In the "relativity busted from the get-go?" thread, I came to the conclusion that if we are to conserve energy and momentum, then when particles are reflected off of a surface, the difference in their speed before and after they are reflected is twice that of the relative speed between them and the reflecting surface. So if we figure the reflecting surface to be travelling through a medium at a speed of v while the average speed of the particles in the medium is c, both relative to the medium itself, then particles will be reflected off of the front with a speed of c+2v and off of the back with a speed of c-2v. Figuring in for the angle of incidence, then, the reflected particles obtain a speed of c+2v*(cos L).

Okay. So let's say we have two equal masses travelling in a circle around each other through the force of gravity, and that gravity is created by some of the particles in the medium being absorbed by the other mass so that a lesser energy density, or negative pressure, is produced in the vicinity of the body where the difference with that of the medium reduces with the square of the distance from the center of the body that absorbs it. That amount of the pressure that is actually absorbed is extremely small. The remaining pressure that is not absorbed simply reflects off of the surfaces of particles in the body in all directions and is the vast majority of it.

So first we have that amount of the pressure that would have normally reached the second body travelling at an average speed of c that has been reflected away. Two bodies of equal mass (and size and shape) should always feel the same force as the other at all times in the same way as they each induce the same forces upon each other. So the current positions of each should be on directly opposite sides of the circle of orbit. But it takes time for the particles reflected off of one (and the "holes" of those that have been absorbed) to travel to the other, so there is a propagation delay. The position each would currently "see" of the other is that where the other was when the particles began to travel from it.

So instead of coordinates of (0,r), one body will see the other at (x,y), where x=(sin 2L)r and y=(cos 2L)r when L is the angle one body sees the other in respect to the origin, so 2L is the angle from the origin itself between that of the current and original positions. The distance travelled between the original position of the first body and the current position of the second is d=ct, where t=(2L/360)*2pir/v. The force that is received is proportional to the number of particles per distance times their energy, but the number per distance is also inversely proportional to their speed, so the force becomes just proportional to their momentum. The force of gravity toward the barycenter, then, is proportional to (y+r)/d^3. In the line of motion, it is x/d^3.

Now, to find the angle for the original position of the first mass to the second, we solve for distance using d=ct and d^2=x^2+(y+r)^2. So

ct=sqrt[x^2+(y+r)^2]
c*[(2L/360)*2pir/v]=sqrt[((sin 2L)r)^2+((cos 2L)r+r)^2]
c*[(2L/360)*2[i]pi/[i]r/v]=r*sqrt[2+2(cos 2L)]
(c/v)*[(2L/360)*2pi]=sqrt[2(1+(cos 2L)]
(v/c)=[(2L/360)*2pi]/sqrt[2(1+(cos 2L)]

We have to find the angle the hard way with this but a simple computer program can find it easily and quickly as precisely as desired.

This is for the amount of gravity that is kept from travelling to the second body by the first, relative to where the particles would have normally been relative to the medium. For those that are reflected, their speed is now relative to the body as well, so d=[c-2(cos (90-L))v]t and the angle is found by the same method as before, resulting in

v/c=1 / [ sqrt[2(1+(cos 2L)] / [(2L/360)*2pi] + 2(cos (90-L)) ]

This resulting angle is different from the other, because the particles travel also with the speed of the body during propagation, as if from the current position of the body if the body had continued in a straight-line path. We will designate these two angles as L1 and L2 (and 1 and 2 for all values in each set). That pressure which is originally blocked at an angle of L1 pulls the second body in that direction because of the greater pressure acting from the other side of the second body, but that pressure which is reflected at L2 helps to push it away, so the overall effect is the difference between the two, and in proportion to the speed of the particles in each case. So for the force in the line of motion, this becomes proportional to

x1/d1^3 - (x2/d2^2)*[1-2*(cos (90-L1))v/c)]

and toward the barycenter is

(y1+r)/d1^3 - (y2+r)/d2^3*[1-2*(cos (90-L2))v/c)]

I have made a computer program for the computations for this and found that for small v/c (anything less than about .1c), the force in the line of motion becomes proportional to -(v/c)^5 /3 and that toward the barycenter is (v/c)^2 /2. Figuring in also for the vectors when considering aberration due to the motion of the second body is rather complex, but I get -(v/c)^5 *6/41 in the line of motion and that toward the barycenter remains the same. Because of the complexity of attempting to figure in for the aberration, I have also tried a couple of other possibilities for the vectors involved and receive -(v/c)^5 *3/43 and (v/c)^2 /2, and -(v/c)^6 *2/145 and (v/c)^2 for the results. I will need someone that is good at finding which precise vectors I should use to help me with this.

Well, that is as far as I have gotten so far, but I thought I should go ahead and post it. I'm not even sure where this might lead yet, but I'm still working on it. Maybe I can get some ideas from any questions and responses as to where I should go from here.

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Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

But what "you figure" is not what we observe, so nothing that follows has observational validity.

Einstein explained what we do observe quite nicely, as a matter of fact.

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PW -- Plant Whisperer

Why not? We observe a null result with the MM experiment, that is all, and so this is what would be required, as far as I can tell, if we are to conserve the energy and momentum of the system. It just means that the motion of the mirrors themselves and how light reflects off of them must also be considered. The speed of the light relative to the mirrors remains the same before and after the reflection, and if the source moves with them, it would just be c. Also, an observer watching a light clock in motion, for example, would not see the beams travelling at some angle to the mirrors as it moves, but still straight up and down with the mirrors. The individual pulses will travel at an angle, but they will always line up in a beam that remains perpendicular to the mirrors at all times (if that is the way it is initially projected), regardless of the relative speed of the observer.

Oh, sure, but it is the result of manipulating the geometry of the system instead of using the laws of conservation to find the answers for us and leaving our definitions and/or concepts of time and space alone. It would still work out much of the time, since it turns out that such manipulations do work surprisingly often, actually, as kind of a shortcut through the reality of a situation, like a shorthand version of mathematics, but not always. I don't want to have to continually sift through those that work out and those that don't. I want to get down to the nitty-gritty. Why does it work in some situations? At what point does it break down and quantum mechanics takes over? What is the common mechanism by which these things operate? How does the energy density of a medium compare to the fabric of space-time? How would a difference in pressure relate to a warping of space-time? These are some of the things I want to examine. Unfortunately, quantum mechanics isn't quite as neat a package as relativity, and will probably get quite messy at times, but this is what I'll have to learn to deal with if I am to get through it. In my opinion, quantum mechanics and relativity are basically just two sides of the same coin, but I'm thinking quantum mechanics is bound to be more precise in the grand scheme of things, since it deals with what is taking place on a more fundamental level, instead of just changing the overall geometry to suit, and so will win out over relativity in the long run.

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Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Not to argue, but this is one of the few ATM threads still unlocked for postings

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PW -- Plant Whisperer

Yeah, I'll post here, too, just for posterity's sake: to prove that I once existed.

Let the lock-down begin!!!

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My invisible elf ??? Why he is made of dark matter and lives off of dark energy !!!
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Grav, I see a few problems with your reasoning.

First, you made an approximation in your reasoning about the reflection. The twice the center of mass velocity change assumes that the stationary mass has an infinite mass. In this case, the change in KE and momentum of the stationary mass is 0 and the reflecting particle must have the same velocity leaving as it had coming. If you do the equations out fully, and use real masses, there will be at least a small change in the stationary masses' momentum.

Second, if you are going to use relativistic particles, you must use the relativistic masses. This should make your infinite mass approximation even worse.

Third, if your particle is actually travelling at c, then the reflected velocity will be c, both from the fron and the back.

Fourth, if you do the special relativistic equations, then the conservation laws hold. Einstien was looking for a geometry where there was a fixed c that would obey conservation.

Fifth, QM is inherently less certain than relativity. Uncertainty is a basic property of QM. It is also a property of the geometry of the system too.

Sorry I didn't reply to this sooner. I became enthralled in another thread and guess I forgot to check back on this one. You have some good insights in this post. Your comments should be helpful to me with this endeavor.

That is exactly what I did with this, you're right. I combined the laws of momentum and energy together in such a way that I could solve for the final speed of the particle and came up with

v2'=(2m1*v1+m2*v2-m1*v2)/(m1+m2), where m2 and v2 are the initial mass and speed of the particle.

I then figured the mass of the particle to be practically insignificant to that of the stationary mass, so used zero for the mass of the particle, giving me

v2'=(2m1*v1+0-m1*v2)/(m1+0)=2*v1-v2=2v-c for the speed of the particle after the reflection, where c is the initial speed of the particle. Reflected at an angle, this becomes v2'=2(cos L)v-c.

Now, as long as the speed of the particle is great and the mass is very small compared to that of the stationary body, the difference should be insignificant to the particle. But it would make a very slight difference on the stationary body, and the formula would then continully change as the body begins to decelerate in the line of motion, which is also what I am finding in this thread so far. I am only considering that for the instanteous velocities so far, though.

Quantum mechanics would (or should) just be another way of doing the same thing as relativity, but by considering the mechanism and without changing the geometry. So what you are implying would be sort of a relativity on top of relativity. If masses really tend to infinity with greater velocities, then quantum mechanics should show this, and it would be relative to the medium in this case. For instance, take an object falling through the atmosphere due to the acceleration of gravity, but against the atmospheric pressure. At some point, the speed of the object will level off at a terminal velocity due to air resistance and will no longer accelerate. At this point, one might say that the mass of the object becomes infinite, because even though it is supposed to be constantly accelerating, it does not do so, but continues along at its present rate. The same thing occurs when we attempt to accelerate an object with something like a high pressure water hose. The initial acceleration of the object depends only upon its rest mass when stationary to the hose. But as the speed of the object increases, the relative velocity between the object and the water molecules decreases, so less pressure is applied. The speed of the object will approach that of the speed of the molecules, but never quite reach it, as it accelerates to less and less a degree, becoming zero when the speeds are the same, since the water molecules can never quite catch up to it, and so it would appear that the object obtains an infinite mass and can be pushed no faster.

Yes, that's correct. Whatever the speed of the particle is initially is what it would be upon reflection. But that is relative to the body doing the reflecting. To an observer with some relative velocity to the body, they would see the body travelling at v, so the difference in the initial and final speeds of the particles would appear to be off by 2v to them.

The laws of conservation do not appear to hold for all instances in relativity, even to the point that some seem to be beginning to suspect (as far as I've read on this forum, anyway) that the laws of conservation themselves are incorrect in the way they are presented, that relativity supercedes them as well, or that they must be modified somehow, by incorporating relativity into the formulas. I do not hold such a view.

That's true. It's kind of a catch-22 in a way. It would be more precise, include a mechanism, and more could be determined from it, but it would also be much more complicated. One must decipher the positions and speeds of all of the particles coming form every point in space at any given time. This does tend to get a bit sloppy, and of course, the uncertainty principle messes with that as well. But with such large numbers of points and particles, we can eliminate some of that mess with the laws of averages and by integrating over the geometry involved.

So yes, relativity is probably a much neater way of doing things in many circumstances, since an overall assessment through some type of simple manipulation of the geometry would be much quicker and easier. But that also involves some methods that may not always be feasible, which sometimes includes using complex numbers as kind of a shortcut through the geometry that can often be misinterpreted. This way of doing things can be useful and often convenient, but I prefer to take the long way around. Even if it ends up being much more difficult than I can handle, I still believe it is the way to go.

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Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Well, I'm not sure myself whether this would be helpful or not. I guess we'll see. I had intended to go back through some of my threads and add to them as I figured out more related details, but I guess I could start a new thread and point to them if I figure out something important enough to warrant it. I'm not sure how I would just add simple calculations or insights that I happen to come across willy-nilly about the same subject, though, without bugging the moderators for permission to temporarily re-open it or something.

If the threads are limited on time, then I guess I will have to start thinking the subject through more thoroughly before I start a thread about it. That might help me become a more rigid thinker, but it will also take half of the fun out of it, for me, anyway. Part of what makes it interesting is to start with some general thoughts and see where it might lead. The replies I get usually help me to mold and shape those ideas into something tangible. That's how I make progress, especially when I'm stuck. Besides, if I have to think a subject all the way through before I post, then I would probably seldom post, since I actually rely on the help I get from other posters in most cases to think it through in the first place. A month does seem to be about as long as my threads run anyway, though, but my main concern is that someone might come along after that with some great insight to a problem that could change the shape of things, but they don't bother to post it because the thread is locked.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

That isnt the problem. When you are using relativistic velocities, the forms of the energy and momentum equations change to take relativity into account. At its most simple, m2 would grow larger, and if very close to c, the small mass approximation is no longer valid.

no the bounced particles would have a velocity of c

That is about the same arguement used by those who didnt want newtonian dynamics altered by relativity. My relativity isnt really good enough to say yeah or nay tho. I am more just pointing out the basic flaws in your arguement.

and at that point it is no longer quantum, pretty much by definition.

I guess what I should say here is that I am not using relativity at all in this. I am attempting to use an "alternative approach", purely through quantum dynamic means, which may or may not produce the same results as relativity. My knowledge on this is limited, however, and I cannot seem to find many good references on the internet. I am mostly working with the Maxwell-Boltzmann formulas for speed distribution in gases and the Planck wavelength distribution formulas, but I'll probably need more than that. Does anybody know of any good easily accessible and easily understandable sites with other useful related formulas when considering a medium?

Well, I have a lot more work to do on this, and only about twenty-something odd days to do it in, so I'd better get at it. As it is, I might not make it. The clock's a' tickin'.

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Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

grav. The coincidences between the Rome and Maryland bar gravitational wave detectors, and the IMB, Kamiokande, Mont Blanc and Baksan neutrino detectors seen during SN1987a, and published ~ 28 times indicate that they differed by ~ 2-3 seconds, and the clocks were not all on universal time.
Distance to the Large Magellanic Cloud is ~ 170,000 light-years. seconds in a day..86,400 times 365.24 days times 170,000 minus 3 over 86,400 times 365.24 times 170,000 yields a gravitational effect and a neutrino velocity that are ~ 99.9999999999 c and 100% c. After 170,000 years they are no more than 3 seconds apart, and if somebody had their wristwatch Universal time synchronized,(IMB), they both might be perfectly c.
There are those in the world of physics who believe those two velocities are close enough to call them both indistinguishable from c. That means there is no propagation delay whatsoever. Pete

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A third rate theory forbids.
A second rate theory explains after the fact.
A first rate theory predicts.
A. Lomonosov

You arent using a quantum approach either. To be honest, there is no easy way to learn QM, and I really doubt you will find a good reference on the internet for how collisions work in QM.

Thank you. I'm not exactly sure what measurements you are referring to with gravitational wave detectors, but it sounds like the same thing. The propagation delay I am referring to here is actually just that gravity travels at c, not instantaneously, but that it appears to be almost instantaneous the way the vectors work out. For the gravity that acts directly between the current positions of the masses, it appears to be off by a factor of (v/c)^2, causing an apparent delay of 1-(v/c)/2. This is almost precisely the same as gamma in relativity for small v. It acts as if it travels at a speed just under c, off by only about one part in a billion for the speed of Earth's orbit around the sun (but for two Earths orbitting each other at that speed), for example. I'm not sure about the Magellanic clouds, but it would probably be even closer to c due to a lesser relative speed. So one could say almost equally that the speed of gravity is just under c over the actual distance travelled, or that it is precisely c, as with relativity, but the distance has expanded by a factor of gamma, although the distance would actually contract with relativity, I guess, so I suppose this would still be off by a factor of gamma^2, which would still only be about one part in billions or trillions, though, depending on the relative speed.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

I'm sure you're right. I doubt if I am going to get too far with this before the deadline. It'll probably take me a while just to get the basics.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Grav;

I don't get it!!! If this 'particle' is traveling at 'c', if it is baryonic it WON"T reflect and if it is Non-baryonic it WON"T reflect, it will go right through ANY baryonic matter.

Are you thinking that this 'particle' should be the 'carrier of light/photons?

That would be correct, and then you have the 'particle/wave' duality...when the photon hits the mirror (we measure the wave), the infintesimal non-baryonic particle keeps going at "c".

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RussT
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Everything is, as it should be, otherwise, it wouldn't be!

There are quite a few possibilities here, actually. I quess I will have to explore them all and narrow it down. One is that the majority of the particles pass right through a body and keep right on going. Only a very small portion, approximately one part in 10^42, is absorbed by the body. Another is that the same small portion is absorbed, and the rest is simply reflected off of the surfaces of baryons. A third is that all of the particles are initially absorbed, bounced around in the interior for some time, then re-emitted, the difference of 10^42 then either demonstrating how the average density of the pressure in free space is "growing" over the same amount of time or how some of the energy is transferred to baryons in the process. Another is that since the energy density in the vicinity of a body becomes smaller, it creates a sort of refractive density that changes the speeds (and directions) of the particles. Finally, the speeds of the particles after they pass through a body might be different, but as the energy is transferred through the medium, it may cause the speeds to change, speeding up or slowing down in accordance with the average speed in the medium, moving "out of phase" as it passes through bodies, and then slowly back "into phase" again after passing through.

Yes, but it would still require at least a very slight interaction. It could also produce electric, magnetic, and nuclear forces. If the majority of the energy density is reflected off of the surfaces, they would block the entire pressure from reaching another from that direction, but reflect them off the other side with their own difference of energy. An angular momentum of the surface, clockwise or counterclockwise in the line of travel through space might produce positive and negative charges. Most likely, though, there are two types of particles in the medium, it and its anti-particle. Negatively charged particles in a body absorb a small portion of the particle, positive absorbs the anti-particle, creating the "in phase", "out of phase" scenario, and electromagnetic waves. The rest just continues right through. The density of the charged particles would be about 10^42 times greater than that of free space. The density of the electron, for instance is about D=m/(4pi/3)r^3=10^13, using its classical radius, while that of the medium is about D=3P/c^2=10^-29. There should be some correlation there. As far as nuclear forces go, though, if the energy density is simply reflected, two charged particles that are close enough together would not allow the energy density to strike the surface between them to as great a degree, which might reduce their repulsion to where they actually begin to attract when close enough, and nucleons that are completely surrounded by other nucleons wouldn't feel the pressure at all, and become electrically neutral while in the nucleus.

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Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Actually I don't think any of those others really exist.

Just change this to all...and that's it.

Where are you getting this #?
Also, think of all of these going through YOUR body, and they are in absolutely prodigous amounts, every minute of every day of your life!

The photons are 'massless' wave like quanta, riding the infitesimal 10^-33 particles that would be going in straight line motion in SR. With no gravity to curve the 'particles' and the particles traveling at "c" in EVERY direction, the 'particles' are bringing/carrying the photons straight to you, no matter where the light source is emitting them from.

SO, if you shine a flashlight 'straight at a mirror, the particles are carrying that light to the mirror, the particles at "c" keep going through the mirror, the photons are reflected straight back at the flashlight with no scatter. SO, the particles at "c" just keep carrying the light to the mirror, and all just keep going as more at "c" keep coming to carry the light, while the photons just keep being reflected back and forth from the mirror to the glass of the flashlight.

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RussT
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Everything is, as it should be, otherwise, it wouldn't be!

Aha. That last part was bugging me, so I went back through the calculations. The speed of gravity directly between the current positions of the two bodies would be measured as 1+(v/c)^2/2, or approximately c/gamma, not 1-(v/c)^2/2. So for small v, one could say almost equally that the speed of gravity is just greater than c over the distance travelled to the other's current position, or it can be said to be precisely c, but the distance is contracted by a factor of gamma.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

All possibilities must be explored and eliminated. It may very well be that the energy waves of the particles are all we really measure and is reflected by surfaces, while the particles keep going, but that still implies some form of interaction. If the particles keep going, then they are still carrying some energy and momentum relative to the body they pass through or they would have been reflected as well, absorbed, or otherwise just stopped in their tracks. So the energy that is reflected would probably just be the additional energy of the wave, as compared to that of the average energies of the particles and/or density of the medium. Also, c would just be the average speed of the particles.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

See, right here in the first sentence [energy waves of the particles]...we have to be SOOOOOO careful with our wording.

The electrons "EMIT" the waves at the source (with a defined enegy level ie; eV, GeV, TeV) and those 'waves' are 'riding on', 'are being carried by' the Planck particles that are themselves traveling at "c", which is the mechanism for why light speed is "c".

All of the interaction is in the 'energy level of the photons'!!!

Grav, remember way back when I said.........when does E NOT =mc^2

This is it, again. E does not = mc^2 FOR the 'base particle' that is the carrier of photons UNTIL those photons are being emitted at TeV energy level...pair production!!!

At any lower energy level NOTHING of the energy of the 'base particle' is interacting with anything at all, that's why they are simply going right through ALL baryonic matter, including your body, and everything else in the room you are in, the earth...everything!!!!

The Energy of Non-baryonic Dark Matter is "INERT" UNTIL TeV energy 'releases' the locked in energy and then you get pair prodiuction of E=mc^2 that is in that tiny infitesimal 'base particle'.

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RussT
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Everything is, as it should be, otherwise, it wouldn't be!

Moderators,

Could this thread be closed until I am ready to apply more research to it and then re-opened for the remaining time? It may take me a while to explore this further.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Done (if a little late) - 10 days used, 20 to go.