Can't avoid the topic any more. I always wanted to get with just the absolute value of bandwidth, and ignore the phase shift across the spectrum. But it now matters because the bubbles solved a cubic root with complex numbers, that means phase delay, and that means charge.
A minimum quant of Phi^17 is one wave of phase delay in the electron, compared to the quarks (or is it 2/3 a wave as Mr. Planck claims). It is the length of the spectrum of the orbitals. It must be both the fundamental unit of mass and the unit of charge. Phase delay is the quants counting with relative delay between them. That is how the Higgs worked, making a sphere slightly odd shaped is associated with charge and motion. Somehow they discovered that a sphere with different quants of the two phase types would be misshapen and move, move just fast enough to escape the phase balancing mechanism. The whole idea of a difference in the exchange delay between the two bubbles is what makes light work, it provides a line of symmetry.
Is the electron negative inside the sphere or outside?
They never said in physics class, they just tell me its negatively charged. I interpret that to mean that it looks, from the outside, like it has a surplus of negative bubbles (small bubbles?), so therefore, it must have a surplus of positive (big?) bubbles in side? No?
Anyway, these numbers need straightening. Phi^17 is very small relative to the center frequency. I have a scale factor to deal with, there is a limit on the size of anything counted, a kind of Planck. The very top of of the spectrum, Phi^N is the size of a quant, and that is huge, but that conforms to one cycle of a gamma ray, more or less, measured in seconds. I never figured out what a second was, in units of bubbles, but the inverse of that shouldn't be greater than Phi^17, think. That is only about one wave number at that level so still fits.
The rest mass of quarks is 20 times the mass of the electron, they say. And they hold about the same amount of charge. But in raw numbers, the electron and the gluon are about 10^19 bubbles apart, by subtraction. So, there is likely a rescaling going on, the first upgrade to quant size happened early, it happend just above the vacuum.
When I checked to see if Plank scaled the same as my two quant ratios, I noticed that my quant ratios counted Phi^18 many counts as Plank, which is a ratio. So this all fits.
And, when I scaled the longest wavelength that the physics have in their planck curve, and the shortest, at the end of my spectrum, they matched. So clearly we all think that bubbles are the right size, (physicists think space is flat). Thus something happened to change quant ratios at about quant Phi^17, definitely, a change which happened above the vacuum layer.
Physicists have the Plank charge at about 10 e-18. That is on the order of the raw difference between the gluon and the electron, on my chart. So, the 3500 bubbles, at Phi^17, is within range, that is 10e-18 seems to be one unit of Phi^17 things.
Wednesday, June 18, 2014
The exchage rate of phase
Phase quantization seems to crap out around Phi^17, according to my spread sheet. That is about 3500 null bubbles. So the phase error of light must be about 1/3500 units of phase per bubble, at the quant rate of Phi, in the first Lagrange, thus, I think, this is linear, and phase error is quadratic.
There are Phi^74 of these globs of null bubbles in the proton, times two if you like, that is 1836 times the mass of the electron. So at least we know the phase error of light relative to a unit of engineering mass. So we have the standard unit of mass. What this number really is in the number of counts required for 3/2 to power through one period of round off error and match Phi. It is sort of the beat frequency between Phi and the ratio 3/2.
In reality, what would have happened is a stationary Gaussian distribution of phase and Nulls about Phi^17. The ratio of the first and second Lagrange is about 1.28. That is about a quart of a standard deviation away from 3500 bubbles. The old and new quant rates separate because the old quants have mutual interference. The new quants are root(2) plus or minus 3.5 (there abouts), their noise separation has doubled from 1/2.
This does not last, of course, eventually the second Lagrange runs its course and the third moment is added when the third Lagrange takes over. That is when we get charge. The part I do not quite get is that the phase error does not go away. At least, not unless the bubbles change characteristics. The noise separation can change by increasing quant size, or adding degrees of freedom, but the phase error stays, I would think.
There are Phi^74 of these globs of null bubbles in the proton, times two if you like, that is 1836 times the mass of the electron. So at least we know the phase error of light relative to a unit of engineering mass. So we have the standard unit of mass. What this number really is in the number of counts required for 3/2 to power through one period of round off error and match Phi. It is sort of the beat frequency between Phi and the ratio 3/2.
In reality, what would have happened is a stationary Gaussian distribution of phase and Nulls about Phi^17. The ratio of the first and second Lagrange is about 1.28. That is about a quart of a standard deviation away from 3500 bubbles. The old and new quant rates separate because the old quants have mutual interference. The new quants are root(2) plus or minus 3.5 (there abouts), their noise separation has doubled from 1/2.
This does not last, of course, eventually the second Lagrange runs its course and the third moment is added when the third Lagrange takes over. That is when we get charge. The part I do not quite get is that the phase error does not go away. At least, not unless the bubbles change characteristics. The noise separation can change by increasing quant size, or adding degrees of freedom, but the phase error stays, I would think.
Computing progatation functions
I recommend we don't, mainly we need only determine the quant sequence in the digit systems of each body. Motion of a unit circle is simply the quantized change in its surface shape, which I know nothing about at the moment. But that motions should go as fourth order, I think, though do not quote me.
After the quant sequence is determined, just go though some calculation on baud rate relative to any other baud rate and we have propagation. In our case, Higgs is the baud rate.
Motion and Compton Bandwidth
If we define Compton bandwidth as the ratio of Nulls to phase in the unit circle, then we have a definition of mass. Mass being the resistance to motion, and it increases as the third moment of that ratio. The more internal phase imbalance that is tolerated, the more spherical the body remains and the less motion. Wave shift raises the quantization ratio of wave and makes the unit circle larger.
Here the hyperbolic and trig functions are multiplied by root(2).
But why would there be a higher null/phase ratio as a result? Because of this:
r^2 = r+2. Two units of separation with two degrees of freedom with the large quants. Larger separation between wave quants, inside the unit circle, means less interference.
The quarks have 4 to 10 times the mass of the electron, and over all, three have about 21 times the mass of the electron. The gluon brackets the bandwidth of the whole system, but likely has three quants of spectrum of the form, 1/f and f. But they are really functions of the action in the three quarks. So the quarks are wide band. So somehow the system has managed to dump much of the quark phase into the orbitals and into the gluon. The wave shift for the quarks seems larger, and it seems they have a different cubic root system then the electron.
After the quant sequence is determined, just go though some calculation on baud rate relative to any other baud rate and we have propagation. In our case, Higgs is the baud rate.
Motion and Compton Bandwidth
If we define Compton bandwidth as the ratio of Nulls to phase in the unit circle, then we have a definition of mass. Mass being the resistance to motion, and it increases as the third moment of that ratio. The more internal phase imbalance that is tolerated, the more spherical the body remains and the less motion. Wave shift raises the quantization ratio of wave and makes the unit circle larger.
But why would there be a higher null/phase ratio as a result? Because of this:
r^2 = r+2. Two units of separation with two degrees of freedom with the large quants. Larger separation between wave quants, inside the unit circle, means less interference.
The quarks have 4 to 10 times the mass of the electron, and over all, three have about 21 times the mass of the electron. The gluon brackets the bandwidth of the whole system, but likely has three quants of spectrum of the form, 1/f and f. But they are really functions of the action in the three quarks. So the quarks are wide band. So somehow the system has managed to dump much of the quark phase into the orbitals and into the gluon. The wave shift for the quarks seems larger, and it seems they have a different cubic root system then the electron.
Is there unquantized empty space?
The behavior of bubbles makes sense if flat space is non existent. There is no such thing as empty? The universe behaves as if empty is an impossibility. Or, empty can only exist in bubble form. Light seems to have the job of removing empty, in any possible form.
Bubbles jump up the Markov tree in order to have enough Nulls to keep light noise contained in the unit sphere. That is what mass means in physics terms. So don't we get a deficit of Nulls in free space? I dunno, it seems to be that if free space is so bad, then packing Nulls does not likely help, a deficit of nulls in free space means empty space might appear. What then? More bubbles?
Bubbles jump up the Markov tree in order to have enough Nulls to keep light noise contained in the unit sphere. That is what mass means in physics terms. So don't we get a deficit of Nulls in free space? I dunno, it seems to be that if free space is so bad, then packing Nulls does not likely help, a deficit of nulls in free space means empty space might appear. What then? More bubbles?
Thinking like a bubble in 3D is hard
I am just now starting to train my brain, but being a bi-symmetric being, it is hard. I have five fingers, but that is DNA and only gives me a fifth root decoder. Not enough CPU units in me.
But, so far. The bubbles have thing called r which adjust the unit sphere. Each r must be place one noise unit away from any other r, so it becomes an r+1 thing. In 3D, any other r will be one volume away, so I have, r^3 = 1+r. So am training myself to place little spheres just inside the unit sphere, tangent to its surface and tangent to each other.
The separation by one unit, by the way, is minimum redundancy, or optimum noise separation; entropy theory. What I mean is that this thing is already encoded, I need the decoder. So, I know the form of values will be (1+r)^(1/3), which gives me r. When the decoder is orthogonal to noise then its decoding graph is given by: log3((1+r)^(1/3)). That gives me a counting system that counts three things for each digit and has the quant log3(1+r).
What happened to pi?
Good question, I have not gotten that far! They are likely ellipsoids. Wait, won't Markov give me ellipsoids? Likely, and I might discover that with a little work. Are Markov sum of squares really the differential of ellipsoids? Good point, maybe.
What about this equation? Does not this give me what I want? The hyperbolic function tanh is a solution to this, likely a good spot to look. But I wonder, what are the ellipsoids that correspond to the Markov triples, do they follow some form like this?
Wiki has a section on hyperbolic differential equations and says:
The tanh gives the slope of the unit circle where the conic has touched. When that value changes, it point in the direction of minimum phase, toward some other point on the unit circle. There is only one quantized solution in that direction because we are at maximum entropy in a fixed bandwidth system. The motion will be a rotation about some center radial of symmetry.
But, so far. The bubbles have thing called r which adjust the unit sphere. Each r must be place one noise unit away from any other r, so it becomes an r+1 thing. In 3D, any other r will be one volume away, so I have, r^3 = 1+r. So am training myself to place little spheres just inside the unit sphere, tangent to its surface and tangent to each other.
The separation by one unit, by the way, is minimum redundancy, or optimum noise separation; entropy theory. What I mean is that this thing is already encoded, I need the decoder. So, I know the form of values will be (1+r)^(1/3), which gives me r. When the decoder is orthogonal to noise then its decoding graph is given by: log3((1+r)^(1/3)). That gives me a counting system that counts three things for each digit and has the quant log3(1+r).
What happened to pi?
Good question, I have not gotten that far! They are likely ellipsoids. Wait, won't Markov give me ellipsoids? Likely, and I might discover that with a little work. Are Markov sum of squares really the differential of ellipsoids? Good point, maybe.
Wiki has a section on hyperbolic differential equations and says:
The solutions of hyperbolic equations are "wave-like." If a disturbance is made in the initial data of a hyperbolic differential equation, then not every point of space feels the disturbance at once. Relative to a fixed time coordinate, disturbances have a finite propagation speed.These sound like fixed bandwidth solutions to me. They have a finite propagation bandwidth. I always eliminate the t thing.
The tanh gives the slope of the unit circle where the conic has touched. When that value changes, it point in the direction of minimum phase, toward some other point on the unit circle. There is only one quantized solution in that direction because we are at maximum entropy in a fixed bandwidth system. The motion will be a rotation about some center radial of symmetry.
Tuesday, June 17, 2014
So the bubbles need charge to make spheres?
Evidently. They cannot pack sphere until the three degrees of freedom. The two phase chains are different in size, they have a slight difference in exchange rate, not enough difference to slip a cycle, but enough to make sharp turns nearly impossible. So, in Lagrange mode one, they can make large bracelets, but not pack sphere. Hence, they alternate, and become a bipolar clock. This makes for the first spin mode but not much else. So the correct name for charge is a sphere packing shift in wave number to compute cubic roots.
What about Lagrange mode 2? Make cylinders? Probably. They are part of the equation. Relative to every third cubic root in space, the wave has two choices to place Nulls. This is likely part of the quarks strange and charm, and likely the cause of fermion spin. There are two types of spin and I have not sorted them out.
What about Lagrange mode 2? Make cylinders? Probably. They are part of the equation. Relative to every third cubic root in space, the wave has two choices to place Nulls. This is likely part of the quarks strange and charm, and likely the cause of fermion spin. There are two types of spin and I have not sorted them out.
OK, then, what about the fine structure constant?
It look like it is defined as the square root of free space impedance, which is the power spectra of the orbitals. Its square root is the first moment, the bandwidth in units of quants, of deviation, which seems to be about 19. They tell me the fine structure is about 137 = 1/(alpha)., or about .36 of the total bandwidth, in units of power, or 30% of the total width in quants (first moment). That indicates to me, using my simple statistics, that fine structure is simply tells us how much of the bandwidth (in units of quants) is outside the standard one deviation point. That is, assume the power spectra is Gaussian, one standard deviation is about 65% of that, leaving 35% for the sqrt(fine spectra). The Plank charge must be the standard deviation of the bandwidth, therefore.
So, how much of the bandwidth is really caused by charge? Probably about 65% of it, or one Plank's charge. The total available is determined by the bandwidth (in quants) of the gluon, 2*(Phi^n - Nulls^m), in the gluon wave. I mean, the gluon is the center of action, it sets center quant and quant span for orbital kinetic energy.
So take Mr. Plancks 11 bits of quant deviation, divide by 3, it has three degrees of freedom, and we get about 3.5 units of phase shift over the whole 19 units of wave. Of that, about 1 are likely due to spin, motion, and other quark modes, leaving us with about 2.5 units of wave shift for charge, a number that seems reasonable. It gives me the original 17 plus my 2.5, and square that I get power spectra at about 377.
The notable differences. This is not gaussian spectra it is spectra in a third degree polynomial, composed of one, two and three degrees of freedom. The whole system seems to be a 16 bit wave bandwidth (in quants), counting relative to the center frequency, with packing gain due to the third order polynomial. Motion, then, seems to be the fourth order polynomial produce of the surface anomaly in the unit one times the third order bandwidth spectra. It should be treated as noise in the channel. I do not think it will show up in the spectra of EM light.
So, how much of the bandwidth is really caused by charge? Probably about 65% of it, or one Plank's charge. The total available is determined by the bandwidth (in quants) of the gluon, 2*(Phi^n - Nulls^m), in the gluon wave. I mean, the gluon is the center of action, it sets center quant and quant span for orbital kinetic energy.
So take Mr. Plancks 11 bits of quant deviation, divide by 3, it has three degrees of freedom, and we get about 3.5 units of phase shift over the whole 19 units of wave. Of that, about 1 are likely due to spin, motion, and other quark modes, leaving us with about 2.5 units of wave shift for charge, a number that seems reasonable. It gives me the original 17 plus my 2.5, and square that I get power spectra at about 377.
The notable differences. This is not gaussian spectra it is spectra in a third degree polynomial, composed of one, two and three degrees of freedom. The whole system seems to be a 16 bit wave bandwidth (in quants), counting relative to the center frequency, with packing gain due to the third order polynomial. Motion, then, seems to be the fourth order polynomial produce of the surface anomaly in the unit one times the third order bandwidth spectra. It should be treated as noise in the channel. I do not think it will show up in the spectra of EM light.
Degrees of freedom in the quarks
They say the mass is relativistic, but what they mean is that the gluon is a tub of Nulls that 1 to 3 waves short of a packed Null, so no unit circle. It makes waves with a few bits of high frequency motion. The various masses, then, must be composed of separate three bits, each with one degrees of freedom. Right? The unit circle is the first Lagrange, no? Especially since I do not think we trade quark pairs until we break something and have to renormalize. So, for each normal configuration, all the bits are centered about the one mass bit.
So the gluons likely set the bandwidth of the combined orbitals to something like:
Phi^17 is the largest quant and they can be decomposed into units of 1/Phi^17. Guaranteed never to make a unit circle. They would be somewhat smaller than or equal to the smallest unit of null as 1/Phi^17.
In base 2, the proton has about 109 units of variation devoted to mass, and it varies little so they are likely the most significant of our 16 total bits. They are something near the top four bits at Lagrange 1 and define the four unit circles. The electron seems to be about 1/17 of the rest masses of the quarks.
So what do we make of the unit of charge in wave motion? It is one value that can appear in three locations, all rotations about a radial from the center of the unit circle. But the relative angle of that radial is determined by motion of the unit circle. The entire system is four gear systems, comprising 16 bits/ 4 (each, I think), designed to move the four unit circles, all centered within the bandwidth of the gluon center.
What causes motion?
Deformation of the unit spheres by the action of the fractional bits, the 1/2^16, (I think) within the unit spheres, the inverse of the hyperbolic wave motion outside. These fraction changes of phase variation inside the unit spheres are actually designed to restore curvature. They will have the same degrees of freedom.
I am not sure about much of this, like are there four yardsticks with one Unit one each, or one yardstick with four unit ones? We only have about 16-19 total, so consider the electron. One for mass at Lagrange 1, one for spin at Lagrange 2 and one for charge at Lagrange 3. Including degrees of freedom, (which are bits, really, it has 6 bits. Higher orders of Lagrange are more minimum redundant, you get more bit action per degree than the lower orders.
Other clues:
Treated as four independent rulers, we have to identical quarks, which therefore must be positioned mostly orthogonal to each other. Second, all three spheres have the same degress of freedom, and must have the same number of equivalents bits (entropy). And they are all equally subject to bandwidth restrictions, both in limits an separation of bits. So I am at least at 20 bits, but some of that is compaction because of Lagrange optimization. It is four, including all bases, then we fit into the 156 bit system and still fill the bandwidth.
What about motion?
That seems to be another amount of entropy, no doesn't it? I have not figured that one out.
Monday, June 16, 2014
Graphs and entropy
The idea that a signal of the form x^(1/b), can always be represented in a base b system of finite size digits should be easy to prove. b is integer, make it simple. make logb(x) finite in base b. So there exists a finite, balanced decoding graph with b branches at each node. From there it is simple to show a base b counter can represent all paths through the tree, up to the precision of my finite log. Shannon did no different, except he started with noise.
Computing the quantization error is simple. If the number of base b digits is n, then add up the inverses of the finite set 1/b^n, to get the finite log base b, then compare that to the Taylor series of the real log. The last term is the error term.
Also, there will exist an encoding that is minimum redundancy. I know there is a general information theory about this in Wiki. If b is not integer, then make it so.
Why do I care? Because we are going to deal with cubic roots. So instead of 0,1,2 as my coefficients, I can use r1,r2,r3; I think, which would be degrees of freedom, say for the l numbers in atomic orbitals. I can rotate though the roots as I count the digits. The L numbers in the orbitals seem to be base three. I think the 'one' spots on the unit oval will rotate about by cubic roots.
What about digits for cyclic graphs? Dunno, I just though about it. Digits on the interior of a sphere? They would not be ordered, necessarily. But more likely three independent wave actions since quarks are three.
The quants on the orbitals: Principal, angular, and magnetic; all seem to be the Lagrange degrees of freedom, in base 1,2 and 3; So they are interleaved in any digits system. The principal quant does not have bu one value per digit, it is the first Lagrange. But it does a wave shift with energy, shortening and leaving the charge to be counted by l in 2 degrees. Then they all shorten for the magnetic, it is almost a shift of the decimal point. More degrees of freedom added out from the electron as energy increases. But those quantum numbers are not completely orthogonal, I can tell. They also do not include the quarks.
Also, I notice, when the principal quant is at its lowest energy level, my picture of the hyperbolasays the radial charge is a straight line, and the electron mass is flat. No, small Lagrange, and small q in the denominator of the minimum error. So error cone from the electron is wide, the packed nulls nearly spherical. But for principal quant 2, there are two spheres, where did the extra degree of freedom come from? Spin? No. The first few bits must be powers of the first Lagrange. Still a bit confused, I am. But it looks like they assign Markov numbers straight from the list, starting with the most least significant bit. I guess the spin took all of the first Lagrange.
Computing the quantization error is simple. If the number of base b digits is n, then add up the inverses of the finite set 1/b^n, to get the finite log base b, then compare that to the Taylor series of the real log. The last term is the error term.
Also, there will exist an encoding that is minimum redundancy. I know there is a general information theory about this in Wiki. If b is not integer, then make it so.
Why do I care? Because we are going to deal with cubic roots. So instead of 0,1,2 as my coefficients, I can use r1,r2,r3; I think, which would be degrees of freedom, say for the l numbers in atomic orbitals. I can rotate though the roots as I count the digits. The L numbers in the orbitals seem to be base three. I think the 'one' spots on the unit oval will rotate about by cubic roots.
What about digits for cyclic graphs? Dunno, I just though about it. Digits on the interior of a sphere? They would not be ordered, necessarily. But more likely three independent wave actions since quarks are three.
The quants on the orbitals: Principal, angular, and magnetic; all seem to be the Lagrange degrees of freedom, in base 1,2 and 3; So they are interleaved in any digits system. The principal quant does not have bu one value per digit, it is the first Lagrange. But it does a wave shift with energy, shortening and leaving the charge to be counted by l in 2 degrees. Then they all shorten for the magnetic, it is almost a shift of the decimal point. More degrees of freedom added out from the electron as energy increases. But those quantum numbers are not completely orthogonal, I can tell. They also do not include the quarks.
Also, I notice, when the principal quant is at its lowest energy level, my picture of the hyperbola
Light and its accuracy
So why is the relative sample rate of light not constant?
The accuracy of the null bubbles, as powers of 3/2, matching powers of Phi stop, somewhere around Phi^107. Mainly because, I think, the ratio that matches Phi better comes from:
Phi^n = a * Phi + b
At some point, that ratio gets better faster than the error term from (3/2)^m can cycle. But does that mean the light is not constant? Light rate is based on the exchange rate, and those are Null bubbles, so I would not expect light to be more accurate.
There is a point where the rational approximation of Phi, at Phi^17, does indeed drop to the approximation error of Nulls at (3/2)^127. So there are packings too small and packings too large. Three points seem to match. Phi, estimating itself, needs a quant of (3/2)^17, which more or less matches the electron matching Phi with (3/2)^89, which matches the error at Phi^107, and in between is the nearly perfect match at the proton peak. So, (3/2)^17 seems to be the smallest thing Higgs can handle. So, something like 1/(3/2)^17 looks like the smallest fraction. The number I throw out are seriously rounded, I have not worked the spread sheet in this. But (2^17-1) makes a 16 bit system at the peak of the proton, so I am merely guessing that that is the number. The smallest thing is packed nulls, that does not include wave kinetic energy.
Bubble size?
There must be a difference, but the two wave bubbles would not change size to allow better packing a low quants. Trying to make light rate variable would remove all the stability of packing. I would guess that the phase error in exchanges corresponds to the minimum packing ratio, simply on minimum redundancy grounds of how the bubbles are set in the quasar.
The accuracy of the null bubbles, as powers of 3/2, matching powers of Phi stop, somewhere around Phi^107. Mainly because, I think, the ratio that matches Phi better comes from:
Phi^n = a * Phi + b
At some point, that ratio gets better faster than the error term from (3/2)^m can cycle. But does that mean the light is not constant? Light rate is based on the exchange rate, and those are Null bubbles, so I would not expect light to be more accurate.
There is a point where the rational approximation of Phi, at Phi^17, does indeed drop to the approximation error of Nulls at (3/2)^127. So there are packings too small and packings too large. Three points seem to match. Phi, estimating itself, needs a quant of (3/2)^17, which more or less matches the electron matching Phi with (3/2)^89, which matches the error at Phi^107, and in between is the nearly perfect match at the proton peak. So, (3/2)^17 seems to be the smallest thing Higgs can handle. So, something like 1/(3/2)^17 looks like the smallest fraction. The number I throw out are seriously rounded, I have not worked the spread sheet in this. But (2^17-1) makes a 16 bit system at the peak of the proton, so I am merely guessing that that is the number. The smallest thing is packed nulls, that does not include wave kinetic energy.
Bubble size?
There must be a difference, but the two wave bubbles would not change size to allow better packing a low quants. Trying to make light rate variable would remove all the stability of packing. I would guess that the phase error in exchanges corresponds to the minimum packing ratio, simply on minimum redundancy grounds of how the bubbles are set in the quasar.
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