There is no "pull" of gravity, only the PUSH of flowing ether!
in [Relativity]
|
Prev: Was Einstein Guilty of Scientific Fraud?
Next: Question about energy eigenvalues of a Hamiltonian, in general
From: Paul Stowe on 6 Apr 2010 20:03 On Apr 5, 8:03 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Mon, 5 Apr 2010, PaulStowewrote: > > On Apr 4, 3:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Apr 5, 2:19 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > ... > > > > > > I wanted to do a quick calculation of heating, so looked at your > > > > > numbers: > > > > > > "the mass attenuation coefficient [u] (m^2/kg) is ~3.147E-06. The > > > > > "momentum flux (Q) is ~6.74E+00 kg/m-sec^2." > > > > > > Are these correct? Given the typical density of matter around here, is > > > > > this compatible with > > > > > > > > > The 'linear attenuation coefficient is on the order of 1E-20 1/m, > > > > > > As I understand it, the intercepted momentum flux, if completely > > > > > transferred to some object, gives the maximum force possible, if there > > > > > was complete shielding on the other side. So, the maximum > > > > > gravitational force on an object (presumably super-dense) of some > > > > > given cross-sectional area. Meanwhile, an object of "normal" mass and > > > > > density, completely shielded from one side, experiences a force of > > > > > Qu*mass. For gravitational attraction, the fluxes from each side, > > > > > accounting for shadowing are Q(1-u*term dependent of mass and geometry > > > > > of shading object) from the shady side, and Q from the other side, for > > > > > a net transfer of momentum of Qu^2 * f(m1,r)*m2. > > > > > In the case of heating it's power flux, not momentum flux. The power > > > > flux (W) is Qc/4pi => ~1.6E+08 Watts/m^2. Then for any body w = W(2GM/ > > > > rc^2), conbining all 'constants' we have (2WG/c^2)(M/r). Thus, > > > > > k = (2WG/c^2) ~= 2.38E-19 m/sec^3 and, > > > > > w = kM/r > > > > > Where M & r are the mass and radius of the body. > > > > > Then, > > > > > w = W - W' > > > > > where W is as defined above and W' the amount that makes it out. > > > > Then, > > > > > w = W(1 - e^-2lr) > > > > > Now, solve for l... (the linear attenuation coefficient) > > > > So, w = W(1 - exp(-2lr)) is approximately w = 2lrW for weak > > > attenuation, or k M/r = 2lrW. > > > > We have M = 4*pi*r^3/3 * rho where rho is the mean density.With your > > > numbers, this gives > > > > l = 2e-20 * rho * r, > > > > for l = 1e-16 m^-1 using rho, r for the Earth. Why does this value > > > depend on r? > > > The flux (Q) is constant on a per area basis. Thus, the total > > momentum/power impinging upon a mass is QA - WA. As area goes to zero > > the intercepted flux goes to zero. If the mass is constant the > > density increase as a function of r^3 and the linear attenuation > > remains constant (in the weak region) BUT! the total area is > > decreasing as function of r^2. This differential results in a 1/r > > reduction in the total attenuated quantities. Thus the form M/r... > > OK, so this is the decrease in flux as it moves into a spherical body? > I'd suggest calling it something different; this is not what is usually > meant by a "linear attenuation coefficient". > > > > > > > > > > What have I gotten wrong here? From the above, I have maximum force > > > > > possible per kg is only Qu = 2e-5N, which is unrealistically low. > > > > > Meanwhile, for a block a steel, the linear attenuation coefficient > > > > > would be 0.02. > > > > You don't comment at all on the calculation of force. Using the values > > > you give for Q and u, I get a maximum force per kg of 2e-5N. This is > > > rather less than we observe terrestrially, and in turn, this must be > > > far less than the maximum possible since shielding is negligible. > > > Since you have repeated the correctness of your values for Q and u, > > > why does the above calculation give a value of l very different from > > > rho*u? > > > As for the force of gravity, this is why big G is Qu^2, NOT! qu. The > > Newtonian is, technically in LeSage's weak region, > > > F = Q(uM)(um)/r^2 > > > The total 'potential', or what we call acceleration due to gravity, > > is, > > > a = Qu(Mu/r^2) > > > BUT! you MUST have that second mass m to realize this 'potential'. > > Given that u is independent of m this 'potential' is considered mass > > indenpendent. This is WHY! big G is only truly valid for multibody > > problems. Trying to get the right answer with only Qu just doesn't > > cut it. > > > Look at Qu (kg/m-sec^2)(m^2/kg) => m/sec^2, it's not a force, it's the > > field's attenuation potential. > > I didn't say that Qu was a force. I said that Qu*mass is a force, the > force you'd get on a mass if you had complete screening, with the momentum > flux Q incident from one side only. > > If you want to use a cube of some size, of some density, you get the same > result, as long as you're still in the weak attenuation limit. > > Isn't saying that the momentum flux Q is 6.74 (kg.m/s)/m^2/s the same > as saying that the force that would act on a perfectly absorbing square 1m > by 1m in area, completely shielded from one side, would be 6.74N? > > If not, why not? Surely the momentum flux of the corpuscles must be much > greater than the gravitational forces we see, since a tiny reduction in > the flux from one side, resulting in an equally tine anisotropy in the > flux, with a tiny absorption, is meant to lead to the gravitational forces > we see. > > If the maximum force we could get on an object of 1m^2 in cross-section > were only 6.74N, how can we see forces larger than this, in real life, on > objects smaller than this? > > -- > Timo
From: Paul Stowe on 6 Apr 2010 20:46 On Apr 5, 8:03 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Mon, 5 Apr 2010, PaulStowewrote: > > On Apr 4, 3:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Apr 5, 2:19 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > ... > > > > > > I wanted to do a quick calculation of heating, so looked at your > > > > > numbers: > > > > > > "the mass attenuation coefficient [u] (m^2/kg) is ~3.147E-06. The > > > > > "momentum flux (Q) is ~6.74E+00 kg/m-sec^2." > > > > > > Are these correct? Given the typical density of matter around here, is > > > > > this compatible with > > > > > > > > > The 'linear attenuation coefficient is on the order of 1E-20 1/m, > > > > > > As I understand it, the intercepted momentum flux, if completely > > > > > transferred to some object, gives the maximum force possible, if there > > > > > was complete shielding on the other side. So, the maximum > > > > > gravitational force on an object (presumably super-dense) of some > > > > > given cross-sectional area. Meanwhile, an object of "normal" mass and > > > > > density, completely shielded from one side, experiences a force of > > > > > Qu*mass. For gravitational attraction, the fluxes from each side, > > > > > accounting for shadowing are Q(1-u*term dependent of mass and geometry > > > > > of shading object) from the shady side, and Q from the other side, for > > > > > a net transfer of momentum of Qu^2 * f(m1,r)*m2. > > > > > In the case of heating it's power flux, not momentum flux. The power > > > > flux (W) is Qc/4pi => ~1.6E+08 Watts/m^2. Then for any body w = W(2GM/ > > > > rc^2), conbining all 'constants' we have (2WG/c^2)(M/r). Thus, > > > > > k = (2WG/c^2) ~= 2.38E-19 m/sec^3 and, > > > > > w = kM/r > > > > > Where M & r are the mass and radius of the body. > > > > > Then, > > > > > w = W - W' > > > > > where W is as defined above and W' the amount that makes it out. > > > > Then, > > > > > w = W(1 - e^-2lr) > > > > > Now, solve for l... (the linear attenuation coefficient) > > > > So, w = W(1 - exp(-2lr)) is approximately w = 2lrW for weak > > > attenuation, or k M/r = 2lrW. > > > > We have M = 4*pi*r^3/3 * rho where rho is the mean density.With your > > > numbers, this gives > > > > l = 2e-20 * rho * r, > > > > for l = 1e-16 m^-1 using rho, r for the Earth. Why does this value > > > depend on r? > > > The flux (Q) is constant on a per area basis. Thus, the total > > momentum/power impinging upon a mass is QA - WA. As area goes to zero > > the intercepted flux goes to zero. If the mass is constant the > > density increase as a function of r^3 and the linear attenuation > > remains constant (in the weak region) BUT! the total area is > > decreasing as function of r^2. This differential results in a 1/r > > reduction in the total attenuated quantities. Thus the form M/r... > > OK, so this is the decrease in flux as it moves into a spherical body? > I'd suggest calling it something different; this is not what is usually > meant by a "linear attenuation coefficient". > > > > > > > > > > What have I gotten wrong here? From the above, I have maximum force > > > > > possible per kg is only Qu = 2e-5N, which is unrealistically low. > > > > > Meanwhile, for a block a steel, the linear attenuation coefficient > > > > > would be 0.02. > > > > You don't comment at all on the calculation of force. Using the values > > > you give for Q and u, I get a maximum force per kg of 2e-5N. This is > > > rather less than we observe terrestrially, and in turn, this must be > > > far less than the maximum possible since shielding is negligible. > > > Since you have repeated the correctness of your values for Q and u, > > > why does the above calculation give a value of l very different from > > > rho*u? > > > As for the force of gravity, this is why big G is Qu^2, NOT! qu. The > > Newtonian is, technically in LeSage's weak region, > > > F = Q(uM)(um)/r^2 > > > The total 'potential', or what we call acceleration due to gravity, > > is, > > > a = Qu(Mu/r^2) > > > BUT! you MUST have that second mass m to realize this 'potential'. > > Given that u is independent of m this 'potential' is considered mass > > indenpendent. This is WHY! big G is only truly valid for multibody > > problems. Trying to get the right answer with only Qu just doesn't > > cut it. > > > Look at Qu (kg/m-sec^2)(m^2/kg) => m/sec^2, it's not a force, it's the > > field's attenuation potential. > > I didn't say that Qu was a force. I said that Qu*mass is a force, the > force you'd get on a mass if you had complete screening, with the momentum > flux Q incident from one side only. Yes, and I've been trying to tell you that QuM is not a proper expression in the model. > If you want to use a cube of some size, of some density, you get the same > result, as long as you're still in the weak attenuation limit. In the weak attenuation limit there is more spacing than attenuation sites (that what matter is in the model). As such each attenuator is exposed to the field flux and has a cross-sectional area. It a 3D problem, and it is the sum of ALL of the attenuating areas that produce the total effect. That WHY I expressed the equation the way I did uM => total cross-sectional area. Note that uM for the Earth is MUCH greater than its physical area. I could be a modernist and just say 'shut up & calculate' :) but that not the way to understanding. > Isn't saying that the momentum flux Q is 6.74 (kg.m/s)/m^2/s the same > as saying that the force that would act on a perfectly absorbing square 1m > by 1m in area, completely shielded from one side, would be 6.74N? Yup, that an interesting problem, isn't it... The exposure of more attenuating area ican increase the effect per unit volume because its a 4 pi omni-directional flux and, being constant per unit volume has some counter intuitive aspects. > If not, why not? Surely the momentum flux of the corpuscles must be much > greater than the gravitational forces we see, since a tiny reduction in > the flux from one side, resulting in an equally tine anisotropy in the > flux, with a tiny absorption, is meant to lead to the gravitational forces > we see. > > If the maximum force we could get on an object of 1m^2 in cross-section > were only 6.74N, how can we see forces larger than this, in real life, on > objects smaller than this? Well, in the end, do the math using the terms 6.744 for Q and 3.147E-6 for u. You'll find that it yields Newton's equations. Why is a very interesting physical process... As I mentioned earlier, this leads to a path of unification between EM- SM and gravity. If Q is, in fact, the momentum flux of the universal aether medium,and given Maxwell own definition of c^2 = 1/u'z Where u' -> compressibility and z the density, then these terms are fundamental also to Q. In fact, Q = a^2/u'(2pi) = (e^2/2h)^2/z2pi = 6.744 kg/m-sec^2 Where a is alpha (1/137.0358) and e elemental charge (1.60217E-19 kg/ sec) So, we have here a direct connection between EM-SM and gravity.... As for the 3kT well if v = c and the kinetic energy is 1/2mc^2 = 3/2kT then mc^2 = 3kT. Given mc^2 = h(nu) well, what more do you want? You really think that given Maxwell's definitions of u' & z and thus leading to e/m being a frequency that it is a coincidence that h(e/m)/ 3k = 2.8 Degrees K??? Hell of a one if it is. Paul Stowe
From: Timo Nieminen on 6 Apr 2010 22:44 On Tue, 6 Apr 2010, Paul Stowe wrote: > On Apr 5, 8:03 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Mon, 5 Apr 2010, PaulStowewrote: > > > On Apr 4, 3:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > On Apr 5, 2:19 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > What have I gotten wrong here? From the above, I have maximum force > > > > > > possible per kg is only Qu = 2e-5N, which is unrealistically low. > > > > > > Meanwhile, for a block a steel, the linear attenuation coefficient > > > > > > would be 0.02. > > > > > > You don't comment at all on the calculation of force. Using the values > > > > you give for Q and u, I get a maximum force per kg of 2e-5N. This is > > > > rather less than we observe terrestrially, and in turn, this must be > > > > far less than the maximum possible since shielding is negligible. > > > > Since you have repeated the correctness of your values for Q and u, > > > > why does the above calculation give a value of l very different from > > > > rho*u? > > > > > As for the force of gravity, this is why big G is Qu^2, NOT! qu. The > > > Newtonian is, technically in LeSage's weak region, > > > > > F = Q(uM)(um)/r^2 > > > > > The total 'potential', or what we call acceleration due to gravity, > > > is, > > > > > a = Qu(Mu/r^2) > > > > > BUT! you MUST have that second mass m to realize this 'potential'. > > > Given that u is independent of m this 'potential' is considered mass > > > indenpendent. This is WHY! big G is only truly valid for multibody > > > problems. Trying to get the right answer with only Qu just doesn't > > > cut it. > > > > > Look at Qu (kg/m-sec^2)(m^2/kg) => m/sec^2, it's not a force, it's the > > > field's attenuation potential. > > > > I didn't say that Qu was a force. I said that Qu*mass is a force, the > > force you'd get on a mass if you had complete screening, with the momentum > > flux Q incident from one side only. > > Yes, and I've been trying to tell you that QuM is not a proper > expression in the model. Why isn't it? A momentum flux, a mass absorption coefficient, and a mass should together give the rate of transfer of momentum to said mass. That is, a force exerted on the mass. If it was what you were trying to say, you could have said it, instead of saying that Qu isn't a force (which it obviously isn't). > > If you want to use a cube of some size, of some density, you get the same > > result, as long as you're still in the weak attenuation limit. > > In the weak attenuation limit there is more spacing than attenuation > sites (that what matter is in the model). As such each attenuator is > exposed to the field flux and has a cross-sectional area. It a 3D > problem, and it is the sum of ALL of the attenuating areas that > produce the total effect. That WHY I expressed the equation the way I > did uM => total cross-sectional area. Note that uM for the Earth is > MUCH greater than its physical area. I could be a modernist and just > say 'shut up & calculate' :) but that not the way to understanding. I see a serious problem here, a very serious problem for the theory. Le Sage must operate in the weak absorption limit for gravitation by the Earth, due to limits imposed by lack of observable shielding. Problem 1: In the weak absorption limit, uM is the total interaction cross-section, and the interaction cross-section is much less than the geometric cross-section. If uM >> geometric cross-section, then you aren't in the weak absorption limit, and the theory has failed. For uM much greater than geometric cross-section, you have very strong absorption, and the interaction cross-section will be close to the geometric cross-section (but still less, unless absorption is total, which you don't get with exponential absorption). > > Isn't saying that the momentum flux Q is 6.74 (kg.m/s)/m^2/s the same > > as saying that the force that would act on a perfectly absorbing square 1m > > by 1m in area, completely shielded from one side, would be 6.74N? > > Yup, that an interesting problem, isn't it... The exposure of more > attenuating area ican increase the effect per unit volume because its > a 4 pi omni-directional flux and, being constant per unit volume has > some counter intuitive aspects. Problem 2: Not that counter-intuitive. 6.74N is what you'd have if the flux were unidirectional. Not being unidirectional will reduce the maximum force possible, and make things worse. Not much worse, but it certainly doesn't help. > > If not, why not? Surely the momentum flux of the corpuscles must be much > > greater than the gravitational forces we see, since a tiny reduction in > > the flux from one side, resulting in an equally tine anisotropy in the > > flux, with a tiny absorption, is meant to lead to the gravitational forces > > we see. > > > > If the maximum force we could get on an object of 1m^2 in cross-section > > were only 6.74N, how can we see forces larger than this, in real life, on > > objects smaller than this? > > Well, in the end, do the math using the terms 6.744 for Q and 3.147E-6 > for u. You'll find that it yields Newton's equations. Why is a very > interesting physical process... Le Sage absorption gives the Newtonian result for terrestrial gravitation only in the weak absorption limit. So, with no restriction on Q and u other than one must remain in the weak absorption limit, one obtains Newtonian gravitation, and G=Qu^2. Problem 3: Weak absorption means that u*density*thickness must be small, much less than 1. u*density being approximately 0.01 means that any thickness of, say, greater than 10m, is already past weak absorption. Since the thickness of the Earth is much greater than 10m, the necessary condition is not satisfied. This is a real problem. It might look at first like the maths tells you that any Q, u that satisfy G=Qu^2 is OK, but u must also be such that you stay in the weak absorption limit. Taken together, these problems are immediately fatal. Not necessarily fatal to a le Sage theory, but fatal to your choice of Q and u. (Well, I think that each point is immediately fatal, but this doesn't change the conclusion when they're taken together.) If the values of Q and u that follow from G and heating are such that u means you no longer have weak absorption, the theory, with this Q and u, is dead. Perhaps your assumption that the corpuscle speed is c is wrong? > As I mentioned earlier, this leads to a path of unification between EM- > SM and gravity. If Q is, in fact, the momentum flux of the universal > aether medium,and given Maxwell own definition of c^2 = 1/u'z Where > u' -> compressibility and z the density, then these terms are > fundamental also to Q. In fact, > > Q = a^2/u'(2pi) = (e^2/2h)^2/z2pi = 6.744 kg/m-sec^2 > > Where a is alpha (1/137.0358) and e elemental charge (1.60217E-19 kg/ > sec) > > So, we have here a direct connection between EM-SM and gravity.... > > As for the 3kT well if v = c and the kinetic energy is 1/2mc^2 = 3/2kT > then mc^2 = 3kT. Given mc^2 = h(nu) well, what more do you want? You > really think that given Maxwell's definitions of u' & z and thus > leading to e/m being a frequency that it is a coincidence that h(e/m)/ > 3k = 2.8 Degrees K??? Hell of a one if it is. If (1/2)mc^2 is the energy, and h(nu) is the energy, I would want a little more than mc^2 = h(nu). If you search for coincidences, you'll find them. Numerologists have been making money for many centuries. In any event, I'd make sure that the gravity part of the theory works properly before getting too excited about such unification. -- Timo
From: Timo Nieminen on 7 Apr 2010 00:57 On Wed, 7 Apr 2010, Timo Nieminen wrote: > On Tue, 6 Apr 2010, Paul Stowe wrote: > > > Yup, that an interesting problem, isn't it... The exposure of more > > attenuating area ican increase the effect per unit volume because its > > a 4 pi omni-directional flux and, being constant per unit volume has > > some counter intuitive aspects. > > Problem 2: Not that counter-intuitive. 6.74N is what you'd have if the > flux were unidirectional. Not being unidirectional will reduce the maximum > force possible, and make things worse. Not much worse, but it certainly > doesn't help. In Edwards (pg 187), you have max flux = Q * pi * (r/R)^2 for a spherical body that absorbs completely. Given that Q is defined as the incident flux per unit area, it is odd that shielding from one side gives a net flux larger than this, but I assume this is in the definition of "net flux" as a different kind of flux. OK, with this, you get a maximum force, next to a completely shielding body, of 6.74 * pi = 21 N per kg. (Eqn (19), on pg 188, is wrong; this should be F = Qu m2 pi (r/R)^2, not F = Qu^2 m2 pi (r/R)^2. Just repeat the weak limit calculation, replacing the weak limit net flux with the strong limit net flux, and this is what you get. In Edwards, pg 188, it's clearly stated (in words, rather than as a mathematical expression) that F = Q_net u m. So why have you been trying to say that Qu * mass isn't a force? -- Timo
From: NoEinstein on 7 Apr 2010 08:41
On Apr 5, 8:52 pm, spudnik <Space...(a)hotmail.com> wrote: > Dear Spudnik: You've got a lot going for you. In the case of the present reply of yours, "a lot" is too much book to read. If any portion of your treatise relates to the mechanism of gravity, and especially, if you take issue with my assertion that ETHER FLOW is the mechanism of gravity, give-me-your-best-shot at proving me wrong. To date no one has proven me wrong about anything. I invite any of the readers to do that, too, but please stick to one issue at a time. Spudnik's book-size replies will go unread. NoEinstein > > nah; we should blame Pascal for discovering, > experimentally, his "plenum," which he thought was perfect. I mean, > it's always good to have a French v. English dichotomy, > with a German thrown-in for "triality." > > > of Newton's "action at a distance" of gravity, > > via the re-adumbration of his dead-as- > > a-doornail-or-Schroedinger's-cat corpuscle, > > "the photon." well, and/or "the aether," > > necessitated by "the vacuum." > > --Light: A History!http://21stcenturysciencetech.com > > --NASCAR rules on rotary engines!http://white-smoke.wetpaint.com > > thus: > Death to the lightcone -- > long-live Minkowski!... yeah; and, > the photon is *still* dead, > no matter what herr Albert said about it! > > > > <<pseudoscientists rarely revise. The first edition of > > > Principia Mathematica, a product of a committee, > > > the Royal Society, after "the MS burnt in an alchemical > > > process that set the trunk in which it was resting, afire," > > > has had several editions, the latter of which take pains > > > to omit mention of Robert Hooke. The sole calculus is > > > is a rectangle, dxdy, in Book 2, Section 2, Paragraph 2.>> > > thus: > as a student of Bucky Fuller -- an army of one, I say -- you've bit- > off > more than you should want to chew, with the n-hole spin on fullerenes; > and that is my clue, because a fullerene should have a very large > manifestation of polarization, not unlike in a game of futbol. I > mean, > just becaus the ball went through only one slit, why wouldn't it be > affected by the total symmetry of the instrumentation?... > all of it, down to teh electronics etc. > > my main thing was, though, that you should at least *try* > to consider the theory of light using only waves, > which can still be pieced-together from almost any "undergrad" > textbook, post-Copenhagen, especially older ones. > > or, just stick with Einstein's refurbishment of Newton's crappy > "theory," > nothing of which is needed for relativity & so on. anyway, > one simply does not need to analyze a phenomenon > by *both* its wavey & bullety aspects -- at the same time; > once you have proven a theorem in projective geometry e.g., > you do not have to give the "2nd column proof," unless > you're just learning it, for the first time! > > >http://en.wikipedia.org/wiki/Louis_de_Broglie > > any moving particle or object had an associated wave." > > thus: > a-ha, I was correct: > say "half," with respect to the beamsplitters, please (as > I comprehend, they generally split the "photon" > into "two photons" of half the energy, I think > of a different frequency, not amplitude -- although > the "photon" is really more akin to a phonon, > such as the audible "click" of the geiger-counter. the *proviso* > with these experiments is that the waves are highly modified > in the LASER apparatus, so that some folks more easily think > of them as "rocks o'light." > > it could have been worse; > lots of more-or-less literate folks use "of" > in the place of "have" -- to be or not to be owned, > that is this particualr question! > > >http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser#The_experi... > > thus: > if you let go of the empty notion of "photon," > there isn't any difficulty, at all, with a geometrical picture. > Death to the lightcone -- long-live the lightcone-heads (because, > Minkowski was only one of them, by haphazard default/death). > yes, I know, that *photonics* is a whole field of engineering; > thank you, herr doktor-professor E., > for unburying Newton's bogus corpuscle and attendant "theory," > that Young had successfully popped! > > thus: > on the wayside, if > you are really going to set so much store in a two-hole procedure > for fullerenes, maybe you shouold read the original article, and > try to question its purpose. as it is, I'd guess that > English is not your mother-tongue, > which can sometimes prove difficult in *using* it; so, > that's why I always suggest Shakespeare, becuase > *no* one can *begin* to comprehend English, > til he *tries* to read the bard. (he also had a hand > in translating the KJV of the Bible .-) > > thus: > NB, quaternions are not "quadrays" (for an amateur attempt > at homogenous co-ordination), but you can "do" special rel. > with them (according to Lanczos .-) > > thus: > The "cap & trade" omnibus bill -- what Waxman-Markey should > be known as, being so fundamental to the Stupid, economy -- is > at least as old as Waxman's '91 bill to ameliorate acid rain. One > must really stop and consider, just who really opposes this "last > hurrah" for Wall Street (like-wise, the healthcare bill, also > under Waxman's House committee, and which, > after all, is geared toward funding a smaller aspect of the S-- > the economy, already tremendously leveraged by the "voluntary" > cap & trade, which the bill would essentially mandate, > a la the much-larger, market-making EU scheme). > > Not so long ago, there was a guest-editorial in the WSJ, > which mentioned that a carbon tax would achieve the same thing, > more or less, as the total "free" trade approach of cap & trade; oh, > but, there're certain, so-called Republicans, who refer to the bill > as "cap & tax!" > > Well, before any "reform" of the financial system, why > would one put all of one's eggs into such a casino -- especially > considering that the oil companies have not bothered > to release the carbon-dating "fingerprints" that they use, > to determine whether two wells are connected, underground; so, > guys & gals, how old is the stuff, on average, anyway? > > Surely, the green-niks who lobby for "renewable" energy, do not think > that oil comes only from dinosaurs, and their associated flora -- > all, from before the asteroid supposedly offed them (I refer them > to the recent issue of Nature -- several articles that may be > related!) > > Finally, note that, in a sense, the whole world is going a) > nuclear, and b) into space, while we are essentially frozen > into '50s and '60s techniques in these crucial frontiers. (While some > folks dither about Iran's nuke-weapons policy, they are rapidly > achieving a full-scale nuke-e and process-heat capbility > for industry & infrastructure.) > > --yr humble servant, the Voting Rights Act o'65 > (deadletter since March 27, 2000, > when Supreme Court refuzed appeal in LaRouche v. Fowler ('96)) |