Essay Abstract

The information or knowledge we can get from a quantum state, depends on the interaction of this state with the measurement process itself. As we know, this involves an uncertainty in the amount of information and accuracy of this knowledge. Now we ask: is there an information content intrinsic and independent of the observer, in quantum reality? Our answer is a resounding, yes. More precisely, we show that this information is encoded on surfaces, and specifically in circular compactifications. That is, there is a holograph on surfaces. In this encoding of quantum information will call as the strong holographic principle. Due to the application of this principle, it will show as the Higgs vacuum value implies a subtle correction by entropic uncertainty. The connections of quantum information, to cosmology, are clearly shown when the values of the dark energy density approxeq = ln2;and matter density Omega_{m}= Omega_{c} Omega_{b}, approxeq = 1-ln2. It showed that the equation of the energy-momentum has five solutions by factoring by two components which appear in terms of mass and imaginary momentum. Far from being a mere mathematical artifice, we see that these states should exist, and that our interpretation of them is that every particle appears to be a mixture of two states, one of them unobservable, having an imaginary component. In other words: these involve imaginary states faster than the speed of light, without contradicting Special Relativity, as we shall see. Failure to observe Cherenkov radiation forces us to determine which are virtual states. These five states also appear to be related to the minimum number of microstates which generate the group E8. Finally: all these results lead us to postulate a particle candidate to the dark matter, of approximately 9,2 Gev.

Author Bio

Independent researcher. Some articles published in vixra

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Angel

"The information or knowledge we can get from a quantum state, depends on

the interaction of this state with the measurement process itself"

No it does not. Whatever the state was it has already existed. That is, not only the physical state as such, but the physically existent representation thereof which you receive (eg light). The sequence order, apart from what measuring physically involves, render this concept impossible.

All any measurement does is decree a point at which the sequence is stopped, conceptually. Physically, all that happens is that receipt of the physical input, ie the interaction involved, causes the physically existent representation to cease existing in that form.

Paul

    For those who read the essay, two small errors.Page 1

    [math][(1-i)/\sqrt{2}]\cdot[(1+i)/\sqrt{2}]c^{2}

    [/math]

    Page 2

    [math]x_{7}=1e_{1}+2e_{2}+3e_{3}+5e_{4}+8e_{5}+11e_{6}+4e_{7}\:\:,\:||x_{7}||=240

    [/math]

    Thanks

    Dear Angel,

    Congratulations with your Higgs vacuum lattice attack.

    You wrote:

    6.Since the Higgs boson itself, acquires mass from Higgs vacuum, by subtracting energy,

    Could it also be the other way around?

    That the Higgs boson itself is massless and the Higgs vacuum Has oscillating Higgs energy able to transfer mass to all Fermioins?

    Best regards,

    Leo Vuyk

      • [deleted]

      • Post #6

      Dear Leo Vuyk, to your first question: it is possible

      As for your second question: are you absolutely right.

      I'm still studying, as it breaks the symmetry of the Higgs vacuum, for, as you well said, oscillate and give mass to the fermions.

      I've made ​​some progress, which in summary, is the expose, for if your interest.

      What is clear, that is for sure is that the sum of the spins, as I have argued in the essay, it should be considered only the spin 1/2, of the fermions. Since, at the moment focusing on the six leptons, the sum of the spins:

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s=(9/2)}

      [/math]

      Now, as you well said, the Higgs vacuum oscillation; corresponds, taking into account the electric charges, all (note the equivalences, very important), and the contribution due to Higgs boson:

      [math]

      10\cdot\bigl\{[\ln(mv(H)/m_{e})-{\displaystyle \sum_{s}(s+1)s}]/7\bigr\}=O_{1}

      [/math]

      [math]10\equiv|3(\frac{4}{3})|+|3(\frac{2}{3})|+|3(\frac{-1}{3})|+|3(\frac{1}{3})|+|1|+|-1|\equiv2{\displaystyle \sum_{s}s\equiv10d}

      [/math]

      Clearly, three, by the three colors QCD

      [math]7=3(\frac{4}{3})+3(\frac{2}{3})+3(\frac{-1}{3})+3(\frac{1}{3})+1\equiv7d[/math]

      1)

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s+O_{1}=(9/2)+O_{1}\simeq\ln(m_{\mu}/m_{e})}

      [/math]

      2)

      [math]2\cdot6\cdot{\displaystyle \sum_{s=1/2}(s+1)s-O_{1}=2\cdot(9/2)-O_{1}\simeq\ln(m_{\tau}/m_{e})}

      [/math]

      The previous two equations must be met, these two oscillations Higgs vacuum, to preserve the symmetry of the sum of the spins, compared to the sum of the spines (module), the quarkcounting three color charges.

      3)

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s+O_{1}+2\cdot6\cdot{\displaystyle \sum_{s=1/2}(s+1)s-O_{1}=3_{C}\cdot6}}\sum_{s=1/2}(s+1)s

      Dear Leo Vuyk, to your first question: it is possible

      As for your second question: are you absolutely right.

      I'm still studying, as it breaks the symmetry of the Higgs vacuum, for, as you well said, oscillate and give mass to the fermions.

      I've made ​​some progress, which in summary, is the expose, for if your interest.

      What is clear, that is for sure is that the sum of the spins, as I have argued in the essay, it should be considered only the spin 1/2, of the fermions. Since, at the moment focusing on the six leptons, the sum of the spins:

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s=(9/2)}

      [/math]

      Now, as you well said, the Higgs vacuum oscillation; corresponds, taking into account the electric charges, all (note the equivalences, very important), and the contribution due to Higgs boson:

      [math]

      10\cdot\bigl\{[\ln(mv(H)/m_{e})-{\displaystyle \sum_{s}(s+1)s}]/7\bigr\}=O_{1}

      [/math]

      [math]10\equiv|3(\frac{4}{3})|+|3(\frac{2}{3})|+|3(\frac{-1}{3})|+|3(\frac{1}{3})|+|1|+|-1|\equiv2{\displaystyle \sum_{s}s\equiv10d}

      [/math]

      Clearly, three, by the three colors QCD

      [math]7=3(\frac{4}{3})+3(\frac{2}{3})+3(\frac{-1}{3})+3(\frac{1}{3})+1\equiv7d[/math]

      1)

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s+O_{1}=(9/2)+O_{1}\simeq\ln(m_{\mu}/m_{e})}

      [/math]

      2)

      [math]2\cdot6\cdot{\displaystyle \sum_{s=1/2}(s+1)s-O_{1}=2\cdot(9/2)-O_{1}\simeq\ln(m_{\tau}/m_{e})}

      [/math]

      The previous two equations must be met, these two oscillations Higgs vacuum, to preserve the symmetry of the sum of the spins, compared to the sum of the spines (module), the quarkcounting three color charges.

      3)

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s+O_{1}+2\cdot6\cdot{\displaystyle \sum_{s=1/2}(s+1)s-O_{1}=3_{C}\cdot6}}\sum_{s=1/2}(s+1)s

      Dear Leo Vuyk, to your first question: it is possible

      As for your second question: are you absolutely right.

      I'm still studying, as it breaks the symmetry of the Higgs vacuum, for, as you well said, oscillate and give mass to the fermions.

      I've made ​​some progress, which in summary, is the expose, for if your interest.

      What is clear, that is for sure is that the sum of the spins, as I have argued in the essay, it should be considered only the spin 1/2, of the fermions. Since, at the moment focusing on the six leptons, the sum of the spins:

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s=(9/2)}

      [/math]

      Now, as you well said, the Higgs vacuum oscillation; corresponds, taking into account the electric charges, all (note the equivalences, very important), and the contribution due to Higgs boson:

      [math]

      10\cdot\bigl\{[\ln(mv(H)/m_{e})-{\displaystyle \sum_{s}(s+1)s}]/7\bigr\}=O_{1}

      [/math]

      [math]10\equiv|3(\frac{4}{3})|+|3(\frac{2}{3})|+|3(\frac{-1}{3})|+|3(\frac{1}{3})|+|1|+|-1|\equiv2{\displaystyle \sum_{s}s\equiv10d}

      [/math]

      Clearly, three, by the three colors QCD

      [math]7=3(\frac{4}{3})+3(\frac{2}{3})+3(\frac{-1}{3})+3(\frac{1}{3})+1\equiv7d[/math]

      1)

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s+O_{1}=(9/2)+O_{1}\simeq\ln(m_{\mu}/m_{e})}

      [/math]

      2)

      [math]2\cdot6\cdot{\displaystyle \sum_{s=1/2}(s+1)s-O_{1}=2\cdot(9/2)-O_{1}\simeq\ln(m_{\tau}/m_{e})}

      [/math]

      The previous two equations must be met, these two oscillations Higgs vacuum, to preserve the symmetry of the sum of the spins, compared to the sum of the spines (module), the quarkcounting three color charges.

      3)

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s+O_{1}+2\cdot6\cdot{\displaystyle \sum_{s=1/2}(s+1)s-O_{1}=3_{C}\cdot6}}\sum_{s=1/2}(s+1)s

      Sorry, no passing, to insert the remaining equations, sorry for the repeats.

      I get to see so

      [math]6\cdot{\displaystyle \sum_{s=1/2}(s+1)s+O_{1}+2\cdot6\cdot{\displaystyle \sum_{s=1/2}(s+1)s-O_{1}=3_{C}\cdot6}}\sum_{s=1/2}(s+1)s

      Angel,

      Please include me out of the "we" that pretend to obtain "the information obtained from a quantum state that depends on the interaction of this state with the measurement process itself."

      As a realist, I know I am a completed unique being. I know there are such things as completed unique elephants. I know there are completed unique stars and completed unique planets. I know that only one completed unique object, or event, real or imagined can only occur once in reality.

      To believe that just because scientists can build machines that produce particles and energy waves, that that exclusive unnatural, unrealistic type of behavior is superior to natural reality is incorrect. Each discrete quantum package is just as unique as each unique snowflake is and each quantum package can only occur once. You can repeatedly write down your seemingly identical mathematical symbolic squiggles as much as you like, the more they perfectly abstractly mean to you, the less they represent sane reality.

        Mr. Doz,

        I left this link on the viXra blog but thought you might like to check it out (http://cdms.berkeley.edu/CDMSII_Si_DM_Results.pdf). It's a paper detailing the latest results from the Cryogenic Dark Matter Search collaboration and relates to your particle candidate for the dark matter . . .

        I haven't read your paper yet so can't comment on it . . . although even when I do get around to it there's a good chance I probably won't thoroughly understand it anyway . . .

        With regards,

        Wes Hansen

          Sorry sir, eliminate we both seem to have bothered you.

          What you say you are squiggles mathematical, that according to you, do not represent reality, are precisely those that allow you and I We'll use two computers, and we exchange conversation, via a network. This is because these squiggles mathematical despises it seems

          Please check their knowledge of physics, because bosons, according to Bose-Einstein statistics, are indistinguishable, why can occupy the same quantum state.

          And now, based on these mathematical squiggles, that as you do not represent reality, I propose a question:

          Is the Riemann hypothesis, it is possible that this relates his demonstration, with electrical charge?

          Here's a hint with these mathematical squiggles:

          First Imaginary part of the first zero of the Riemann hypothesis:

          s= 1/2 + 14.134725141734693i; Z1= 14.134725141734693

          http://oeis.org/A058303

          [math]\exp Z_{1}-(m_{h}/m_{e})(\ln^{-1}(\varphi)-1)=m_{pk}/\sqrt{\pm e^{2}/G_{N}}

          [/math]

          Gn = Newton Constant (S.I ) = 6.67428 *10^-11 , Mpk = Planck mass

          e= electric charge ( S.I ) = 1.602176565 *10^-19 ; mh= Higgs boson mass

          me = electron mass; (mh/me)=246924 ; Phi= golden number = (sqr(5)+1)/2

          2)

          [math]

          \ln(m_{pk}/\sqrt{\pm e^{2}/G_{N}})+\ln^{-1}({\displaystyle \sum_{F_{n}/240}F_{n}^{2}}+(1+\Omega_{c})^{2})=14.1347251004\simeq Z_{1}[/math]

          Omega(c) = dark matter density, 240= non zero roots group E8

          Regards

          Thank you very much, dear Hansen, I will watch the results indicated.

          Do not worry, I do not quite understand many things

          Regards

          Angel,

          All of your so-called knowledge of physics is abstract guesswork. Sure, your perfect abstract invisible bosons according to Bose-Einstein perfect abstract statistics are indistinguishable and can occupy the same perfect abstract quantum state, but each real snowflake, each concoction of real DNA, and each real fingerprint is unique and only occurs once and can only occupy the real place it is located in once. There are no statistics needed for reality.

          As I thoughtfully pointed out in my essay BITTERS, Einstein's guess that perfect abstract energy was perfectly equal to perfect abstract mass times perfect constant speed of perfect abstract light perfectly squared was utter real nonsense. Real energy consists of integrated differing unique amounts of real light, real heat, and real material motion.

          Excuse me, sir, but my great ignorance, I can tell you that commits serious mistakes resulting from their ignorance of quantum mechanics: First, "your perfect abstract unseen bosons"

          Not my bosons, are bosons that exist and are observable: photons of light, for example, with which you can see, through the interaction of the cells of the retina of the eye, with these bosons that you call abstract.

          Second error: existing and observable reality of bodies of matter, composed of a large number of quantum particles, it becomes a non-interlaced macrostate, which depends on a critical value, as a factor of the Planck mass.

          Third mistake: what you call real is the result of the actual existence of quantum microstates, which interact. Any physicist knows this first-year university.

          I see you have not answered my question, on the possible demonstration of the Riemann hypothesis, by quantum physics, and specifically in relation to the electric charge.

          I'll give the last track, this way, to prove the Riemann hypothesis, by quantum mechanics.

          For the rest: greetings

          [math]\pm e/\sqrt{m_{0}^{2}\cdot G_{N}\cdot n}=1

          [/math]

          [math]{\displaystyle \sum_{n=1}^{\infty}}\pm e/\sqrt{m_{0}^{2}\cdot G_{N}\cdot n}={\displaystyle \sum_{n=1}^{\infty}}\frac{1}{n^{1/2}}

          [/math]

          Renormalization: The vacuum is neutral, the electric charge.

          And this can only happen, if all non-trivial zeros, the Riemann zeta function; are values ​​of (1/2+it). t = real number

          Fulfilled, moreover, that:

          [math]{\displaystyle \sum_{n=1}^{\infty}}\pm e/\sqrt{m_{0}^{2}\cdot G_{N}\cdot n}={\displaystyle \sum_{n=1}^{\infty}}\frac{1}{n^{(1/2+it)}}=0

          [/math]

          Be, all the imaginary parts of (1/2 + it), that satisfies:

          [math]

          \sum_{n=1}^{\infty}\frac{1}{n^{s}}=0\:;\: s=\frac{1}{2}+it[/math]

          [math]\frac{m(VH)}{m_{e}}(1+{\displaystyle \sum_{q}}\sqrt{q^{2}})=O{\displaystyle [\sum_{n=1}^{\infty}\exp(-Im(s))]^{-1}}[/math]

          Where, m(VH) is the value, in mass, the Higgs vacuum, and ,me, is the mass of the electron

          And, electric chargues:

          [math]{\displaystyle \sum_{q}}\sqrt{q^{2}}=\sqrt{(4/3)^{2}+(2/3)^{2}+(-1/3)^{2}+(1/3)^{2}+1^{2}}=\sqrt{31/9}[/math]

          Regards

          Angel,

          Nobody has ever seen a boson because bosons are abstractions and invisible. Everybody with eyes has the ability to see light. How many photons are there in the light emitted from a glow worm's backside? I mean are there a specific number of perfect photons per square millimeter in every light no matter what the source of the light is, or its unique intensity? How are perfect photons arranged? Are they lined up, side by side, an equal distance apart? Or are they stacked neatly one above the other? What shape is a typical photon? Is it a perfect circle Angel? Or is a photon a perfect sphere? It cannot be a perfect cube right?

          One last question Angel, Einstein was reputed to be the greatest squiggler of all time. His guesswork squiggles supposedly explained how the abstract Universe behaved. How would you rate your squiggles? Are they superior to Einstein's? Or are they about the same? I do hope you are not going to honestly admit that your squiggles are inferior to the ones produced by Einstein.

          It is not I that needs any useless lessons in unrealistic, unnatural abstract physics. No lessons are needed for my understanding of reality. No experiments are needed. And mercifully, no squiggles are ever required to be produced by me.

          • [deleted]

          • Post #19

          You contradicts himself: "Nobody has ever seen a boson because bosons are abstractions and invisible. Everybody with eyes has the ability to SEE light. How many photons are there in the light emitted from a glow worm's backside?

          Ergo, you have seen the bosons.

          Look, sir: As one of the largest Einstein scientists of all time, to try to compare me with Einstein myself, would be an unforgivable act of vanity. I am who I am, and point. And finally this exchange of ideas, I leave my last scribble for you.

          [math]\frac{\sqrt{(m_{pk}\cdot\exp-[\ln(m_{pk}/m_{e})/{\displaystyle 2\sqrt{{\displaystyle \sum_{q}q^{2}}}])^{2}\cdot G_{N}}}}{(1\frac{1}{\pi\sqrt{240-\ln^{2}[m(VH)/m_{e}]}})}=\pm e=1.602176565\cdot10^{-19}C[/math]

          Regards

          • [deleted]

          • Post #20

          How can the scientific method's testability rule be satisfied? ... if a particle has 2 states, but one is unobservable, how do we know it has real existence?