Mass-Energy Equivalence
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From: Paul Stowe on 25 Apr 2010 14:15 On Apr 24, 11:21 am, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 24, 1:05 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > If observing a force response (inertia) is an indirect inference > > rather than relying on a calculation well, > > An energy response and momentum response are measurable. For example, > the latter is directly related to the curvature of a trajectory in a > magnetic field. This is certainly no less complicated that deriving an > acceleration from a trajectory, and then using the acceleration to > derive a mass. You misunderstood my answer (it wasn't clear). I meant like using an old fashion balance scale... 'Assuming' that force is always equal to mass times acceleration and 'calculating it IS NEVER! a direct observation. It is AN ASSUMPTION, not a direct measurement. As Majorana's test showed, this may in fact NOT be true in all cases. How does one 'know' FOR SURE the mass of any astronomical body? Answer, THEY DON'T! The 'assume' that the equations we've developed are correct 'in all cases' even when they have never been 'directly' tested in the domains utilized. Thus my question that nobody has ever been able to answer, how do you know the difference between, F = kA^2/r^2 and F = GM^2/r^2 from long distance observation alone? I say, you can't! If you can't the is no uniqueness and the answer is uncertain. > > we certainly have very > > different definitions of indirect... As for your argument, by your > > definition, photons have mass... > > The definition I was using earlier was m^2 = E^2 - p^2 (in natural > units). In this definition, the photons have zero mass. Now this is rather the point, don't'cha think, you've defined it the way you want... IF E = kmv^2 it is alway true, or it is simply not a fundamental expression. It cannot be both. > > > > > > I have come to the realization that mass is an electromagnetic > > > > > > phenomena. That is to say, the fields that constitutes 'matter' will > > > > > > create counter EMF effects when perturbed from equilibrium. > > > > > > I'm sorry, I'm afraid I don't understand this claim. So if something > > > > > is proceeding at constant momentum in the absence of an > > > > > electromagnetic field present, and there is a change in a > > > > > gravitational field that affects this thing, the inertia that governs > > > > > the acceleration response to this gravitational field is somehow > > > > > electromagnetic in origin? > > > > > As you 'should' know all so-called material systems actually consist > > > > of charged entities, Quarks. > > > > Electrons are not quarks and they are a key constituent in matter. > > > They are electrically charged, though. > > > > But to your point, quarks are distinct from electrons not just by the > > > value of their charge, but by the fact that they have a charge that > > > electrons do not exhibit at all: strong charge, or "color". And in > > > fact, one is hard pressed to understand how it is that nucleons > > > inhabit a nucleus that is 10^-15 times smaller in volume than the > > > atom, if one only considers the interaction of the electric charge. > > > The behavior of the interaction due to strong charge is much, much > > > different than the interaction due to the electric charge, something > > > we know from extensive studies. > > > Yes, and because of this fact the apparent inertia of hadrons is > > different than leptons. I am currently investigating this aspect and > > 'think' it has to do with the QM base harmonic of the entity. I've > > made some progress but am not there yet. > > Let us know when you get there. You're competing against QCD, which > has a number of successes already. I don't see it as a competition, nor conflict, but we'll see where it leads, if anywhere. > > As for gravity being an > > offshoot of EM, that would yield unification, now wouldn't it? As it > > looks to me now, gravity is a second order effect of changes in > > velocity, as such, the sign force (+/-) disappears, leaving only one > > opposing vector. It looks now that it is directly related to Grad E, > > where E is the electric potential. > > Then you should be able to derive the electric field of the solar > system, since the gravitational effects are so well mapped. This seems > like a straightforward exercise. Ah, if it were only that simple. It's NOT the resultant E, it's each individual charge's E field that responds. These are only of the 10^21 Nt/dv per charge but given the number of said charges per cc of matter it adds up to a significant response (inertia). > > > > Last I check charges manifest electric > > > > 'fields' that are believed to be infinite in extent. In an > > > > equilibrium state (what you called constant momentum) these fields > > > > configuration are consistent. Perturb the equilibrium, and the fields > > > > are 'forced' to change, when electric fields are accelerated they > > > > create a reactive counter EMF. Hal Puthoff and others have realized > > > > this also. > > > > > > > That EMF is the source of inertia, thus, by definition, 'mass'. > > > > > > > Regards, Paul Stowe
From: Paul Stowe on 25 Apr 2010 14:25 On Apr 24, 11:21 am, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 24, 1:05 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > If observing a force response (inertia) is an indirect inference > > rather than relying on a calculation well, > > An energy response and momentum response are measurable. For example, > the latter is directly related to the curvature of a trajectory in a > magnetic field. This is certainly no less complicated that deriving an > acceleration from a trajectory, and then using the acceleration to > derive a mass. You misunderstood my answer (it wasn't clear). I meant like using an old fashion balance scale... 'Assuming' that force is always equal to mass times acceleration and 'calculating it IS NEVER! a direct observation. It is AN ASSUMPTION, not a direct measurement. As Majorana's test showed, this may in fact NOT be true in all cases. How does one 'know' FOR SURE the mass of any astronomical body? Answer, THEY DON'T! The 'assume' that the equations we've developed are correct 'in all cases' even when they have never been 'directly' tested in the domains utilized. Thus my question that nobody has ever been able to answer, how do you know the difference between, F = kA^2/r^2 and F = GM^2/r^2 from long distance observation alone? I say, you can't! If you can't the is no uniqueness and the answer is uncertain. > > we certainly have very > > different definitions of indirect... As for your argument, by your > > definition, photons have mass... > > The definition I was using earlier was m^2 = E^2 - p^2 (in natural > units). In this definition, the photons have zero mass. Now this is rather the point, don't'cha think, you've defined it the way you want... IF E = kmv^2 it is alway true, or it is simply not a fundamental expression. It cannot be both. > > > > > > I have come to the realization that mass is an electromagnetic > > > > > > phenomena. That is to say, the fields that constitutes 'matter' will > > > > > > create counter EMF effects when perturbed from equilibrium. > > > > > > I'm sorry, I'm afraid I don't understand this claim. So if something > > > > > is proceeding at constant momentum in the absence of an > > > > > electromagnetic field present, and there is a change in a > > > > > gravitational field that affects this thing, the inertia that governs > > > > > the acceleration response to this gravitational field is somehow > > > > > electromagnetic in origin? > > > > > As you 'should' know all so-called material systems actually consist > > > > of charged entities, Quarks. > > > > Electrons are not quarks and they are a key constituent in matter. > > > They are electrically charged, though. > > > But to your point, quarks are distinct from electrons not just by the > > > value of their charge, but by the fact that they have a charge that > > > electrons do not exhibit at all: strong charge, or "color". And in > > > fact, one is hard pressed to understand how it is that nucleons > > > inhabit a nucleus that is 10^-15 times smaller in volume than the > > > atom, if one only considers the interaction of the electric charge. > > > The behavior of the interaction due to strong charge is much, much > > > different than the interaction due to the electric charge, something > > > we know from extensive studies. > > > Yes, and because of this fact the apparent inertia of hadrons is > > different than leptons. I am currently investigating this aspect and > > 'think' it has to do with the QM base harmonic of the entity. I've > > made some progress but am not there yet. > > Let us know when you get there. You're competing against QCD, which > has a number of successes already. I don't see it as a competition, nor conflict, but we'll see where it leads, if anywhere. > > As for gravity being an > > offshoot of EM, that would yield unification, now wouldn't it? As it > > looks to me now, gravity is a second order effect of changes in > > velocity, as such, the sign force (+/-) disappears, leaving only one > > opposing vector. It looks now that it is directly related to Grad E, > > where E is the electric potential. > > Then you should be able to derive the electric field of the solar > system, since the gravitational effects are so well mapped. This seems > like a straightforward exercise. Ah, if it were only that simple. It's NOT the resultant E, it's each individual charge's E field that responds. These are only of the 10^-21 Nt/dv per charge but given the number of said charges per cc of matter it adds up to a significant response (inertia). > > > > Last I check charges manifest electric > > > > 'fields' that are believed to be infinite in extent. In an > > > > equilibrium state (what you called constant momentum) these fields > > > > configuration are consistent. Perturb the equilibrium, and the fields > > > > are 'forced' to change, when electric fields are accelerated they > > > > create a reactive counter EMF. Hal Puthoff and others have realized > > > > this also. > > > > > > > That EMF is the source of inertia, thus, by definition, 'mass'. > > > > > > Regards, Paul Stowe
From: mpc755 on 25 Apr 2010 17:10 On Apr 24, 8:40 pm, glird <gl...(a)aol.com> wrote: > On Apr 23, 10:22 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > In E=mc^2, mass is conserved. > > Yes, the mass(amount of matter) IS conserved; even if some of it may > have no weight (in grams) after it is released. > > glird What do you weigh the aether with? What do you use to weigh the lowest common denominator of matter? You can't weigh the aether. I wouldn't use the term 'no weight'. Saying the matter has no weight, but still has mass, after it is 'released' is confusing. It is better conceptually to describe aether and matter as different states of the same material. The material has mass. Aether and matter have mass. 'DOES THE INERTIA OF A BODY DEPEND UPON ITS ENERGY-CONTENT? By A. EINSTEIN' http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf "If a body gives off the energy L in the form of radiation, its mass diminishes by L/c2." The mass of the body does diminish, but the matter which no longer exists as part of the body has not vanished. It still exists, as aether. As the matter transitions to aether it expands in three dimensions. The effect this transition has on the surrounding aether and matter is energy. In E=mc^2, mass is conserved.
From: BURT on 25 Apr 2010 17:14 On Apr 25, 2:10 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Apr 24, 8:40 pm, glird <gl...(a)aol.com> wrote: > > > On Apr 23, 10:22 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > In E=mc^2, mass is conserved. > > > Yes, the mass(amount of matter) IS conserved; even if some of it may > > have no weight (in grams) after it is released. > > > glird > > What do you weigh the aether with? > > What do you use to weigh the lowest common denominator of matter? > > You can't weigh the aether. > > I wouldn't use the term 'no weight'. Saying the matter has no weight, > but still has mass, after it is 'released' is confusing. It is better > conceptually to describe aether and matter as different states of the > same material. > > The material has mass. > > Aether and matter have mass. > > 'DOES THE INERTIA OF A BODY DEPEND UPON ITS ENERGY-CONTENT? By A. > EINSTEIN'http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf > > "If a body gives off the energy L in the form of radiation, its mass > diminishes by L/c2." > > The mass of the body does diminish, but the matter which no longer > exists as part of the body has not vanished. It still exists, as > aether. As the matter transitions to aether it expands in three > dimensions. The effect this transition has on the surrounding aether > and matter is energy. > > In E=mc^2, mass is conserved. Weight is one aether being creating by another. Mitch Raemsch
From: spudnik on 25 Apr 2010 17:17
there is always some missing mass in any fission or fusion event, taht's the energy; of course, only matter-antimatter reactions approach a complete conversion; true? > Yes, the mass(amount of matter) IS conserved; even if some of it may > have no weight (in grams) after it is released. thus: all "photons" are readily absorbed by the correct tuner, generally a change of orbital of an electron, I suppose; all "photons" "go" at the speed of propogation of lightwaves, depending upon the index of refraction of the medium (given that there is really no vacuum "a la Pascal"). > Can we see or detect any -99.9999%c photons? thus: but, dood, what in Hell do *you* mean, by "aether & matter are different states of the same material" -- why do atoms and electrons & antiatoms need "an other state" of themselves? thus: Skeptics were just another Greek cult under the Roman Empire; Peripatetics, Gnostics, Stoics, Epicureans etc. ad vomitorium. I recall also recently reading that Justice Kennedy had come out for WS in some moot court, but that he later came to Oxford ... most likely, because it serves his oligarchical worldview (or, it was Scalia). I know of at least three "proofs" that WS was WS, but I recently found a text that really '"makes the case," once and for all (but the Oxfordians, Rhodesian Scholars, and others brainwashed by British Liberal Free Trade, capNtrade e.g.). > On the contrary, others include Justices Scalia, OConnor, > Blackmun and Powell, as the WSJ article noted. Only two current > Justices (Breyer and Kennedy) openly support the Stratford man. thus: what ever it says, Shapiro's last book is just a polemic; his real "proof" is _1599_; the fans of de Vere are hopelessly stuck-up -- especially if they went to Harry Potter PS#1. http://www.google.com/url?sa=D&q=http://entertainment.timesonline.co..... --Light: A History! http://wlym.com |