Why GPS Atomic Clocks Run More Slowly Before Launch
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From: mpc755 on 14 Jun 2010 00:20 On Jun 13, 9:37 pm, Mitchell Jones <mjo...(a)21cenlogic.com> 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 analogy of a boat and its associated bow wave accurately reflects the observed behaviors in a double slit experiment. If you execute a boat double slit experiment the boat would enter and exit a single slit and the associated bow wave would enter and exit multiple slits. The bow wave creates interference upon exiting the slits which alters the direction the boat travels. Detecting the boat (i.e. placing buoys at the exits to the slits in order to detect the boat) causes decoherence of the associated bow wave (i.e. turns the wave into chop) and there is no interference. 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. When an object moves with constant velocity the associated aether pressure is exerted equally to each and every part of each and every nuclei which the object consists of. When an object is accelerating the associated aether pressure is not exerted equally to each and every part of each and every nuclei which the object consists of. > 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. > The faster the satellite moves relative to the aether the more aether the satellite displaces. The aether is not at rest when displaced and 'displaces back'. The 'displacing back' is the pressure associated with the displaced aether. The faster the satellite moves with respect to the aether the more pressure exerted by the displaced aether towards and throughout 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: > Two satellites the same distance from the surface of the Earth are moving with two different constant momentums with respect to the aether. The atomic clock in the satellite traveling faster with respect to the aether 'ticks' slower because that satellite is displacing more aether and that aether is exerting more pressure towards and throughout the satellite. > (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. > You are mistaking the pressure associated with aether when discussing a body moving with constant momentum with respect to the aether with the aether pressure associated with a body accelerating with respect to the aether. When an object moves with constant momentum with respect to the aether the pressure exerted by the aether is equally applied to each and every part of the object. This is what occurs when a particle moves through a frictionless superfluid. When an object accelerates with respect to the aether the pressure exerted by the aether towards the front of the object is greater than the aether pressure exerted towards the back of the object. > 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. > What physically occurs in nature to cause curved 'space-time', E=mc^2, conservation of mass, conservation of energy, gravity, double slit experiments, Higg's background field, what physically occurs in nature to cause atomic clocks to 'tick' at different rates, Einstein's train gedanken, how the torsion in teleparallelism manifests itself as the pressure associated with the displaced aether, MMX 'null' results, Casimir Effect, and on and on and on are the list of things which are easily understood in the theory of Aether Displacement. 'Ether and the Theory of Relativity by Albert Einstein' http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html "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. Aether and matter are different states of the same material. The material is maether. Maether has mass. Aether and matter have mass. Aether is uncompressed maether and matter is compressed maether. Aether is displaced by matter. The aether is not at rest when displaced and 'displaces back'. The 'displacing back' is the pressure exerted by the aether. Gravity is pressure exerted by displaced aether towards matter. 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 and there is no interference. '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. Mass does not convert to energy. Matter converts to aether. As the maether transitions from matter to aether it increases in volume. The physical effect the increase in volume has on the neighboring matter and aether is energy. The physical effect of maether decompressing is energy. Mass is conserved. 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 ìs/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 ìs/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 ìs/day faster than clocks on the ground." (quoted text from http://en.wikipedia.org/wiki/Effects_of_relativity_on_GPS). 'The Need to Understand Mass' By Roger Cashmore Department of Physics University of Oxford, UK. http://www.phy.uct.ac.za/courses/phy400w/particle/higgs2.htm "There is, however, one very clever and very elegant solution to this problem, a solution first proposed by Peter Higgs. He proposed that the whole of space is permeated by a field, similar in some ways to the electromagnetic field. As particles move through space they travel through this field, and if they interact with it they acquire what appears to be mass. This is similar to the action of viscous forces felt by particles moving through any thick liquid. the larger the interaction of the particles with the field, the more mass they appear to have. Thus the existence of this field is essential in Higg's hypothesis for the production of the mass of particles." The "action of viscous forces felt by particles moving through any thick liquid" is the particles interaction with the aether. The force is the pressure exerted by the displaced aether towards the particle. The "thick liquid" is the aether behaving as a frictionless superfluid 'one something'. "the larger the interaction of the particles with the field, the more mass they appear to have." The faster the particle moves with respect to the aether, the greater the pressure exerted by the displaced aether towards the particle. 'Politics, Solid State and the Higgs' By David Miller Department of Physics and Astronomy University College, London, UK. http://www.phy.uct.ac.za/courses/phy400w/particle/higgs3.htm "1. The Higgs Mechanism In three dimensions, and with the complications of relativity, this is the Higgs mechanism. In order to give particles mass, a background field is invented which becomes locally distorted whenever a particle moves through it. The distortion - the clustering of the field around the particle - generates the particle's mass. The idea comes directly from the physics of solids. Instead of a field spread throughout all space a solid contains a lattice of positively charged crystal atoms. When an electron moves through the lattice the atoms are attracted to it, causing the electron's effective mass to be as much as 40 times bigger than the mass of a free electron." The distortion of the background field is the displacement of the aether by the moving particle. The 'clustering' of the field around the particle is the 'displacing back'. The 'clustering' of the field is the pressure exerted by the displaced aether towards the particle. "The idea comes directly from the physics of solids." The aether behaves as a frictionless superfluid 'one something'. In the Casimir Effect, the aether displaced by each of the plates extends past the other plate, forcing the plates together.
From: mpc755 on 14 Jun 2010 00:20 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.
From: Xan Du on 14 Jun 2010 00:38 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
From: mpc755 on 14 Jun 2010 01:02 On Jun 14, 12:38 am, Xan Du <xan...(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 The other poster is discussing an entrained aether which I will let them discuss in particular. Astronauts are practicing for a space walk. The astronauts practice over and over again in a huge tank of water. When the astronauts are actually performing the space walk think of the astronauts as being in a huge tank of aether which extends forever. In Aether Displacement, the aether is displaced by matter. You throw a bowling ball off of a jetty into the ocean. The bowling ball displaces all of the water which is the ocean. Can you detect this displacement on the other side of the ocean? No. Someone tosses in a train engine next to the bowling ball. The bowling ball still displaces the water but the effect of the bowling ball will be overshadowed by the train engine once you get far enough away from the bowling ball itself. Someone tosses in the Empire State Building next to the train engine. The effects of the train engine on the displaced water will soon be overshadowed by the displaced water of the Empire State Building. This is what occurs to the aether. 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. This means the aether is displaced by the matter which is the Earth. The aether is displaced far past the Moon. 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. Gravity is pressure exerted by displaced aether towards matter. Once you get far enough away from the Earth the state of the aether as determined by its connection with the matter and the state of the aether in neighboring places is the aether's state of displacement as determined by the Sun. Once you get far enough outside of the solar system 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 as determined by the Milky Way. There are no voids. Aether and matter are different states of the same material. The material is maether. Aether is uncompressed maether and matter is compressed maether. Maether is the substance of space.
From: mpc755 on 14 Jun 2010 01:48
On Jun 14, 12:38 am, Xan Du <xan...(a)yahoo.thom> wwote: > 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 The othew poftew ith dithkuthing an entwained aethew whitth I wiww wet them dithkuth in pawtiquwaw. Aftwonautth awe pwaktithing fow a thpathe wawk. The aftwonautth pwaktithe owew and owew again in a huge tank of watew. When the aftwonautth awe aktuawwy pewfowming the thpathe wawk think of the aftwonautth ath being in a huge tank of aethew whitth ekthtendth fowewew. In Aethew Dithpwathement, the aethew ith dithpwathed by mattew. You thwow a bowwing baww off of a jetty into the othean. The bowwing baww dithpwatheth aww of the watew whitth ith the othean. Kan you detekt thith dithpwathement on the othew thide of the othean? No. Thomeone totheth in a twain engine nektht to the bowwing baww. The bowwing baww ftiww dithpwatheth the watew but the effekt of the bowwing baww wiww be owewthadowed by the twain engine wonth you get faw enough away fwom the bowwing baww itthewf. Thomeone totheth in the Empiwe Thtate Buiwding nektht to the twain engine. The effektth of the twain engine on the dithpwathed watew wiww thoon be owewthadowed by the dithpwathed watew of the Empiwe Thtate Buiwding. Thith ith what okuwth to the aethew. The ftate of the aethew ath detewmined by itth konnektionth with the mattew and the ftate of the aethew in neighbowing pwatheth ith the aethew'th ftate of dithpwathement. Thith meanth the aethew ith dithpwathed by the mattew whitth ith the Eawth. The aethew ith dithpwathed faw paft the Moon. The aethew ith not at weft when dithpwathed and 'dithpwatheth bathk'. The 'dithpwathing bathk' ith the pwethuwe exewted by the dithpwathed aethew towawdth the mattew. Gwawity ith pwethuwe exewted by dithpwathed aethew towawdth mattew. Onthe you get faw enough away fwom the Eawth the ftate of the aethew ath detewmined by itth konnektion with the mattew and the ftate of the aethew in neighbowing pwatheth ith the aethew'th ftate of dithpwathement ath detewmined by the Thun. Onthe you get faw enough outthide of the thowaw thyftem the ftate of the aethew ath detewmined by itth konnektionth with the mattew and the ftate of the aethew in neighbowing pwatheth ith the aethew'th ftate of dithpwathement ath detewmined by the Miwky Way. Thewe awe no woidth. Aethew and mattew awe diffewent ftateth of the thame matewiaw. The matewiaw ith maethew. Aethew ith unthompwethed maethew and mattew ith kompwethed maethew. Maethew ith the thubftanthe of thpathe. |