Sensei Posted April 7, 2015 Posted April 7, 2015 Thanks for defining the value of Phi. here is ground potential defined to three decimal places. http://www.wolframalpha.com/input/?i=0.511%3D((938.272-x)%2F2)*√(1-(x^2%2F938.272^2)) Not sure why I have to justify the proton mass experiment in this thread, it is well known. It's well know for us, but not to you. I believe so, you don't know where this 938.272 MeV came from... Alternative question: show convertion of electrovolts to kWh.. I am just asking to see whether you know basics. If you know, then show us. If you don't know how to do it, then you should ask. As least you will learn something..
beejewel Posted April 9, 2015 Author Posted April 9, 2015 Once again for the sceptics, Ground Potential Theory (GPT) begins with the following thought experiment; "take two paralell metal plates at some fixed distance apart, and charge one plate negative and the other plate positive, so that there exists an electrical potential between the plates, now take this experiment to the absolute limit, so one plate consists purely of electrons and the other of protons" It shall then be obvious that the potential has reached a limit, and adding more protons to the anode or more electrons to the cathode can not further increase the potential between the plates. If we also know how many electrons there are in total, we can divide the overall energy in eV stored in the anode by the number of electrons in the cathode (assuming same number) to get the net surface potentials. We also understand that removing the same number of electrons and protons from each plate does not change the potential, so let's take this experiment to the limit, so our plates in the end contain one electron and one proton, i.e Hydrogen. With this simple gedanken experiment, I hope to have shown that electrical potential does indeed have a limit, and that this limit is equal to the surface potential of the proton. In other words, the mass of Hydrogen is stored in the potential energy field between the electron and the proton. Finding the surface potential of the proton in Volts is a simple matter of taking its rest mass in whatever unit you have and converting it to electron volts and dividing it by 1 electron. This will give you the current value of a Proton: 938.272046(21) MV* Source: The NIST Reference on Constants, Units and Uncertainty We now have two valuable clues from which to begin the construction of GPT; 1) the constancy and limit of the speed of light and 2) the constancy and limit of the proton potential. Further our new understanding about the upper limit of electrical potential, let's us realise that ground potential must lie somewhere between zero and the surface potential of the proton, and obviously we would like to know what GP is, so we can see where we are in the scheme of things. So in GPT I postulate that the protons surface potential is a physical constant in the same way as the speed of light, and assign the symbol [latex]\Phi[/latex] We can now show that there is a linear relationship between [latex] v \propto \Phi[/latex] In the above sketch I have set [latex] c = \Phi = 1 [/latex]. Note that the velocity axis is the four-velocity, and so an observer at rest shall be at some potential between zero and [latex]\Phi[/latex] and likewise somewhere between zero and c on the velocity axis. Now let's have two observers at rest, Jack and John, Jack owns a fuel station and John owns a sports car. John buys a litre of gas from Jack, and speeds off down the highway at increasing velocity until his fuel runs out, at this point his velocity relative to Jack is [latex]\Delta v[/latex] and his potential has perceivably increased by [latex]\Delta \phi = \frac{mv^2}{2}[/latex] All this proves is that [latex]\frac{v}{c} \propto \frac{\phi}{\Phi}[/latex] and I will be able to use this later to show that the electron is the proton's anti particle. Keeping this to the current thread, it also proves that four-velocity and real velocity for that sake must increase with potential, which brings us back to the revolving planets, and I have not yet been convinced, I think Kepler was wrong. Steven
ajb Posted April 9, 2015 Posted April 9, 2015 (edited) Once again for the sceptics, Saying it again does not make it right this time! "take two paralell metal plates at some fixed distance apart, and charge one plate negative and the other plate positive, so that there exists an electrical potential between the plates, now take this experiment to the absolute limit, so one plate consists purely of electrons and the other of protons" It shall then be obvious that the potential has reached a limit, and adding more protons to the anode or more electrons to the cathode can not further increase the potential between the plates. If we also know how many electrons there are in total, we can divide the overall energy in eV stored in the anode by the number of electrons in the cathode (assuming same number) to get the net surface potentials. But adding more charge to the plates would change the potential difference. This is irrespective of just considering the unphysical situation of electron and proton plates. (Q = CV right?) So in GPT I postulate that the protons surface potential is a physical constant in the same way as the speed of light, and assign the symbol \Phi This is the real unfounded jump. You are now trying to use your potential in analogy to the speed of light as used in special relativity. The difference is that Einstein did not have to postulate 'c=constant' from scratch; it was already written into Maxwell's equations. Einstein's leap was to realise the significance of this and not really the fact that it is constant (in all inertial frames). Your potential seems to lack this 'hook' and none of us really have a clue what you are trying to do. Several big holes in the theory have appeared as well as many mathematical ones. Edited April 9, 2015 by ajb 1
beejewel Posted April 9, 2015 Author Posted April 9, 2015 (edited) Saying it again does not make it right this time! But adding more charge to the plates would change the potential difference. This is irrespective of just considering the unphysical situation of electron and proton plates. (Q = CV right?) This is the real unfounded jump. You are now trying to use your potential in analogy to the speed of light as used in special relativity. The difference is that Einstein did not have to postulate 'c=constant' from scratch; it was already written into Maxwell's equations. Einstein's leap was to realise the significance of this and not really the fact that it is constant (in all inertial frames). Your potential seems to lack this 'hook' and none of us really have a clue what you are trying to do. Several big holes in the theory have appeared as well as many mathematical ones. Just going over it again for those who may not have read earlier posts.. I don't agree that adding more charge to a plate consisting of only protons will incease it's potential, Q=CV but you can't increase Volts when your anode is 100% protons, which is my whole point here because volts is a ratio of electrons to protons. [edit] I agree adding more charge increases the stored energy, but it does not increase the potential.[/edit] The jump to my postulate is only unfounded if the above claim is unfounded. Lucky for Einstein that someone else had already proved it, but all you have to do is open your eyes and look around, all protons are the same size, all protons have the same charge, there are no stable particles with a higher potential than a proton, there appear to be roughly as many electrons as protons, and the list goes on. But fine, let's debate the charged plates if you feel that is the sticking point, go one step further and gather every electron and every proton in the universe on two plates, then divide by the number of electrons, and you still come back to that same constant. Steven Edited April 9, 2015 by beejewel
Strange Posted April 9, 2015 Posted April 9, 2015 I don't agree that adding more charge to a plate consisting of only protons will incease it's potential, Q=CV but you can't increase Volts when your anode is 100% protons Say your plate consists of 100 protons. (Disregarding the fact this impossible and therefore the entire argument is meaningless). Now add another one: there are now 101. Repeat. 102 protons. Repeat ... However many protons you add, it will be 100% protons. (And nonexistent.)
beejewel Posted April 9, 2015 Author Posted April 9, 2015 Say your plate consists of 100 protons. (Disregarding the fact this impossible and therefore the entire argument is meaningless). Now add another one: there are now 101. Repeat. 102 protons. Repeat ... However many protons you add, it will be 100% protons. (And nonexistent.) Am I missing something here? Strange just confirmed my argument, you can't increase potential! Steven
ajb Posted April 9, 2015 Posted April 9, 2015 I don't agree that adding more charge to a plate consisting of only protons will incease it's potential, Q=CV but you can't increase Volts when your anode is 100% protons, which is my whole point here because volts is a ratio of electrons to protons. But the electric field is proportional to the charge density. As you change the change the potential. Or I am missing something? I don't see that it matters exactly what your plates are made of as long as they hold charge. Use the basic equations for parallel plate capacitors to show you are right. The jump to my postulate is only unfounded if the above claim is unfounded. You have not convinced anyone yet... Lucky for Einstein that someone else had already proved it, but all you have to do is open your eyes and look around, all protons are the same size, all protons have the same charge, there are no stable particles with a higher potential than a proton, there appear to be roughly as many electrons as protons, and the list goes on. This is anecdotal evidence. You need to really calculate something in your theory and show it matches nature well. So far you have not been able to do this. But fine, let's debate the charged plates if you feel that is the sticking point, Under some simple assumptions the potential between charged plates is well understood and part of school physics. ...go one step further and gather every electron and every proton in the universe on two plates, then divide by the number of electrons, and you still come back to that same constant. This is different to what you have said in the above. Maybe in that sense the number of +ive charges and -ive charges in the observable Universe does put a bound on potential difference. But does this point to something significant about electromagnetic theory? Does this really point to the surface potential of a proton (which is the potential difference between the centre of the proton and its surface when we think of it classically?) Am I missing something here? Strange just confirmed my argument, you can't increase potential! Strange did not say anything about potential, just that if I have 100 protons so a charge of 100e, then 101 protons with have a charge of 101e. The charge has changed, that was his point.
beejewel Posted April 9, 2015 Author Posted April 9, 2015 Under some simple assumptions the potential between charged plates is well understood and part of school physics. The problem here is we are dealing with 938 million volts, and for safety reasons school physics experiments mainly deal with less than 30 volts, and at University undergraduate level one is usually limited to 30 kV. None of these effects would show up at these potentials. Steven
ajb Posted April 9, 2015 Posted April 9, 2015 The problem here is we are dealing with 938 million volts, and for safety reasons school physics experiments mainly deal with less than 30 volts, and at University undergraduate level one is usually limited to 30 kV. None of these effects would show up at these potentials. The theory is the same, provided we are not near the breakdown potential of the capacitor. You just need to argue using high school physics that the potential difference between two plates is limited by some physical bounds that are irrespective of the material used to create the dielectric. In practice we would expect a real capacitor to fail well below this limit. Anyway, your 938 million Volts is the potential needed for an electron from rest to gain the energy equal to that of the rest mass of a proton. Is that right?
beejewel Posted April 9, 2015 Author Posted April 9, 2015 Definition of Volts from hyperphysics [latex]V=\frac{WorkDone}{charge}[/latex] therefore on an anode consisting of 100% protons the work done in adding each extra charge is 938 million volts, which means you add one more unit of work to the numerator and add another one charge in the denominator, school maths tell you that nothing has changed. Steven Anyway, your 938 million Volts is the potential needed for an electron from rest to gain the energy equal to that of the rest mass of a proton. Is that right? My understanding is that the rest mass of a proton is calculated or more likely measured in the ionised form, probably using an accellerated beam through a mass selection magnet, so it is the mass of the proton at it's full potential which acording to GPT is the potential at which the electron mass is near enough to zero. So expressing the mass as energy according to the mass energy equation we choose the unit of eV because it's an SI standard and when we divide by one electron we have the all familiar unit for potential. You may ask why we divide by 1 electron, and that is because it is the other plate in our capacitor which we in reality call for Hydrogen. The entire 938 million eV is the work done in separating a proton from an electron, hence it is equivalent to the stored potential. When we ionize hydrogen, the electron is sufficiently far away from the proton, that we can safely ignore any work required to move it the last distance from say 100 meters to infinity. Steven
ajb Posted April 9, 2015 Posted April 9, 2015 You have a proton. You have another at infinity (we define this to be at zero potential) and then you look at the work done to bring them close to each other. You see we have a 1/r dependence here and that is divergent. So how close do you want to bring them together? The equivalent of 938 million volts? I don't think this is the same as the potential you are discussing. You have two plates and a uniform electric field between them. The potential is then defined in terms of work done on a test particle in that field. Changing the charge density on the plates has an effect on the field between the plates.
beejewel Posted April 9, 2015 Author Posted April 9, 2015 You have a proton. You have another at infinity (we define this to be at zero potential) and then you look at the work done to bring them close to each other. You see we have a 1/r dependence here and that is divergent. So how close do you want to bring them together? The equivalent of 938 million volts? I don't think this is the same as the potential you are discussing. You have two plates and a uniform electric field between them. The potential is then defined in terms of work done on a test particle in that field. Changing the charge density on the plates has an effect on the field between the plates. I boldly claim that the whole concept of charge is a misinterpretation of the real world, there are NO coloumb forces! All the energy contained in matter is the stored potential energy between the electron and the proton. What everyone generally thinks of as the Coulomb force is according to GPT [latex] \Delta v [/latex], which briefly means you the observer at ground potential (930 MV) experience the proton (938 MV) moving at a speed of [latex]v=c(\frac{8}{938})= ~2500 km/s[/latex] , and it will never stand still, you can confine it, but that just means it jiggles even more, but let's not derail this thread with that argument just yet, convincing you guys to forget Coulomb is a challenge I am keeping for another day Steven
ajb Posted April 9, 2015 Posted April 9, 2015 I boldly claim that the whole concept of charge is a misinterpretation of the real world, there are NO coloumb forces! That is a big claim considering that the standard notion(s) of charge and Coulomb force are put to good use all the time. I assume you are aware of the rather general notion of charges and currents via Noether's theorems?
swansont Posted April 9, 2015 Posted April 9, 2015 Definition of Volts from hyperphysics [latex]V=\frac{WorkDone}{charge}[/latex] therefore on an anode consisting of 100% protons the work done in adding each extra charge is 938 million volts, which means you add one more unit of work to the numerator and add another one charge in the denominator, school maths tell you that nothing has changed. No, that's not right. F = -kq1q2/r2 and work is the integral of the force and the distance. So work depends on how much charge you have. The work done in adding the first proton is zero. Let's assume the work done in adding the second is 938 MeV. The work done in adding the third will be 1876 MeV, because there's already 2e of charge there. The work for each additional charge will grow. The way you've described it, all capacitors must have the same voltage rating, and that's nonsense. 2
beejewel Posted April 9, 2015 Author Posted April 9, 2015 I assume you are aware of the rather general notion of charges and currents via Noether's theorems? I am aware of it now, looks like something that might take some time to digest. No, that's not right. F = -kq1q2/r2 and work is the integral of the force and the distance. So work depends on how much charge you have. The work done in adding the first proton is zero. Let's assume the work done in adding the second is 938 MeV. The work done in adding the third will be 1876 MeV, because there's already 2e of charge there. The work for each additional charge will grow. The way you've described it, all capacitors must have the same voltage rating, and that's nonsense. F = -kq1q2/r2 Well there is that coloumb force again, lets get to that later. We generally think of capacitors as charged plates or spheres separated by some dielectric. These conductors are at least from a classical view made of some kind of metal, which in turn is made from electrons and protons with a few neutrons chucked in, further when the plates are neutral there is said to be the same number of electrons as protons in each plate hence believed to have no charge. Charge is added by transferring some electrons from anode to cathode, and when we do the actual numbers we realise that the number of electrons in proportion of the total are actually insignificantly few. Richard Feynman once used an example that even 1% difference in charge between to people would be equal to a force strong enough to lift the Earth, he seemed like the kind of guy who would have done the numbers. My argument here is that it is unlikely that anyone has physically experimented with any charges high enough to notice any potential limit, as was the case with special relativity. Had Einstein not pointed out that high relative speed caused time dilation, it would have been very unlikely that we would have discovered it anytime soon, as the effect is so small. Please read over my description again, if it is not clear I shall edit it, but I think it's okay, work done per unit charge approaches an asymptote as the ratio of protons over electrons approaches infinity. Steven
ajb Posted April 9, 2015 Posted April 9, 2015 My argument here is that it is unlikely that anyone has physically experimented with any charges high enough to notice any potential limit, This still does not explain your reasoning in terms of basic capacitors and more importantly you have been unable to match any of this with nature. In particular I have no idea what this has to do with dark matter!
beejewel Posted April 9, 2015 Author Posted April 9, 2015 (edited) This still does not explain your reasoning in terms of basic capacitors and more importantly you have been unable to match any of this with nature. In particular I have no idea what this has to do with dark matter! After 90 replies I don't blame you fortunately I still remember. According to GPT the planets in our solar system, as well as the stars in galaxies move backwards, this means their speeds relative to you the observer actually increases with increasing radius, when plotted this forms a rotation curve excactly like those found in nature, and not like predicted by todays science. Simply google "galaxy rotation curve" and click on images, and you will find hundreds of rotation curves that look like my prediction and absolutely none that look like Keplers. Dark Matter was invented by scientists to explain the anomalous rotation curve, that in itself is a concern. Steven Edited April 9, 2015 by beejewel
Strange Posted April 9, 2015 Posted April 9, 2015 absolutely none that look like Keplers. Not surprising as Kepler's law is not applicable - for rather obvious reasons (that you refuse to address). Dark Matter was invented by scientists to explain the anomalous rotation curve, that in itself is a concern. You think they should just ignore the evidence and not attempt to explain it? By the way, how does your theory explain the gravitation lensing caused by dark matter? Even when separated from a galaxy...
ajb Posted April 9, 2015 Posted April 9, 2015 According to GPT the planets in our solar system, as well as the stars in galaxies move backwards, This was a sticking point. It is not clear what you mean by backwards, and especially unclear when you said something about backwards in time. ...this means their speeds relative to you the observer actually increases with increasing radius, This is not very clear at all. You mean their orbital speeds or really the speeds relative to the Earth. The orbital speed is probabily more useful. ...when plotted this forms a rotation curve excactly like those found in nature, and not like predicted by todays science. Exactly? I mean you have taken an estimate for the mass of a galaxy using the visible matter + a little bit of baryonic dark matter and matched the observed rotation curves well? How many galaxies have you managed to match and to what sort of range of parameters (eg masses of the galaxies) do you need? Dark Matter was invented by scientists to explain the anomalous rotation curve, that in itself is a concern. Not just the rotation curves, the cosmological significance cannot be ignored.
beejewel Posted April 9, 2015 Author Posted April 9, 2015 (edited) By the way, how does your theory explain the gravitation lensing caused by dark matter? Even when separated from a galaxy... Good question, allthough I have done no work on the gravitational lensing that you refer to, GPT is quite clear on the amount of energy stored by space itself, and it's excactly half. Allthough I do not have watertight proof yet, I believe the frequent number two in the denominator which we see in most of our equations is there, because we are trained to only work with solid matter. GPT says that for every 1 gram of matter there must be 1 gram of antimatter, and as I am claiming that the electron which is only 0.5 MeV is the other half of the proton, there is a discrepancy of 929.5 MeV and that's the mass of space itself. When you integrate the energy required to separate a mass "m" from the Schwartzchild radius to infinity (taking redshift into account) and after some rediculous algebra you get the exact answer 2mc^2, that pretty much says it all. Steven Edited April 9, 2015 by beejewel
ajb Posted April 9, 2015 Posted April 9, 2015 ...I believe the frequent number two in the denominator which we see in most of our equations is there, This sounds like numerology now! GPT says that for every 1 gram of matter there must be 1 gram of antimatter, Another 'random' prediction. So where is all this missing antimatter? and as I am claiming that the electron which is only 0.5 MeV is the other half of the proton, The big difference between the proton and the electron have been pointed out to you several times. There is just no way they are two halves of the same thing. ...there is a discrepancy of 929.5 MeV and that's the mass of space itself. Is that a prediction of the cosmological constant? Should that be mass per unit volume or something?
beejewel Posted April 9, 2015 Author Posted April 9, 2015 Is that a prediction of the cosmological constant? Should that be mass per unit volume or something? Now that you mention it, yes, I worked out the mass per volume of space several years ago, before GPT, but I believe it still holds. [latex] V = \frac{8Gm\pi}{3H^2}[/latex] and it works out that one cubic meter of space weighs about 1.118 x 10^-26 kg anyway this is off topic and it's late here now, and I'm off to bed soon, so don't forget to google those rotation curves. Steven Here is a very basic youtube explanation of Galaxy rotation curves for anyone interested. https://youtu.be/Hcc0dToHf18 Steven
swansont Posted April 9, 2015 Posted April 9, 2015 My argument here is that it is unlikely that anyone has physically experimented with any charges high enough to notice any potential limit, as was the case with special relativity. We can cram 80 or 90 protons together in a small sphere. That's a pretty trivial exercise.
Sensei Posted April 9, 2015 Posted April 9, 2015 So you don't know how to calculate proton invariable mass-energy to be 938272046 eV and you don't even know how to convert one unit of energy to other unit of energy (!) electron volts to kWh. Otherwise you would answer questions from #76 post.. Let's start with how to convert from eV -> kWh 1 kWh = 1000 W * 3600 s = 3,600,000 W*s = 3,600,000 J P=I*U [1 W = 1A * 1V) but P=E/t Analyze units more closely, if you don't know why J = kg*m^2/s^2 W = kg*m^2/s^3 so we can substitute: E/t=I*U E=I*t*U (energy is ampere * time * voltage) Other equation is: I*t=Q so again substitute, and we have: E=Q*U (energy is charge * voltage) But Q (in Coulombs) is quantized, it can have 1e, 2e, 100e, .... integer multiply of elementary charge e (http://en.wikipedia.org/wiki/Elementary_charge) so we have at the end: E=e*U 1.602176565*10^-19 J = 1.602176565*10^-19 C * 1 Volt That's kinetic energy of electron in 1 Volt potential difference. So if we have 1 kWh energy = 3,600,000 J it's 3,600,000 / 1.602176565^-19 = 2.247e+25 eV Second question from #76 post was about mass-energy of proton 938272046 eV Convert eV -> J, like above showed: 938272046 eV * 1.602176565^-19 = 1.5032774836958E-010 J divide by c^2: 1.5032774836958E-010 J / 299792458^2 = 1.67262177649828E-027 kg and there is mass in kilograms unit!
beejewel Posted April 9, 2015 Author Posted April 9, 2015 So you don't know how to calculate proton invariable mass-energy to be 938272046 eV and you don't even know how to convert one unit of energy to other unit of energy (!) electron volts to kWh. Otherwise you would answer questions from #76 post.. Let's start with how to convert from eV -> kWh 1 kWh = 1000 W * 3600 s = 3,600,000 W*s = 3,600,000 J P=I*U [1 W = 1A * 1V) but P=E/t Analyze units more closely, if you don't know why J = kg*m^2/s^2 W = kg*m^2/s^3 so we can substitute: E/t=I*U E=I*t*U (energy is ampere * time * voltage) Other equation is: I*t=Q so again substitute, and we have: E=Q*U (energy is charge * voltage) But Q (in Coulombs) is quantized, it can have 1e, 2e, 100e, .... integer multiply of elementary charge e (http://en.wikipedia.org/wiki/Elementary_charge) so we have at the end: E=e*U 1.602176565*10^-19 J = 1.602176565*10^-19 C * 1 Volt That's kinetic energy of electron in 1 Volt potential difference. So if we have 1 kWh energy = 3,600,000 J it's 3,600,000 / 1.602176565^-19 = 2.247e+25 eV Second question from #76 post was about mass-energy of proton 938272046 eV Convert eV -> J, like above showed: 938272046 eV * 1.602176565^-19 = 1.5032774836958E-010 J divide by c^2: 1.5032774836958E-010 J / 299792458^2 = 1.67262177649828E-027 kg and there is mass in kilograms unit! So what's your point, how is it relevant to this discussion?
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