answers on my end. I am just saying there is something fishy about an electron and positron having the exact same mass.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 24, 2016, 04:05:08 pm
content: Quote from: John Heath on April 24, 2016, 03:41:43 pmThe Earth is pretty damn near neutral, with, I'd guess, at most a few billion volts potential with respect to the interplanetary medium. Which isn't much electric field on a planetary scale (there's as much voltage in the atmosphere itself, occasionally building up to breakdown, causing teeny little sparks..). Probably a physicist has ran the numbers and come up with a more reasonable (and even smaller) quantity; for example, photoemission isn't going to push more than x eV of voltage; the highest energies coming from the Sun are in the 10MeV range (coronal electrons), which would simply get trapped if there were much more field nearby.In any case, energy is stored in the field, not the particle (except as a matter of bookkeeping), so once you've performed the work to put the particles inside a given region, the particles themselves don't "know" anything further.Tim

author: orolo
author_rank: Frequent Contributor
timestamp: April 24, 2016, 04:20:05 pm
content: You can easily introduce a positron inside an arbitrarily charged positive farady cage: just introduce a ~1MeV photon inside, and let it decompose into an electron-positron pair, and then make the electron hit the cage (just a gentle push, there is no E field), changing the cage's charge by a trivial amount, while you retain the positron. Now the positron is inside, no problems at all. This might seem contrived, but makes perfect physical sense. Besides, any excess of mass attributed to positrons should be also attributed to electrons, reversing the experiment with a negatively charged cage. Of course the electron and the positron have the same mass, they are the same particle after all: they are both eigenstates of the same Dirac equation.QuoteAssuming a perfectly conducting Earth, it's just the complementarity principle: the field is equivalent to that generated by an opposite charge simmetrically placed with respect to earth.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 24, 2016, 04:59:13 pm
content: Quote from: T3sl4co1l on April 24, 2016, 02:49:42 pmMaybe you are on to something . Mass change means frequency change. Frequency change means a change to either inductance or capacity or both of a L/C resonate circuit. If Uncle Albert has it right then it is the inductance L that changed for the relative inertia of a system to cause time to dilate , lower L/C resonance frequency . On the other hand if capacity changed then it is a different story. A fork in the road. Which way to turn. There is a possible solution found in the work of Oliver Heaviside to calculate impedance. If impedance is higher then a reflected pulse will be the same polarity. If the impedance is lower then the reflection is reversed in polarity. The math works out that you know if it was L or C that changed if the impedance change is known provided the speed of light is constant. Speed of light becomes 1/resonate frequency for vacuum properties 8.8 p f and 1.28 u Henry and impedance Z 376 ohms. If relativity is right then the impedance should be Z 377 for an increase in relative inertia of a system or the alternative 375 Z ohms for an increase in the relative size of the system , capacity C. If the 10 K volt Faraday cage door could be left open a jar then a test could be done to see if a EM pulse reflects polarity the same or reversed to test for relative inertia of a system L vs relative size of a system C for a L/C resonate time dilation.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 24, 2016, 08:25:02 pm
content: Quote from: John Heath on April 24, 2016, 12:49:50 pmI have thought about the same thing long ago, and concluded that that there IS such a thing as a an 'absolute voltage', though an electrometer won't measure it if inside a Faraday cage at the same potential. In such a case, there will be no electric fields anywhere within the cage, because the cage and all within are at the same potential, so the electrometer won't register, but Coulomb's law is still not violated. The gold leaves are still being repelled by each other, but see an equal and opposite force on account of the charged cage. I'll think further about that, but a first glance that appears to be the case.This discussion leads naturally to asking, then what is the absolute voltage of the Earth, and what keeps it in check, and from memory the answer to that lies in the 'solar wind'. I do a lot of work calculating the electric fields in complex geometries in particle accelerators and the like, so have some familiarity with this kind of stuff. I've gotta go for now, but will think further about your interesting questions.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 25, 2016, 01:47:50 am
content: Quote from: orolo on April 24, 2016, 04:20:05 pmYour thoughts brought up an interesting idea. If I pair separate +e and -e out of the vacuum with your 1 MeV spark plug the positron will not last long before it annihilates with the plentiful electrons in the Faraday cage. I will make the Faraday cage more and more negative until the life expectancy of the positron equals the life expectancy of the electrons. That could be more and more positive depending on how you think of it. In any event when pair separated -e and +e reaches equilibrium 50/50 for life expectancy the voltage of the Faraday cage should be the zero voltage frame of reference of the universe ,, maybe. In short the laws of physics are not the same for all voltage frames of reference if Mr positron has a very short life compared to the electron. The positron being anti matter in equilibrium with electron , matter , makes this an experiment best done at a distance with sun block

author: John Heath
author_rank: Frequent Contributor
timestamp: April 25, 2016, 04:05:09 am
content: Quote from: ruffy91 on April 24, 2016, 01:25:44 pmIf the leaf touches the inside Faraday cage it will not discharge. If it will not discharge then there can not be a voltage difference between the inside walls and the charged leaf. The person inside the cage can not witness a separated gold leaf if he measures 0 voltage difference between the leaf and the Faraday cage walls as his laws of physics must be the same as ours. It gets worse. If he walks outside the Faraday cage he may discharge the gold leaf to the outside of the Faraday cage. As strange and counter intuitive as that sounds it appears to be true. I have a video of such an experiment by Professor Eric Rogers. He has a propensity to electrocute little girls but his physics is cool.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 25, 2016, 04:22:25 am
content: Whoa whoa, slow down there, John!I'm not sure what sense to make of your other post, so about this one,Quote from: John Heath on April 25, 2016, 01:47:50 amAll the positron sees, inside the cage, is the electric field near it. If there is a net force, it accelerates in that direction.If the field inside a Faraday cage weren't independent of outside fields (which is the definition of such a cage!), for example the drift region in any CRT would not work.The last grid/anode in a CRT electron gun (in the neck) is the highest voltage anode. After that point, the electric field is nearly zero -- the entire volume of the picture tube. Most CRTs would not work if this weren't the case!(Tektronix scope CRTs, for example, cheat by accelerating the beam with a relatively low potential (~2kV), deflecting it, then passing through a pair of screens which accelerate at a much higher potential (10-15kV). Some of the beam strikes the screens, reducing beam current and scattering it, making the spot size larger, so it comes at a cost. Still, the final drift region, between second screen and phosphors, is essentially a Faraday cage.)So in conclusion, the net force on a positron (or electron) in the middle of a Faraday cage, is zero. If the particle has nonzero velocity, it will eventually hit a wall, so this isn't a very effective trap.If you can perform this measurement, with extremely large fields outside the cage, and do measure a difference (that cannot be attributed to leakage of fields through the cage), you will have some interesting physics papers to write!Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: April 25, 2016, 11:36:54 am
content: I see your point., Once the electron passes the neck of the CRT its in a Faraday cage and the voltage is not changing so the electron is coasting from that point on.

author: orolo
author_rank: Frequent Contributor
timestamp: April 25, 2016, 02:38:51 pm
content: Quote from: John Heath on April 25, 2016, 04:05:09 amHi. If the gold leaf is charged, it will discharge when contacting the cage, regardless of the charge already stored in the cage: due again to Gauss' law, all the charge in the leaf will migrate to the outside of the cage. So even if the leaf discharges, it will tell nothing about the total charge of the cage.This is beautifully demonstrated in aVan de Graaff generator. The VdG generator also exemplifies pumping positive charges inside an already positively charged (quasi perfect) faraday cage.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 25, 2016, 10:13:59 pm
content: Quote from: DiligentMinds.com on April 25, 2016, 06:58:55 pmThe jest of this thread is measurement. How to measure one's voltage frame of reference compared to the universe. Metrology seems like the place to bring it up.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 25, 2016, 10:46:35 pm
content: Quote from: orolo on April 25, 2016, 02:38:51 pmIf you watch the video by Professor Eric Rogers it can be seen that the charged gold leaf will not discharge inside the Faraday Cage . outside yes but not inside. It is towards the end of the video.Van De Graaf generator. I have a funny story about this that turned out to be a learning experience. I bought a kids Van De Graaf just to try it out. It was cheap and I could see that copper wires at the bottom were not even toughing the belt. Good thing it was in my hands as I sorted that problem out making sure the belt had good contact with the copper brushes. This was not enough as it still did not seem to work properly. In a moment of desperation I broke the cardinal rule. I am not proud of this but I read the manual. I know I know but I really wanted to see it work. It said in rather large print " THE VAN DE GRAAF COPPER WIRES MUST NOT TOUCH THE BELT DIRECTLY" . I made the adjustments and it started to work , ha. Turns out Mr Van De Graff is in the business of moving protons not electrons and the path of least resistance for a proton is ionized air not copper. As the saying goes you are never too old to learn something new.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 26, 2016, 03:43:43 am
content: I have thought further about this problem. Previously I said that an electrometer within the charged Faraday cage will not register that the cage is charged, and that is correct, but the explanation I originally gave is not. I'm still thinking about exactly what is going on, and the best way to explain it. I'll be back.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 26, 2016, 12:02:48 pm
content: Let me add to your thoughts with some out of the box ideas. Electron mass is known to be .511 MeV and proton 938.272 MeV. In a Faraday cage charged to - 1 MeV is this still true? To conserve energy I will say electron .5 MeV and proton 939 MeV mass. Let us go more and more negative in the Faraday cage until the electron mass equals the proton mass at about 450 MeV each. This could be the point of absolute zero voltage frame of reference making the electron and proton the compliment of the same particle but opposite charge. That was a Paul Dirac idea. He later tossed it in the trash bin when the positron was discovered. I sometimes wonder if he was too quick to give up on this?

author: Zeranin
author_rank: Regular Contributor
timestamp: April 26, 2016, 12:15:42 pm
content: Before embarking on an explanation, I need to explain what has been bothering me personally, about this problem. In a hand-waving way, a gold leaf electroscope measures how ‘charged’ an object is. If an object has an excess of charge (AKA the object is charged) then when connected to an electroscope, the gold leaf will rise through electrostatic repulsion of like charges. We all know that. OK, so we charge up a hollow sphere or Faraday cage, to say 10kV above ground, and connect our electroscope to the outside of the sphere, and of course the gold leaf flips up, confirming the fact that the sphere is charged. The entire sphere is at the same potential, because it is a metallic conductor. We then place an uncharged electroscope inside the charged sphere, and connect the electroscope terminal to the inside surface of the sphere, and it registers nothing, yet the inside and outside surfaces of the sphere are both at the same potential. How so? It does not need to be an electroscope, human hair standing on end performs the same experiment. Apparently we cannot detect or measure the charge on the sphere, if the observer or experiment is placed within the sphere, at least with an electroscope.I believe that much of the confusion (or at least my confusion) stems from a misunderstanding of what and how an electroscope actually measures. As we know, an electroscope is a single terminal device. If the gold leaf is uncharged (equal number of +ve and –ve charges), then there is no repulsive force on the leaf, and no deflection of the leaf. If we push charge into the electroscope so that the leaf is ‘charged’, then the leaf will deflect.That said, an electroscope can also function as a crude and insensitive voltmeter, which I would argue is how it is actually being used in these discussions. How so, when an electroscope has only one input terminal, while a voltmeter is a differential device with two input terminals, that measures potentialdifference? As I understand the original posting, the Faraday cage is charged up to 10 kV, meaning that a 10kV power supply is connected between Earth, and the cage. A wire is then taken from the cage to an electroscope, which will deflect. The deflection is proportional to the 10 kV, and the capacitance from the electroscope to Earth. The charge that will flow into the electrometer (which is what makes it deflect) is given by Q=CV, so for a given 10kV, the electrometer deflection will be proportional to this capacitance. That is why when an electroscope is used in this way, the case is grounded, to increase this capacitance. The net effect of all this is that the electroscope is actually measuring adifferentialvoltage between it’s input terminal and case, just as with a conventional differential voltmeter.Now we consider the case where the electroscope is placed inside the Faraday cage, which for convenience can be a spherical conducting shell. We connect the electroscope input terminal to the inside surface of the sphere/cage, but to what do we connect the electrometer case? The only thing we can connect it to is the sphere/cage, but if we do that then there will be zero volts between the input and case as they will be connected to the same thing, and we will get zero reading. We could leave the case unconnected, but that would be even worse, decreasing the capacitance from the electroscope to the cage. It’s best to forget about using an electroscope voltmeter at all, and just bring your favourite voltmeter into the cage, but you will have no more success in detecting that the cage is charged and by how much, for the only way to do that is to measure voltage between the sphere and ground, which you fundamentally cannot do if you and your equipment are inside the cage. Getting back to the electroscope, the only way that you can push charge into the electroscope and make it deflect, is if you have a potentialdifference, but there are no potential differences or electric fields to be found inside the cage, so the electroscope is never going to deflect.So what about the suggestion that we could detect the mass of excess charge on a metallic object within the sphere, that had touched the inside surface. That won’t work either, because all of the excess charge is distributed on the outside surface of the sphere/cage, so can’t be ‘seen’ or detected from inside the cage. To put that another way, as your mate outside the cage starts cranking up the 10kV power supply, at all times there are no electric fields anywhere inside the cage, so no driving force to push charge around so as to create an excess charge on anything inside the cage. Or deflect an electroscope or voltmeter inside the cage. I believe the OP is right. From within the cage, there is no way of knowing whether the cage is charged or not, or by how much.From outside the cage it’s easy to detect if the cage is charged. If a large, electrically neutral body like Earth is conveniently nearby then use a voltmeter. If the cage is in deep space, then observe the motion of nearby charged particles.Interesting topic, even if the link to metrology is tenuous.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 26, 2016, 11:28:20 pm
content: Some fine thoughts. Enjoyed the read. Yes the electroscope has only one lead plus the stray capacity from the outside of the gold leaf to whatever is round. Let us make some differential voltage measurements then apply Coulomb's law outside the Faraday cage. From a grounded water tap the gold leaf measures 10 K volts. From the same water tap the chair , ceiling , floor and walls measure 0 volts. Let us apply Coulomb's law. Q x Q / R^2 = force. 0 X 10 k volt = 0 so the attractive force between the room and the gold leaf is 0. What about the leaf/ It measures 10 K volt on either side so 10K X 10 K / R^2 is an impressive repelling force.What if we lived on the leaf and our water tap ground reference was the leaf itself. Without bias we will let the numbers tell us armed only with Coulomb's law. Relative to the leaf the walls and ceiling are - 10 K volts but they are far away and in every direction. Q x Q / R^2 so there is a small attractive force outward in all directions. What about the Leaf relative to the leaf?. The leaf relative to the leaf the voltage is 0 volts therefore no force to separate the leaf.In summary we in the kitchen with our Coulomb's law book predict a strong outward force on the gold leaf. Our friends that live on the gold leaf armed with the same Coulomb's law book predict a weak outward force on the leaf.There can not be two different prediction from the same theory of Coulomb's law yet there it is depending on the observer being in a kitchen or an observer living on a 10 K volt gold leaf. This is the part that keep me pacing the floor at night.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 27, 2016, 02:42:09 am
content: Quote from: John Heath on April 26, 2016, 11:28:20 pmLet me summarize your situation, just to be sure I understand. You have a gold-leaf electroscope on your kitchen table, and you have a 10kV power supply connected between your water tap and the electrosope. You analyse the separating force on the gold leaf according to Coulomb's Law, from the perspective of sitting on your kitchen chair, and from the perspective of living on the gold leaf, and you get two different answers. Yes, that would keep me awake at night also.You seem to be mixing up ‘charge’ and ‘voltage’, in Coulomb’s Law, and be not paying attention to what the electroscope actually measures. You also seem to be forgetting that while voltage is nearly always described as a relative quantity, AKA potential difference, excess charge (the Q in Coulombs Law) is absolute, not relative. You can say that Earth is at some particular large voltage with respect to something else in the Universe, but this in no way alters the fact that the Earth is electrically neutral, Q=0. I’m also uncomfortable with your term ‘the leaf relative to the leaf’ . Both leaves are always at the same potential, because they are metal and connected to each other, but I’m sure you know that, so on that point I’m probably just arguing with semantics.Fundamentally, the electroscope responds to charge, more specifically, excess charge on the leaf. It can be made to respond to a voltage, but only if and because such a voltage has the effect of producing excess charge on the leaf.Here is how I analyse your example, in a way that produces consistent results regardless of the position and perspective of the observer. Assume the electroscope is of the type that uses a pair of gold leaves, that hang down vertically. There will be a stray capacity, ‘C’, from each gold leaf to earth. You could increase C significantly by connecting your metal electroscope case to the water pipe, but either way, you have a particular stray capacity C from each leaf to ground. The fixed-voltage power supply is also connected from ground to leaves, so the power supply charges up this capacitor. Assuming that mother Earth is electrically neutral, the total excess charge that is pushed onto each gold leaf is given by Q=CV. The gold leaves then respond according to F = Q x Q / R^2, where Q is the charge on each leaf, and R is the separation between the leaves. This analysis and result is exactly the same, regardless of the observer, isn’t it? The stray capacity depends only on the geometry, which is observer independent, as is the magnitude of the power supply voltage, 10kV.I’m not sure if the above explains what was bothering you, but if it does not, get back to me and we’ll thrash it out.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 27, 2016, 04:22:45 am
content: I like you model better. Let us make it clearer by floating in the middle of space somewhere. Stray capacity is no longer an issue. I will replace gravity with a spring on the gold leaf. If charged the leaf spring opens. Take the charge away and the leaf closes. Both us as observers from a distance and the ones that live on this gold leaf apply Coulomb's law as an accounting issue of literally counting the number of + charges and _ charges , protons / electrons , and applying Coulomb's law accordingly. This resolves the conflict between the distant observer and the one that lives on the gold leaf as they will now arrive at the same answer for the gold leaf.I do not mean to be a trouble maker here but there is one fly left in the soup. Replace the spring gold leaf with a Faraday cage that has 0 charge. We place one electron in the middle somewhere. As the cage has 0 charge then Q X Q / R^2 or 0 X 1 electron = 0 force. The electron is free to float around not effected by a Coulomb force. I am going to throw a wrench in the works by sprinkling some electrons on the Faraday cage. let us say the cage now has a charge of -1 and the electron floating inside has a charge of -1 just to give it a number. If Q X Q / R^2 is true then the electron should stay in and exact center of the cage as R^2 means closer to any side of the cage will increase the Coulomb force repelling it. The problem is the electron could care less and will still just float around indifferent to R^2 Coulomb force from the sides of the Faraday cage. This was demonstrated in the video by Professor Eric Rogers ironically using the inverse law R^2 to explain it. Maybe I misunderstood it.He hints that it is similar to gravity at the center of the world where one weighs 0 pounds. That can be understood. However if the earth were a egg shell and we were inside the egg shell gravity would not pull one towards the shell just like the 1 electron floating around in a positive charged Faraday cage for this example would not feel a Coulomb force to the cage walls . How can an electron not want to go to a positive charged Faraday wall? In an oscilloscope the electron will go toward a positive deflection plate but not in the Faraday cage despite R^2.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 27, 2016, 06:28:56 am
content: Quote from: John Heath on April 27, 2016, 04:22:45 amI think we are getting close to complete resolution. The question is, if you charge up a Faraday cage, meaning the cage has an excess charge, then why does an electron inside see no force, at ANY position within the cage. Professor Rodgers explained this, but you may have missed his point.Assume for simplicity that the cage is spherical. At the centre of the cage, we know from symmetry there can be no force on the electron. You might think intuitively that if you move the electron off centre and closer to the inside wall, the net Coulomb force would increase, as it gets closer to the charged wall, especially as the force goes as 1/R^2.The trick is to realize that the electron is being pulled literally in EVERY direction, and these forces always cancel just as long as the inverse square law is true. Look again at the video. If you move the electron off centre, then you still have to consider ALL the Coulomb forces, not just the forces from the part of the wall nearest to the electron, which is what you are doing. If you cleverly divide the inside surface of the sphere into little ‘patches’, you find that the ‘patches’ on the wall where the wall is nearest to the electron have less area, in inverse proportion to R^2, compared to the larger ‘anti-patches’ on the part of the wall that is further away. The force from any such patch is proportional to it’s area, and therefore inversely proportional to it’s distance R from the electron. Each patch area scales as R^2, but the Coulomb force scales inversely as 1/R^2, with the result that the magnitude of the Coulomb force is the same for every patch, but the forces cancel out because each pair of patches pull in an opposite direction. It’s messy to explain verbally, but if you look again at the video then I’m sure it will be clear, not to mention rather nifty.You can reach the same conclusion trivially easily by application of Gauss’s Law to an imaginary Gaussian closed surface, chosen to be a concentric sphere inside the spherical Faraday shell. As there is no charge enclosed in the spherical Gaussian surface, the electric field must be zero everywhere on this surface, and thus everywhere inside the spherical Faraday shell. If there is no electric field, then there is no force on an electron, as it is an electric field (units Newtons/Coulomb) that provides electrostatic force on charges. However, this explanation is cheating because you have to take Gauss’s Law on trust, whereas Rodger’s explanation in the video is intuitive and understandable by anyone. Although spherical Faraday cages are easiest to analyse, in fact the electric field is zero everywhere inside ANY closed, conducting shape. For a sphere, the charge excess charge distributes itself evenly on the surface, as it must from symmetry considerations. For other shapes, the surface charge density is not constant, but distributes itself in such a way that the E-field is everywhere zero inside, and it turns out that in order to do so, the charge density is much higher at corners and edges.Yet another way we know that the E-field must be zero inside a hollow conductor, is to note that the entire conductor must be equipotential, at the same voltage. This we know from Ohm’s Law. At DC, any two point on a conductor are at the same voltage, unless there is a current flowing, which there is not in an electrostatic problem. If every point on the inside surface is at the same voltage, then intuitively one can see that there can be no electric field created, units of V/m. You can’t produce volts per meter, unless you have voltage difference(s), and there can be no voltage differences anywhere on a closed equipotential surface.Quicky getting back to your intuitive feeling that the electron would see a stronger Coulomb force when nearer to the inside wall, consider the case of an electron above a large, charged flat plate. You may intuitively think that Coulomb force of attraction to (or repulsion from) the plate would be stronger at closer distance to the plate, yet in fact the force is a constant, independent of distance, because as the electron moves further from the plate (force dropping off as R^2), it ‘sees’ more of the plate, increasing by R^2.Enough raving. I enjoy solving mysteries and paradoxes, or at least trying to, so keep them coming.

author: RIS
author_rank: Regular Contributor
timestamp: April 27, 2016, 10:21:55 am
content: to say that the field does not exist because it has no effect on electron inside is ridiculousand yes the charge of the cage can be detected from insideyour just need a good hammer,not for breaking but to measure the energy oscillations.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 27, 2016, 12:14:53 pm
content: Quote from: RIS on April 27, 2016, 10:21:55 amActually, an electric field isdefinedin terms of the force it exerts on a charged particle.No electrostatic force on electron means no electric field, end of story.If you disagree then it might be best if we agree to disagree, and let others here judge for themselves.

author: RIS
author_rank: Regular Contributor
timestamp: April 27, 2016, 12:54:36 pm
content: I agree with this philosophical issues.but if we put man at the center of gravity the it will not be a philosophical issue,it will be a matter of life and death.These are just my rough thoughts.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 28, 2016, 11:22:23 am
content: Quote from: Zeranin on April 27, 2016, 06:28:56 amSorry for the delay. Had to watch some videos on Gauss's law to wake up some distant college memories. I see your point. Can not have a voltage gradient within a closed Faraday cage for evenly distributed charges on the Faraday walls. Will set the net Faraday cage charge at +1 . On the outside the electric field will fall off at R^2 but uniform inside the cage. Will now place 1 electron at the center of the Faraday cage . Mr Electron's -1 charge should fall off at R^2 from within the Faraday cage being a single point charge , yes/no? If the electron were very close to one of the Faraday walls it will run into infinity problems with R^2 for electron - Coulomb force causing it to stick to the Faraday walls. I could make the same argument for gravity. Say the earth is an egg shell and we are inside this egg shell. The gravity gradient inside is 0 in the same way the voltage gradient in the Faraday cage is 0. However I am a single point mass generating my gravity field that should drop off at R^2 from within the egg shell. I should stick to the inside of the egg shell much like the - electron should stick to the + Faraday wall.I am confident everything I said is false , incorrect. The question is what went wrong and can anything be leaned from this? I will toss one in your court for consideration.A] Coulomb force comes from the properties of the vacuum around the electron only not the Faraday cage walls. The implication of this is an electron accelerating in a CRT gun is pulling on a vacuum not the positive charge of the second HV anode. To conserve momentum the vacuum must have accelerated in the other direction. The TV as a total must move backwards while the electron is accelerating forward to conserve momentum. Can the vacuum around the accelerating electron pull on the TV set or do we have to wait for the electron to hit the phosphor to settle the score? NASA has ion accelerators to propel space ships using a Coulomb force so apparently the vacuum around an accelerating electron can pull a TV set backwards. Not sure if this is going anywhere , just brain storming.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 29, 2016, 12:24:13 am
content: Quote from: John Heath on April 28, 2016, 11:22:23 amHmm. I thought my previous posting showed clearly that when viewed from any angle, a electron inside a Faraday cage will not be attracted to the cage wall.Can not have a voltage gradient within a closed Faraday cage for evenly distributed charges on the Faraday walls.Absolutely correct. You can’t have a voltage gradient (imposed from outside the cage) within a Faraday cage,period, because the inside surface is equipotential. You can set up non-symmetric charge distributions with resultant voltage gradients outside the cage if you wish, yet still you will not create any voltage gradient (AKA electric field) inside the cage, because the inside surface is equipotential. This is the principle of electrostatic shielding. Of course, you can set up electric fields within the cage if you wish, by placing charged particles or electrodes inside the cage.On the outside the electric field will fall off at R^2 but uniform inside the cage.You say the electric field inside is ‘uniform’, but the correct description is that the field inside is ZERO.Mr Electron's -1 charge should fall off at R^2 from within the Faraday cage being a single point charge , yes/no?Yes, but after that your reasoning goes astray. To calculate the electrostatic force on said electron, we multiply the electron’s charge in Coulombs, by the strength of the E-field (units N/C, or V/m, they are the same unit) in absence of the electron. As I think we agree, the E-field strength in the absence of Mr electron is zero, and thus the force on the electron is zero. Period. The field strength is zero right up to the inside wall, and even inside of the wall, so the force on the electron is zero, right up to the wall. I can’t make it any clearer. ?The implication of this is an electron accelerating in a CRT gun is pulling on a vacuum not the positive charge of the second HV anode.That is not correct. The reaction force is most certainly on the electrostatic structure that created the electric field, and thus accelerates the electron. The simplest accelerating structure is a charged sphere, and you KNOW that there is an electrostatic force between two charges, one being the mechanically fixed sphere and the other an electron. Coulomb’s law is symmetric, and automatically predicts the same force on both the charges. As a practical example of this, we design and manufacture ion-thrusters for NASA, for interplanetary space craft. These nifty ‘rocket engines’ work by accelerating charged ions out the back of the engine. If your proposal was correct, these ion-thrusters would not work. The reason ion-thrusters are used is twofold. Firstly, the power for the engine can come from solar panels. Secondly, we need to get the maximum possible amount of thrust per unit mass of ejected fuel, because we can’t afford to carry ridiculous masses of fuel. This means we need the highest possible exhaust velocity. With an ion-thruster, we can achieve exhaust velocities approaching the speed of light, far higher than is possible with a conventional fuel-burning rocket. As described, the spacecraft would charge up, so we developed a ‘helicon double layer’ design where the ions and electrons are both ejected into the exhaust. Getting back to your TV set, if we take the case of accelerating asingleelectron, then while the electron is being accelerated, the TV and electrons accelerate in opposite directions, from the equal and opposite force. When the electron eventually hits the phosphor, the ‘score is settled’, and both end up with no velocity. In reality we have a continuous beam of electrons, so the forces are at all time balanced.I enjoy your ideas, and you really exercised my brain on the topic of why an electroscope will not register that a sphere is charged, when placed inside the sphere. I initially though I knew that an electroscope simply measured excess charge, but was forced to think more carefully. My excuse is that I don’t use electroscopes, so had never thought deeply about what they actually measure.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 29, 2016, 05:15:37 am
content: I have an analogy that tracks Q X Q / R^2. Will replace the vacuum with a 2D matrix of 100 1 M resistors somewhat like a screen door. Will solder a copper wire along the outside of the matrix of the resister grid. This will be the Faraday cage walls. Now it works. If I place 1 electron in the middle of this matrix of resistors it will have no reason to move up , down , left or right as there is no voltage gradient in the matrix to cause the electron to move even if it is almost touching the Faraday wall. A vacuum is an insulator not a resistor so it is not the best analogy but at least it is tracking Coulomb's law inside a sphere.Yes I can see how an ion accelerated to .99 c would give you a big bang for your buck for fuel mass. .99 c would be a mass increase of 7 . There are some tiny satellites they plan to launch using laser photons to accelerate them to our neighboring galaxy. That would be 0 fuel mass , ha.As to your pondering on a possible absolute 0 voltage frame of reference. I have given that very question some considerable thought. I have a 1 foot square Faraday cage that I charge up to 10 K volts then back down to ground every 10 seconds. While that is going on a fiber cable is sampling a 10 MHz clock inside the cage but the frequency counter is outside the cage. The 10 MHz oscillator is in a difference voltage frame of reference than the frequency counter. I believe the key to measuring a change caused by a different voltage frame of reference is to not be in the frame of reference when measuring. The predicted target is 50 uHz faster per second for a 10 MHz oscillator. To put that in perspective a GPS clock would be faster by 5 mHz. That is not as bad as it sounds as the oscillator only has to be stable for 10 seconds. Noise and jitter is a problem but net average stability for a 10 second period seems okay. 10 mHz is about as close as I can get so far. Need 3 more zeros of resolution. I sometimes wonder if there is a better way to do this. In any event a measurable change caused by a change in a voltage frame of reference could point to the direction of what is an absolute 0 voltage frame of reference.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 29, 2016, 10:53:50 am
content: Quote from: John Heath on April 29, 2016, 05:15:37 amYes!Your analogy is smack on the money, and more valid and powerful than you could imagine. What you are describing is known as FEA, Finite Element Analysis. Of course, while you describe a 2D array of 'resistors', the full solution is a 3-D matrix of such 'resistors', AKA finite elements. As you say, a vacuum is not literally resistive, but it turns out that the math for solving the electric potential at all points in space, for any structure of conductive objects in 3D space, is identical as if you were solving the voltage at every point in the situation where the vacuum is occupied by a solid resistive material such as carbon, which in turn is (in effect) modelled as a matrix of resistors, exactly as you describe. As you would realize, solving for the voltage at every node in a huge 3D array of resistors, where there could be millions of nodes, is computationally intensive, though conceptually simple.The FEA method is extraordinarily powerful. I invented and developed it 25 years ago for solving magnetic problems, writing my own FEA code to run on the first 8086 PC's, years before commercial magnetic modelling applications were generally available. I developed an iterative algorithm for efficiently solving the huge matrices, being the FEA 'engine', and then added a (primitive by today's standards) graphical user interface so I could view the 3D model from any angle, and display 2D slices on any plane. My application at that time was for designing ultra-high-efficiency brushless electric motors, and my FEA software allowed me to calculate the flux density at every point within an electric motor consisting of windings, magnetic materials and air. By observing the change in flux as the armature is rotated, one can calculate torque, speed, efficiency etc. The fantastic thing about FEA is it's complete generality, so soon I was using the same software to model and design complex systems of magnetic coils used in physics research.Next I modified the code to solve electrostatic problems as well, exactly as you describe, and went on to design all manner of 'ion optics' for physics research, specialized electron guns, ion lenses and accelerators and so on. The commercial electrostatic FEA modelling package was (and still is) called SIMION, with similar capabilities. As FEA gives you the electric potential at every point in 3D space, it also gives you the E-field at every point in space, being just the difference in potential at 2 points on the matrix, divided by the distance between these 2 points. Then I added to capability for 'ray tracing' of charged particle(s), which is actually quite easy, as the force on the particle at any point is just the E-field times the charge, so it is just a matter of iteratively applying Newton's laws of motion to trace out the path of a charged particle(s). With Simion, the user inputs a particular physical design of electrodes, and can then ray-trace to observe the result. The user can then manually modify the design check the result, and repeat so as to hone and optimize the design, a slow and laborious process. I added more code to my FEA application to automate that process, so that many thousands of designs can be tried with the design being progressively optimized automatically, a process that would take hundreds of man years with the commercially available software. As a result, I have designed some of the most perfect ion imaging lenses ever built for the physics research group that I work with. FEA is a lot of fun.FEA can solve almost any 3D field problem, or any similar problem where a physical quantity 'flows' as a result of a driving force. Another example where I have applied my code to good effect is thermal modelling. The math for all this FEA stuff is essentially the same, its just a matter of adding different front ends to the FEA engine to solve different types of problem. I can model any arrangement of thermal materials, define the heat sources, and then calculate and display the temperature at every point in the model, a lot of fun.Another application is calculating the resistance of any arbitrary 3D shape, consisting of any number of different materials. I collaborate with electronic manufacturers on lots of stuff, one very large resistor manufacturer is interested in my code for calculating the resistance of complicated shapes of resistive foil, for example in the very large shunt that they built for me. At present, they 'guestimate' the effect of square bends etc in the zig-zag shape, then fine tune the design later to get exactly the right resistance. With FEA, you can accurately calculate the resistance of ANY shape.Another really big application is calculation of stress and strain in mechanical engineering structures, and it goes on and on.Some of your ideas may not work out, but your 'resistor matrix' ides is more relevant and powerful than you could have imagined.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 29, 2016, 02:17:56 pm
content: This is indeed good news. It has been nothing but good news from the time I came to this group only a few months ago. Have received a wealth of information on precise frequency counting from fine minds that have been there done that. Then to find out that you are a specialist in 3D metric tensors. I have a pile of questions. Off to work.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 30, 2016, 10:55:15 am
content: I have a 1 cubic meter of vacuum on my kitchen table . It is 8.8 p farad by 1.2 Henry for a propagation speed of c and impedance of 376 . If I tap this 1 meter cube of vacuum it should resonate at about 50 MHz. I like to pace around this cube to better understand it. In the middle of this cube there is a sun that spits out bursts of charged particles now and then moving freely outward through the vacuum. You see the problem? The vacuum is both a insulator and conductor of charges at the same time ? Both can not be true. We have to say that a vacuum is a conductor of charges with low resistance. Mr copper wire bathed in our conducting vacuum just got a lot more complicated. Electrons are free to move inside copper and they are free to move in a vacuum. This means it is the transition between copper and a vacuum that is an insulator not the vacuum itself as both copper and the vacuum conduct electrons. The only reason to bring this up is to say your use of a resistor matrix to represent the vacuum inside a Faraday cage is justified as the vacuum should act this way being a conductor of charged particles.My questions. What does a 1 K resistor measure in a X , Y 2 dimensional matrix that goes off into infinity? Secondly how do I terminate the outside of this resister matrix so that it appears to go off into infinity. The outside resisters would have to be lower than 1 K to make it seem to be continuous yes / no ?. Thirdly have you tried inducters Henry and condensers Farad matrix for a vacuum then terminate the outside with resisters. That to myself would be the ultimate vacuum model so you must have entertained this at some point.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 30, 2016, 01:45:57 pm
content: Quote from: John Heath on April 30, 2016, 10:55:15 amThese are not simple questions.The only reason to bring this up is to say your use of a resistor matrix to represent the vacuum inside a Faraday cage is justified as the vacuum should act this way being a conductor of charged particles.The use of a 'resistor matrix' is certainly justified, but not for the reason you give. The first step is to convince yourself that a 'resistor matrix' is a valid way to model and solve for the voltages and current at all points inside a block of electrically conductive material, such as carbon, with metal surfaces/shapes/electrodes at fixed potentials embedded in the carbon. That said, what we ACTUALLY want to model is to find the electric potential at all points in space, where the carbon is actually vacuum, and in this REAL case there is nothing physical that actually 'flows'. That said, if you look at the equations relating permittivity, electric flux, electric flux density, electric field and potential, you find anexact mathematical analogy. The math and the method is therefore identical, and it's just a matter of re-naming the familiar quantities such as electrical resistance in terms of the equivalent electrostatic quantity. In the same way, you find an exact mathematical analogy with magnetic quantities, so you can equally well do magnetic modelling, essentially just by renaming the quantities (eg resistance and EMF) into the equivalent magnetic quantities (eg reluctance and magneto-motive-force)What does a 1 K resistor measure in a X , Y 2 dimensional matrix that goes off into infinity?I don't exactly understand the question, though maybe my answer to your next question covers it.Secondly how do I terminate the outside of this resister matrix so that it appears to go off into infinity. The outside resisters would have to be lower than 1 K to make it seem to be continuous yes / no ?.Yes, this is an important point in electrostatic or magnetic FEA. In theory, you should build your 3D FEA model out to infinity, but of course that is not practical. Alternatively, you can model out 'well beyond' the volume of interest, with the result that you will get 'near enough' to the right answer in the volume of interest. However you look at it though, your FEA model stops at a 'boundary', and you ask exactly the right question, how do you fudge this boundary to get the most accurate results, by making the boundary behave as if the resistors extended out to infinity. For starters, you need to assign a potential (voltage) at the boundary, else the matrix is insoluble. If you use the same resistance value at the boundary as elsewhere, then that is equivalent to having the vacuum extend outward, terminating at an equipotential surface, and in many cases that is exactly what you want - you are modelling inside of a shielding can. However, if you want to model vacuum out to infinity, then by choosing the boundary resistor values 'just right' then the matrix behaves as if it extended out to infinity. The 'just right' value of resistance is higher than the other resistors representing the vacuum, and the optimum value depends on the overall size of the model, the further out you model empty space, the larger the optimum resistor value. My program lets the user choose they type of boundary, fixed potential, infinite resistance, or 'magic elements' that behave as if the matrix extended forever. When I last looked at the commercial SIMION package, it did not have this level of sophistication, and gave erroneous results near the model boundary as a result, when modelling vacuum out to infinity.Thirdly have you tried inducters Henry and condensers Farad matrix for a vacuum then terminate the outside with resisters. That to myself would be the ultimate vacuum model so you must have entertained this at some point.That is similar to how you model a transmission line, but not for solving electrostatic problems.Wonder if anyone else is interested in this stuff? The link to metrology is getting ever more tenuous.

author: John Heath
author_rank: Frequent Contributor
timestamp: April 30, 2016, 09:37:14 pm
content: At 1000 views in just 5 days it is safe to say this Metrology group has an interest in what you are saying. The only difference between Metrology and FEA is 2D measurement vs 3D measurement. Cern particle detectors , basically FEA , is the leading edge of the limits of measurements made in our day. To dip your toes in those waters would be a dream come true for a Metrology nut.In ion len shape modeling does FEA adjust for ion mass? In other words would a proton charge +1 and a gold atom with charge + 1 end up in the same place at the same time or does FEA account for this mass difference?

author: Zeranin
author_rank: Regular Contributor
timestamp: May 01, 2016, 08:20:11 am
content: Quote from: John Heath on April 30, 2016, 09:37:14 pmA good question. Strictly speaking, the FEA finds the electric potential and E-field at all points in space, while modelling the movement of charged particles through this space is a separate subsequent process.So to your question, electrostatic systems produce ion trajectories that are mass independent, provided all the particles have the same KE, which is usually the case, and provided the particles have the same charge of course. In ion-optic systems, usually all charged particles are accelerated to the sameenergy, because all are accelerated through the same voltage, the gain in energy given by the charge multiplied by the change in potential. Therefore your proton and positively charged gold atom will follow exactly the same trajectory through any ion-optic system, and end up at the same place, though not at the same time, because the heavier particles will be travelling slower for the same energy. The reason that different masses follow the same path is as follows. At first glance it may seem as if the heavier particle would be deflected less, because the electrostatic deflecting force is the same, but the heavier particle 'takes more' to deflect. However, the heavier particle is travelling slower as discussed, so the electrostatic deflecting force has longer to act on the particle, with the net result that the path taken is mass independent. Cute, eh? In my program, and SIMION, you can of course specify the particle mass, and even set up a variety of masses, and then observe all the masses to take identical paths, but different transit times. By contrast, different masses DO take different paths through magnetic fields. In general, electrostatic fields change the particles energy, and of course deflect the particle, while magnetic fields deflect but do not change the energy, because the magnetic force is always at right angles to the velocity, so no work is done.Maybe I'll attach some screen shots from the program, showing an electrostatic or magnetic structure, with the associated field strength displayed as coloured contours and the field direction displayed as arrows. These coloured contour plots are often quite beautiful and artistic, independently of the information they convey.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 01, 2016, 07:21:52 pm
content: I would like to see some windows screen shots. If memory servers you said the first version use a 100 bus system so it must have been a printout of numbers armed only with CPM and the brand new Dr John G. Kemeny's BASIC programming language. Liberated from machine code at last. I am old enough to remember flipping those switches for a cold cold boot.The magnetic field force is at right angles to the movement of the charge therefore no energy. That is a profound statement. I see your point. If I place an electron charge next to a magnet it will have no effect or force. It is only when the electron is moving or the strength of the magnetic field changers that a force is seen. Brings to mind the first paragraph of the electrodynamics of moving bodies where a distinction is made between a moving wire and a moving magnet. The distinction can not be made as movement of the wire or movement of the magnet is a relative concept depending on the wire's point of view or the magnet's point of view. The wire and the magnet can not have different laws therefore Maxwell must be wrong. If a magnet were to accelerate a charge it would be in clear violation of energy conservation laws as I could simply place a permanent magnet to accelerate charges. However the notion that the magnetic force on a moving charge is at right angles therefore no energy is a new way of thinking of it. Must stew on this some more.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 03, 2016, 04:48:23 am
content: Here is a screen shot from my ancient 3D FEA magnetic modelling program. This is about the simplest model I could make, being a cubic NdFeB permanent magnet (orange) sitting in empty space (olive). The magnet is 13x13x13mm, while the modelling boundary is a cube measuring 40x40x40. The grid spacing is 1mm, and each FEA ‘element’ is a 1x1x1 mm cube, chosen here to be coarse so you can see the individual elements. The boundary is shown in dark blue, and the permeability of the boundary is less than that of free space (vacuum), to make the model behave as if the boundary was at infinity, as discussed in previous posting.The view is a 2D slice through the 3D model, with the magnet being magnetized in the vertical direction. The direction of the field at every point is space is shown by a tilted line inside every element. As you see, the FEA shows the classic magnetic field lines for a permanent magnet, often illustrated by placing the magnet under a piece of paper, on top of which is sprinkled iron filings.Although not visible in the screen shot, in the real program, the user can move the cursor anywhere on the model, and the (x,y,z) coordinates and magnetic values are displayed at the bottom of the screen. For example, looking at the bottom of the screen reveals that when this screen shot was taken, the cursor was at (20,27,20), which is centrally just above the top pole piece of the magnet. As shown, at this point the cursor is in ‘air’, the flux density is 0.4701 Tesla, and the relative permeability at this position (air) is 1.0. By moving the cursor, and changing which 2D slice is being viewed, one can therefore read out the flux density (or any one of a number of other magnetic quantities) at any point within the 3D model. FYI, the flux density is higher in the centre of the magnet (0.783 tesla), while vertically above the magnet, at the boundary, the field has dropped off to only 0.053 Tesla.Alternatively, one can produce coloured contour plots of any magnetic quantity, on any chosen 2D slice, and overlay the vectors if desired. These are often quite pretty. I’ll follow up with another posting showing the coloured contour plots of mod(B), Bx, By, Bz and magnetic potential, on the same 2D slice. One can include in the model current carrying coils, and permeable magnetic materials (eg steel), and can also handle ‘real’ non-linear magnetic materials providing the B-H curve is known. Even this very simple model of 8000 elements takes about half a second to solve, so for large, complex models the processor is kept very busy indeed.All lots of fun, if you are into this sort of stuff.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: May 03, 2016, 05:14:36 am
content: It is quite difficult to solve these problems on intuition. Newton and Leibnitz developed the calculus to better define and understand gravitational problems. Gauss, Maxwell, Hilbert and others developed other mathematical tools to concretely describe what is going on. If you want to understand Faraday cages, field shielding and other related things in detail you will need to spend quite a bit of time developing your math skills. These tools are quite accurate and provide great insight, but work best on a limited set of geometries. Things like spheres and rectangular boxes.Finite element methods are a brute force method of solving differential and integral equations. They work for any geometry you are willing to set up, though they have limitations at corners and some other types of discontinuities. The computer science course I took in the early 70s described how to do this and there were homework assignments. The fact that these tools were widely known fifty years and more ago does not in any way detract from the achievement of applying these solutions to particular problem fields, adapting them to the relatively limited capabilities of early personal computers and particularly in generating graphical interfaces as a way to present the enormous amounts of data involved. The peculiarities of boundary conditions and various non-linearities in different problem fields makes any particular implementation a work of great merit.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 03, 2016, 11:23:12 pm
content: Quote from: Zeranin on May 03, 2016, 04:48:23 amNicely done ! I can see the lines tilting just like iron filings on paper. You said you pinched off the blue boundary permeability to a little less than a vacuum to simulate infinity so the Tesla net numbers near the outside should be about as real as you can get.On looking at the screen shot my mind wonders to what a forced mono pole would look like ? By forced mono pole I mean gluing a bunch of magnets together forcing all south poles to face center therefore all north poles facing out in the shape of a golf ball. That is about as close as we are going to get to a mono pole. I would like to run this golf ball on your program just to get a intuitive feel of what mono pole would look like. Could your simulation multitask two such mono poles at the same time? Would the two identical mono poles repel each other just like two positive charges? Is a mono pole magnet the true nature of a charge? If it walks like a duck and sounds like a duck then maybe it is a duck. Your simulator would know if two identical golf ball mono poles are sounding and walking like two + charges. If so then flipping a down quark to a up quark could take on a whole new meaning.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 04, 2016, 01:45:30 am
content: Quote from: John Heath on May 03, 2016, 11:23:12 pmBefore moving on to magnetic monopoles, I'll send more screen shots of the rectangualr bar magnet, being coloured contour plots of |B|, Bx,By,Bz and magnetic potential.Then I'll build you a 'monopole'. A spherical monopole as you describe would be a pain to model, but I can very easily do just as well, by building you a 'monopole cube', where each of the 6 faces is magnetized through the thickness of the face, with every face having the south pole on the outside surface. I’ll make each face something like 20x20x5 mm, so they will be magnetized in the 5mm direction. I’ll make ‘em in NdFeB, so they’ll be nice and powerful, too. Of course I can make two of them if you want, but suggest we build one first.How could this NOT behave as a monopole, you ask? Every exterior face is a south pole, no denying that. As Catalina said, the results for this type of problem are not always intuitive, but it’s always fun to model it and see what happens.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 04, 2016, 03:41:26 am
content: Looking first at image Bar_modB, this is a plot of the magnitude of the flux-density vector, AKA modB or |B|, sometimes called the magnetic field strength. As in the plot in the previous posting, the magnetization is in the vertical direction, that from here on I will refer to as the Y-direction. There is a vertical ‘colour-key’ on the RH side of the plot, so bright yellow represents the highest flux density, fading out to black at zero flux density. It is seen that the flux density is higher inside the magnet than outside. Perhaps counterintuitively, the highest flux density is not in the centre of the magnet, but at the edges of the magnet. Outside of the magnet, the flux density is seen to fall off rapidly, first showing as a dim red halo, then fading into black at the model boundary. The operator can ‘zoom’ the colour sensitivity, to bring out the detail in areas of low field strength, in which case the magnet would ‘saturate’ in bright yellow.Image Bar_modB_plan_view is a plan view, in the X-Z plane, with the 2D slice taken at the vertical centre of the magnet, Y=20mm. Here it is seen that the flux density is highest of all at the vertical edges of the cubic magnet, again, not an intuitive result. This screen shot was taken with the cursor exactly at the centre of the screen, at (20,20,20), as displayed at the bottom left. The permeability (relative to air) is always displayed at the cursor position, here showing a value of 1.069, this being the relative permeability of this grade of NdFeB. It is counterintuitive that a magnetic material should have such a low permeability, almost the same as vacuum, but such is the case for NdFeB.Next posting will show the coloured plots for Bx and magnetic potential, then I’ll move on to building something more exciting, a magnetic monopole.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 04, 2016, 08:54:31 am
content: Image file Bar_Bx shows the strength of the X-component of the flux density vector, where the X-direction is left-to-right, and the magnetization is vertical. Observe two black ‘bars’ passing through the center of the model, one horizontal and one vertical, indicating that the X-component of the field is zero here, or put another way, the field is vertical. If you look back to the first plot a few postings back, showing the field direction at all points in this plane as tilted lines, then you will observe that the field is indeed vertical along these black ’bars’. There is nothing unexpected or exceptional about this plot - I included it mainly because it is rather pretty. Call me strange, but I happen to think that models are often very pretty, and I certainly enjoy playing with them. The fact that I get paid for mucking around with pretty models is icing on the cake.Image file Bar_mag_potential is a plot of the “magnetic potential’ AKA magneto-motive-force, a concept possibly not familiar to most. It is the magnetic analog of electro-motive-force, the magnetic driving force that drives the magnetic flux (Webers) around the magnetic circuit, and is in units of ampere-turns. A permanent magnet is modelled as a large number of ampere-turns, distributed along the length of the magnet in the direction of magnetization. As with potentials in general, we are really only interested in potential differences, and can choose the point of zero magnetic potential for our convenience. The horizontal black ‘bar’ across the centre of the model indicates zero magnetic potential, and the potential built up by the magnet is +ve above the magnet, and symmetrically –ve below it. The colours represent only magnitude, so the coloured plot looks identical above and below the dark bar where the potential is zero. In the screen shot, the cursor was placed just above the top pole piece of the magnet where the potential is greatest, and you can see from the readout at the bottom of the screen that the magnetic potential at this point is 2275 ampere-turns (AT). Just below the lower face of the magnet, the potential is -2275 AT, so the equivalent total number of ampere turns for this physically small (13x13x13mm) NdFEB magnet is an astonishing 2275x2 = 4550 ampere turns. There is no way that you could build a single turn 13x13mm current carrying coil that could carry 4550 amps, so the conclusion here is that at least in smaller sizes, NdFeB permanent magnets will easily outperform any electromagnet than you can build, which is why they are used.Surely the OP must be bored by now?

author: John Heath
author_rank: Frequent Contributor
timestamp: May 05, 2016, 02:22:50 am
content: Quote from: Zeranin on May 04, 2016, 01:45:30 amThe square is fine . In only 3 generations of distance from the center the square will be a golf ball. A surprisingly fast transition from square to a circle but I have empirical evidence that this will happen. If you measure I B I Tesla diagonally 2 pi will jump off the page and bite you on the nose. This comes from the Dave spirit of " don't talk about it , build it and measure ".Pay close attention to the last 1/3 of the video where 2 pi is right in your face from a decidedly square matrix. How cool is that !You have put too much on the table to respond to your other posts. Prefer to read with the benefit of time before responding. Do not want to push my luck but we need either the right or left blue boundary impedance to a magnetic field set very low to simulate a fridge wall. It is a given that two monopoles will act like two + charges but if it sticks to a fridge all is lost. The monopole must not stick to a fridge. Would not ask you to compromise your scientific integrity however there is a possibility I could jump off my balcony if the simulated monopole sticks to a fridge. It's only the second floor but hey I could sprain an ankle.

author: IanB
author_rank: Super Contributor
timestamp: May 05, 2016, 03:12:10 am
content: This is a long and wordy thread, and very difficult to read and comprehend everything that has been said. However, if I may jump in at the end, and forgive me if this has already been said, but I see one presumption which I think is (or may be) wrong. The presumption is that an electrometer indicates charge. It doesn't, it indicates a electric field gradient. Under prescribed circumstances an electrometer may be used to measure charge, but that's not what it indicates.By analogy, consider a spring scale for measuring weight (or force). The pointer on the scaleindicatesdisplacement of the spring, which we hopemeasuresthe weight (mass) of the thing being weighed, but only under appropriate circumstances. If you hang a 1 kg weight on the scale it will indicate 1 kg, but if you submerge the weight in a bucket of water it will no longer indicate 1 kg. The mass of the weight hasn't changed, but the indication on the scale has.So coming back to the analogy, if we take a gold leaf electrometer in free space and apply a charge to it, the charge will produce a non-uniform electric field around the device, and the field gradient will cause the electrometer leaves to separate. However, if we bring the surroundings to the same potential as the electrometer there will no longer be a field gradient and the electrometer will indicate nothing.This I think is the solution to the conundrum. The electrometer can indicate the local electric field gradient produced if the electrometer has a different potential from its surroundings. The difference in potential can be produced by adding or removing charge from the electrometer, but the charge itself is just the agent of change, it is not itself the thing that is indicated.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 05, 2016, 05:26:53 am
content: Quote from: IanB on May 05, 2016, 03:12:10 amGreat to have you in on the discussions. Just to be clear about what you are saying, do you mean theelectric field gradient, in units of V/m/m, or do you mean theelectric potential gradient, AKA the electric field, in units of V/m?Regardless, I maintain that fundamentally a gold-leaf electroscope measures the excess charge in Coulombs on each leaf, and that the repulsive force between the leaves is proportional to Q1xQ2/R^2, as per Coulomb's Law, where Q1 and Q2 are the excess (equal) charge on each leaf, and R is the leaf separation.That said, your way of looking at it is really just the same thing, for you cannot have excess charge on the leaves without an electric field extending outward from the leaves. We are talking about 2 sides of the same coin.You refer to the electroscope being at a different potential to it's surroundings in order to deflect, but in one sense that is not true. If the electroscope was the only object in the Universe, in which case there are no physical 'surroundings', the electroscope will still operate correctly, and it will indicate excess charge on the leaves, just as it always does. That said, there will also be an electric field surrounding the electroscope, dropping off as 1/R^2 at distances large compared to the dimensions of the electroscope, but most people (me included) would say that this field is a consequence of the electroscope having excess charge.If there are physical surroundings at a particular potential (volts) with respect to the electroscope, then the excess charge on the electroscope is given by Q=CV, where C is the capacitance between the electroscope and the surroundings. Then, the force on the leaves can be calculated from Coulomb's law, knowing this excess charge. Yet again, it seems to me that what is fundamentally creating the force on the leaves is the excess charge on the leaves, which in this case has been 'pushed' there by the potential difference between electroscope and surroundings.Keep in mind that we can create electric fields and field gradients by all manner of electrostatic apparatus, such as the accelerating stack in a CRT, or a particle accelerator. You can place an electroscope smack in the middle of such a field or field gradient, but it won't register anything unless there is excess charge on the leaves, adding weight to my argument that fundamentally the force on the leaves is as a result of excess charge on the leaves.Mostly, you and I are looking at 2 sides of the same coin, but you may prefer looking at the other side of the coin. Are we in agreement?

author: Zeranin
author_rank: Regular Contributor
timestamp: May 05, 2016, 06:56:09 am
content: OK JohnHeath, it’s time to build (inside my PC) your magnetic monopole, by creating a cube, where each face is a NdFeB permanent magnet, magnetized with the South pole facing outward on every face, evidently producing a magnetic monopole.First though, I need to talk to you sternly about the Law. It seems to me that you are a habitual Law Breaker, or at least you try your best to be. You need to know that if you break the Law, then there will be consequences. As you probably know, there is a Law against creating magnetic monopoles, actually one of Maxwell’s Laws. You have convinced me to try to beak this Law, so we should expect consequences. Perhaps the FEA program will crash, or perhaps grind on forever trying to reach a solution. Or maybe my PC will explode in a ball of fire, or disappear in a puff of smoke. Or maybe the program will correctly model the monopole, but the result will be unlike what you expect. I have never before attempted to model a magnetic monopole, but I’m expecting trouble, as every Law breaker should.As a precursor to creating the cube, I have modelled a single face of the cube, so we can see how much magnetic flux emanates from this single face. Then I’ll assemble 6 such faces to form our monopole cube.The face is 2mm thick (2 grid squares), magnetized in the direction of the 2mm. The attached image file is a 2D slice through this face, so you can see it is indeed 2mmm thick, and 23mm across. The edges are beveled so that the 6 faces can fit together to form a cube.The plot is exactly what you would expect, with the magnet driving the magnetic flux upwards, with the flux then spreading out and eventually returning to the bottom of the magnet. At a distance 2mm above the top surface, the average flux density is about 0.12 Tesla. Keep that number in the back of your mind. For a successful monopole, we need to have a magnetic flux of this order emanating outward from every face of the cube.This evening I’ll model the full cube, and set the FEA program running. Assuming no injury from an exploding PC, I’ll then report the result back here. I also look forward to viewing the infinite resistor matrix video that you attached.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 05, 2016, 12:15:42 pm
content: I see. Build one side first then add the other sides later. This way if the program freezes you can always remove one side of the monopole to see what went wrong. There is a trick I used solve Koide formulas that can often freeze windows.Set timer to 100On timer end programThis way you do not have to reset your compute if Mr monopole runs into infinity problems. Look forward to your test results and thanks for taking the time for this.

author: monkeysuncle
author_rank: Contributor
timestamp: May 05, 2016, 01:28:31 pm
content: Lots of great discussion of the physics here. I wanted to share a philosophical thought.The OP wanted to know if there was such a thing as a "voltage frame of reference," analogous to a relativistic frame of reference for motion.My father, who was a physics professor at an engineering college, once looked at me very darkly when I asked him a similar question. "Analogy has no place in science," he told me.I objected that that wasn't true. Everyone who has taken freshman physics knows that an LC tank circuit is analogous to a mass-spring system.He explained that that "analogy" works only because the dynamic equations that describe the two systems happen to have identical form. If you want to assert that two physical systems are analogous, you must be absolutely clear what you are asserting in mathematical terms. Otherwise you haven't asserted anything that has any physical meaning.With respect to the current discussion, it is true that a physical system will have the same behavior if the potential of the entire system is raised or lowered by an arbitrary amount (relative, let us say, to zero at infinity). That includes not only the potentials on the components of the system, but also the potential function that exists in space surrounding the components.The principle at work here, however, is not the principle of relativity, but the principle of linear superposition.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 06, 2016, 01:21:08 am
content: Well, here it is, a perfectly constructed ‘magnetic monopole cube’, each face magnetized such that the South pole is on the exterior surface. The attached plot is a 2D cross section through the cube, so you can see 4 of the faces in cross section. Each face is plotted in a different colour, signifying that the direction of magnetization is different for each face. The FEA program chugged away and found the solution, meaning it solved for the magnetic flux density, B, at every point in space. No drama of any sort. To see the image in sufficient detail, you may need to click on the filename and open it and view it full size in a proper jpeg viewer, rather than just clicking on the small icon of the image.Overlaid on the same attached plot are the magnetic field direction vectors at every point in space on the 2D slice that is being viewed. In my previous posting I attached an identical plot showing the field vectors when only one face (the top orange face) is present, so you already know from that plot that each side is a powerful and effective magnet, that really pushes the magnetic flux through the space around the magnet, achieving a flux density of 0.12 tesla a few mm from the surface.OK. Looking at the attached plot of the cube, we straight away we can see that something odd is going on here, as predicted, because we are attempting to build something that Maxwell says cannot exist. Rather than plot field vectors, the program has instead plotted a matrix of little crosses, actually ‘+’ signs. Why so? Why has the program refused to plot the field vectors? It’s a fairly smart program, and it will not (indeed cannot) plot something that does not exist. The FEA program has correctly calculated the Bfield at every point in space, no problems, and found that the said field strength is precisely zero, everywhere. You can’t plot the direction of a zero length vector, so the program instead plots little ‘+’ signs, informing the user that the field strength is zero. Although we are viewing a single 2D slice through the center of the model, I can choose any and every 2D slice within the model and the magnetic field strength, B, is zero everywhere, both outside the cube, inside the cube, and inside the NDFeB magnetic material.If you were to actually build such a cube of magnets, or a sphere, it would behave as if made fromunmagnetizedNdFeB. It would not stick to your fridge, and two such ‘monopoles’ would neither attract nor repel each other, any more than two plastic golf balls. You could get your best magnetometer probe and wave it around everywhere, including touching the surface of the magnets, and measure nothing at all. Truth stranger than fiction.Maybe we should not be surprised at the result, we did after all attempt to break the Law, but it is nonetheless an unusual and unintuitive result. Anyone who has ever played with NdFeB magnets knows how incredibly powerful they are, to the point where it can be very difficult to separate such a magnet from a piece of steel, once attached. And yet, if you take 6 such magnets and arrange them into a ‘monopole’ cube, they will become completely ‘dead’. Very cute, I reckon. If you were to build such a cube, demonstrate to your mate that your metal cube is magnetically as dead as a dodo, and then tell him it was made from powerful NdfeB magnets, he probably wouldn’t believe you. On a practical note, if you were to actually build such a cube, then if you later disassembled the cube, you may find that the magnets arepermanentlydemagnetized to some extent, because when assembled in this configuration, each magnet experiences a massive demagnetizing field provided by the other five. Actually, if you were to build the cube with one side missing, then in terms of the field produced, this ensemble of 5 magnets would behaveidenticallyto the single missing side. Very cool.For anyone that feels tempted to build such a cube, be aware that the commonly available rectangular shaped magnets won’t do the trick, because they won’t properly fit together at the edges of the cube, you would need properly beveled edges. Also, for total field cancellation, all 6 magnets would need to be ofidenticalstrength, and in the real world they won’t be. The FEA result is correct, and tells us that the more ‘perfectly’ we build such a ‘monopole’, the more it will converge to producing zero magnetic field at all points in space. Of course, if we build it imperfectly, then it will behave as a (weak) conventional magnet, not as a monopole. The magnetic monopole remains as elusive as ever.In the next posting I’ll include a plot of the magnetic potential produced by our ‘monopole’ cube, and discuss why this cube behaves in the way that it does, using a number of independent arguments.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 06, 2016, 06:42:03 am
content: Surprise and fascinating. I have a question.Will the cube be stressed outward trying to be a sphere rather than a square? There are many more questions but that is the important one.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 06, 2016, 06:58:06 am
content: Quote from: John Heath on May 06, 2016, 06:42:03 amHmm. Good question. Intuition tells me 'Yes', and we know that the first few magnets will repel one another like crazy as we bring the North poles together to form the interior surface. So I'll say yes, when the cube is assembled, all faces will experience an outward force. If I change my mind, I'll let you know.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 06, 2016, 12:12:41 pm
content: I have attached a plot of ‘magnetic potential’ for our ‘monopole cube’ just for completeness. Boring plot, isn’t it. Outside of the cube is black, meaning the magnetic potential is uniformly zero. Inside of the cube, the magnetic potential is very high at 1782 ampere-turns (yellow colour), but equally uniform. The uniformity of magnetic potential tells us that there will be no flux ‘flowing’ from one place to another, and thus the flux density will be zero. There is, however, a large difference in magnetic potential (AKA magneto-motive-force) between the inside and outside of the cube. The magnets can be seen as a sort of ‘magnetic battery’ that provides this difference in magnetic potential from one end of the magnet to the other, so we certainly expect to see the inside of the cube at a different magnetic potential to the outside. If there was a hole in the cube, this considerable difference in potential would drive flux through the hole, creating a ‘fringing leakage flux’ from one pole back to the other. However, as there are no such holes in this cube, or a sphere, there is simply no path for flux to return from outside to inside, and the flux density is zero. The coercive force of the NdFeB material that I have modelled is 891000 ampere-turns per meter length of magnet, in the direction of magnetization. Therefore, as these magnets are 2mm thick in their direction of magnetization, we should expect them to produce a magneto-motive-force of 891000 x 0.002 = 1782 AT. When I took this screen shot, the cursor was in the center of the cube, and at the bottom of the screen it shows that the magnetic potential at this point (or anywhere within the cube) is indeed 1782 AT. This FEA stuff really does work, and I hope some have found it of interest.The simplest explanation of why a ‘monopole’ cannot exist is to examine one of Maxwell’s Equations, sometimes known as Gauss’s Law for magnetics, which states that the total net flux leaving a closed volume must equal zero. By definition, a magnetic monopole would have a net flux emanating from it’s surface, so a magnetic monopole would violate this Law. FEA programs make use of this Law, so it is impossible to create an FEA magnetic model of a structure that violates this law, and so it is impossible right from the outset to create a magnetic monopole with my FEA program. That said, I wanted to see what happened when I tried, so I went ahead and modelled it anyway.However,it’s quite OK to have equal amounts of flux leaving and entering the volume, as the Law in no way says that the flux densityon any given partof a closed surface must be zero. What my FEA program actually predicts for the cube is that the flux density, B, will be zeroat all pointsin space, but the Law does not in itself require that to be true, for the Law would still be satisfied if the flux left the cube at some places, such as in the centre of the faces, and returned at other places such as near the corners. In the simpler case of a permanent magnetsphere, it MUST be the case that B is zero everywhere, because symmetry dictates that there can be no place(s) on the sphere where the flux would preferentially choose to leave or enter, but you can’t use that argument for the cube.Here is an elegant explanation of why our cube produces zero magnetic field everywhere. You will need to take my word for it that a thin, planar permanent magnet can be modelled as a current-carrying loop around the perimeter. Assume for the moment that the permanent magnet faces of the cube are ‘very thin’, so we replace each of them with a square loop of very-small-cross-section current-carrying wire, a single-turn square winding if you like. So the cube is formed by 6 such square loops of wire. Each side of the cube will therefore consist of 2 lengths of wire, one from each adjacent loop, but here is the trick. If the direction of current in each loop is chosen so that the resultant magnetic fields all point outward, then you will find (easily drawn as a sketch on a piece of paper) that on every edge, the current flows in opposite direction in the two wires, and thus no magnetic field is produced, because we all know that the net magnetic field from 2 counter-current , very thin, co-located parallel conductors is zero.In practice, the sides of the cube are not infinitely ‘thin’ actually 2mm thick in my FEA model, but this does not change the argument. All you need to do is stack a large number of ‘very thin’ magnets together to get the thickness you want, and the field from each and every one of them still cancels, as already described. To get the sides to fit exactly you need to bevel the edges, but that does not alter the argument either, just means that each ‘thin’ square face is a slightly different dimension. Thus we see that our square ‘monopole’ cube is expected to produce zero fieldeverywhere, exactly as shown by the FEA, even though the Law would have permitted flux to leave the square faces in one place and return in another. This ‘square-winding-model’ also explains why if you remove one face, then the remaining 5 behave identically to the one that was removed, except of opposite sign. Try it on a sketch. It’s true, and cute as.I find the above ‘square-winding’ argument to be rather beautiful, and it gives me great pleasure to find that the FEA model produces the same result, of zero field everywhere.Interestingly, the same argument predicts that we can make our cube of different width, length and height, and still it will produce zero magnetic field everywhere. Anyone care to agree or disagree, or do I need to ‘stretch’ my FEA modelled cube to find out.?Taking it a step even further, I’ll stick my neck out and say we can build a permanent magnet ‘monopole’ ofany shape at all, and in all cases it will produce zero field at all points in space, just provided that the magnetization vector is always at right angles to the surface, and every face is the same thickness. You agree with that, John Heath?

author: John Heath
author_rank: Frequent Contributor
timestamp: May 07, 2016, 02:59:00 am
content: I am getting a sense you want myself to disagree with you. You have a nagging doubt. Hey join the club. I will do my best to make a counter argument for what its worth. In an earlier screen shot of a square magnet north top south bottom the corners have the highest magnetic density becoming weaker towards the center of the square magnet. This is sounding somewhat similar but opposite to Mr Monopole. I will now use your symmetry argument as my counter argument. Your elegant analogy of a thin wire square loop is indeed exactly the opposite to the current edge of the other monopole edge but this ignores the center of the monopole plate. The thin wire square loop should spiral round and round inward to the center of the monopplole wall to justify a uniform magnetic density. However the center of the monopole wall is far away from the edge where it's opposite counterpart lives. For this reason the magnetic field in the center of the wall should express itself unlike the edge of the wall that can not as the other walls cancel it out. This would be my symmetry counter argument.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 07, 2016, 11:01:50 am
content: Quote from: John Heath on May 07, 2016, 02:59:00 amI have no 'doubts', nagging or otherwise, just find the topic interesting. It is true that I've never modelled a system of magnets that produces zero magnetic field everywhere, but then again, I never tried to model an apparent monopole either.My method of modelling permanent magnets is correct, but I understand what you are saying as to why it does not seem intuitively right. The following argument is both elegant and intuitive.In reality, you would expect that a thin permanent magnet is in effect composed of a very large number of tiny current-carrying coils uniformly occupying the entire area of the magnet, not just on the outside as my model appears to be. Agreed.OK, get a piece of paper and sketch it. Draw a square, say 100x100mm representing a thin, square magnet, magnetized through the thickness. Now draw a matrix of small, square current-carrying coils, for example a 5x5 matrix, so each of the 25 small squares will be 20 x 20mm. Of course, the current around each square loop will be in the same direction, say clockwise. Draw arrows showing the direction of current, on each side of each square. About now, a light bulb will be turning on inside your head. All of the current-carrying conductors inside the square cancel out,except for the sides that form the outside of the 100x100 mm square, and this will always be true regardless of how small we make each current loop, right down to the size of atoms. Nature is very beautiful, is she not? So now you can see why it is valid to model a permanent magnet as a thin sheet of current travelling around the outside surface. Quite literally, if you were to build the matrix of 25 squares and pump a current I through all of them, then the field produced would in every way be identical to the field produced by passing the same current through a single square 100 x 100 loop.The FEA model and result are correct, I guarantee it. Just for fun, I might punch a small hole through the centre of one face on our FEA 'monopole' cube, and we should see a torrent of flux squirting through the hole, from the inside of the cube to the outside, as a result of the large magneto-motive-force between inside and outside, as shown on one of my previous plots. That will convince you that the permanent magnet faces are well and truly alive and kicking, including in the center of the face, and yet plug up the hole and you get zilch, nothing, no flux density anywhere and everywhere, even for shapes that lack the perfect symmetry of a sphere. Nature is beautiful.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 07, 2016, 11:41:42 am
content: Quote from: Zeranin on May 07, 2016, 11:01:50 amAnd here it is, a real 'flux cannon', sort of like punching a hole in a hollow cube with high pressure gas inside. If you remember back to where I modelled the top face in isolation, the flux density a couple of millimetres above the face (pole piece) was 0.15 Tesla.When this screen shot was taken I had the cursor right in the middle of the hole, so as to read out the flux density at that point. If you look at the bottom of the plot you can see that the flux density inside the hole is 0.49 Tesla, much higher than 0.15 Tesla the top face was able to produce on it's own. Very cute. The flux density does fall off very rapidly beyond the hole though.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 07, 2016, 01:59:44 pm
content: Quote from: Zeranin on May 07, 2016, 11:01:50 amYou misunderstand. There can be no doubt to the screen shot model as that is based on known laws of physics. The doubt comes from the counter intuitive picture the known laws of physics paints. You seemed frustrated so I presented a counter argument as a possible way out that being Q X Q / R^2 with focus R^2.Your rebuttal that every square in the screen shot may be considered 1 atom with electron rotating clockwise is more precise than my wire current spiraling towards the center of a monopole wall. However there is a price to be payed for this analogy. If I have an infinitely large 2 dimensional wall where every atom is oriented the same way with all electron rotating clockwise there can not be a magnetic field as all edges of all atom electron obits cancel out the neighboring atoms as per your argument. There in is the rub ,, I think.A hole in a monopole " flux cannon " , well named

author: John Heath
author_rank: Frequent Contributor
timestamp: May 07, 2016, 09:03:18 pm
content: Just happened to have a few neodymium iron boron magnets around ear marked for a Faraday rotator. In the spirit of Dave's " don't talk about it , build it and measure " it would suffice to say I now have a 2 inch , 50 mm , square monopole cube. If I open the front door to the cube the outward force is great enough to keep the door open under it's own gravity weight when facing up. Your guess that there is an outward force is confirmed. However it is not a perfect monopole as the magnets are smaller 20 mm on a 50 mm perf board. I think the real monopole would only be a small 5 m cube in the center of the box. Would like to FEA model this box to see how close it comes to a real monopole. Will take pictures and specifications for modeling , if you have to time.

author: IanB
author_rank: Super Contributor
timestamp: May 07, 2016, 10:36:36 pm
content: Quote from: Zeranin on May 05, 2016, 05:26:53 amI think I mean the electric field, or potential gradient.QuoteThe question now becomes excess charge relative to what? After all, both leaves are at the same potential and so neither can have excess charge relative to the other. I think it is only possible for the whole instrument to have an excess of charge relative to something outside of it, essentially "the surroundings".QuoteThis for me is the crucial point. In a sense, Coulumb's Law is an outcome, not a first principle. The first principle is that a charged particle within an electric field experiences a force. If two charged bodies are close to each other, then body A produces an electric field and body B within this field experiences a force. Likewise, body B produces a field and body A within B's field experiences a force. If the two bodies are in free space with no interference from other charged bodies or external fields nearby then Coloumb's Law describes what happens. However, if there is a third charged body nearby, or if there are external electric fields, then the situation becomes more complicated.QuoteI think that if the electroscope were the only thing in the universe, then the "surroundings" are infinite space, which has a potential of zero at infinity. Then the field drops off towards zero as it extends away from the electroscope. Since both leaves are at the some potential there is no potential gradient and thus no field between them, only a field around them dropping away towards infinity. Therefore the leaves don't respond to each other, but each responds independently to the unsymmetrical field surrounding them which they both have created.If we were to bring the surroundings to the same potential as the electroscope then there would be a flat field all around it with no potential gradients. In this case the leaves would experience no force and would not separate.QuoteI guess I do prefer looking at the other side of the coin. I think this is because in any arbitrarily complex system the best way to solve it will be to compute the shape of the electric field as a vector field and then compute the force vector acting upon each charged body within this field.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 08, 2016, 12:05:53 am
content: Quote from: IanB on May 07, 2016, 10:36:36 pmNo!Excess charge is excess charge, and is not relative to anything. An object with no excess charge (net charge is you prefer) has equal numbers of +ve and -ve charges, for example, equal numbers of electrons and protons. This is purely an accounting exercise, and is not relative to anything. Similarly, an electron has an absolute net charge, that can be expressed in Coulombs. I can address your other points as necessary, but some may now have been 'neutralized', if you'll pardon the pun.

author: IanB
author_rank: Super Contributor
timestamp: May 08, 2016, 12:31:09 am
content: Quote from: Zeranin on May 08, 2016, 12:05:53 amYes, an electron has a charge of -1 unit. I do not dispute physics.But I still believe that the electrometer does not measure absolute charge, it measures relative charge when compared to its surroundings. If we charge up an electrometer to an absolute potential of 100 V and then place it in surroundings which are uniformly at a potential of 100 V, then I do not think the electrometer will show any deflection. If the electrometer showed a deflection then I think we would have a way of measuring absolute potential, and I do not think this is possible.(Similarly, if we consider any two adjacent electrons in a sea of electrons, where all electrons are equidistant from each other, than no adjacent pair of electrons will experience any repulsive force.)

author: Zeranin
author_rank: Regular Contributor
timestamp: May 08, 2016, 12:37:38 am
content: Quote from: John Heath on May 07, 2016, 01:59:44 pmThe real problem is that you are constantly trying to break the Laws, but far from being 'frustrated' by this I regard it as agood thing. Good scientists are always questioning the Laws and trying to break them, that is how science progresses. It does seem unlikely that very well established Laws such as Maxwell's Equations could be broken by playing around with a few permanent magnets on the kitchen table, but it sure is fun trying.QuoteThere is no problem here. What I actually said, is that the net effect of all those current loops in the centre is the same as a single current loop around the outside, and therefore you are wrong to saythere can not be a magnetic field. The large current loop around the outside most certainlydoescreate a magnetic field, and it does so at all points within the loop. FEA is one method of calculating the field produced by a current carrying loop, but the traditional and more fundamental method is to use the Biot Law. Basically, all magnetic fields are produced by moving charge. The Biot Law describes the field produced by a single moving charge, or if you prefer, an infinitely short length of current carrying conductor. If you want to know the field produced by a current carrying loop of any shape, then all you need to do is add up the field contributions from all the short lengths of conductor that make up the loop. Sometimes the resultant integrals are easily soluble, and sometimes not. I also have a program that numerically calculates the magnetic field in 3D space from any arrangement of coils and current carrying conductors, using the Biot Law, and it always gives the same results as my FEA program. However, the Biot Law is no longer applicable when permeable magnetic materials are present, whereas FEA can handle the more general case of any structure consisting of current carrying coils and permeable magnetic materials.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 08, 2016, 01:06:14 am
content: Quote from: John Heath on May 07, 2016, 09:03:18 pmSure, I'm happy to FEA model it. Presumably it's constructed from rectangular shaped NdFeB magnets? The great thing about FEA is that you can build something without actually building it, the best of both worlds.My 'guess' of an outward force on each face was based on good physics. If we use the 'square-current-carrying-loop' model for our monopole cube, then we agree that on every edge we have a pair of currents flowing in opposite directions. When you have 2 parallel current carrying conductors, they exert a force on one another, and this force is repulsive if the currents are in opposite directions, thus the force on each face must be outwards repulsive. Nature, beautiful one day, perfect the next.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 08, 2016, 04:41:29 am
content: Quote from: IanB on May 08, 2016, 12:31:09 amHere you are appear to be mixing up charge and potential. An electroscope does indeed measure the absolute excess charge on the leaves. There is no such thing as 'relative charge compared to surroundings', though it's fine to talk about the electricalpotentialof something relative to something else.When you talk of 'charging an electroscope to anabsolutepotential of 100V', then you need to be clear about exactly what you mean. This refers to the situation where the electroscope is the only object in the universe, and charging it to anabsolute100V means adding excess charge such that that 100 Joules of energy would be either gained or lost when bringing 1 Coulomb of charge from an infinite distance, up to the electroscope, noting that Volts are in units of J/C. Sure, you can charge your electroscope in that way, OR you can charge it to 100Vrelative to it's surroundings, which would typically mean connecting a 100V battery between mother earth and the electroscope terminal, let's call this Case B. In this case, it would take 100 Joules per Coulomb to bring additional charge from earth to the electroscope terminal. In both charging scenarios, the electroscope will receive excess charge, but not by the same amount. In Case B, the amount of excess charge received is given by Q=CV, where C is the capacitance between electroscope and surroundings.Either way, your electroscope will receive excess charge, AKA it will be charged, and the leaves will be deflected. Now remove any connection from the electroscope terminal, so that the amount of excess charge is trapped on the electroscope. Now place the electroscope in a Faraday cage that is at a potential of 100V, ie, one side of a 100V battery is connected to Earth, and the other to the cage. The electroscope deflection will remain unchanged, because the electroscope fundamentally responds to the excess charge on it's leaves, and this has not changed.Now to really confuse you, charge up your electroscope by whatever means you choose, and remove any connection from it's terminal. Now, by whatever means you choose, charge up a sphere on the end of an insulating rod (ie give it excess charge), and bring the sphere close to the electroscope terminal, but not touching it. The electrometer will respond, with the leaves falling or rising higher, depending on the sign of the excess charge on the sphere, although no charge has moved from the sphere to the electrometer. Aha, I hear you say, gotcha there Zeranin, no change in electroscope charge, and yet it responds! But what is actually happening here is called induced charge. If the sphere is positively charged, then it will attract electrons to the electroscope terminal, from the gold leaves. Thus the excess charge on the gold leaves is changed, and the leaves respond accordingly. In every case, the electrometer leaves rise or fall depending on the amount of excess charge on the leaves. That is fundamentally what an electroscope measures, excess charge on the leaves, and everything else gets down to a discussion about different methods of producing an excess charge on the said leaves.QuoteNot true. If you have a dense bunch of electrons in empty space, then they are indeed repelled from each other, and fly apart. This effect is known as 'space charge', and it is a pain in the butt for electrostatic designers such as me that design electron guns and lenses and the like, because it means that when you have a high-current electron beam at low energy (low speed), the beam 'blows apart' rather than staying nicely focussed as you would like. Standard FEA is concerned only with voltages between electrodes, and therefore does not account for space charge effects, but space charge can be accounted for separately in the modelling.

author: IanB
author_rank: Super Contributor
timestamp: May 08, 2016, 06:19:21 am
content: Quote from: Zeranin on May 08, 2016, 04:41:29 amA change in potential is produced by a spacial displacement of charge. The two are inextricably linked, are they not?QuoteOK. I can't do the theoretical derivation to prove this for myself, so I will accept it as correct.QuoteYes, you are right. I was thinking of the electroscope being in free space and far from anything else.QuoteI think this is because inside the Faraday cage there is a uniform electric field, and therefore no external field gradients to influence the electroscope.However, believe that the electroscope inside the Faraday cage forms one plate of a capacitor, the Faraday cage the other. If we raise or lower the potential of the Faraday cage from outside then the potential of the electroscope will change by the same amount, since the potential difference between plates of a capacitor fixed in space remains unchanged unless there is a displacement of charge between them. In order to achieve the desired goal of bringing the electroscope and the Faraday cage to the same relative potential we need to transfer charge between them until their potentials are equalized. Once we do this the electroscope leaves will no longer deflect, even if the system of Faraday cage and electroscope has a huge excess charge and has a high potential difference relative to the surroundings.QuoteNo confusion here.QuoteWhat I meant was a sea of electrons, uniform and infinite in extent. If there is any boundary then uniformity is lost and the equidistant property no longer holds.

author: Rerouter
author_rank: Super Contributor
timestamp: May 08, 2016, 06:38:06 am
content: Hmm at a basis, the original question could likely be solved through indirect measures, a 10KV charged cage would cause ionizing in the surrounding air, so by measuring the difference in gas composition he could come up with a rough basis for the voltage to the outside reference frame, with nothing but the air moving through the holes in the cage,Next up, again probably not that cheap of a way to accomplish it, measure the energy input and wavelength spectra out of a laser, with a higher voltage reference frame the atoms already have higher energy levels, so you should see a tiny amount of higher energy wavelengths compared to the outside measurement, (This one is me guessing from previous knowledge)

author: Zeranin
author_rank: Regular Contributor
timestamp: May 08, 2016, 10:43:59 am
content: Quote from: IanB on May 08, 2016, 06:19:21 amMore correctly, inside the Faraday cage iszeroelectric field, rather than a uniform electric field, but I think you were just being careless here in your wording, rather than in error as such. I think we agree that if the Faraday cage is highly charged (large excess charge), the electroscope sitting inside will not have an excess charge (or deflection) when it's terminal is connected to the cage, because all of the excess charge resides on theoutsideof the cage/sphere, again illustrating my point that the electroscope responds to excess charge on it's leaves.QuoteThe two are linked, yes. However, if I connect a capacitor across my favourite power supply, and then crank up the supply voltage, then most people would say that the movement of charge was as a result of changing the potential, not the other way around. In other cases, such as a Van der Graaf generator I agree that a change in potential can be produced by physically displacing charge.Apart from those minor points, I agree with all your entire posting.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 08, 2016, 02:17:52 pm
content: Quote from: Rerouter on May 08, 2016, 06:38:06 amEnergy input to spectra out of a laser or frequency = energy / h . Cool idea and thanks for suggesting it. I tried it fell fell right on my face. By trying I mean googled the pieces out of lasers. Turns out laser are not like atomic clocks despite the fact that both are atomic events. What sets the high Q bandwidth of a laser is the resonate optical cavity after the photon source. This being the case the optical resonate cavity is subject to imperfection in the reflecting mirrors and expansion temperature coefficient of the cavity itself. At hundreds of THz just looking at it will cause too much of a change in the 1 KHz range. In short too unstable to measure a relativistic change in time for a 10 K volt change in a voltage frame of reference but thanks for bringing it up.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 08, 2016, 03:20:15 pm
content: Great questions coming up from new blood in the thread. Great stuff. Have the monopole cube pictures.The magnets are round however if made square the PEA rendering should be in the ballpark considering they are only 1/2 the size of the cube walls.Spec:A] Magnets are about 3 Gauss at 2 inches , 50 mm. The 6 magnets are within in 15 percent of each other. North faces in with south on the outside.B] Your outside single loop current magnet model would be in the range of a few thousand amps with diameter of 20 m as the magnet diameter is 20 m.When cube door is closed the outside Gauss at 2 inches is 2 Gauss however when door closed the Gauss reading at 2 inches is 1 Gauss +- 20 percent. Not a problem for you however I will have to eat a little crow and take a second look at your thin wire around the edge as the better magnet model. I suspect a open door and closed door modeling of the cube will verify this. If this is the case then a perfect square magnets the same size as the cube walls will have no magnetic field on the outside. This would mean a monopole can not have a magnetic field. Put another way you can not have a north pole without a south pole.A problem popped up. Wghen the c

author: Zeranin
author_rank: Regular Contributor
timestamp: May 09, 2016, 06:50:29 am
content: Quote from: John Heath on May 08, 2016, 03:20:15 pmHey, I love the way you are using a clip-on ammeter to measure the equivalent circulating current of the permanent magnet! Did it drive your ammeter off scale? If you put one of your magnets directly in the jaw-gap of your ammeter with no air gap, I’ll bet it will drive it hard off scale, and possibly damage the ammeter as well!I didn’t know it was possible to build a ‘monopole’ so roughly, but fair enuff just for playing around I suppose. If you really want to approximate a ‘proper’ monopole cube, then you will need to get the magnets much closer together. The magnets are 20mm diameter, right? OK, then make yourself a 20mm cube out of anything convenient, wood would be fine, and then glue or screw one magnet to each face. That would be much better than what you have now, and you would find that the flux density(Gauss) outside of the cube would be quite small, as the closer you get to building the monopole perfectly, the closer the flux produced will approach zero. The conclusion is that you cannot build a monopole, but what you do end up building (if you build it properly) is quite cute in it’s own right, a structure of magnets that produces zero field everywhere, quite novel.How thick are your magnets, and what is the size of the hole through the centre? I’m happy to FEA model it.QuoteI don’t understand. You are saying that when the door is closed (all 6 faces present) you measure 2 Gauss at 2 inches from the cube, but when the door is closed, you measure 1 Gauss +-20% at 2 inches. Um What!!!QuoteLooks like your PC exploded. I did warn you about such things happening when you try to break the Laws of Physics!

author: Zeranin
author_rank: Regular Contributor
timestamp: May 09, 2016, 11:41:09 am
content: The attached images are from FEA modelling John's rather 'rough' cube of magnets that he constructed. I didn't have the exact dimensions, but that doesn't really matter. The plots show clearly the kind of magnetic behaviour that such an assembly of 6 permanent magnets will produce.I have modelled a single side, and the flux density 42mm above the face is approximately 9 Gauss. When the entire cube of magnets is assembled, the flux density 42mm above this same face is only 3.4 Gauss, showing as expected that once you start to assemble a cube of magnets in this way, the fields start to cancel, until in the limiting case of perfectly constructing the cube of magnets, you get zero field everywhere, perfect cancellation.In the image monopole_rough_vectors, you can see clearly the paths and directions that the magnetic flux is taking. Starting out from any of the outward-facing poles, the flux always eventually curves around and returns to the opposite pole on the inside of the cube, but it is only able to do this because the magnets are smaller than the faces of the cube, providing plenty of space for the said flux to find it's way back to the opposite pole on the inside.The image monopole_rough_modB is a coloured contour plot, showing the field strength at every point. As you would expect, the flux density is highest near the magnets, rapidly falling away as you get further from the cube.With the benefit of FEA generated plots such as these, it is actually quite easy to understand and visualize the flow of magnetic flux for any magnetic structure that you wish to dream up. For me, FEA is both powerful and beautiful.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 09, 2016, 12:06:43 pm
content: Ha , yes mother nature does not care for our monopole so she is taking it out on my computer.The magnet has a 3 m hole and 20 m on the outside with thickness of 3 m .The amp meter is saturating as you suspected. Need to calibrate at 1 amp current then move away from the sensor until it reads 10 m amp . By marking that distance as times X100 much like a X10 scope probe. With this in place I should be able to measure the magnetic current.Electrons moving in opposite directions where the two edges of the monopole cube touch should run into each other , scattering , now and then. If this happen the electrons will have to radiate a photon from a change in velocity or direction. That would be free energy. Can not have free energy so something must be stopping it or if it does happen the magnet is demagnitized to even the score energy wise. I remember you hinting at this saying the edges touching could damage the magnets. If it does happen then it should spit out a photon now and then. I wonder if it could be measured?Speaking of energy conservation I have an unrelated question. You design then perfect ion accelerator for NASA making sure the ion stream never hits the electrostatic accelerator electrode. If the accelerator electrodes is never hit by an ion then we do not need a power supply to keep it at 10 K volts . All we need is a condenser charged to 10 K volt and our ion accelerator engine will run for years until we run out of ions. Is this not a violation of energy laws??

author: John Heath
author_rank: Frequent Contributor
timestamp: May 09, 2016, 12:11:32 pm
content: Just received your PEA rendering . Will stew on it . Off to work.

author: IanB
author_rank: Super Contributor
timestamp: May 09, 2016, 03:00:10 pm
content: Quote from: John Heath on May 09, 2016, 12:06:43 pmThat's a bloody big magnet!

author: Zeranin
author_rank: Regular Contributor
timestamp: May 09, 2016, 11:38:02 pm
content: Quote from: John Heath on May 09, 2016, 12:06:43 pmActually that won’t work. You completely change the whole geometry one you move the magnet away from the jaws, or don’t have the jaws properly closed. You are correct in saying that the equivalent surface current of your magnets is around 2000 amps, and that is way too high for your current clamp meter to measure, and that’s all there is to it.QuoteYou misunderstand. What I meant to say was that a permanent magnet (PM)behaves as ifthere was a sheet of current moving around the outside surface, and thus it can be modelled in that way. That said, there is notliterallya sheet of current flowing through the material, if there was then it would get very hot from I^2R heating! Ferrites are modelled the same way, but they are not even electrically conductive in the bulk material. Re demagnetization, if you place any PM in a powerful field that pushes the flux in the opposite direction, then you may demagnetize the PM. NdFeB is fortunately very difficult to demagnetize (and also difficult magnetize, but that’s another story) so in practice I doubt that you would demagnetize the individual magnets if assembled into a perfect ‘monopole cube’, but you would be giving the magnets a ‘hard time’.QuoteAn excellent question, damn you! I need to do some ‘real work’ right now, but I’ll get back to you. I feel like I’m monopolizing the thread – anyone else care to comment on this intriguing question?

author: IanB
author_rank: Super Contributor
timestamp: May 10, 2016, 01:10:07 am
content: Quote from: Zeranin on May 09, 2016, 11:38:02 pmTake the simplest possible charged particle, an electron, and follow it (conceptually) through a closed cycle. The electron starts out at the generator and is accelerated towards the target, picking up energy as it goes. When it hits the target it loses some energy, but it is now in close proximity to the accelerator electrode and thus within its field. Our electron now has to be taken on a journey back to the generator to repeat the cycle. To do this, the electron has to be moved away from the accelerator electrode, against the attractive force that accelerated it on the first part of its journey. Moving the electron in this direction back to the starting point requires work to be done on the electron that balances the energy acquired by the electron when it was accelerated. Thus, energy is conserved.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 10, 2016, 01:43:28 am
content: Quote from: IanB on May 09, 2016, 03:00:10 pmShould have said 3 mm not 3 m . Yes a 30 m magnet would be rather large. Would not fix in a shirt pocket.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 10, 2016, 03:24:10 am
content: Quote from: IanB on May 10, 2016, 01:10:07 amAll true. The classic example is the old CRT based TV. You have an electron 'beam current' that strikes the phosphor and, as you rightly say, the EHT power supply provides the beam current. The power supplied by the EHT supply is the EHT voltage (10's of kV) multiplied by the beam current. All very straightforward. As you say, the EHT current is in a closed loop.However, John is thinking about the much trickier case of an ion-thruster, where the 'ion beam' is shot out the back of the spacescraft, and thus never collected by an electrode, quite unlike the case of a CRT where the beam electrons are collected by the phosphor electrode and returned to the EHT power supply. Much trickier situation indeed.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: May 10, 2016, 04:02:59 am
content: Quote from: Zeranin on May 10, 2016, 03:24:10 amSpacecraft thrusters do shoot a beam out the back. In many (maybe most) cases they shoot an electrically neutral beam consisting of both electrons and positively charged ions. By doing this they avoid accumulating charge on the spacecraft which has a number of benefits. One of which is eliminating the ever increasing "escape velocity" and associated reduction in propulsion efficiency which occurs when you allow charge to accumulate.

author: IanB
author_rank: Super Contributor
timestamp: May 10, 2016, 04:31:44 am
content: As a thought experiment, I would like to imagine an "iron thruster" mounted to a wheeled cart of some kind. This thruster would have a hopper of small iron balls and a system of permanent magnets to accelerate the balls to high velocity. By suitable arrangement of the mechanism, balls would be fed from the hopper, accelerated by the magnets, and ejected out the back of the cart, thus propelling it forwards. Since the thruster is using permanent magnets, no energy input would be required to propel it as long as the supply of balls remains.I do not think such an iron thruster can be constructed. I suspect therefore, that the equivalent "zero power" ion thruster cannot be constructed either.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 10, 2016, 05:56:40 am
content: Quote from: CatalinaWOW on May 10, 2016, 04:02:59 amAll true, but see my reply to IanB.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 10, 2016, 06:01:02 am
content: Quote from: IanB on May 10, 2016, 04:31:44 amA permanent magnet poweredironthruster eh? I like it, and we agree that it cannot be built. However, no one has really answered Johns question. If the ions, or electrons, or both are not actually collected on an electrode (as in a CRT) then how can there be a current flowing on the accelerating electrodes, in which case there is no electrical power needed to run the system. That's what John is asking, and it's a cute question.Let me ask a similar question. As you know, CRTs have a ‘control grid’ so that the electron beam can be turned on and off, indeed, that is how they create an image. OK, start with the electron beam turned off, so there is no electron beam, and no current being drawn from the EHT supply. Now turn on the beam for just a few microseconds, just long enough to release a dense bunch of electrons. The electron bunch will be accelerated towards the phosphor anode, so make no mistake, work is being done in accelerating those electrons. However, the electrons do not touch and are not collected by any electrode, or the phosphor, while they are in transit, so there can be no electrode or EHT phosphor current until the electron bunch actually strikes the phosphor, so during this time there is no energy being provided by the electrical power supplies.Question. Where does the energy come from to accelerate these electrons, while they are in transit, before they strike the phosphor anode? The grey matter is pleasantly tickled.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 10, 2016, 10:01:19 am
content: Quote from: IanB on May 10, 2016, 01:10:07 amI see your point. However in the case of an ion accelerator the ion never returns to put the energy back into the accelerator gun. This means the ion accelerator gun must lose a tiny bit of energy every time an ion is accelerated into space .

author: John Heath
author_rank: Frequent Contributor
timestamp: May 10, 2016, 11:13:19 am
content: My solution for the energy conservation problem for an ion accelerator space ship is to point a finger at where the energy is. If I say the energy is in the concentration of positive charges in the bag of ions to accelerate the ship not the accelerator gun then we are okay. Entropy had to go down to have a bag of ions. If I bust the bag of ions they will spread out in all directions causing entropy to go up. Same applies to a CRT . What business does an electron have going 30 percent c in the middle of a CRT tube if it did not hit the accelerator first anode and is yet to hit the 32 K volt second anode? Neither first nor second anode have been discharged yet Mr electron has the energy to accelerate to 30 percent c. My guess is the energy came from increasing the entropy of the vacuum in the CRT by adding 1 negative electron charge to a vacuum that is decidedly positively charged to 32 K volts.Great PEA rendering . I am getting better at predicting what it will look like. I had the middle part right but goofed up the outside . I was sure the field attenuation would be more round but it turned out to be more of a square with round edges.Mr monopole has been reduced is size from 2 inch square to 1 inch square. A second monopole is on its way that will complement the first , north facing out and south facing out. It is a given that there will not be an attractive force between the + and - monopoles but have to verify.Another unrelated thought. In the 2 dimensional infinite resistor problem 2 pi pops up if a resistors square is measured diagonally. If the outside of the resister matrix is terminated properly to simulate infinity then the diagonal measurement should be pi to the last digit. If true then the formula to calculate the termination resisters would be the first time anyone in the history of man has ever derived ip to the last digit from math alone. This would raise a few eye brows in the mathematicians camp.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 10, 2016, 12:10:02 pm
content: Quote from: John Heath on May 10, 2016, 11:13:19 amLike you, I expect that if we modelled out further that the mod B contours would start to look round.QuoteYes and no. Because it is difficult to accurately build a 'monopole', especially using round magnets, you will not get zero field outside of the cube, though it will be less than for a single magnet alone. The cube with the S poles facing outward will have a total of 6 'South Poles', but will also have 6 North poles - you can see this clearly on the FEA vector plot. Therefore, you will be able to get your to monopoles to weakly attract or repel, depending on their relative orientation.

author: IanB
author_rank: Super Contributor
timestamp: May 10, 2016, 01:29:54 pm
content: Quote from: Zeranin on May 10, 2016, 06:01:02 amIn a CRT a large potential difference is created between the anode and the cathode, producing a strong electric field. It takes a power input to charge up this field, and once in place it stores energy. When electrons travel from the cathode to the anode they are accelerated through the potential gradient created in the field, drawing energy from it, and upon striking the anode they discharge it a little. If we continued without providing a replenishing current between the anode and cathode to keep it charged up the system would become discharged and would stop working. The power input to store potential energy in the electric field is where the energy comes from to accelerate the electrons.

author: RIS
author_rank: Regular Contributor
timestamp: May 10, 2016, 04:39:52 pm
content: Quote from: IanB on May 10, 2016, 01:29:54 pmnow thats a very nice explanation I like it and that would be nice and a closed systembut I wonder Why is the outer glass of the CRT has a charge.I mean that is a technical problem but I somehow see some interaction between the closed and open systems.so What would be a good explanation or opinion for that

author: Zeranin
author_rank: Regular Contributor
timestamp: May 10, 2016, 11:54:17 pm
content: Quote from: IanB on May 10, 2016, 01:29:54 pmYes. We can go further, and calculate the stored energy from E=0.5CV^2, where 'C' is the stray capacitance between the electrodes to which the EHT power supply is connected. So far, so good.QuoteWell yes, but what I actually asked was where the energy came from to accelerate a small bunch of electrons,beforethey strike the anode.QuoteWe all agree with that, but here you are presumably referring to the steady state situation, where electrons are returned to the anode, and then ‘pumped’ back to the cathode by the EHT power supply. My question was, where does the energy come from to accelerate a single bunch of electrons, while they are in transit,beforethey strike the anode.QuoteThat does not really answer the question. Creating the accelerating electric field in the first place is a one-off. Thereafter the electric field strength (V/m) is constant, and does not ‘wind down’ while the bunch of electrons is in transit. Does anyone disagree that theonlyplace that the energy can come from to accelerate our bunch of electrons while in transit is the EHT power supply. Thus the question becomes, how can current be drawn from the EHT power supply during this transit time, when the bunch of electrons does not touch and is not collected by any electrode?Of course, we all understand how the EHT power supply provides the current and power to accelerate CRT electrons when there is a constant electron beam current, but that was never my question.My question relates to the case where the CRT control grid is used to produce a small bunch of electrons, that are then accelerated towards the anode. We all understand that once this bunch of electrons strike the anode, then the EHT power supply will need to provide a small pulse of current (and thus energy) to pump those electrons back to the cathode, but that also has nothing to do with my question.What I am asking, is where does the energy come from to accelerate a single bunch of electrons, while they are in transit, before they strike the phosphor? I have put forward the proposition that the only place this energycancome from is the EHT power supply, which means that current would need to be drawn from the supply during the transit time. Assuming you agree with this, how can such a current exist, given that during transit, the electrons do not touch and are not collected by the anode or any electrode. It’s a fair question, and as yet we have no answer. Where is Catalina to WOW us with the answer?

author: IanB
author_rank: Super Contributor
timestamp: May 11, 2016, 12:44:45 am
content: Quote from: Zeranin on May 10, 2016, 11:54:17 pmFirstly, the electrons left the cathode before they started on their journey. Since the electrons carry charge, they discharged the cathode a little at this point.Next, the electrons are accelerated through the electric field towards the anode, gaining energy as they go. Since the electrons in motion form an electric current, the moving electrons are actually creating a magnetic field. In this process, some potential energy from the electric field is being converted to stored energy in the magnetic field around the electrons. As the electrons are accelerated,the electric field is weakened(by whatever minuscule amount). Given sensitive enough measurements, the potential difference between the anode and the cathode while the electrons are in transit could be observed to be decreasing. It is as if an inductor has been connected in parallel with a capacitor and some energy is transferred from the capacitor to the inductor.It might be imagined that the voltage on the capacitor remains constant during the journey of the electrons and drops suddenly when the electrons strike the anode. But this will not be found to be the case. What will be found is that the potential difference remains the same at the moment the electrons leave the cathode, it will then drop gradually as the electrons are accelerated, and it will stop dropping once the electrons strike the anode.(It might be asked what happens to the stored energy in the magnetic field created by the fast moving electrons? Since there is no recovery path for this energy, it will be dissipated on the collapse of the magnetic field as heat (from eddy currents) and low energy radiation.)

author: Zeranin
author_rank: Regular Contributor
timestamp: May 11, 2016, 03:34:17 am
content: Quote from: IanB on May 11, 2016, 12:44:45 amHi Ian,I’m glad you find the problem of sufficient interest to think about it, and you clearly understand that there is a ‘mystery’ here, that requires explanation.One thing is not clear about your reply. In a CRT system, the cathode-to-anode voltage is fixed, this being the EHT supply voltage, and all my discussions have been on that basis. However, you seem to be considering the case where there is no EHT power supply as such, but only a capacitor to maintain the cathode-to-anode potential, is that correct? Assuming that you are in fact considering the case where there is no EHT power supply, I believe that your posting is largely correct, but still incomplete.It has taken me a while to reach my own conclusion about the exact correct explanation, but FWIW I finally feel confident in my understanding, so will relate it here. I will describe the situation with a fixed-voltage EHT supply between anode and cathode.QuoteUsually the cathode is at ground potential, connected to mother Earth via the mains ground wiring, so the cathode potential won’t actually change after squirting out a bunch of electrons. In any event, we have a fixed voltage between anode and cathode, and this is the only thing that matters as far as the electric fields inside the CRT are concerned, so our analysis need not be concerned with the fact that our bunch of electrons came from the cathode.It is true that moving charge produces a magnetic field, and also true that accelerated charge produces EM radiation, but both these effects are very small, and in my view have nothing to do with the true explanation of what is going on, and consideration of either is certainly notnecessaryto explain what is going on, so I will leave magnetic and EM effects out of the discussion altogether.We agree that energy is required to accelerate our bunch of electrons while in transit, and I proposed that the only place that this energycouldcome from is the EHT power supply, and I maintain this is the case. During transit, current is drawn from the EHT supply, even though the anode-cathode voltage remains constant, and even though the moving electron bunch never contacts the anode or any electrode. Furthermore, there isnota pulse of EHT current when the electrons actually strike the anode.The question then, is how can current be drawn from the supply, when the electrons are in transit, and not touching or collected by the anode or any other electrode?The explanation is subtle, and is concerned with the movement of charge (ie a current) through induction. As the electrons approach the anode, they attract positive charge (or repel electrons, same thing) to the anode, as per Coulomb’s Law. The only way that additional positive charge can get to the anode is from the EHT supply, so that is exactly what happens, current flows from the EHT supply to the anode. As the electrons get closer to the anode, they attract still further positive charge, with the result that there is a constant flow of current from the EHT supply to the anode, for the entire time that the electrons are moving from cathode to anode. The relationship Q=CV is in effect maintained at all times, except that Q includes a contribution from the moving bunch of electrons. As the electrons get closer to the anode, they increasingly ‘neutralize’ some of the +ve charges on the anode, and so additional +ve charge has to flow from the EHT supply to the anode, to compensate. By the time the electrons are only a tiny distance from the anode, about to strike it, the total negative charge that has flowed from the anode to the EHT supply is exactly equal to- ve charge about to strike the anode, as must be the case to maintain Q=CV, where Q is the net charge consisting of the charge on the anode, and the moving charge that is just about to strike the anode. Thus, when the moving electrons actually strike the anode and are absorbed into it, there is no pulse of anode current because Q=CV is already satisfied, and the EHT current to the anode that has been present for the entire electron journey ceases. At all times, all the equations of Physics are satisfied, and at all times during the transit and acceleration of the bunch of electrons, current is being drawn from the EHT supply, to provide the power for accelerating the electrons. Everything adds up. If it did not, then I would not sleep at night.

author: IanB
author_rank: Super Contributor
timestamp: May 11, 2016, 04:43:03 am
content: Quote from: Zeranin on May 11, 2016, 03:34:17 amYes, I realized after the fact that I didn't define the basis of my system model very well. I had in mind the earlier question about the ion thruster, where the electrode once charged up was isolated. By analogy, I considered the case where the anode and cathode of the CRT were charged up by the EHT supply and then the supply was disconnected, so that only the stray capacitance remained to maintain the potential difference (and in my perfect world there are no leakage currents to allow the charge to drain away).The interesting question to me then became, what would happen to the potential difference between anode and cathode during the time the bunch of electrons were accelerated from cathode to anode? I concluded that the potential difference would gradually decrease since the only source of energy to accelerate those electrons is the energy stored up in the electric field of the capacitance between anode and cathode. The electrons acquire energy, so another part of the system must lose energy to balance it. The only source of that energy is the electric field accelerating the electrons, and nature will not be denied; ergo the field energy must be depleted and the potential difference must decrease.It is then a small step to conceptually reconnect the EHT supply and observe that it will replace the lost charge in the stray capacitance by supplying some current.Quote from: Zeranin on May 11, 2016, 03:34:17 amThis I also concluded, as evidenced by my statement here:Quote from: IanB on May 11, 2016, 12:44:45 amThe gradual drop in potential difference across the stray capacitance if isolated would correspond to the gradual supply of current from the EHT supply during transit if it were connected.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 11, 2016, 05:08:38 am
content: We are in complete agreement, IanB.That's the great thing about science and engineering. After sufficient thought and discussion, agreement is always reached.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 11, 2016, 12:45:50 pm
content: Great posts guys. I'm liking the electron slowly changing the high voltage in transit for the CRT.The + and - 1 inch square monopole are complete. They are acting as predicted. They will not stick to each other nor a fridge. This is good.However the complementary monopoles should have acted like an electron and positron. This did not happen? Special relativity predicts this so where is my voltage difference. I feel cheated or more likely I have misunderstood something , story of my life. However I do have an indication that the elusive voltage is there when placing a round magnet on the screen of a TV set. When north or south pole are facing the screen there is a marked black ring where there is a void of electrons hitting the phosphor . Outside and inside the round magnet is okay but not on the edge which is where the electrons should be moving. I have a picture of it.