Why GPS Atomic Clocks Run More Slowly Before Launch
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From: Dirk Bruere at NeoPax on 14 Jun 2010 14:57 On 14/06/2010 18:33, mpc755 wrote: > On Jun 14, 1:17 pm, Dirk Bruere at NeoPax<dirk.bru...(a)gmail.com> > wrote: >> On 14/06/2010 17:23, mpc755 wrote: >> >> >> >>> On Jun 14, 11:24 am, Dirk Bruere at NeoPax<dirk.bru...(a)gmail.com> >>> wrote: >>>> On 14/06/2010 06:54, Mitchell Jones wrote: >> >>>>> ***{It is. The surface of the Earth's pool lies within the region of >>>>> gravitational dominance of the Earth. Above the surface there is a an >>>>> interface zone, where the Earth's pool, which moves with the Earth, >>>>> comes into turbulent contact with the Sun's pool. Since the Earth moves >>>>> around the Sun at 29 kps, those are the relative speeds at which the two >>>>> pools interact across the interface zone. >> >>>>> Above that zone of turbulence, we move into the Sun's pool, which >>>>> includes all regions of the Solar System not within the gravitational >>>>> dominance of lesser bodies, such as planets, moons, asteroids, comets, >>>>> etc. Thus each body within the solar system carries with it its own >>>>> aether pool, in its region of gravitational dominance. >> >>>>> Above the surface of the Sun's pool, we transition into the aether pool >>>>> of the Milky Way galaxy, which includes all regions in the galaxy not >>>>> within the gravitational dominance of the included stars, clusters of >>>>> stars, black holes, etc. >> >>>>> Above the surface of the galaxy's pool lies the local group's pool, >>>>> which includes all regions of the local group not within the >>>>> gravitational dominance of the included galaxies and other structures. >> >>>>> Bottom line: each gravitating body or large scale structure in the >>>>> universe carries within the zone of its gravitational dominance an >>>>> associated aether pool, which extends throughout the region not >>>>> dominated by lesser bodies, whether gravitationally bound to the primary >>>>> or just passing through. >> >>>>> --Mitchell Jones}*** >> >>>>> ***************************************************************** >>>>> If I seem to be ignoring you, consider the possibility >>>>> that you are in my killfile. --MJ >> >>>> And so we have a fluid that exerts pressure but does not do so in a >>>> manner that causes drag? >> >>>> -- >>>> Dirk >> >>>> http://www.transcendence.me.uk/-Transcendence UKhttp://www.blogtalkradio.com/onetribe-Occult Talk Show >> >>> 'On the super-fluid property of the relativistic physical vacuum >>> medium and the inertial motion of particles' >>> http://arxiv.org/ftp/gr-qc/papers/0701/0701155.pdf >> >>> "Abstract: The similarity between the energy spectra of relativistic >>> particles and that of quasi-particles in super-conductivity BCS theory >>> makes us conjecture that the relativistic physical vacuum medium as >>> the ground state of the background field is a super fluid medium, and >>> the rest mass of a relativistic particle is like the energy gap of a >>> quasi-particle. This conjecture is strongly supported by the results >>> of our following investigation: a particle moving through the vacuum >>> medium at a speed less than the speed of light in vacuum, though >>> interacting with the vacuum medium, never feels friction force and >>> thus undergoes a frictionless and inertial motion." >> >>> A particle in the super fluid medium displaces the super fluid medium, >>> whether the particle is at rest with respect to the super fluid >>> medium, or not. A moving particle creates a displacement wave in the >>> super fluid medium. >> >>> A particle in the aether displaces the aether, whether the particle is >>> at rest with respect to the aether, or not. The particle could be an >>> individual nucleus. A moving particle creates a displacement wave in >>> the aether. >> >>> The super fluid medium is not at rest when displaced and 'displaces >>> back'. The 'displacing back' is the pressure exerted by the displaced >>> super fluid medium towards the particle. >> >>> Gravity is pressure exerted by displaced aether towards matter. >> >> Nice to see you relying on relativity theory to prove the Ether exists... >> > > Einstein's 'First' paper was all about the ether. > > 'Albert Einstein's 'First' Paper > http://www.worldscibooks.com/etextbook/4454/4454_chap1.pdf > > Einstein relied on ether in relativity. > > 'Ether and the Theory of Relativity by Albert Einstein' > http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html > > "According to the general theory of relativity space without ether is > unthinkable" .... a nice quote almost 100 years out of date. > In the Casimir Effect, the aether displaced by each of the plates > extends past the other plate, forcing the plates together. Why should it force the plates together? Why not apart, for example? -- Dirk http://www.transcendence.me.uk/ - Transcendence UK http://www.blogtalkradio.com/onetribe - Occult Talk Show
From: Mitchell Jones on 14 Jun 2010 15:39 In article <hv4bo3$7k4$1(a)news.eternal-september.org>, Xan Du <xan747(a)yahoo.com> wrote: > On 6/14/2010 12:20 AM, mpc755 wrote: > > On Jun 13, 9:50 pm, Xan Du<xan...(a)yahoo.com> wrote: > >> On 6/13/2010 9:37 PM, Mitchell Jones wrote: > >> > >> > >> > >>> In article > >>> <37c0330d-8b86-4efc-9c1f-4db7b17f3...(a)u26g2000yqu.googlegroups.com>, > >>> mpc755<mpc...(a)gmail.com> wrote: > >> > >>>> The rate at which an atomic clock 'ticks' is based upon the aether > >>>> pressure in which it exists. In terms of motion, the speed of a GPS > >>>> satellite with respect to the aether causes it to displace more aether > >>>> and for that aether to exert more pressure on the clock in the GPS > >>>> satellite than the aether pressure associated with a clock at rest > >>>> with respect to the Earth. This causes the GPS satellite clock to > >>>> "result in a delay of about 7 s/day". The aether pressure associated > >>>> with the aether displaced by the Earth exerts less pressure on the GPS > >>>> satellite than a similar clock at rest on the Earth" causing the GPS > >>>> clocks to appear faster by about 45 s/day". The aether pressure > >>>> associated with the speed at which the GPS satellite moves with > >>>> respect to the aether and the aether pressure associated with the > >>>> aether displaced by the Earth causes "clocks on the GPS satellites > >>>> [to] tick approximately 38 s/day faster than clocks on the ground." > >>>> (quoted text > >>>> fromhttp://en.wikipedia.org/wiki/Effects_of_relativity_on_GPS). > >> > >>> ***{The use of the term "displacement" brings to mind buoyancy concepts, > >>> and evokes images such as that of a boat displacing the water in which > >>> it is immersed. That image, however, is utterly wrong: in the case of > >>> velocity entrainment, the aether actually flows into the moving object. > >> > >>> The actual situation with a GPS satellite, for example, is that as the > >>> satellite moves around the Earth in a circular orbit at 3,875 m/s, it > >>> moves through, and hence relative to, the vast pool of aether entrained > >>> by the Earth. Thus we can, by switching to coordinates centered on the > >>> satellite, create a picture wherein the satellite is stationary and the > >>> aether is flowing past it at v = 3,875 m/s. > >> > >>> Within that framework, it becomes easy to visualize the situation in > >>> terms of the Bernoulli effect: the aether slows down as it penetrates > >>> into the satellite and into the atomic clock inside the satellite, where > >>> it achieves a speed of essentially zero with respect to the > >>> satellite--which means: as the satellite moves, the aether within it > >>> also moves, in the same direction at the same speed. Result: the energy > >>> that manifests as kinetic energy outside of the satellite is converted > >>> into pressure energy within the satellite. > >> > >>> Let's consider the situation mathematically. > >> > >>> The Earth's pool of entrained aether is a spherical fluid reservoir > >>> which extends from the center of Earth up to a surface that is millions > >>> of meters out in space. For such a reservoir considered in its entirety, > >>> variations in the acceleration due to gravity, g, will be very large, > >>> and cannot be neglected. However, within a sufficiently narrow range of > >>> altitudes, g will be effectively constant, and can be treated as such. > >>> In that case, if D as the distance in meters from the bottom of the > >>> Earth's pool to the surface, there will be horizontal bands within > >>> which, for all volumes of mass m, the following will be true: > >> > >>> mgD = C > >> > >>> In the above, C is a constant. > >> > >>> If h is the height of a location in meters above the bottom of the > >>> Earth's pool, and d is its depth in meters below the surface, then D = h > >>> + d, and it follows that > >> > >>> mgh + mgd = C > >> > >>> The quantity mgh is the potential energy, U, of a stationary mass of > >>> aether at the location of interest, and mgd is its "pressure-energy," > >>> which we will denote by P. > >> > >>> Hence U + P = C. > >> > >>> Therefore we can say that, within such a horizontal band, > >> > >>> U1 + P1 = U2 + P2 > >> > >>> Which expands to: > >> > >>> mgh1 + mgd1 = mgh2 + mgd2 > >> > >>> The above statement compares two separate volumes of aether of mass m > >>> within the horizontal band of altitudes. > >> > >>> Now let's suppose that the second volume is within a GPS satellite which > >>> is in inertial motion at a speed of v = 3876 m/s relative to the aether > >>> within Earth's pool. Since aether penetrates through the interstices of > >>> matter, the interior of the satellite, including the spaces between > >>> atoms, is filled with it. However, the necessity to flow in circuitous > >>> paths around the atoms of matter has the effect of entraining the aether > >>> within the satellite: as the satellite moves, the aether within it also > >>> moves, in virtually the same way as the satellite. Hence the velocity of > >>> the satellite relative to the surrounding aether is also the velocity of > >>> the aether within the satellite. To express that state of affairs, the > >>> preceding equation may be adjusted as follows: > >> > >>> (1/2)mv^2 + mgh1 + mgd1 = mgh2 + mgd2 + (1/2)mv^2 > >> > >>> In the above, (1/2)mv^2 has been added to both sides, so the statement > >>> remains mathematically sound. But what is its physical meaning? If, for > >>> example, the vertical axis is taken to pass through the center of the > >>> Earth and the center of the satellite and to move with the satellite, > >>> the satellite becomes "stationary" by stipulation, and the aether > >>> outside the satellite is then moving past the satellite at the speed of > >>> 3976 m/s. Thus the v on the left side of the equation is easy to > >>> interpret: it refers to the speed of a volume of mass m moving past the > >>> satellite in the retrograde direction at 3976 m/s. But what does v mean > >>> on the right side, given that the volume of mass m inside the satellite > >>> is attached to the satellite, which is stationary by stipulation? > >> > >>> The answer is obtainable only by the use of physical understanding based > >>> on our experiences with the behavior of fluids. We know, for example, > >>> that if we stick an arm out the window as we drive down the road at 100 > >>> kph, we will experience a force pushing the arm toward the back of the > >>> car. Because we are aware of the existence of air, we know that a moving > >>> stream of air is going past the car in the opposite direction, that the > >>> stream slows and divides into an upper and lower stream as it strikes > >>> our arm, then speeds back up again as it converges back into a single > >>> stream behind our arm. Because energy is conserved, the kinetic energy > >>> lost when the stream slows down in front of our arm must go somewhere. > >>> But where does it go? The answer: the drop in speed in the front begets > >>> a rise in pressure in the front, and the rise in speed at the back > >>> begets a fall in pressure there. The force pushing our arm back, in > >>> short, arises because the air pressure is higher at the front than at > >>> the back. Kinetic energy lost when the air slows down at the front does > >>> not cease to exist: it is converted into pressure energy; and pressure > >>> energy lost when the air speeds back up again behind the arm does not > >>> cease to exist, either: it is merely converted back into kinetic energy > >>> again. > >> > >>> Therefore we interpret the above equation in accordance with the > >>> following: > >> > >>> K1 + U1 + P1 = K2 + U2 + P2 > >> > >>> In the above, K1 = (1/2)mv^2; U1 = mgh1; P1 = mgd1; K2 = 0; U2 = mgh2; > >>> and P2 = mgd2 + (1/2)mv^2. > >> > >>> We can emphasize that effect by the way we group the terms: > >> > >>> (1/2)mv^2 + mgh1 + mgd1 = 0 + mgh2 + [mgd2 + (1/2)mv^2] > >> > >>> In short, (1/2)mv^2 on the right side--i.e., within the satellite--is a > >>> measure of the increment to pressure-energy that is necessary to > >>> conserve energy, given the kinetic energy possessed by an equal mass of > >>> fluid on the outside. > >> > >>> Such insights enable us to make sense out of the experimentally derived > >>> equation below: > >> > >>> t' = t[1 - v^2/c^2]^.5 > >> > >>> What it means is that (a) aether pressure rises within regions where the > >>> motion of aether is impeded, relative to surrounding regions where it is > >>> in unimpeded motion, (b) the aether is very nearly incompressible, but, > >>> if the relative motion between an impeded region and its surroundings is > >>> great enough, the gradual rise in pressure within the impeded flow > >>> region will begin to have measurable effects, (c) one of those effects > >>> will be a gradual compression that, in turn, will bring about a slowing > >>> of all movement within the region, and (d) the pattern of that slowing, > >>> based on experimental results, is described by the above equation. > >> > >>> What this all has to do with "aether displacement," on the other hand, I > >>> have no idea. > >> > >>> --Mitchell Jones}*** > >> > >>> ***************************************************************** > >>> If I seem to be ignoring you, consider the possibility > >>> that you are in my killfile. --MJ > >> > >> So, why isn't the aether interplanetary? Interstellar? Intergalactic? > >> > >> -Xan > > > > It is. > > "The Earth's pool of entrained aether is a spherical fluid reservoir > which extends from the center of Earth up to a surface that is millions > of meters out in space." > > That describes a sphere. Did I miss something? > > -Xan ***{Defining the actual shape of the Earth's pool takes a lot of computer number crunching. What you would have to do would be to write a computer program comparing the strength of the Earth's field to that of the Sun, and generate a display of the results. You might, for example, color as red the points where the Earth's field was stronger, and color as green the points where the Sun's field was stronger. We can't do that much number crunching here, but we can get a rough idea of the shape. To that end, consider the radius beginning at the center of the Sun and passing outward through the center of the Earth, and ask yourself how far out from the center of Earth you would have to go on that line to reach each of the two points where the gravitational force exerted by the two bodies was equal. (To reach one point, you go toward the Sun; to reach the other, you go away from it.) The force exerted by the Earth on an object is: Fe = G(Me)m/r^2 The force exerted by the Sun on the same object: Fs = G(Ms)m/(R - r)^2 We are looking for the point where Fe = Fs, so substitution gives us: G(Me)m/r^2 = G(Ms)m/(R - r)^2 Dropping equal factors out of both sides gives: Me/r^2 = Ms/(R - r)^2 Sundry manipulations give: r^2(1 - Ms/Me) - 2Rr + R^2 = 0 Solving the above quadratic equation for r yields up two roots: r1 = {2R + [4R^2 - 4(1 - Ms/Me)R^2]^.5}/2(1 - Ms/Me) = {2R + [4R^2(1 - (1 - Ms/Me))]^.5}/2(1 - Ms/Me) = {2R + [4R^2(Ms/Me)]^.5}/2(1 - Ms/Me) The other root is: r2 = {2R - [4R^2(Ms/Me)]^.5}/2(1 - Ms/Me) In the above, the mass of Earth is Me = 5.974x10^24 kg, the mass of the Sun is Ms = 1.989x10^30 kg, and the mean radius of Earth's orbit is R = 149598261000 m, so we have: r1 = {2(149598261000) + [4(149598261000)^2(1.989x10^30/5.974x10^24)]^.5}/2(1 -1.989x10^30/5.974x10^24) = - 259,713,830 m, or about 260,000 km to the outside of Earth's orbit. r2 = 258,815,185, or about 259,000 km to the inside of Earth's orbit. The reason for the difference is that Solar gravity is slightly stronger on the sunward side of Earth than on the dark side. Since the gravitational force exerted by the Sun only varies slightly within 260,000 km of Earth, it follows that the Earth's pool is spherical to pretty good accuracy, and so we can say that the mean radius--i.e., the depth--of Earth's pool is: D = (259713830 + 258815185)/2 = 259264507 m, which is about 259,300 km. The above calculation, of course, ignores the effect of whatever zone of turbulence may exist near the contact interface between the Earth's pool and the Sun's pool. Note also that the Moon's orbit (at about 384,400 km) is outside the Earth's pool, entirely in the zone of gravitational dominance of the Sun. The reason is that it is the gravitational field of the Sun that holds the Moon (and the Earth) in its orbit around the Sun. The effect of the nearby presence of the Earth is merely to induce a slight wiggle into the path that the Moon takes around the Sun. --Mitchell Jones}*** ***************************************************************** If I seem to be ignoring you, consider the possibility that you are in my killfile. --MJ
From: mpc755 on 14 Jun 2010 15:49 On Jun 14, 2:57 pm, Dirk Bruere at NeoPax <dirk.bru...(a)gmail.com> wrote: > On 14/06/2010 18:33, mpc755 wrote: > > > > > On Jun 14, 1:17 pm, Dirk Bruere at NeoPax<dirk.bru...(a)gmail.com> > > wrote: > >> On 14/06/2010 17:23, mpc755 wrote: > > >>> On Jun 14, 11:24 am, Dirk Bruere at NeoPax<dirk.bru...(a)gmail.com> > >>> wrote: > >>>> On 14/06/2010 06:54, Mitchell Jones wrote: > > >>>>> ***{It is. The surface of the Earth's pool lies within the region of > >>>>> gravitational dominance of the Earth. Above the surface there is a an > >>>>> interface zone, where the Earth's pool, which moves with the Earth, > >>>>> comes into turbulent contact with the Sun's pool. Since the Earth moves > >>>>> around the Sun at 29 kps, those are the relative speeds at which the two > >>>>> pools interact across the interface zone. > > >>>>> Above that zone of turbulence, we move into the Sun's pool, which > >>>>> includes all regions of the Solar System not within the gravitational > >>>>> dominance of lesser bodies, such as planets, moons, asteroids, comets, > >>>>> etc. Thus each body within the solar system carries with it its own > >>>>> aether pool, in its region of gravitational dominance. > > >>>>> Above the surface of the Sun's pool, we transition into the aether pool > >>>>> of the Milky Way galaxy, which includes all regions in the galaxy not > >>>>> within the gravitational dominance of the included stars, clusters of > >>>>> stars, black holes, etc. > > >>>>> Above the surface of the galaxy's pool lies the local group's pool, > >>>>> which includes all regions of the local group not within the > >>>>> gravitational dominance of the included galaxies and other structures. > > >>>>> Bottom line: each gravitating body or large scale structure in the > >>>>> universe carries within the zone of its gravitational dominance an > >>>>> associated aether pool, which extends throughout the region not > >>>>> dominated by lesser bodies, whether gravitationally bound to the primary > >>>>> or just passing through. > > >>>>> --Mitchell Jones}*** > > >>>>> ***************************************************************** > >>>>> If I seem to be ignoring you, consider the possibility > >>>>> that you are in my killfile. --MJ > > >>>> And so we have a fluid that exerts pressure but does not do so in a > >>>> manner that causes drag? > > >>>> -- > >>>> Dirk > > >>>>http://www.transcendence.me.uk/-TranscendenceUKhttp://www.blogtalkradio.com/onetribe-OccultTalk Show > > >>> 'On the super-fluid property of the relativistic physical vacuum > >>> medium and the inertial motion of particles' > >>>http://arxiv.org/ftp/gr-qc/papers/0701/0701155.pdf > > >>> "Abstract: The similarity between the energy spectra of relativistic > >>> particles and that of quasi-particles in super-conductivity BCS theory > >>> makes us conjecture that the relativistic physical vacuum medium as > >>> the ground state of the background field is a super fluid medium, and > >>> the rest mass of a relativistic particle is like the energy gap of a > >>> quasi-particle. This conjecture is strongly supported by the results > >>> of our following investigation: a particle moving through the vacuum > >>> medium at a speed less than the speed of light in vacuum, though > >>> interacting with the vacuum medium, never feels friction force and > >>> thus undergoes a frictionless and inertial motion." > > >>> A particle in the super fluid medium displaces the super fluid medium, > >>> whether the particle is at rest with respect to the super fluid > >>> medium, or not. A moving particle creates a displacement wave in the > >>> super fluid medium. > > >>> A particle in the aether displaces the aether, whether the particle is > >>> at rest with respect to the aether, or not. The particle could be an > >>> individual nucleus. A moving particle creates a displacement wave in > >>> the aether. > > >>> The super fluid medium is not at rest when displaced and 'displaces > >>> back'. The 'displacing back' is the pressure exerted by the displaced > >>> super fluid medium towards the particle. > > >>> Gravity is pressure exerted by displaced aether towards matter. > > >> Nice to see you relying on relativity theory to prove the Ether exists.... > > > Einstein's 'First' paper was all about the ether. > > > 'Albert Einstein's 'First' Paper > >http://www.worldscibooks.com/etextbook/4454/4454_chap1.pdf > > > Einstein relied on ether in relativity. > > > 'Ether and the Theory of Relativity by Albert Einstein' > >http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html > > > "According to the general theory of relativity space without ether is > > unthinkable" > > ... a nice quote almost 100 years out of date. > Correct does not go out of date. Einstein's 'First' paper was all about the ether. 'Albert Einstein's 'First' Paper http://www.worldscibooks.com/etextbook/4454/4454_chap1.pdf Einstein relied on ether in relativity. 'Ether and the Theory of Relativity by Albert Einstein' http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html "According to the general theory of relativity space without ether is unthinkable" "the state of the [ether] is at every place determined by connections with the matter and the state of the ether in neighbouring places, ... disregarding the causes which condition its state." The state of the aether as determined by its connections with the matter and the state of the aether in neighboring places is the aether's state of displacement. > > In the Casimir Effect, the aether displaced by each of the plates > > extends past the other plate, forcing the plates together. > > Why should it force the plates together? Why not apart, for example? > 'Casimir effect' http://en.wikipedia.org/wiki/Casimir_effect "Because the strength of the force falls off rapidly with distance, it is only measurable when the distance between the objects is extremely small." With the distance between the plates being extremely small the aether displaced by each plate extends past the other plate. The aether is not at rest when displaced and 'displaces back'. The 'displacing back' is the pressure exerted by the displaced aether towards the matter doing the displacing. Since the plates are extremely close together and the force of the displaced aether is back towards the matter doing the displacing and since the aether displaced by each plate extends outside of the other plate the pressure exerted by the displaced aether forces the plates together. > -- > Dirk > > http://www.transcendence.me.uk/- Transcendence UKhttp://www.blogtalkradio..com/onetribe- Occult Talk Show
From: Mitchell Jones on 14 Jun 2010 20:25 In article <mjones-B0A1AF.14390414062010(a)newsfarm.iad.highwinds-media.com>, Mitchell Jones <mjones(a)21cenlogic.com> wrote: > In article <hv4bo3$7k4$1(a)news.eternal-september.org>, > Xan Du <xan747(a)yahoo.com> wrote: > > > On 6/14/2010 12:20 AM, mpc755 wrote: > > > On Jun 13, 9:50 pm, Xan Du<xan...(a)yahoo.com> wrote: > > >> On 6/13/2010 9:37 PM, Mitchell Jones wrote: > > >> > > >> > > >> > > >>> In article > > >>> <37c0330d-8b86-4efc-9c1f-4db7b17f3...(a)u26g2000yqu.googlegroups.com>, > > >>> mpc755<mpc...(a)gmail.com> wrote: > > >> > > >>>> The rate at which an atomic clock 'ticks' is based upon the aether > > >>>> pressure in which it exists. In terms of motion, the speed of a GPS > > >>>> satellite with respect to the aether causes it to displace more aether > > >>>> and for that aether to exert more pressure on the clock in the GPS > > >>>> satellite than the aether pressure associated with a clock at rest > > >>>> with respect to the Earth. This causes the GPS satellite clock to > > >>>> "result in a delay of about 7 s/day". The aether pressure associated > > >>>> with the aether displaced by the Earth exerts less pressure on the GPS > > >>>> satellite than a similar clock at rest on the Earth" causing the GPS > > >>>> clocks to appear faster by about 45 s/day". The aether pressure > > >>>> associated with the speed at which the GPS satellite moves with > > >>>> respect to the aether and the aether pressure associated with the > > >>>> aether displaced by the Earth causes "clocks on the GPS satellites > > >>>> [to] tick approximately 38 s/day faster than clocks on the ground." > > >>>> (quoted text > > >>>> fromhttp://en.wikipedia.org/wiki/Effects_of_relativity_on_GPS). > > >> > > >>> ***{The use of the term "displacement" brings to mind buoyancy concepts, > > >>> and evokes images such as that of a boat displacing the water in which > > >>> it is immersed. That image, however, is utterly wrong: in the case of > > >>> velocity entrainment, the aether actually flows into the moving object. > > >> > > >>> The actual situation with a GPS satellite, for example, is that as the > > >>> satellite moves around the Earth in a circular orbit at 3,875 m/s, it > > >>> moves through, and hence relative to, the vast pool of aether entrained > > >>> by the Earth. Thus we can, by switching to coordinates centered on the > > >>> satellite, create a picture wherein the satellite is stationary and the > > >>> aether is flowing past it at v = 3,875 m/s. > > >> > > >>> Within that framework, it becomes easy to visualize the situation in > > >>> terms of the Bernoulli effect: the aether slows down as it penetrates > > >>> into the satellite and into the atomic clock inside the satellite, where > > >>> it achieves a speed of essentially zero with respect to the > > >>> satellite--which means: as the satellite moves, the aether within it > > >>> also moves, in the same direction at the same speed. Result: the energy > > >>> that manifests as kinetic energy outside of the satellite is converted > > >>> into pressure energy within the satellite. > > >> > > >>> Let's consider the situation mathematically. > > >> > > >>> The Earth's pool of entrained aether is a spherical fluid reservoir > > >>> which extends from the center of Earth up to a surface that is millions > > >>> of meters out in space. For such a reservoir considered in its entirety, > > >>> variations in the acceleration due to gravity, g, will be very large, > > >>> and cannot be neglected. However, within a sufficiently narrow range of > > >>> altitudes, g will be effectively constant, and can be treated as such. > > >>> In that case, if D as the distance in meters from the bottom of the > > >>> Earth's pool to the surface, there will be horizontal bands within > > >>> which, for all volumes of mass m, the following will be true: > > >> > > >>> mgD = C > > >> > > >>> In the above, C is a constant. > > >> > > >>> If h is the height of a location in meters above the bottom of the > > >>> Earth's pool, and d is its depth in meters below the surface, then D = h > > >>> + d, and it follows that > > >> > > >>> mgh + mgd = C > > >> > > >>> The quantity mgh is the potential energy, U, of a stationary mass of > > >>> aether at the location of interest, and mgd is its "pressure-energy," > > >>> which we will denote by P. > > >> > > >>> Hence U + P = C. > > >> > > >>> Therefore we can say that, within such a horizontal band, > > >> > > >>> U1 + P1 = U2 + P2 > > >> > > >>> Which expands to: > > >> > > >>> mgh1 + mgd1 = mgh2 + mgd2 > > >> > > >>> The above statement compares two separate volumes of aether of mass m > > >>> within the horizontal band of altitudes. > > >> > > >>> Now let's suppose that the second volume is within a GPS satellite which > > >>> is in inertial motion at a speed of v = 3876 m/s relative to the aether > > >>> within Earth's pool. Since aether penetrates through the interstices of > > >>> matter, the interior of the satellite, including the spaces between > > >>> atoms, is filled with it. However, the necessity to flow in circuitous > > >>> paths around the atoms of matter has the effect of entraining the aether > > >>> within the satellite: as the satellite moves, the aether within it also > > >>> moves, in virtually the same way as the satellite. Hence the velocity of > > >>> the satellite relative to the surrounding aether is also the velocity of > > >>> the aether within the satellite. To express that state of affairs, the > > >>> preceding equation may be adjusted as follows: > > >> > > >>> (1/2)mv^2 + mgh1 + mgd1 = mgh2 + mgd2 + (1/2)mv^2 > > >> > > >>> In the above, (1/2)mv^2 has been added to both sides, so the statement > > >>> remains mathematically sound. But what is its physical meaning? If, for > > >>> example, the vertical axis is taken to pass through the center of the > > >>> Earth and the center of the satellite and to move with the satellite, > > >>> the satellite becomes "stationary" by stipulation, and the aether > > >>> outside the satellite is then moving past the satellite at the speed of > > >>> 3976 m/s. Thus the v on the left side of the equation is easy to > > >>> interpret: it refers to the speed of a volume of mass m moving past the > > >>> satellite in the retrograde direction at 3976 m/s. But what does v mean > > >>> on the right side, given that the volume of mass m inside the satellite > > >>> is attached to the satellite, which is stationary by stipulation? > > >> > > >>> The answer is obtainable only by the use of physical understanding based > > >>> on our experiences with the behavior of fluids. We know, for example, > > >>> that if we stick an arm out the window as we drive down the road at 100 > > >>> kph, we will experience a force pushing the arm toward the back of the > > >>> car. Because we are aware of the existence of air, we know that a moving > > >>> stream of air is going past the car in the opposite direction, that the > > >>> stream slows and divides into an upper and lower stream as it strikes > > >>> our arm, then speeds back up again as it converges back into a single > > >>> stream behind our arm. Because energy is conserved, the kinetic energy > > >>> lost when the stream slows down in front of our arm must go somewhere. > > >>> But where does it go? The answer: the drop in speed in the front begets > > >>> a rise in pressure in the front, and the rise in speed at the back > > >>> begets a fall in pressure there. The force pushing our arm back, in > > >>> short, arises because the air pressure is higher at the front than at > > >>> the back. Kinetic energy lost when the air slows down at the front does > > >>> not cease to exist: it is converted into pressure energy; and pressure > > >>> energy lost when the air speeds back up again behind the arm does not > > >>> cease to exist, either: it is merely converted back into kinetic energy > > >>> again. > > >> > > >>> Therefore we interpret the above equation in accordance with the > > >>> following: > > >> > > >>> K1 + U1 + P1 = K2 + U2 + P2 > > >> > > >>> In the above, K1 = (1/2)mv^2; U1 = mgh1; P1 = mgd1; K2 = 0; U2 = mgh2; > > >>> and P2 = mgd2 + (1/2)mv^2. > > >> > > >>> We can emphasize that effect by the way we group the terms: > > >> > > >>> (1/2)mv^2 + mgh1 + mgd1 = 0 + mgh2 + [mgd2 + (1/2)mv^2] > > >> > > >>> In short, (1/2)mv^2 on the right side--i.e., within the satellite--is a > > >>> measure of the increment to pressure-energy that is necessary to > > >>> conserve energy, given the kinetic energy possessed by an equal mass of > > >>> fluid on the outside. > > >> > > >>> Such insights enable us to make sense out of the experimentally derived > > >>> equation below: > > >> > > >>> t' = t[1 - v^2/c^2]^.5 > > >> > > >>> What it means is that (a) aether pressure rises within regions where the > > >>> motion of aether is impeded, relative to surrounding regions where it is > > >>> in unimpeded motion, (b) the aether is very nearly incompressible, but, > > >>> if the relative motion between an impeded region and its surroundings is > > >>> great enough, the gradual rise in pressure within the impeded flow > > >>> region will begin to have measurable effects, (c) one of those effects > > >>> will be a gradual compression that, in turn, will bring about a slowing > > >>> of all movement within the region, and (d) the pattern of that slowing, > > >>> based on experimental results, is described by the above equation. > > >> > > >>> What this all has to do with "aether displacement," on the other hand, I > > >>> have no idea. > > >> > > >>> --Mitchell Jones}*** > > >> > > >>> ***************************************************************** > > >>> If I seem to be ignoring you, consider the possibility > > >>> that you are in my killfile. --MJ > > >> > > >> So, why isn't the aether interplanetary? Interstellar? Intergalactic? > > >> > > >> -Xan > > > > > > It is. > > > > "The Earth's pool of entrained aether is a spherical fluid reservoir > > which extends from the center of Earth up to a surface that is millions > > of meters out in space." > > > > That describes a sphere. Did I miss something? > > > > -Xan > > ***{Defining the actual shape of the Earth's pool takes a lot of > computer number crunching. What you would have to do would be to write a > computer program comparing the strength of the Earth's field to that of > the Sun, and generate a display of the results. You might, for example, > color as red the points where the Earth's field was stronger, and color > as green the points where the Sun's field was stronger. > > We can't do that much number crunching here, but we can get a rough idea > of the shape. To that end, consider the radius beginning at the center > of the Sun and passing outward through the center of the Earth, and ask > yourself how far out from the center of Earth you would have to go on > that line to reach each of the two points where the gravitational force > exerted by the two bodies was equal. (To reach one point, you go toward > the Sun; to reach the other, you go away from it.) > > The force exerted by the Earth on an object is: > > Fe = G(Me)m/r^2 > > The force exerted by the Sun on the same object: > > Fs = G(Ms)m/(R - r)^2 > > We are looking for the point where Fe = Fs, so substitution gives us: > > G(Me)m/r^2 = G(Ms)m/(R - r)^2 > > Dropping equal factors out of both sides gives: > > Me/r^2 = Ms/(R - r)^2 > > Sundry manipulations give: > > r^2(1 - Ms/Me) - 2Rr + R^2 = 0 > > Solving the above quadratic equation for r yields up two roots: > > r1 = {2R + [4R^2 - 4(1 - Ms/Me)R^2]^.5}/2(1 - Ms/Me) > > = {2R + [4R^2(1 - (1 - Ms/Me))]^.5}/2(1 - Ms/Me) > > = {2R + [4R^2(Ms/Me)]^.5}/2(1 - Ms/Me) > > The other root is: > > r2 = {2R - [4R^2(Ms/Me)]^.5}/2(1 - Ms/Me) > > In the above, the mass of Earth is Me = 5.974x10^24 kg, the mass of the > Sun is Ms = 1.989x10^30 kg, and the mean radius of Earth's orbit is R = > 149598261000 m, so we have: > > r1 = {2(149598261000) + > [4(149598261000)^2(1.989x10^30/5.974x10^24)]^.5}/2(1 > -1.989x10^30/5.974x10^24) > > = - 259,713,830 m, or about 260,000 km to the outside of Earth's orbit. > > r2 = 258,815,185, or about 259,000 km to the inside of Earth's orbit. > > The reason for the difference is that Solar gravity is slightly stronger > on the sunward side of Earth than on the dark side. > > Since the gravitational force exerted by the Sun only varies slightly > within 260,000 km of Earth, it follows that the Earth's pool is > spherical to pretty good accuracy, and so we can say that the mean > radius--i.e., the depth--of Earth's pool is: > > D = (259713830 + 258815185)/2 = 259264507 m, which is about 259,300 km. > > The above calculation, of course, ignores the effect of whatever zone of > turbulence may exist near the contact interface between the Earth's pool > and the Sun's pool. Note also that the Moon's orbit (at about 384,400 > km) is outside the Earth's pool, entirely in the zone of gravitational > dominance of the Sun. The reason is that it is the gravitational field > of the Sun that holds the Moon (and the Earth) in its orbit around the > Sun. The effect of the nearby presence of the Earth is merely to induce > a slight wiggle into the path that the Moon takes around the Sun. > > --Mitchell Jones}*** ***{I would add that the value of D is of crucial importance in telling how to calculate the clock offsets that must be applied, when a clock is at sea level, to ensure that after it has been placed in a particular satellite orbit, it will advance at the same rate as clocks on Earth. The key point is this: the speed that must be used is the speed of the clock relative to the aether in which it is immersed. Thus if the orbit is within the Earth's pool, the value of v that must be plugged into t' = [1 - v^2/c^2]^.5 will be the orbital speed of the satellite. If, on the other hand, the orbit is outside the Earth's pool--at 320,000 km, say--then the value of v that must be plugged in will be the speed of the satellite relative to the Sun's pool, not its speed relative to the Earth's pool. And the difference is huge: 29 km/sec, rather than a few thousand meters/sec. (Note: that's the "aether wind" which Michelson was looking for.) The so called "theory of relativity" provides no understanding of the underlying reality in which the mislabeled "time dilation" equations are to be used, and so the decision regarding when to switch from speeds relative to the Earth to speeds relative to the Sun is based on trial and error: you plug in the smaller numbers first and, if they don't give the right answers, you switch to the larger ones. If, however, you have the understandings conveyed by the gravitationally entrained aether theory, no such trial and error is required. --Mitchell Jones}*** ***************************************************************** If I seem to be ignoring you, consider the possibility that you are in my killfile. --MJ
From: mpc755 on 14 Jun 2010 21:19
On Jun 14, 5:50 pm, Dirk Bruere at NeoPax <dirk.bru...(a)gmail.com> wrote: > > Ignorance is lack of math. > You are its epitome. > Ignorance is not being able to explain what occurs physically in nature in the following and not caring to understand what occurs physically in nature in the following when what is occurring physically in nature in the following is easily understood in Aether Displacement. The slits are the length the C-60 molecule travels in one year. A C-60 molecule is in the slit(s). After 6 months detectors are placed at the exits to the slits. A C-60 molecule is in the slit(s). After 6 months detectors are placed at the exits to the slits and 3 months later the detectors are removed. Perform the experiment above and every time the detectors are left at the exits the C-60 molecule is detected exiting a single slit. When the detectors are placed and removed and the experiment is performed over and over again the non-detected C-60 molecules create an interference pattern. Explain what occurs physically in nature to cause this behavior. Does the C-60 molecule enter one slit or multiple slits depending upon there being detectors at the exits to the slits a year into the future? Of course not. A moving C-60 molecule has an associated aether displacement wave. The C-60 molecule enters and exits a single slit. The aether wave enters and exits multiple slits. The aether wave creates interference upon exiting the slits which alters the direction the C-60 molecule travels. Detecting the C-60 molecule causes decoherence of the aether wave (i.e. turns the wave into chop) and there is no interference. You do realize your inability to answer the above thought experiment is evidence Aether Displacement is more correct than whatever nonsense you choose to believe in which does not allow you to explain what occurs physically in nature, correct? The above too difficult for you? How about explaining what occurs physically in nature to cause the Casimir Effect. And if you are going to require 'virtual' particles then describe how a virtual particle exists out of nothing. Can't explain that either, can you? Each of the plates in the Casimir Effect displaces the aether past the outside of the other plate. The aether is not at rest when displaced and 'displaces back'. The 'displacing back' is the pressure exerted by the displaced aether towards the matter. The pressure exerted by the displaced aether which exists outside of the plates forces the plates together. You have no idea what occurs physically in nature in a double slit experiment or in the Casimir Effect. Enjoy your ignorance. |