Mediating bosons cannot exist without the matter to make them. That they exist doesn't make them real--they are epifenomena. As are other equations from the force equation. Set it equal to ms'' + m's' + m''s and one can get cooking. As long as electrons have restorers, they exert and experience proxy and sunthetic forces (my terms for what scientists call pseudoforces) such as magnetism and exchange. A ball whirling on a sling in a medium where all particles are balls whirling on their slings I'd expect to show the same effects.
Oh? How do you expect the strength to be the same at every distance from the emitter? For the length of the pipe, near the pipe, the strength doesn't fall off much from the end because the charges are over a region. But power is Fc, and F still falls off with distance. Maybe you are confused with the fact that surfaces with variable emissivity send power only in certain directions and fractions. For a perfect pipe, the total power flows in only one direction, for the best meaning of "one" as given by raytracing about an extended region.
The difference between an absorber-emitter and a reflector is only of path. A conventional reflector is an atomic surface absorber; a conventional absorber is an interatomic bulk absorber. I would say that an optic sail is an isatomic bulk absorber. If no energy were transferred to the receiver, then it would be transparent, not reflective or absorptive.
I didn't say there wasn't a need for them, only that they shouldn't be reified.
The so-called emission at A violates Coulomb's law if B is perfectly reflective. The body cannot radiate if there is no outside matter to interact with. If a probe at B is used to tell if there's any energy leaking from A, then the probe counts as an absorber, with its own masses and charges, that can then interact with A by transacting with A. This is possible because a particle is infinitely big, so the electrons in A are already at B, however weakened, and it only takes a while for their field to modulate to B. By presuming perfect reflection, one is also presuming infinite mass-inertia at both ends so that the pulse cannot be lost in transit.
If the receiver is farther and farther from the emitter, its charges are that much farther and the forces and energies transmitted are likewise littler. In optic terms, the receiver subtends a smaller and smaller solid angle and can only receive that much less power (P A/4 pi r^2). The transit frequency isn't as drastic (f l/r). If the emitter's charges start quick, then slow off before 2t whereupon they rise again to an intermediate steady state, then this is only possible because the receiver's charges are taking up the energy. Because a particle is infinitely big, this energy exists in the space between the two ends and is manifest as the motion of the field for every intermediate region. The foton is really the particleicle, or the particuliculum. In real life, the electrons at the receiver aren't subject to infinite forces, so they are able to store the energy from their end during reflection. Their storing train takes time to shuttle. If they were perfect reflectors, they would have infinite restoring forces, and so their field would be infinite everywhere from their lattice positions. So it would take no time for an emitter anywhere to regenerate its radiation. Objects are nominally nonreflectors because their bulk is accelerated by any force gradient from the emitter and any dipole moment from the absorber; they'll fly away from radiation pressure.
BTW, why doesn't/. have miniscripts and extended characters??
I'd been thinking of this problem for awhile because of energy-conservation considerations. The latency effect has been used in some half-baked proofs, such as in Bearden's essay for getting free energy. But it's not irrelevant what happens to either source. This is how I see it:
Source A is a perfectly insulated and reflective directional emitter--a pipe--pointed at B, which is a shutter with reflective and absorptive shutters, being either Source B or Sink B. However, B is in reflective mode. Source A is switched on: Its charges wiggle quickly. Because they can't interact with anything, they can't radiate and thus their energy is regenerated and constant. A's charges probe B's charges; but because B is reflective now and here, A's charges don't lose. A is switched off: Its charges are stopped and return to their former potential. However, the signal still exists and is being transmitted to B, which is in timelike separation to A. Before it gets there, B opens its shutters and tries to get the energy. At A's end, of course B was shut the whole time and so A's energy gauge should be full. Then, the weakened force at B interact's with B's charges and intercepts the energy. At A, A can be off, or gone, or dead, or blown up, yet it still has its energy. So how can B get the energy without it appearing from nowhere?
The trick is to consider that fotons aren't even things; they are changes in the fields from source and sink charges. Radiation is fundamentally a consequence of Coulomb's law between charges; it's a transactional effect between particles that are infinitely big, by mechanics standards. Yes, an electron is infinitely big; that's what I learned from my thinking. (So there's no need for those virtual particles, or action-at-a-distance.) So Sink B does take the energy, but this too is a signal that goes back to A. At 2t, whether or not A is bare or has put up a lens cap to insulate interactions from B, B's charges will tug at A's charges inside or at the cap and hull such that the energy is yanked thence in order to conserve both's energy. So latency and conservation work like a collection agency sending out a loan shark or, more intuitively, a blue spiky Koopa shell that you see on Mario Kart 64. Playing with that game, one finds that if in two-player mode, the player releases this invincible homing shell at the computer kart, and the computer happens to lap the player, that the shell will still take out the nearest kart and will instead take out the player's. Sometimes, if the track is short and the shell is released in some directions, it assumes a fononic crustal ("phononic crystal" in corruptive Latinish) mode and lets the shell loop forever without hitting either kart. However, either kart moving shifts the equilibrium of the three-body sustem and will eventually knock out one or both of the karts. Oh, it seems like I've gone off-topic...
Slashdot Mirror
english -> English By writing, you are talking about grammar. Nonetheless, many English-speakers misspell the same words as above.
Oh? How do you expect the strength to be the same at every distance from the emitter? For the length of the pipe, near the pipe, the strength doesn't fall off much from the end because the charges are over a region. But power is Fc, and F still falls off with distance. Maybe you are confused with the fact that surfaces with variable emissivity send power only in certain directions and fractions. For a perfect pipe, the total power flows in only one direction, for the best meaning of "one" as given by raytracing about an extended region.
The difference between an absorber-emitter and a reflector is only of path. A conventional reflector is an atomic surface absorber; a conventional absorber is an interatomic bulk absorber. I would say that an optic sail is an isatomic bulk absorber. If no energy were transferred to the receiver, then it would be transparent, not reflective or absorptive.
The so-called emission at A violates Coulomb's law if B is perfectly reflective. The body cannot radiate if there is no outside matter to interact with. If a probe at B is used to tell if there's any energy leaking from A, then the probe counts as an absorber, with its own masses and charges, that can then interact with A by transacting with A. This is possible because a particle is infinitely big, so the electrons in A are already at B, however weakened, and it only takes a while for their field to modulate to B. By presuming perfect reflection, one is also presuming infinite mass-inertia at both ends so that the pulse cannot be lost in transit.
If the receiver is farther and farther from the emitter, its charges are that much farther and the forces and energies transmitted are likewise littler. In optic terms, the receiver subtends a smaller and smaller solid angle and can only receive that much less power (P A/4 pi r^2). The transit frequency isn't as drastic (f l/r). If the emitter's charges start quick, then slow off before 2t whereupon they rise again to an intermediate steady state, then this is only possible because the receiver's charges are taking up the energy. Because a particle is infinitely big, this energy exists in the space between the two ends and is manifest as the motion of the field for every intermediate region. The foton is really the particleicle, or the particuliculum. In real life, the electrons at the receiver aren't subject to infinite forces, so they are able to store the energy from their end during reflection. Their storing train takes time to shuttle. If they were perfect reflectors, they would have infinite restoring forces, and so their field would be infinite everywhere from their lattice positions. So it would take no time for an emitter anywhere to regenerate its radiation. Objects are nominally nonreflectors because their bulk is accelerated by any force gradient from the emitter and any dipole moment from the absorber; they'll fly away from radiation pressure.
BTW, why doesn't /. have miniscripts and extended characters??
http://groups.google.com/group/comp.sys.mac.advoca cy/browse_frm/thread/b842efbe82707af5/ab80aa357a7e 97da#ab80aa357a7e97da
Yes it is. The universe did it already.
I'd been thinking of this problem for awhile because of energy-conservation considerations. The latency effect has been used in some half-baked proofs, such as in Bearden's essay for getting free energy. But it's not irrelevant what happens to either source. This is how I see it:
Source A is a perfectly insulated and reflective directional emitter--a pipe--pointed at B, which is a shutter with reflective and absorptive shutters, being either Source B or Sink B. However, B is in reflective mode. Source A is switched on: Its charges wiggle quickly. Because they can't interact with anything, they can't radiate and thus their energy is regenerated and constant. A's charges probe B's charges; but because B is reflective now and here, A's charges don't lose. A is switched off: Its charges are stopped and return to their former potential. However, the signal still exists and is being transmitted to B, which is in timelike separation to A. Before it gets there, B opens its shutters and tries to get the energy. At A's end, of course B was shut the whole time and so A's energy gauge should be full. Then, the weakened force at B interact's with B's charges and intercepts the energy. At A, A can be off, or gone, or dead, or blown up, yet it still has its energy. So how can B get the energy without it appearing from nowhere?
The trick is to consider that fotons aren't even things; they are changes in the fields from source and sink charges. Radiation is fundamentally a consequence of Coulomb's law between charges; it's a transactional effect between particles that are infinitely big, by mechanics standards. Yes, an electron is infinitely big; that's what I learned from my thinking. (So there's no need for those virtual particles, or action-at-a-distance.) So Sink B does take the energy, but this too is a signal that goes back to A. At 2t, whether or not A is bare or has put up a lens cap to insulate interactions from B, B's charges will tug at A's charges inside or at the cap and hull such that the energy is yanked thence in order to conserve both's energy. So latency and conservation work like a collection agency sending out a loan shark or, more intuitively, a blue spiky Koopa shell that you see on Mario Kart 64. Playing with that game, one finds that if in two-player mode, the player releases this invincible homing shell at the computer kart, and the computer happens to lap the player, that the shell will still take out the nearest kart and will instead take out the player's. Sometimes, if the track is short and the shell is released in some directions, it assumes a fononic crustal ("phononic crystal" in corruptive Latinish) mode and lets the shell loop forever without hitting either kart. However, either kart moving shifts the equilibrium of the three-body sustem and will eventually knock out one or both of the karts. Oh, it seems like I've gone off-topic...
illiterate retard:
Your -> You're
it's -> its
String-cuts propagate at the local speed of sound. Illiterate retards pull scientific proclamations out of their arse and call it truth.