Monday, April 21, 2014

Compton wave length and space impedance

1836 15.6175392619  log(1/2+sqrt(5)/2)
377   14.6307168455   log(3/2)

If I have the theory right, then then light should sample the mass of the proton at the same rate as mass samples the space impedance.

They are off by one wave number, mainly because there are adjacent mass quants at the point where the electron orbitals start.  How does the proton know all this? Because when one of these phase imbalances began to quantize the proton nulls, the proton nulls rumble, they tell the proton that phase imbalance is about to go Shannon.

The proton is going to be sensitive to space impedance because the charge it, and the electron share is related to keeping wave from quantizing a chunk of the proton nulls.  

I am looking at spectral lines from the atoms.  Now all of these should go as natural log, they are emitted before they go Shannon.  Their frequency of emission and the space impedance of 377 should be should be aligned, for the same reason. The frequency they emit at is energy, and the mass they match plus that frequency causes the same rumble in the proton phase imbalance.

So, you put a heavy muon in the middle of those orbitals, the rumble point changes.
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Sunday, April 20, 2014

Thinking like a proton

The thing that makes the atom work is efficient packing by the quarks.  The mass quant at the proton is 108 = 2*2 *3*3*3.  No matter how wave moves, the quarks find an efficient packing. That makes the proton attractive to phase because of its excess capacity to balance phase variance. So we get a dearth of free nulls in the outer ring, and kinetic energy is supported, hence the atom.  How would the proton use this capacity? By allowing phase to move in and out in response to energy changes.

There are seven principle quant numbers, and about 14 missing mass quants from the electron to the proton.  There are some 11 missing wave numbers corresponding to those missing mass quants. There are an additional 7 angular momentum orbital slots. The angular and principal split the slots available. Count up the vectors of principal and angular. Some eight.  These sets of angular and principle are the contours within which the proton can maintain constant precision, and none of the wave numbers in the atomic shell will materialize.

 These lines lie along an integer defined manifold of phase variance in the proton

That is the plan, find the curvature in which that happens.

And these integers, the first three quantum numbers, should be separable into an integer set, i,j,i*j which define the equal potential contours in the proton. There exists some integral equations that tells me so, and proton precision should be allocated so all those integrals converge uniformly with integer summation. These phase potentials should all be equal gradients, and they should have simple errors with respect to the main quant wave numbers up the quant chain from the atom. The proton seeks to maximize instability of the Nulls and maintain stability of phase by accepting phase imbalance.

Simplifying units

I am looking at atomic orbitals, and the necessity of dealing with the units of physics that keep coming up. I need to simplify, because everything is really, just phase and null and how much of those we have with respect to two quantization ratios.  So I am dumping the units of physics and replacing them with units of things counted, as follows:

Primary units:

m is the quantization ratio of Nulls
f is the quantization ratio of Phase

Secondary units:

j*m is m multiplied by an integer
k*f is f multiplied by an integer

m^j that is m taken to an integer power
f^k  and f taken to an integer power

There, that is simpler, and eliminates charge and magnetism and particles and mass and time and energy and power and force and space.  As long as I can write equations with these, in some symmetrical frame, I can count them up and let physicists add their semantics later.

I add some secondary conjectures.

2^(j*m) and 2^(k*f)  define a digit system when operations on m and f are Shannon separable, and
e^(j*m) and e^(k*f) define the digits system when operations are not Shannon separable.

A symmetric frame is one in which either conjecture always holds. Symmetry is determined by the density of phase and nulls samples. To obtain density I construct two operators:

D1 = 1/2(m^k - f^k) and D2 = 1/2(m^k + f^k) k integer

When the operators are Shannon equal to one, we have a binary system, otherwise a natural log system.

I suspect my symmetric frames will be defined by integers:  
{k, j, j * k}; k, j   integer.

The trick to the orbitals is to find the set j,k that define non-Shannon frames. I can find them in two ways; 1) Cheat and look at the centuries of work done by physicists, or 2) Use some methods of group theory. Once the integer set is found, I count them out and draw the natural log results.

Phase theory predicts the Hamiltonian is always the sum of variance in Phase + variance in Nulls, Vp +  Vn Physics is the art of finding these two quantities. In the Shannon equation we have the term: log2(1+ S/N) which now gets converted to:
log2((Vp+Vn)/Vp).

Lets call that the Shannon operator. I think we can fix the Shannon sample rate to the value 2. So then the Shannon operator will be a function of (j*k,j,k). The Shannon condition not met when the predicted quant value some j,k is < 1. The trick then becomes finding the form of the Shannon operator.

We know why we have the j,k axis of symmetry, because the proton is made of a threesome of quarks which would thus have separable axis of symmetry.  Group theory likely tells us then, we have a third axis, j*k. I think that is how it is going to work.  We make a simplification by treating the proton as a sphere along the three axis, and the value Vp and Vn will be mappings of their values inside the proton along the three axis. We are damn near done! We know the orbitals are constructed in, our spectral charge, between two points, one where the m,f for magnetic is Shannon and one where m,f for the electron is Shanon. We know how much proton precision is available, so we just take our spread the precision among the density operators along the three axis. We scale up the precision to the point where Shannon is almost met, take the 'int' of j an k and presto.

At any given point in the proton, Vn is  (D1+D2)^2, Vp = (D1-D2)^2 Computed over the three axis, up to the precision of the proton.  Precision is:
sum(j*k*p + j*p + k*p); where p is a scale factor and must scale each term evenly. We need only find the integer set where the manifold over the three axis converge uniformly, the manifold curvature being the limit of proton precision. That manifold curvature makes the Shannon Uncondition in the Shannon operator.

Charge falls out because that curve component prevents EM light from forming. Spin falls out because two adjacent mass in our spectral chart.  The j*k axis defines null density that prevents mass quantization,  The magnetic is really the phase angle of the phase imbalance in the orbital, and when that is to straight, the electron flies away.  

The spacial size of a Lagrange point in gravity and the primordeal Neutron

Wiki has a full description of the lagrange points around the solary system, interesting. Lagrange points around the solar system

Orbits tend to be unstable a bit around these points. Borrowing from Einstein, we get the same equations whether we think the Lagrange points move around or whether the potential is too close to zero. We can make the same analogy if we believe in quantum gravity; either the quants can change ratios easily or they move around quite a bit. But the important thing is these Lagrange points have size of the order of 1E6 meters.

So take that as the Compton wavelength and then move the idea to a Neutron start where gravity half wave are only 10e6, so their Lagrange points only have 10E3 size, and that makes the Compton 'mass' of these fields very large in galactic standards, the mass of a neutrino, which we think is small, but compared to gravity nulls they are quite hefty. The proton is not supported, it cannot hold charge gradient in these intense fields. So what happens at the center of these Neutron stars? Maybe they have negative protons, negatrons.

I jest. But in a Black Hole, matter at the center is not even supported. The gravity nulls are huge, nearing a size greater than even the electron. They would have absorbed most of the Nulls in the center, the center would be devoid of packed Nulls, and most of free space would be a phase gradient, gravity having collected all the available nulls. If you continually grow gravity, what would you get? Gravity Nulls approach the size of a Neutron, you get the primordeal Neutron.

But the conservation of energy tells us that is impossible, we just cannot recyle mass and energy continually. Unless the Null quant ratio grew a bit. Such a thing would be possible because the vacuum needs to generate noise to keep its sample rates constant. But in a region where nulls and phase have become so widely separated, they would not get the opportunity. You get these bizzare regions of space where the fine grained accuracy of the proton is not supported, just these dull neutron like gravity nulls held in groups with a distance of a few hundred meters.  The Compton equivalence is off, Plank is way off. Nothing interesting, just dull void.

If I were a group theorist, I would change the mass/wave quant ratios, and see if there are spots just above the Proton where we can get a near integer 'one.  Somewhere near 107, up from 91, in wave number. I think the Higgs wave number is about 107. Reset the mass quant ratio to match that, see how much grouping is supported. I set the mass ratio at 5/3 and get a very lose match with wave number 138, keeping the sample rate of light at Fibonacci.

The 5/3 world

The idea is a dull world with two type of packed nulls, huge useless neutron like things, and gravitrons slightly bigger than an electron.  Very little kinetic energy, all precarious potential energy. This is the pre-big bang world. If this is what a quasar is, then they would be easily disturbed by the normal vacuum and occasionally shoot out huge quantities of 3/2 matter at high kinetic energy.  I am not sure quasars are sucking in matter but maybe shooting it out.

Saturday, April 19, 2014

OK, a simple attempt at a four body Kepler

I take four masses, pair then up :

Mass pairs (1,2,3,4) , taken two at a time
2  3  4  6  8 12

My signal, counting radii up by some small increment and computing gravitational force (G=1), totalled across the six radii. So I compute a signal of 1/r^2 dividing each of those mass pairs. Assume a total energy. Play around with variable until I get a smooth spread of quant ratios.

Compute SNR for a total energy E if the noisecounting radii by d in N steps

My signal:
Sig <- function(d,E,N) {
     foo.sig = NULL
     for(i in 1:N){
         foo.sig[i] = sum(m/(i*d)^2)/E +1}
     foo.sig
 }



Here is the SNR +1. These are the bauds, some eight of them that count out six radii in steps. I have really done a spectral decomposition. The result tells me that those signals divided by that number E, is best counted with these bauds. Nothing more.

So I selected a possible function. There were six radii so I suspect they counter out powers of 2 modulo six. So I count out the function, using these quants in an eight digit sequence.  and generate the signal of six radii counting.  My counter was my suspected function is counter squared modulo 6. I was sure to do this:
2^(B[j]*k)/(2^k), divide by the bit power, to normalize the digits to an eight bit twos; I divided by the bit power.

I got this plot.


Six radii counting out a quadratic form.  Is it useful? No, not without some better knowledge of Kepler.  If I set two quant rates, and counted out a polar R, and a theta, then picked the ellipse as a possible function, then yes, useful.  As for this thing on the right, it could just as easily been someone juggling six balls in the air.

But here is the point.  If we know the generating force and the total energy, then we can find the best quants before applying some law of physics in the Schrodinger equation. Do the Schrodinger last.  Do group separation first.

Do I have wave and null packing in reverse order?

I am a lousy physicist, and do not trust myself, I hesitate.

It should take less phase to pack a null because phase moves at light speed.  Nulls should represent potential energy.  Which order? Is it Null at quant k+1 and wave at k? Or the other way around?

Readers really cannot trust me to get this right the first go around. But I trust the proton, and it comes in at wave then null.  I am going with that for now.

Consider the multi-body Kepler problem

We want to compute the orbits numerically for some fixed energy level. The fixed energy level is the noise. There are N! equations of mass on mass with G, the summation of these equations at any step is the signal. You step all radius with some common step value such that the divergence along the way does not break Shannon separability. Check the -iLog(i) along the way with you computer, or just scale up.

The amplitude is some 2^(k* quant). You want to compute the quants for each step from k=0 to k = Nmax. Compute the mass on mass signal at each step, use Shannon to get the quant. What does this get you? You have simply counted up the group with respect to some fixed point of symmetry. You have created the Kepler group. Does this find you the central Lagrange, or is there one? I have not done this, but if you lay out the radii on a complete sequence, they should count out the potential energy.  The kinetic energy makes that phase flat.  When the potential energy is maximum they are closest together, and visa versa. So, use a little ingenuity and fins a common Lagrange.

For example: On the complete symmetry group of the classical Kepler system

What is in those electron orbitals?

Wave(k) does not have Null(k), it has Null(k-1), it can't pack the proper quant of null.   Wave(k) has dumped excess phase into the orbital, and the Null(k-1) is now a Null(k) but half packed with phase.

So, what remains in the orbital? Up to half a quant of free Nulls. And the electron does indeed operate like a spread out bunch of nulls. I do not think is it is a probability of finding an electron, it is the actual electron is spread around. There are not enough Nulls to pack the next quant up, so add energy to that orbital, and the current quant of electrons mixes and matches with the 1/2 quant of free nulls. It really does spread around. Phase is unbalanced and held by the proton charge, phase can't fly away.  So, up to a point, those orbitals are Null soup, nothing but a mish mash of unbalanced phase and free Nulls.

This came up because I was looking at how the hyperbolics would accumulate imbalance as they march energy up the orbitals. The functions are ideal, for the purpose, they keep fractions and count out the orbitals with few operations.  I substitute the twos base for the natural log, and fractions appear to accumulate smoothly.

Anyway, I went through the Schrodinger stuff for the hydrogen atom, I took that class and I get what is happening.  But here is the thing. We know we step energy levels by Plank.  We know we are at maximum entropy. So we have to sample at twice Plank, and we have to encode the spherical charge function and mass function.  We know we are symmetric about the nucleus. So, simple enough in the bitstream version.  Energy levels count as Plank, they are the signal. Amplitude, quant size, count as log Plank.  I think your noise is actually the spheroidal functions you need to quantize. What I mean is that the spheroidal charge is a disturbance to the system, and it is resolved by quantization.

So, just plug is in and sample away. The hyperbolics should count out properly. You may have no idea what the quants mean, but they should be accurate. So give them meaning after the fact, count up first, then apply physics second. The same way I did the Plank black body. Applied force, or applied heat, or applied gravity, or whatever are disturbances to the quantum system, they should normally be noise in the Shannon equation. But the S/N is always an energy ratio. It took me a while to get that. So, we want to know the orbitals for a certain energy level. The hyperbolics count out the kinetic and potential, as you step through the Plank energies. It is just a sampled data version of the Hamiltonian.

I tire of thinking like a vacuum

So, rather than write a quantizer code, I am going directly to the hyperbolic. I am taking a short cut.  Its a simple trick in which I pick the order I want to work with, and pack it with force, obeying the rules of Fibonacci packing while accumulating quantization error in the proton.

My approach is to scale some region of the quant chart to the precision I want. Then make two binary numbers, one for null quants and one for wave quants.  I want everything packed in wave/null pairs. These two numbers are my Eigenvectors, and the hyperbolic functions my Eigenfunctions.

I will shift and add the nulls/wave Eigenvalues to make them match, in wave/Null pairs. and shift and subtract the same amount from the proton. Thus, each null quant gets the proper wave number, and the error in the proton will be kinetic energy. I make my new binary number all ones, but they are not all matched in quants.  The matching error is in the proton. (Rule 1) All actions can be considered as actions of wave/null pairs.

(Rule 2) I know the vacuum quantizes the maximum order it can in a wave/null pair to stabilize.  So I can just fill in the blanks until I have what I need quantized, then extract the proton accumulated error and see what I have to play with.

Later, I will assign even and odd hyperbolic functions to the wave and null quants, make the thing automatic, then convert from bits to qubits for pretty pictures.

When I scale I get two binary numbers. The one bits indicate the real quantization, the actual vacuum samples. The numbers look like:

Wave                          Null
1000100100010010   0010010100101

(Rule 3) The wave get the higher order bit (Rule 4) The wave ends in zero, the null ends in one.

Then I shift the nulls left to match the first wave, accumulate it in my output sequence, and subtract from the proton. As I do that, I am sure I will discover more rules, but these four are a good start. I guess the main rule is Fibonacci packing.

What about charge and spin?
I think charge is a fractional overlap in wave/null pair for the electron, and the proton should be initialized with the value. Spin is a 1/2 allowable variation in the null quant number, as near as I can tell. I have no idea how to work these, yet. I am not a physicist.

Looking at one of my previous posts, we discover that the hyperbolics manage fractions quite well.

Friday, April 18, 2014

Nucleon accuracy and the shoe industry

When the inaccuracy of packing nucleons equals the inaccuracy of making orbital slots. Quantum accuracy is all about maintaining the integer 'one' for the group, and that is limited by the nucleon. If the integer 'one' is violated then the conservation of baryon has broken.

Let Qw be (1/2+sqrt(5)/2)*91 and Qn be (3/2)*108.

These are the quant values of the proton Wave/Null pair.

Lets use log two because our computers use it.
The log difference, log2(Qw/Qn) = log2(Qw) - log2(Qn), is 5e-5.

At full accuracy, the difference should be:
 log(2^(Nmax) /2^(Nmax-1)) or log(2) = 1,

But we make that 1/2 to keep Nyquist sampling. Nmax is a twos bit equivalent of the atom that is maximally Fibonacci packed.

At that point the physicist is dealing with the twos binary version of an atom that is Fibonacci packed according to the real wave/particle quantization.  The Fibonacci packing is more efficient than the binary, but more difficult to do math. But the actual standard model, including the Feynman diagrams acts just like a Fibonacci packing. The top wave/Null quants (91,108) come as a pair, and mark the Finonacci integer, the other wave/null quants are separated, using the actual particle quants and their matching wave quants. The standard model, and the atom,  is a Fibonacci packing.

Thus scaling to get our twos binary version of the Fibonacci atom, so we can do math:

 1e4 * [log2(Qw)-log2(Qn)],  must equal 1/2 (with Nyquist sampling).  So, at scale, the twos binary is a 10000 digit integer, a whopper.

Making Fractions

But wait, that's not all.  Somewhere these stupid vacuum elements figured out how to use the proton atom to make crystals, liquids, solids, rocks, galaxies and shoes. I have no clue how they figured this out. But they use different null quant ratios, evidently. So the proton, in its generosity, as agreed to carry fractions, and has a decimal point somewhere up its binary quant chain of 10000 digits. This decimal point allows the proton to hold fractional errors imposed upon it in the making of shoes.

Where is this decimal point? I have little clue, except:

We have one clue, the magnetic null quant is unavailable, and that seems to be 18 orders down from the electron, out of 91.  Plus we have vacuum noise. So 20 out of 91 times 10000 makes a big fraction, in binary digits. The most likely scenario is that the magnetic kept kicking protons free, and with enough free protons, magnetic null quants failed. But about half the digits gained from kicking out the magnetron were needed to manage charge.

Consider the electron, about 14 orders down from the Proton, or 1/6 of 91. That becomes a 1600 digit binary number. Can we describe everything we know about chemistry as a series of subgroups making up 1600 digit binary number?  The periodic table needs 8 digits to count the atoms, another 10 to put them in their place on the chart, another 15 digits to describe their oribital quants. We are at 32 bits, with 1500 left over to describe the chemical elements. Easily done, I think.

 The vacuum phase, using the proton, as invented the decimal point.