Logical theories have two main origins. The theory is either, predicted before the evidence is found, as was clearly the case when Fizeau succeeded in confirming what he had intended, that Freznel's much earlier prediction was valid, or the theory evolves as evidence becomes available.
Freznel's prediction was a consequence of ether theory, but the nuclear atom theory is predicted by nothing whatever in nature or anywhere else. It stands completely alone. So it's unlikely that the theory boldly preceded the evidence. How could one fully commit to such a theory before supporting evidence began to emerge?
If new evidence emerges through experiment, the successful experimenter imposes his or her theories regarding the new evidence on the physics community, and it's accepted as reality if there is no other apparent explanation for what is observed. The nuclear atom is bizarre to the point of being ridiculous, but was still accepted by the physics community, and that was highly irresponsible. We discard our intuition because it's in conflict with an intuitively absurd prediction based on nothing at all. How can that be justified? Why not vice versa? Intuition has not been proven wrong by any means.
There is obviously some other answer, we just don't know what it is.
The nuclear atom consists of a nucleus that houses almost the entire mass of the atom, and that is surrounded by an appropriate number of electrons drifting about in vast expanses of nothing that extends to a radius which is around 11,000 times that of the nucleus. The atomic radius is maintained when the electrons are present, but if they are stripped from the atom, the whole outer shell of emptiness around the nucleus disappears. The remaining nucleus can then traverse the vast expanses of space housed within other nuclear atoms that have not shed their electrons.
Does that seem logical?
If this pixel '.' represents one hydrogen atom nucleus in a jar of liquid hydrogen, its nearest neighboring nucleus is 22000 pixels away, or around 20 meters depending on screen size. And there are only two electrons that will pass through the void between them. That's what is proposed as the foundation of solids.
This paper, published almost a century ago, describes the original scattering experiments. Paragraph 3 in the final experiment listed, gives a 1mm air equivalent stopping power for a 2.1e-7 meter thick gold foil. Table V11 in the experiment listed prior to that gives 5.5mm as the stopping range in air for the probe particle when no mica sheets are in place. So the probe is stopped completely by 5.5 gold foils, 2.1e-7 meters thick.How can that be, when chances of a collision is so extremely remote through 1 foil? Coulomb forces can't be the cause once the probe has entered the foil because there are then just as many such forces pulling it forward as there are holding it back.
If an alpha particle is stationary, its charge can be detected in the surrounding dimension. If the charge was to continually switch on and off, a recording of that fact would be noted moving outward at light speed into the surrounding dimension. If the charge is annulled and nothing is changing, the presumption is that nothing else is shifting outward from the source at light speed. But a direct consequence of the zero origin concept is that every fundamental component of the entire mass of the particle is leaving behind a similar recording.
Everything that exists in the zero origin universe is made up of two fundamental components, being absolute opposites and point sized. They are of course electrons and positrons. From the mass difference between an electron and proton, the number of fundamental components in the proton is 1835 (near enough). That's 917 electrons and 918 positrons.
It would be expected that such components would all collect together in a point and disappear into oblivion. But they don't ever cease to exist.
It should be obvious that a gamma ray has no mechanism with which to make an electron or positron. How can it possibly know what an electron or positron is? Does it carry some kind of code built into the frequency that can be physically extracted in stages as the pair slowly become aware of their existence? How can a dumb ray be so amazingly clever? And why only electrons and positrons? And why does the creation process stop when the charge on each has reached the required value even if the ray carries much more energy than the combined value of the two?
It could be postulated that the creation process is entirely controlled by the emerging electron and positron as they become self aware because they are, without doubt, fundamental forces of nature and the parameters for their development are set by the laws of nature. They will always end up being exactly the same as any other electron or positron in the universe. It all sounds convincing enough, but it's no more realistic than any other postulate.
The only other way the combined pairs could be detected is through the minute distortion in spacetime generated by their presence (gravity).
In a sea of these pairs, they would all be driven toward any location where the concentration is highest, and since the minimum distance between them is zero they will begin to interact beyond the individual pairs. And there's no reason why the interaction would extend beyond the point.
The enclosed dimension created by a coulomb force field is vastly different to the enclosed dimension of a gravity induced black hole. That too should be obvious. The coulomb force field generated by an electron positron pair in a less than immediate association can't restrain the light speed expansion of its force field because the charge origin is a point and the coulomb force field is 100% effective only at that point. Evidence to that effect arises in every pair "creation" event.
As pairs continue to group together in a point and interaction energy increases, individual particles could be temporarily ejected from the group at times and that will cause a pulse of instability to resonate throughout the remaining charges of the group, extending the size of the group beyond the point. The Coulomb force field would momentarily extend into the surrounding dimension. The limit of such re-occurring cycles will be reached when the momentary forces extending beyond the point have become strong enough to enclose dimension around the expanded contents, from which an electron (normally) has been momentarily ejected. The electron then remains trapped in the world outside the newly formed event horizon.
Regardless of the number of charges involved in the interaction in a point, the residual field emanating from "within" can only be zero. But now the charges are separated by the contained instability and that sets the residual field strength showing at the proton surface. So how much will it be?
If the electron is by some means returned to the inside, the interaction radius could perhaps again reduce to zero. But the internal environment is back to the stage where particle ejection was occurring. The cycle of particle ejection would be expected to continue indefinitely, maintaining a permanent separation between everything within. The mass of the system has increased by 1 electron, and the interaction intensity will have also increased. The radius of the field (neutron radius) could be vastly less than the proton radius.
The residual field at the surface of a neutron could also be substantially less than that for a proton. If that's so, then neutron probes used in scattering experiments should suggest a much greater radius for the nucleus of the nuclear atom.
Evidence that the proton is constructed in the manner described emerges through deep inelastic scattering, and apparently from maths as well, which would all seem to be a remarkable achievement. The internal charge structure is broken up and its form exposed. Whatever combinations of particles eventuate from any destructive collision, it's essential that the positive charge of the proton be represented in full somewhere in the ruins.
Top and Bottom quarks carry fractional charges. The charge on the up quark is 2/3 that of the positron charge while the down quark carries a 1/3 electron charge. That gives a charge ratio of 2 to 1. So, if the positive charge is to be fully represented, other particle combination of the same charge imbalance must also emerge at the same time. If there is 1 unit positive charge, there are two fractional charges of .5e+ and .5e-. And that would make sense considering how the field is maintained.
According to the mass difference between the proton and electron, the proton contains a total of 1835 e+ and e- charges. But the residual charge that extends beyond the immediate interaction within is apparently only .000545 of that amount (1/1835).
The positive charge excess within a proton can exist only at the horizon from the viewpoint in the outside world. Beyond that, nothing exists. Assuming that the charge is centered at a point which is almost infinitely removed from the outside world cannot be correct because the charge at any part of the horizon could only be zero. In a sense the horizon is the point origin of dimension around which the proton charge is represented in full, over the zero thickness horizon. There's no other way. The same applies for the equal quantities of residual charges of opposite signs from within.
Atoms that are fixed relative to each other won't be attracted to or repelled from each other. The residual e- components represented at the event horizon of one atom are drawn to the residual e+ components represented at the event horizon of the other, and vice versa. The force direction is outward from each horizon. And each residual e- component represented at the event horizon of one atom is forced away from all residual e- components represented at the event horizon of the other, and vice versa. The same applies for the residual e+ components. The force direction is then pointed toward each horizon. But a field imbalance is generated if they move toward or away from each other.
If a probe particle is moving at light speed toward the atoms of a scatter material the opposite charges would be moving toward each other at the rate at which the attractive forces are updated, and the force will be zero. But the repulsive force will have doubled because it points in the opposite direction. The total resultant force is then equal to the combined residual coulomb forces of the whole system that emerge through the event horizons. That's .0545% (not 100% as before) of the value of 1835 repulsive charges acting between every relatively moving proton or neutron in the system. The maximum motion generated charge for each proton or neutron is 1.602e-19 coulombs.
If the probe is moving away at light speed, the attractive forces would double and the repulsive forces would be zero.
f = v^2/c^2*Q1*Q2/(4*pi*e0*r^2) gives the resultant force. Q1 and Q2 represent .0545% of the entire charge of each proton and neutron that make up the probe and anything else that the probe is moving relative to. The radii 'r' to the scatter material atoms directly in line with the probe's trajectory are naturally the most significant.
The probe's motion is restrained to some degree as soon it leaves the source. As it nears the target, the nearest depression in the field around the gold atoms to which it's heading will begin to become apparent and the probe will be shifted off course toward it. The possibility of an atom in the target surface being exactly aligned through the center of mass along the trajectory of the approaching probe is extremely remote, as is demonstrated in the attached Qbasic program.
The rolling process involved in the manufacture of scatter foils demands that the atoms on both surfaces all be oriented in the plane of the surface, and the atoms that bridge the distance between the two surfaces would be expected to be compressed together as much as possible.
The arrangement of atoms shown in this animation is about as close together as they could get. And this is not supposed to be an accurate representation of the elastic movement. The atomic structure will shift wherever it can.
When the probe has negotiated the first layer, it's confronted by an atom in the second layer that's directly in its path. It will deflect in the direction where the repulsive forces are weakest. Chances of an exact alignment with the centers of mass along the trajectory would be zero in that environment. It's certainly not stable because the probe just crashed through the surface. The rest of the journey is obvious. The restraining force applied as the probe pushes forward is almost the same as the driving force applied as the structure closes back in behind it. As the probe approaches the final layer, it deflects downward and then along the plane of the original trajectory. The atomic configuration on both sides of the foil is the same. How it goes in is how it comes out.
The configuration of atoms shown in the animation would appear to limit access to any pathway through the material to just the one straight line perpendicular to the surface, but the 3D picture allows direct access for the probe to negotiate any path through the material at any angle to the surface.
The probe's movement through the foil would obviously follow a crooked path, but the momentum of the probe would maintain a reasonably straight line on average through the foil.
The two frames of this animation were generated from the attached Qbasic program. Note that a probe speed of around .06c was used in the initial Rutherford scattering experiments.
In each frame the trajectory of the probe has been shifted to point at a different spot on the target atom. The consequences are clear enough.
The force direction between the probe and target atom is along the line directly between them, while the trajectory is at an angle to that line. The force acting perpendicular to the trajectory is determined by the offset distance over the radius from the start point.
The next two frame animation shows the result from two different probe speeds. Clearly, the lower speed probe would align itself closer to the initial trajectory when it finds its way between the two relevant atoms than would the higher speed probe. Meaning that the lower speed probes would scatter less than the latter. And that doesn't comply with observation.
The problem is apparently in the equation
f = v^2/c^2*Q1*Q2/(4*pi*e0*r^2). Changing it to this,
f = (v^2/c^2) ^2 *Q1*Q2/(4*pi*e0*r^2) gives this result.
Justifying the change shouldn't be too difficult either because the action involved in the emergence of the residual charges is two fold. In one half of the process, the attractive forces are reducing, while in the other half, the repulsive forces are increasing. In the case of gavity, v^2/c^2 applies because only one such action can occur at any time. So (v^2/c^2) is squared again, which is of course v^4/c^4.
As the probe moves toward the upper target atom, it will be driven back down again, until it finally finds its way to the weakest part of the local charge field. The final apparent nuclear radius will be substantially greater than the 1e-21 used in the above animation, and substantially less than the atomic radius.
For a probe speed of .9c, the required trajectory offset from the target center to achieve a similar result to the above is 1/100 of the offset required for a .06c speed. The higher the probe speed, the smaller is the presumed radius of the measured component.
From this number graph, it's blatantly obvious what's going on.
Trajectory offset from Probe target atom center. speed .00000000000000000000001 (1e-23) .9c .000000000000000000001 (1e-21) .06c .000000000000001 Proton radius. (1e-15 .9c .00000000000001 Nuclear radius. (1e-14) .06c .0000000001 Atomic radius. (1e-10)
When all of the target atoms local to the probe's trajectory are taken into account, the required trajectory offsets will need to be substantially greater than were used in the animations. 1e-15 meter offset for a .9c speed and 1e-14 meter offset for a .06c speed is somewhere near what would be needed to deflect the probe past an atomic radius sized nucleus.
Electron probes used as a means of measuring the proton radius would be affected in exactly the same manner. Only one set of residual charges would emerge through the electron's motion because the electron charge is constant, but its motion toward the target atoms will have much the same consequences. A light speed advance removes the e- e+ attraction completely, and doubles the repulsive force. The force equation would need to be altered to, v^3/c^3 etc. If all charges were permanent, v^2/c^2 would apply.
The permanent e+ charge carried by an alpha particle probe also needs to be included in the program. But it wouldn't change what is blatantly obvious.
Copy-paste the program directly off this screen.
'-----Start-------------------------
'Escape exits program.
SCREEN 12
CLS
pi = 3.1416
c = 3E+08
e0 = 8.8462E-12
LOCATE 19, 1
INPUT " Radius multiplier (1 = 1e-9 meter radius)"; mul
INPUT " Probe speed (c=1)"; v
PRINT
IF v = 0 OR mul = 0 THEN END
v = v * c
e1 = 1.602E-19
'Total residual charge within each proton and neutron is 1.
' (.5e+ and .5e-)
e = e1 * 4 'Alpha particle residual charge.
ep = e1 * 197 'Gold atom residual charge.
m = 1.67E-27 * 4 'Alpha particle mass.
ma = 1.67E-27 * 197 'Gold atom mass.
r = 1E-09 'The graphics are based on this radius.
r = r * mul
rr = r 'Graphics.
mul = 1 / mul 'Graphics.
b = 340 'Graphics.
unit = 1E-11 'meter base unit.
'----These figures can be altered--------
ds = 10 'unit multiplier.
shift = 1E-21
'Sets the trajectory offset on the target
'atom. Zero offset from center of target
'atom results in zero deflection.
'----------------------------------------
shift = shift / 100 'Sets the initial shift rate.
dx = unit * ds
aa: ON ERROR GOTO ab
LINE (50, 250)-(390, 250)
LOCATE 17, 14: PRINT "<--- "; rr; "meters "
LOCATE 2, 1
PRINT r; "radius from impact zone. "
f = (v ^ 4 / c ^ 4) * e * ep / (4 * pi * e0 * r ^ 2)
PRINT f; "newtons. "
accel = f / m
PRINT accel; "m/sec^2 acceleration rate. "
time = dx / v
PRINT time; "seconds over"; dx * (v / c); "meter segment."
'-----------------
shiftx = shiftx + (shift * (v / c * mul)) * ds
'The shift multipliers set the trajectory to always point at the
'chosen offset on the target atom. The force is applied directly
'along the line through the centers of the masses, while the
'trajectory points at an angle to that line. The sideways force
'is directly proportional to the lengths of the opposite and
'adjacent of that triangle (shiftx / r).
'-----------------
PRINT shiftx; "The chosen offset on the target is tracked."
shifty = accel * time * (shiftx / r)
shift2 = shift2 + shifty
shift3 = shift3 + shift2
PRINT shift2; "meter shift in"; time; "seconds. "
PRINT shift3; "total shift. "
b = r * 3.4E+11 * mul 'Graphics
a = shift3 / m * 2.2E-15 'Graphics
u = shift3 / ma * 2.2E-15 'Graphics
CIRCLE (390 + mul * 34, 250 - mul * 68 - u * mul), 34 * mul
CIRCLE (390 + mul * 34, 250 + u * mul), 34 * mul
CIRCLE (390 + mul * 34, 250 + mul * 68 + u * 2 * mul), 34 * mul
CIRCLE (390 - b - mul * 34, 250 - a * mul), 34 * mul
CIRCLE (390 - b, 250 - a * mul), 1
LINE (390 - b, 250 - a * mul)-(390 - bb, 250 - aa * mul)
bb = b: aa = a
IF r < dx * unit THEN GOTO ab
DO: s$ = INKEY$: LOOP UNTIL s$ <> ""
IF s$ = CHR$(27) THEN GOTO ab
CIRCLE (390 + mul * 34, 250 - mul * 68 - u * mul), 34 * mul, 0
CIRCLE (390 + mul * 34, 250 + mul * 68 + u * 2 * mul), 34 * mul, 0
CIRCLE (390 + mul * 34, 250 + u * mul), 34 * mul, 0
CIRCLE (390 - b - mul * 34, 250 - a * mul), 34 * mul, 0
r = r - dx * (v / c)
GOTO aa
ab: COLOR 12
LOCATE 23, 1
PRINT "As the probe shifts more toward the upper target"
PRINT "atom, it will be pushed back toward the weakest"
PRINT "point in the field, obviously."
END
'----------end-------