Let's settle the argument When gravity has no strength mpc.
in [Relativity]
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From: BURT on 10 Jun 2010 15:05 On Jun 10, 7:11 am, mpc755 <mpc...(a)gmail.com> wrote: > On Jun 10, 2:38 am, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > On Jun 9, 1:31 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > On Jun 6, 2:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On Jun 5, 8:00 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > Dear Burt: Where did you get the notion that circular orbits have no > > > > gravity? If that were so, then, how are those telecommunications > > > > satellites held in orbit? I've got gravity nailed as: Flowing ether, > > > > replenished by photon exchange. Nothing that you've ever said changes > > > > those facts. NE > > > > There is a round curve of gravity for energy in a circular orbit. But > > > there is no strength of gravity to change the motion of circular > > > speed. The strength of gravity does not lie in the curve but in space > > > flow. A circular orbit has zero gravity strength but a pre speed > > > through the round curve. You can quantify the prespeed in space for > > > the circular orbit. Pre-speed is the motion through space independant > > > of the strength of gravity pushing it faster or slower. Gravity gives > > > and takes from pre-motion of falling energy in elliptical orbit. But > > > pre speed is always a preserved quantity in time orbit. > > > > MItch Raemsch > > > Mpc? A circular orbit of energy follows the curve by its premotion. > > And there is no gravity strength to accelerate or decelerate the speed > > of energy. > > > Mitch Raemsch > > There is still an outstanding question you have not answered. > > 'Interpretation of quantum mechanics > by the double solution theory > Louis de BROGLIE'http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001.pdf > > 'I called this relation, which determines the particle's motion in the > wave, "the guidance formula". It may easily be generalized to the case > of an external field acting on the particle.' > > 'The particle when in motion on its wave, thus has its vibration > constantly in phase with that of the wave. This result may be > interpreted by noticing that, in the present theory, the particle is > de¯ned as a very small region of the wave where the amplitude is very > large, and it therefore seems quite natural that the internal motion > rythm of the particle should always be the same as that of the wave at > the point where the particle is located. A very important point must > be underlined here. For this interpretation of the guidance to be > acceptable, the dimensions of the minute singular region constituting > the particle ought to be very small compared to the wavelength of the > v wave.' > > The 'particle' occupies a very small region of its associated wave. > The external field acting on the particle is the aether. > > A moving particle has an associated aether wave. > > http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory > > "In de BroglieBohm theory, the wavefunction travels through both > slits, but each particle has a well-defined trajectory and passes > through exactly one of the slits." > > You said you agreed with Bohm. Bohm states the particle has a well- > defined trajectory and passes through exactly one of the slits. > > Do you agree with Bohm or don't you?- Hide quoted text - > > - Show quoted text - I can do without your question. But can you deal with mine? Mitch Reamsch
From: mpc755 on 10 Jun 2010 15:29 On Jun 10, 3:05 pm, BURT <macromi...(a)yahoo.com> wrote: > On Jun 10, 7:11 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Jun 10, 2:38 am, BURT <macromi...(a)yahoo.com> wrote: > > > > On Jun 9, 1:31 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > On Jun 6, 2:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On Jun 5, 8:00 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > Dear Burt: Where did you get the notion that circular orbits have no > > > > > gravity? If that were so, then, how are those telecommunications > > > > > satellites held in orbit? I've got gravity nailed as: Flowing ether, > > > > > replenished by photon exchange. Nothing that you've ever said changes > > > > > those facts. NE > > > > > There is a round curve of gravity for energy in a circular orbit. But > > > > there is no strength of gravity to change the motion of circular > > > > speed. The strength of gravity does not lie in the curve but in space > > > > flow. A circular orbit has zero gravity strength but a pre speed > > > > through the round curve. You can quantify the prespeed in space for > > > > the circular orbit. Pre-speed is the motion through space independant > > > > of the strength of gravity pushing it faster or slower. Gravity gives > > > > and takes from pre-motion of falling energy in elliptical orbit. But > > > > pre speed is always a preserved quantity in time orbit. > > > > > MItch Raemsch > > > > Mpc? A circular orbit of energy follows the curve by its premotion. > > > And there is no gravity strength to accelerate or decelerate the speed > > > of energy. > > > > Mitch Raemsch > > > There is still an outstanding question you have not answered. > > > 'Interpretation of quantum mechanics > > by the double solution theory > > Louis de BROGLIE'http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001.pdf > > > 'I called this relation, which determines the particle's motion in the > > wave, "the guidance formula". It may easily be generalized to the case > > of an external field acting on the particle.' > > > 'The particle when in motion on its wave, thus has its vibration > > constantly in phase with that of the wave. This result may be > > interpreted by noticing that, in the present theory, the particle is > > de¯ned as a very small region of the wave where the amplitude is very > > large, and it therefore seems quite natural that the internal motion > > rythm of the particle should always be the same as that of the wave at > > the point where the particle is located. A very important point must > > be underlined here. For this interpretation of the guidance to be > > acceptable, the dimensions of the minute singular region constituting > > the particle ought to be very small compared to the wavelength of the > > v wave.' > > > The 'particle' occupies a very small region of its associated wave. > > The external field acting on the particle is the aether. > > > A moving particle has an associated aether wave. > > >http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory > > > "In de BroglieBohm theory, the wavefunction travels through both > > slits, but each particle has a well-defined trajectory and passes > > through exactly one of the slits." > > > You said you agreed with Bohm. Bohm states the particle has a well- > > defined trajectory and passes through exactly one of the slits. > > > Do you agree with Bohm or don't you?- Hide quoted text - > > > - Show quoted text - > > I can do without your question. But can you deal with mine? > > Mitch Reamsch There is still an outstanding question you have not answered. 'Interpretation of quantum mechanics by the double solution theory Louis de BROGLIE' http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001.pdf 'I called this relation, which determines the particle's motion in the wave, "the guidance formula". It may easily be generalized to the case of an external field acting on the particle.' 'The particle when in motion on its wave, thus has its vibration constantly in phase with that of the wave. This result may be interpreted by noticing that, in the present theory, the particle is defined as a very small region of the wave where the amplitude is very large, and it therefore seems quite natural that the internal motion rythm of the particle should always be the same as that of the wave at the point where the particle is located. A very important point must be underlined here. For this interpretation of the guidance to be acceptable, the dimensions of the minute singular region constituting the particle ought to be very small compared to the wavelength of the v wave.' The 'particle' occupies a very small region of its associated wave. The external field acting on the particle is the aether. A moving particle has an associated aether wave. http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory "In de BroglieBohm theory, the wavefunction travels through both slits, but each particle has a well-defined trajectory and passes through exactly one of the slits." You said you agreed with Bohm. Bohm states the particle has a well- defined trajectory and passes through exactly one of the slits. Do you agree with Bohm or don't you?
From: BURT on 10 Jun 2010 15:46 On Jun 10, 12:29 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Jun 10, 3:05 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > On Jun 10, 7:11 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > On Jun 10, 2:38 am, BURT <macromi...(a)yahoo.com> wrote: > > > > > On Jun 9, 1:31 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > On Jun 6, 2:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On Jun 5, 8:00 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > Dear Burt: Where did you get the notion that circular orbits have no > > > > > > gravity? If that were so, then, how are those telecommunications > > > > > > satellites held in orbit? I've got gravity nailed as: Flowing ether, > > > > > > replenished by photon exchange. Nothing that you've ever said changes > > > > > > those facts. NE > > > > > > There is a round curve of gravity for energy in a circular orbit. But > > > > > there is no strength of gravity to change the motion of circular > > > > > speed. The strength of gravity does not lie in the curve but in space > > > > > flow. A circular orbit has zero gravity strength but a pre speed > > > > > through the round curve. You can quantify the prespeed in space for > > > > > the circular orbit. Pre-speed is the motion through space independant > > > > > of the strength of gravity pushing it faster or slower. Gravity gives > > > > > and takes from pre-motion of falling energy in elliptical orbit. But > > > > > pre speed is always a preserved quantity in time orbit. > > > > > > MItch Raemsch > > > > > Mpc? A circular orbit of energy follows the curve by its premotion. > > > > And there is no gravity strength to accelerate or decelerate the speed > > > > of energy. > > > > > Mitch Raemsch > > > > There is still an outstanding question you have not answered. > > > > 'Interpretation of quantum mechanics > > > by the double solution theory > > > Louis de BROGLIE'http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001..pdf > > > > 'I called this relation, which determines the particle's motion in the > > > wave, "the guidance formula". It may easily be generalized to the case > > > of an external field acting on the particle.' > > > > 'The particle when in motion on its wave, thus has its vibration > > > constantly in phase with that of the wave. This result may be > > > interpreted by noticing that, in the present theory, the particle is > > > de¯ned as a very small region of the wave where the amplitude is very > > > large, and it therefore seems quite natural that the internal motion > > > rythm of the particle should always be the same as that of the wave at > > > the point where the particle is located. A very important point must > > > be underlined here. For this interpretation of the guidance to be > > > acceptable, the dimensions of the minute singular region constituting > > > the particle ought to be very small compared to the wavelength of the > > > v wave.' > > > > The 'particle' occupies a very small region of its associated wave. > > > The external field acting on the particle is the aether. > > > > A moving particle has an associated aether wave. > > > >http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory > > > > "In de BroglieBohm theory, the wavefunction travels through both > > > slits, but each particle has a well-defined trajectory and passes > > > through exactly one of the slits." > > > > You said you agreed with Bohm. Bohm states the particle has a well- > > > defined trajectory and passes through exactly one of the slits. > > > > Do you agree with Bohm or don't you?- Hide quoted text - > > > > - Show quoted text - > > > I can do without your question. But can you deal with mine? > > > Mitch Reamsch > > There is still an outstanding question you have not answered. > > 'Interpretation of quantum mechanics > by the double solution theory > Louis de BROGLIE'http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001.pdf > > 'I called this relation, which determines the particle's motion in the > wave, "the guidance formula". It may easily be generalized to the case > of an external field acting on the particle.' > > 'The particle when in motion on its wave, thus has its vibration > constantly in phase with that of the wave. This result may be > interpreted by noticing that, in the present theory, the particle is > defined as a very small region of the wave where the amplitude is very > large, and it therefore seems quite natural that the internal motion > rythm of the particle should always be the same as that of the wave at > the point where the particle is located. A very important point must > be underlined here. For this interpretation of the guidance to be > acceptable, the dimensions of the minute singular region constituting > the particle ought to be very small compared to the wavelength of the > v wave.' > > The 'particle' occupies a very small region of its associated wave. > The external field acting on the particle is the aether. > > A moving particle has an associated aether wave. > > http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory > > "In de BroglieBohm theory, the wavefunction travels through both > slits, but each particle has a well-defined trajectory and passes > through exactly one of the slits." > > You said you agreed with Bohm. Bohm states the particle has a well- > defined trajectory and passes through exactly one of the slits. > > Do you agree with Bohm or don't you?- Hide quoted text - > > - Show quoted text - What you are doing isn't even a question. But I think that what you are putting up is sciences first shot. And that never is right all of the way. In fact most of the time only one thing is right and the rest wrong. I see everybody doing this. Trying to support something even with all its error. Why? Because science must be right. The short history of science demonstrates that it is built mostly of mistakes. So we need intellectual honesty here and hopefully Einsteinian objectivity to make the corrections. Mitch Raemsch
From: mpc755 on 10 Jun 2010 18:14 On Jun 10, 3:46 pm, BURT <macromi...(a)yahoo.com> wrote: > On Jun 10, 12:29 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Jun 10, 3:05 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > On Jun 10, 7:11 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Jun 10, 2:38 am, BURT <macromi...(a)yahoo.com> wrote: > > > > > > On Jun 9, 1:31 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > On Jun 6, 2:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On Jun 5, 8:00 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > > Dear Burt: Where did you get the notion that circular orbits have no > > > > > > > gravity? If that were so, then, how are those telecommunications > > > > > > > satellites held in orbit? I've got gravity nailed as: Flowing ether, > > > > > > > replenished by photon exchange. Nothing that you've ever said changes > > > > > > > those facts. NE > > > > > > > There is a round curve of gravity for energy in a circular orbit. But > > > > > > there is no strength of gravity to change the motion of circular > > > > > > speed. The strength of gravity does not lie in the curve but in space > > > > > > flow. A circular orbit has zero gravity strength but a pre speed > > > > > > through the round curve. You can quantify the prespeed in space for > > > > > > the circular orbit. Pre-speed is the motion through space independant > > > > > > of the strength of gravity pushing it faster or slower. Gravity gives > > > > > > and takes from pre-motion of falling energy in elliptical orbit.. But > > > > > > pre speed is always a preserved quantity in time orbit. > > > > > > > MItch Raemsch > > > > > > Mpc? A circular orbit of energy follows the curve by its premotion. > > > > > And there is no gravity strength to accelerate or decelerate the speed > > > > > of energy. > > > > > > Mitch Raemsch > > > > > There is still an outstanding question you have not answered. > > > > > 'Interpretation of quantum mechanics > > > > by the double solution theory > > > > Louis de BROGLIE'http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001.pdf > > > > > 'I called this relation, which determines the particle's motion in the > > > > wave, "the guidance formula". It may easily be generalized to the case > > > > of an external field acting on the particle.' > > > > > 'The particle when in motion on its wave, thus has its vibration > > > > constantly in phase with that of the wave. This result may be > > > > interpreted by noticing that, in the present theory, the particle is > > > > de¯ned as a very small region of the wave where the amplitude is very > > > > large, and it therefore seems quite natural that the internal motion > > > > rythm of the particle should always be the same as that of the wave at > > > > the point where the particle is located. A very important point must > > > > be underlined here. For this interpretation of the guidance to be > > > > acceptable, the dimensions of the minute singular region constituting > > > > the particle ought to be very small compared to the wavelength of the > > > > v wave.' > > > > > The 'particle' occupies a very small region of its associated wave. > > > > The external field acting on the particle is the aether. > > > > > A moving particle has an associated aether wave. > > > > >http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory > > > > > "In de BroglieBohm theory, the wavefunction travels through both > > > > slits, but each particle has a well-defined trajectory and passes > > > > through exactly one of the slits." > > > > > You said you agreed with Bohm. Bohm states the particle has a well- > > > > defined trajectory and passes through exactly one of the slits. > > > > > Do you agree with Bohm or don't you?- Hide quoted text - > > > > > - Show quoted text - > > > > I can do without your question. But can you deal with mine? > > > > Mitch Reamsch > > > There is still an outstanding question you have not answered. > > > 'Interpretation of quantum mechanics > > by the double solution theory > > Louis de BROGLIE'http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001.pdf > > > 'I called this relation, which determines the particle's motion in the > > wave, "the guidance formula". It may easily be generalized to the case > > of an external field acting on the particle.' > > > 'The particle when in motion on its wave, thus has its vibration > > constantly in phase with that of the wave. This result may be > > interpreted by noticing that, in the present theory, the particle is > > defined as a very small region of the wave where the amplitude is very > > large, and it therefore seems quite natural that the internal motion > > rythm of the particle should always be the same as that of the wave at > > the point where the particle is located. A very important point must > > be underlined here. For this interpretation of the guidance to be > > acceptable, the dimensions of the minute singular region constituting > > the particle ought to be very small compared to the wavelength of the > > v wave.' > > > The 'particle' occupies a very small region of its associated wave. > > The external field acting on the particle is the aether. > > > A moving particle has an associated aether wave. > > >http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory > > > "In de BroglieBohm theory, the wavefunction travels through both > > slits, but each particle has a well-defined trajectory and passes > > through exactly one of the slits." > > > You said you agreed with Bohm. Bohm states the particle has a well- > > defined trajectory and passes through exactly one of the slits. > > > Do you agree with Bohm or don't you? > > What you are doing isn't even a question. But I think that what you > are putting up is sciences first shot. And that never is right all of > the way. In fact most of the time only one thing is right and the rest > wrong. I see everybody doing this. Trying to support something even > with all its error. Why? Because science must be right. The short > history of science demonstrates that it is built mostly of mistakes. > So we need intellectual honesty here and hopefully Einsteinian > objectivity to make the corrections. > > Mitch Raemsch In a double slit experiment, does the moving C-60 molecule travel a single continuous path through three dimensional space as a particle with an associated external wave?
From: BURT on 10 Jun 2010 18:20
On Jun 10, 3:14 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Jun 10, 3:46 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > On Jun 10, 12:29 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > On Jun 10, 3:05 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > On Jun 10, 7:11 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > On Jun 10, 2:38 am, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > On Jun 9, 1:31 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > > On Jun 6, 2:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > On Jun 5, 8:00 pm, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > > > Dear Burt: Where did you get the notion that circular orbits have no > > > > > > > > gravity? If that were so, then, how are those telecommunications > > > > > > > > satellites held in orbit? I've got gravity nailed as: Flowing ether, > > > > > > > > replenished by photon exchange. Nothing that you've ever said changes > > > > > > > > those facts. NE > > > > > > > > There is a round curve of gravity for energy in a circular orbit. But > > > > > > > there is no strength of gravity to change the motion of circular > > > > > > > speed. The strength of gravity does not lie in the curve but in space > > > > > > > flow. A circular orbit has zero gravity strength but a pre speed > > > > > > > through the round curve. You can quantify the prespeed in space for > > > > > > > the circular orbit. Pre-speed is the motion through space independant > > > > > > > of the strength of gravity pushing it faster or slower. Gravity gives > > > > > > > and takes from pre-motion of falling energy in elliptical orbit. But > > > > > > > pre speed is always a preserved quantity in time orbit. > > > > > > > > MItch Raemsch > > > > > > > Mpc? A circular orbit of energy follows the curve by its premotion. > > > > > > And there is no gravity strength to accelerate or decelerate the speed > > > > > > of energy. > > > > > > > Mitch Raemsch > > > > > > There is still an outstanding question you have not answered. > > > > > > 'Interpretation of quantum mechanics > > > > > by the double solution theory > > > > > Louis de BROGLIE'http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001.pdf > > > > > > 'I called this relation, which determines the particle's motion in the > > > > > wave, "the guidance formula". It may easily be generalized to the case > > > > > of an external field acting on the particle.' > > > > > > 'The particle when in motion on its wave, thus has its vibration > > > > > constantly in phase with that of the wave. This result may be > > > > > interpreted by noticing that, in the present theory, the particle is > > > > > de¯ned as a very small region of the wave where the amplitude is very > > > > > large, and it therefore seems quite natural that the internal motion > > > > > rythm of the particle should always be the same as that of the wave at > > > > > the point where the particle is located. A very important point must > > > > > be underlined here. For this interpretation of the guidance to be > > > > > acceptable, the dimensions of the minute singular region constituting > > > > > the particle ought to be very small compared to the wavelength of the > > > > > v wave.' > > > > > > The 'particle' occupies a very small region of its associated wave. > > > > > The external field acting on the particle is the aether. > > > > > > A moving particle has an associated aether wave. > > > > > >http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory > > > > > > "In de BroglieBohm theory, the wavefunction travels through both > > > > > slits, but each particle has a well-defined trajectory and passes > > > > > through exactly one of the slits." > > > > > > You said you agreed with Bohm. Bohm states the particle has a well- > > > > > defined trajectory and passes through exactly one of the slits. > > > > > > Do you agree with Bohm or don't you?- Hide quoted text - > > > > > > - Show quoted text - > > > > > I can do without your question. But can you deal with mine? > > > > > Mitch Reamsch > > > > There is still an outstanding question you have not answered. > > > > 'Interpretation of quantum mechanics > > > by the double solution theory > > > Louis de BROGLIE'http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001..pdf > > > > 'I called this relation, which determines the particle's motion in the > > > wave, "the guidance formula". It may easily be generalized to the case > > > of an external field acting on the particle.' > > > > 'The particle when in motion on its wave, thus has its vibration > > > constantly in phase with that of the wave. This result may be > > > interpreted by noticing that, in the present theory, the particle is > > > defined as a very small region of the wave where the amplitude is very > > > large, and it therefore seems quite natural that the internal motion > > > rythm of the particle should always be the same as that of the wave at > > > the point where the particle is located. A very important point must > > > be underlined here. For this interpretation of the guidance to be > > > acceptable, the dimensions of the minute singular region constituting > > > the particle ought to be very small compared to the wavelength of the > > > v wave.' > > > > The 'particle' occupies a very small region of its associated wave. > > > The external field acting on the particle is the aether. > > > > A moving particle has an associated aether wave. > > > >http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory > > > > "In de BroglieBohm theory, the wavefunction travels through both > > > slits, but each particle has a well-defined trajectory and passes > > > through exactly one of the slits." > > > > You said you agreed with Bohm. Bohm states the particle has a well- > > > defined trajectory and passes through exactly one of the slits. > > > > Do you agree with Bohm or don't you? > > > What you are doing isn't even a question. But I think that what you > > are putting up is sciences first shot. And that never is right all of > > the way. In fact most of the time only one thing is right and the rest > > wrong. I see everybody doing this. Trying to support something even > > with all its error. Why? Because science must be right. The short > > history of science demonstrates that it is built mostly of mistakes. > > So we need intellectual honesty here and hopefully Einsteinian > > objectivity to make the corrections. > > > Mitch Raemsch > > In a double slit experiment, does the moving C-60 molecule travel a > single continuous path through three dimensional space as a particle > with an associated external wave?- Hide quoted text - > > - Show quoted text - The wave is a continuous form that needs no particle for absorption. It also behaves immatterial when it moves through the two slit partition inbetween the holes. What do you say to that? Mitch Raemsch |