Advanced concepts in black hole cosmology

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by
U.V.S.Seshavatharam
Honorary faculty, I-SERVE, Alakapuri, Hyderabad-35, AP, India
QA-Spun division, LANCO Industries Ltd, Srikalahasti-517641, AP, India
E-mail: seshavatharam.uvs@gmail.com
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Abstract
Based on the big bang concepts- in the expanding universe, ‘rate of decrease in CMBR temperature’ is a measure of  the cosmic ‘rate of expansion’. Modern standard cosmology is based on two contradictory statements. They are – present CMBR temperature is isotropic and the present universe is accelerating. In particle physics also, till today laboratory evidence for the existence of ‘dark matter’ and ‘dark energy’ is very poor. Recent observations and thoughts supports the existence of the ‘cosmic axis of evil’. In this connection an attempt is made to study the universe with a closed and growing model of cosmology. If the primordial universe is a natural setting for the creation of black holes and other non-perturbative gravitational entities, it is also possible to assume that throughout its journey, the whole universe is a primordial (growing and rotating) cosmic black hole. Instead of the Planck scale, initial conditions can be represented with the Coulomb or Stoney scale. Obtained value of the present Hubble constant is close to 71 Km/sec/Mpc.
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333 comments to Advanced concepts in black hole cosmology

  • Wladimir Guglinski

    JR wrote in July 24th, 2013 at 8:03 AM
    ——————————————————
    I am not, and never was, ‘inverting causality’ to make any argument. I answered a fairly vague question in a way that did not provide what you really wanted, and when you pointed this out, I provided the answer you asked for.
    ——————————————————-

    COMMENT
    Dear JR, you not only had inverted the causality, but you are also a lier.

    Look what you had written in July 18th, 2013 at 9:50 AM, so that to explain why 8Be is no stable:

    ——————————————————-
    Why isn’t 8Be bound?

    In a very simple picture, a nucleus is bound if it’s mass is below that of all combinations of constituents (and it’s binding energy relative to those states is this mass difference – mass of constituents minus mass of bound state). 4He is bound because it’s mass is less than that of 2 proton plus 2 neutrons AND below that of 2 deuterons. If it were not bound with respect to 2p+2n or 2 deuterons, it would just fall apart into those constituents.

    8Be is not bound with respect to two 4He nuclei, so it has a ‘negative binding energy’ if you want to think of it that way, and it just falls apart
    ——————————————————-

    Dear JR,
    such argument of yours:
    ——————————————————-
    a nucleus is bound if it’s mass is below that of all combinations of constituents
    ——————————————————-
    is just the definition mass defect

    Look at the definition in the link bellow:
    ——————————————————-
    mass defect :
    The amount by which the mass of an atomic nucleus is less than the sum of the masses of its constituent particles.
    http://www.thefreedictionary.com/mass+defect
    ——————————————————-

    Therefore, dear JR,
    you had used the mass defect so that to explain why 8Be is no stable.

    However, the CAUSE of 8Be to be no stable is NOT the mass defect as you had claimed.

    The CAUSE of the 8Be to be no stable is because the nucleons in 8Be are not bound

    The mass defect of 8Be is only a CONSEQUENCE of instability of 8Be.

    In order to explain why 8Be is no stable, we have to start from the CAUSE: we have to explain why the nucleons are not bound.

    So, you had used indeed the causality inversion.

    Therefore, you have not only a dishonest way of argumentation.
    You are also a lier.

    regards
    wlad

  • JR

    Dear Wladimir,

    I am well aware of how the binding energy of the deuteron is measured. I’m also saying that there are many calculations of the binding energy which start from the n-p interaction and are able to successfully reproduce the deuteron binding energy. That’s why I say that there’s no mystery as to how the deuteron is bound.

    I am not, and never was, ‘inverting causality’ to make any argument. I answered a fairly vague question in a way that did not provide what you really wanted, and when you pointed this out, I provided the answer you asked for.

  • Wladimir Guglinski

    JR wrote in July 23rd, 2013 at 7:59 AM

    ——————————————————–
    So what’s the point of this other paper which apparently uses some completely unconventional theory to get the same answer. I’m not saying that it’s impossible to get ~2 MeV of binding for the deuteron in other theories,
    I’m just saying that it’s perfectly well understood in conventional theory so there’s no reason to invent new theories to explain the deuteron binding.

    ——————————————————-

    Dear JR,
    you say that 2MeV of deuteron binding energy is perfectly well understood in conventional theory because you actually have not understanding of the fundamental questions in Physics.

    You had written in July 22nd, 2013 at 3:12 PM :
    ——————————————————–
    By the way, if the deuteron “cannot have a binding energy weaker than 8.56 MeV, according to nuclear models”, then why does basically every calculation performed over the last 20 years yield a smaller binding energy? With the exception of your calculation, of course.
    ——————————————————–

    So, you cannnot understant that those “basically every calculation performed over the last 20 years” actually ARE NOT THEORETICAL.

    The value of 2MeV that you claim to be theoretical actually is obtained by EXPERIMENTS

    Look the masses :
    ———————————————-
    Proton: Mp = 1.007276466812 u (by EXPERIMENTS)

    Neutron: Mn= 1,00866491600 u (by EXPERIMENTS)

    Deuteron: Md = 2.01355 u (by EXPERIMENTS)

    The mass defect due to the packing loss is:

    D = 0.00240 u (therefore it is an EXPERIMENTAL value)

    Using Einsteins`s equation E = mc^2 we calculate the equivalent energy for that mass defect:

    mass = 0.00240 * 1.66053886 * 10^-27
    mass = 3.98529326 * 10^-30 Kg

    E = mc^2
    E = 3.98529326 * 10^-30 * 299792458^2
    E = 3.58180296 * 10^-13 J

    E = 3.58180296 * 10^-13 * 6.24150974 * 10^18 = 2,2MeV

    Therefore, the binding energy calculated in conventional theory which you suppose to be theoretical is actually obtained via EXPERIMENTAL way.

    For the calculation of a THEORETICAL value of the deuteron the theorists have take as a point of departure the following data:
    * the mass of proton
    * the mass of neutron
    * the strong force in a distance of about 2fm

    From this point of departure they have to calculate the mass defect from the theoretical way (and not getting the mass defect via experiments, as it is made in the conventional theory).

    —————-
    CONCLUSION:
    —————-
    You are not able to understand the following two points, dear JR:

    1-
    ——————————————————-
    The mass defect is NOT the cause of the binding energy…
    ——————————————————-
    … as you had wrongly supposed (that`s why I said that you use the inversion of causality when you tried to explain why 4Be8 is no stable).

    2-
    Unlike…
    ——————————————————
    …the binding enery is actually the cause of the mass defect
    ——————————————————
    (and so you cannot use the mass defect detected in the experiments so that to explain why 4Be8 is no stable).

    You proved, dear JR,
    that you cannot even to understand easy fundamental questions in Physics.
    ——————————————————–
    So, you are not able to understand why the current models are wrong, because you are not able to detect when the authors of a paper use an arbitrary unacceptable assumption as a point of departure in their calculations.
    ——————————————————–

    regards
    wlad

  • JR

    Dear Wladimir,

    What on earth are you trying to say in your latest post? Nothing you said has anything to do with the question we were discussing. The question was whether or not traditional nuclear physics can explain the deuteron binding. The answer is that it can and does. It is in no way any kind of a challenge for conventional theory.

    So what’s the point of this other paper which apparently uses some completely unconventional theory to get the same answer. I’m not saying that it’s impossible to get ~2 MeV of binding for the deuteron in other theories, I’m just saying that it’s perfectly well understood in conventional theory so there’s no reason to invent new theories to explain the deuteron binding.

  • Wladimir Guglinski

    JR wrote in July 22nd, 2013 at 3:12 PM

    ——————————————————
    By the way, if the deuteron “cannot have a binding energy weaker than 8.56 MeV, according to nuclear models”, then why does basically every calculation performed over the last 20 years yield a smaller binding energy?
    ——————————————————-

    COMMENT

    Let us see a theoretical calculation made in the paper linked bellow:
    http://prof.usb.ve/ggonzalm/invstg/pblc/deuteron1.pdf

    In the Conclusion the authors say:
    —————————————————–
    From a theoretical point of view it is interesting to know that this “strong” SU(2)Q electromagnetism, without the help of any other force, generates sufficient attractive short range potentials to provide the binding energy of light nuclides, composed of protons and electrons.
    ——————————————————

    ha, ha , ha ,ha

    The nuclear theorists claim that proton-neutron interaction is via strong force.

    But in the paper above the author simply do not consider the strong force between the proton and neutron in the structure of the deuteron.

    Instead of, they consider a kind of “strong” electromagnetism.

    Then of course that they did get a good result in their calculation, because they did not consider the strong nuclear force in the calculation. But they had to consider !!!!!!! (according to nuclar models).

    ———-
    CONCLUSION
    ———-
    So, while the nuclear theorists claim that protons and neutrons are bound via strong force within the nuclei, however for the calculation of the binding energy of deuteron they do not consider the strong force.
    ————————————————–
    They actually consider a new kind of force: the “strong” electromagnetic force
    ————————————————–

    Dear JR
    those authors have to share the Nobel Prize of Physis in 2014 with you !!!!
    —————————————————–
    While you had discovered the strong nuclear force of REPULSION as the 5th fundamental force of nature…
    ,,, they had discovered a new 6th fundamental force of nature: the strong electromagnetism
    ——————————————————

    ha , ha , ha

    My God,
    it seems a joke !!!!

    regards
    wlad

  • Andrea Rossi

    Frank Acland:
    It is among our priorities in this period.
    Warm Regards,
    A.R.

  • Frank Acland

    Dear Andrea,

    You mention that the IP for the E-Cat may be made useless by strong industrialization of E-Cat products. The only way for this to be the case would be for your partner to produce E-Cats on a massive scale very cheaply.

    Is this a priority for you and your partner?

    Many thanks,

    Frank Acland

  • JR

    Dear Wladimir,

    I’m not claiming that conventional theory is perfect and can explain absolutely anything, although I think it’s done a very good job so far. Since I’m not claiming anything, there’s nothing for me to prove.

    You’re claiming that there’s a long list of problems that can’t be explained, but every example you’ve explained in sufficient detail to understand is wrong, and I’ve pointed you to works that demonstrate that. All that’s left are the ones where you just assert that measurement X can’t be explained by nuclear theory, so since you’re trying to prove something, it’s up to you to give evidence.

    If your only claim is that YOU don’t understand how nuclear models can explain the measurements of pear-shaped nuclei in 220Ra and 224Rn, and therefore YOU believe that something is wrong, then I certainly agree with that claim. But if you’re trying to convince anyone else that there are problems, then you should probably try to provide actual explanations or examples, rather than just claiming that it’s obvious that conventional physics can’t explain this. If it were obvious, then you wouldn’t need to keep going on about it, so if you think you have something worth saying, try explaining what you mean.

  • Wladimir Guglinski

    JR wrote in July 22nd, 2013 at 9:31 AM

    —————————————————
    Dear Wladimir,
    You assert many things about what other people think: that conventional theory requires all even-even nuclei to be spherical, that the fact that the strong force is bound means that everything has to be bound, etc….

    I believe that many of these statements are incorrect and given your track record with unsupported assertions, I don’t count you as a credible source without some kind of additional support. So if you believe that these statements are correct, please provide some kind of evidence that supports that idea that these are currently held beliefs based on conventional theory.
    —————————————————–

    COMMENT
    No, dear JR

    actually YOU and the nuclear theorists have to show that the current nuclear models are able to have PHYSICAL causes so that to contribute for the shape of nuclei detected by experiments.

    A big nucleus as 92U238 has 238 nucleons, and so that it behaves as a ball filled with 238 particles.
    The shape of the distribution of the nucleons must have STATISTICAL distribution, and that`s why it has to take a SPHERICAL shape(according to nuclear models) no matter if it excited or not.

    What we expect from a ball filled with particles following STATISTICAL laws is to keep a spherical shape.

    You and the theorists have to show what is the PHYSICAL cause existing that makes the excited 238U to become ellipsoidal, since it is IMPOSSIBLE to have such PHYSICAL cause according to the nuclear models.

    Concerning the pear-shaped nuclei Ra and Rn, the theorists recognize that there is need a PHYSICAL cause capable to give them the pear shape.

    They are speaking about an internal axis within the nuclei:
    ——————————————————-
    “The pear shape is special,” said University of Michigan physicist Tim Chupp, who participated in the study, in a press release. “It means the neutrons and protons, which compose the nucleus, are in slightly different places along an internal axis.”
    ——————————————————-
    http://www.csmonitor.com/Science/2013/0509/Why-hasn-t-everything-been-annihilated-yet-Pear-shaped-atomic-nuclei-could-hold-answer

    From current nuclear models is IMPOSSIBLE to find a PHYSICAL cause responsible for the existence of such an internal axis.

    Therefore, something is missing in the current nuclaar models, and therefore the nuclear models are WRONG.

    No matter what is the PHYSICAL cause responsible for the existence of the internal axis within the nuclei, the consequence is obvious: it`s impossible to find such physical cause by considering the current nuclear models, and therefore THE NUCLEAR MODELS ARE WRONG

    There is need to change the nuclear models, so that to explain how they can have an internal axis responsible for the pear shape of 224Ra and 220Rn.

    As there is need to change the nuclear models, it means that they are wrong, since it is missing something in them.

    regards
    wlad

  • JR

    Dear Wladimir,

    You appear to be confused by the nomenclature that physicists use. Just because it’s called the “tensor interaction” doesn’t mean that it’s referring to a mathematical tensor – that’s just what they call it. That’s like not believing in imaginary numbers because they’re called “imaginary”.

    There IS a repulsive part of the strong interaction, and in fact it’s very, very strong. It’s commonly referred to as the ‘short-range repulsive core’ and it’s not some new, mysterious 5th force. You can search for it on google an learn all about it, if you like.

    By the way, if the deuteron “cannot have a binding energy weaker than 8.56 MeV, according to nuclear models”, then why does basically every calculation performed over the last 20 years yield a smaller binding energy? With the exception of your calculation, of course.

    p.s. just a little clarification. When I said “there is no rule that says all even-even nuclei need to be bound”, what I really mean is “that statement, while apparently obvious to you, is simply wrong”. I’m not claiming that all arguments need to involve “rules”, but I am suggesting that they should include correct statements of fact. I didn’t think that would be confusing.

  • Wladimir Guglinski

    JR wrote in July 22nd, 2013 at 7:57 AM

    ————————————————
    There is also no puzzle about why the deuteron is bound but the di-proton and di-neutron are not. You can look at figure 1 in the following paper:
    http://iopscience.iop.org/1402-4896/2013/T152/014021/article
    It shows a calculation of the ‘central’ part of the N-N interaction and the ‘tensor’ part.
    ————————————————–

    COMMENT
    BLA-bla-BLA again

    Two protons are bound with energy of 10MeV via strong force of attraction, according to current nuclear models

    Therefore only a stronger force of REPULSION capable to yield an energy repulsion stronger than 10MeV so that to avoid the two protons to form the diproton.

    A tensor is a mathematical entity. A tensor cannot create a FORCE OF REPULSION.

    The same we can say about the dineutron.

    And the deuteron cannot have a binding energy weaker than 8,56MeV, according to nuclear models.

    2-
    ——————————————————
    All nucleons feel the central part, which has a slight attraction near 1fm separation, and a strong repulsion at shorter distances.
    —————————————————–

    COMMENT
    WHAT REPULSION ?????????????????????????

    The only repulsion can be due to Coulomb interaction.
    But as you had already claimed, strong nuclear force (of attraction) is 100 times stronger than the Coulomb repulsion.

    Then it is IMPOSSIBLE to exist a repulsion able to win the force due to strong nuclear force

    Unless you are proposing a new 5th fundamental force of nature: the strong nuclear force of repulsion

    In such case, I will recommend AGAIN your name to the Nobel Comission, so that to give you the Nobel Prize in 2014, for the discovery of the 5th fundamental force of nature

    The mainstream newspaper will announce with triumph the new discovery of the Modern Physics:
    —————————————————–
    Dr. JR discovers the 5th fundamental force of nature
    —————————————————–

    3-
    —————————————————–
    For a pair of protons or neutrons, the net binding is not enough to yield a pp or nn bound state. The n-p interaction also includes the ‘tensor’ part, which adds significantly more binding, giving just enough for the deuteron to be bound by 2.2 MeV.
    —————————————————

    COMMENT
    bla-bla-bla again

    A tensor cannot create a force of repulsion, since a tensor is an abstract concept of mathematics.

    All this sort of DESPERATE and stupid explanations are similar to that ABSURD proposal of isospin by Heisenberg. The isospin cannot create a force of repulsion so that to avoid two neutrons to form the dineutron, as a tensor cannot either.

    Dear JR,
    I am tired to hear so much stupid arguments.

    You are happy cheat yourself with stupid arguments.
    So, be happy.

    But unfortunatelly I cannot share your happyness, since I do not enjoy to make me stupid to myself believing in stupid arguments.

    regards
    wlad

  • JR

    Dear Wladimir,

    You assert many things about what other people think: that conventional theory requires all even-even nuclei to be spherical, that the fact that the strong force is bound means that everything has to be bound, etc….

    I believe that many of these statements are incorrect and given your track record with unsupported assertions, I don’t count you as a credible source without some kind of additional support. So if you believe that these statements are correct, please provide some kind of evidence that supports that idea that these are currently held beliefs based on conventional theory. Not 80 years ago, but in the context of modern theory.

    In the meantime, for those topics where you did provide sufficient detail, I explained why I believe that you are incorrect and gave arguments and references to support these arguments. I’m not going to try and counter assertions that I believe to be incorrect (and which you don’t explain) if you won’t provide any kind of support or explanation to back up your claims. Especially since you tend to ignore my specific comments on your claims and just jump to other topics.

  • Wladimir Guglinski

    To: Steven N Karels and the readers of Rossi`s blog

    Dear readers

    In a previous comment some days ago Mr. JR had tried to convince us that 8Be is not stable by using the unacceptable argument of inversion causality

    Now he try to convince us that 8Be is not stable by the use a new unacceptable argument:
    ————————————————–
    The need of A RULE so that to point out the obvious
    ————————————————–

    So, we realize that Mr. JR is happy by deceiving himself with stupid arguments.
    He enjoys to believe in stupid arguments, sure that they can save the current theories in which he trust.

    But I dont understand is his desire:
    —————————————————-
    Mr. JR wants that we share his happyness.

    He wants we feel ourselves happy by enjoying to deceive ourselves, believing in stupid arguments
    —————————————————–

    I dont understand his desire.

    My trouble with Mr. JR is not because he enjoys to deceive himself with stupid argument,

    My trouble is because I am not agree to be happy by deceiving myself with stupid arguments.

    If someone is happy deceiving himself, there is no problem, I desire to him a good life plenty of happyness.

    But I not agree to share such sort of happiness with him.

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in July 21st, 2013 at 10:50 PM

    1 –
    —————————————————
    The paper you cited as “experiments published in 2012″ does not show any experimental results, it is a calculation.
    —————————————————

    COMMENT
    The calculations have been made because experiments made in 2011 have shown that light nuclei with Z=N=pair have NON-spherical shape.

    Again you exhibit silly arguments.

    All the evidences had been suggesting to the nuclear theorists, along 80 years, that light nuclei with Z=N=pair have spherical shape.
    For instance, all nuclei with Z=N=pair have Q(b)=0. As Q(b)=0 requires a spherical shape, the theorists had concluded (along 80 years) that those nuclei have spherical form.

    In 2009 (before the experiments which detected the non-spherical form of those nuclei with) the theorist Martin Freer had published a paper where he had proposed that 8O16 has spherical shape:
    http://www.scholarpedia.org/article/Clusters_in_nuclei

    HOWEVER IN 2012 Martin Freer had changed his opinion about the shape of those nuclei.
    WHY ????????

    If the experiments in 2011 had not detected the non-spherical shape of the nuclei with Z=N=pair, then WHY?????>/B> would the authors of the paper published in Nature to propose that those nuclei have non-spherical shape ?????

    After all, the conjecture that light nuclei with Z=N=pair have non-spherical shape seems to contradict some nuclear properties, like the quadrupole moment.

    I have also to remember that the journal Nature had published a PLAGIARISM of my argument published in the page 137 of my book (the same argument published in Nature, so that to explain why those nuclei have non-spherial shape, in spite of they have null quadrupole moment).

    2 –
    —————————————————
    238U is not an N=Z nucleus
    —————————————————

    CORRECTION of my mistake:
    Excited nuclei with Z=pair and N=pair as 92U238 have to continue with spherical shape, by considering the current nuclear models

    3 –
    —————————————————
    220Rn and 224Ra are not N=Z nuclei.
    —————————————————

    CORRECTION of my mistake:
    Nuclei with Z=pair and N=pair as 220Rn and 224Ra cannot be pear-shaped, according to current nuclear models

    4 –
    —————————————————
    There is no rule that says that all even-even nuclei need to be bound, simply because it’s “the strong force”.
    —————————————————

    COMMENT
    Actually there is no need any rule so that to point out the OBVIOUS

    Only a silly person needs the need of a rule to point out the obvious.
    Because since there is not any force with the magnitude of the strong force (according to current Nuclear Physics), then obviously it is IMPOSSIBLE to exist any force of repulsion so that to avoid the bound of 8Be via strong force.

    If you need a rule to point out the obvious, then sorry, something is wrong with your head.

    Dear JR,
    again you try to convince the people with your stupid arguments, because:

    1- As the strong force is the stronger force existing in nature (its agnitude is 100 times stronger than any other force existing in nature).

    2- As the protons and neutrons are bound via strong force in the 8Be

    3- Then it is IMPOSSIBLE from nuclear models to explain why the nucleons bound via strong force do not succeed to form a stable 8Be.

    Sure that any person with judgment can understand it very easily.

    But I can understand if a person without judgment cannot understand it.
    So of course such a person needs a ”RULE so that to point out the obvious to him.

    regards
    wlad

  • JR

    Dear Wladimir,

    There is also no puzzle about why the deuteron is bound but the di-proton and di-neutron are not. You can look at figure 1 in the following paper:
    http://iopscience.iop.org/1402-4896/2013/T152/014021/article
    It shows a calculation of the ‘central’ part of the N-N interaction and the ‘tensor’ part. All nucleons feel the central part, which has a slight attraction near 1fm separation, and a strong repulsion at shorter distances. For a pair of protons or neutrons, the net binding is not enough to yield a pp or nn bound state. The n-p interaction also includes the ‘tensor’ part, which adds significantly more binding, giving just enough for the deuteron to be bound by 2.2 MeV.

    One other issue is that the binding energy is not equal to the interaction strength of the force, which is what you seem to be arguing. In a bound system, the particles are moving and have significant kinetic energy, so the attractive force has to be enough to hold the system together even when the particles have a lot of energy. This is illustrated in the table shown in the paper I cited above: while the binding energy is 2.2 MeV, this comes from ~22 MeV of interaction between the nucleons which has to overcome ~19.8MeV of kinematic energy due to their motion. So it doesn’t make sense to just look at the energy scale of the interaction and assume that would be the binding energy.

  • Andrea Rossi

    Roberto Rampado:
    We have not to change our patent.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    ERRATA:

    In my last post the correct is:
    ————————————————-
    The attempts so that to conciliate the nuclear properties of the deuteron have failed to get good predictions. One reason is obvious: if the nucleons were bound via strong force the binding energy of the deuteron would have to be at least E= 10MeV, but its binding energy is only 2,2MeV.
    ————————————————–

  • Wladimir Guglinski

    To Steven, Joe, and the readers of the JNP:

    ——————————————————-
    PUZZLE : why does diproton no exist ????
    ——————————————————-

    Here we will compare three nucleons: diproton, dineutron, and deuteron

    1- Diproton
    In the page 84 of my book Quantum Ring Theory it is calculated that the Coulomb repulsion energy between two protons in a distance of 2fm is:
    ——————
    Er = 1,44MeV
    ——————

    The energy interaction between two protons vai strong force is between 10MeV and 100MeV:
    ——————————
    10MeV < Ea < 100MeV
    ——————————

    Therefore if the nucleons were bound via strong force , two protons would have to form a diproton with a binding energy of at least:
    ———————————–
    E= 10 – 1,44 = 8,56MeV.
    ———————————–
    However, diprotons do not exist in Nature. WHY ??????

    2- Dineutron
    The case of dineutron is worst.
    As there is no Coulomb repulsion between two neutrons, and as they are bound with an energy between 10MeV and 100MeV , then if the nucleons were bound via strong force dineutrons would have to exist with a binding energy of at least 10MeV.

    3- Deuteron
    The interaction proton-neutron via strong force is between 10MeV and 100MeV.
    As there is no Coulomb repulsion between them, then if the nucleons were bound via strong force the binding energy of the deuteron would have to be at least E= 10MeV.
    However the experiments show that the binding energy of deuteron is only 2,2MeV. WHY ????

    4- Diproton, dineutron, deuteron

    In the case of the diproton and dineutron, as they do not exist, the nuclear tables do not give information on the nuclear properties they should have to have, as binding energy, spin, magnetic moment, quadrupole moment, etc.

    So, in the case of diproton and dineutron there is no way to investigate the interaction proton-proton in the diproton and neutron-neutron in the dineutron.

    But as the deuteron exists, its nuclear properties are investigated.
    The attempts so that to conciliate the nuclear properties of the dineutron have failed to get good predictions. One reason is obvious: if the nucleons were bound via strong force the binding energy of the deuteron would have to be at least E= 10MeV, but its binding energy is only 2,2MeV.

    As we already have explained here, along 50 years 34 models had been proposed, so that to calculate the quadrupole moment of the deuteron. And the reason is because when the author fits his model so that to calculate the quadrupole moment, his model does not fit to other properties, as we had seen in the last paper published in 2005 by the Turkish Journal of Physics:
    ————————————————————————–
    These thirty-three potential models have different deuteron properties, such as deuteron quadrupole moment QD, D-state probability PD, asymptotic D-state amplitude AD and asymptotic ratio .
    —————————————————————————
    http://journals.tubitak.gov.tr/physics/issues/fiz-05-29-3/fiz-29-3-1-0408-3.pdf

    ———————
    CONCLUSIONS
    ———————
    If the nucleons were bound via strong force we would have to have:

    1- Diprotons with binding energy of at least 8,56MeV

    2- Dineutrons with binding energy of at least 10MeV

    3- Deuterons with binding energy of at least 10MeV

    4- The deuteron nuclear properties would have to be successful calculated, if the nucleons were bound via strong force ( as the nuclear theorists use in their calculations ).
    However, as along 50 years of attempts with 34 different models the theorists did not succeed to conciliate the nuclear properties of the deuteron, it is hard to believe that the proton and the neutron are bound via strong force in the deuteron..

    But perhaps I am wrong, and 50 years and 34 attempts is not enough yet.
    Then we have to be patient.
    Perhaps along 100, 200, 2000 years they will finally succeed.

    So, let us be patient

    Regards
    wlad

  • Roberto Rampado

    Caro Dr Andrea Rossi,
    Le teorie con cui spiegate attualmente il funzionamento dell’E-Cat, sono poco o molto diverse da quelle esposte nella domanda dei vostri brevetti?
    Grazie e buon lavoro.

    Dear Dr. Andrea Rossi,
    Current theories which explain the operation of the E-Cat, are little different or very different from those in your patents’ demand?
    Thanks and good work.
    Roberto Rampado

  • JR

    Dear Wladimir,

    In response to your examples that you claim show inconsistencies between conventional nuclear theory and experiment:

    1 – The paper you cited as “experiments published in 2012” does not show any experimental results, it is a calculation.

    2 – 238U is not an N=Z nucleus

    3 – 220Rn and 224Ra are not N=Z nuclei. Also, as previously noted, the paper compares the measured transition strengths to conventional calculations and they do a pretty reasonable job; this suggests to me that the pear-shape is predicted in these calculations, although it’s not a direct comparison.

    4 – There is no rule that says that all even-even nuclei need to be bound, simply because it’s “the strong force”. I pointed to the calculations which do show that 8Be is not bound, although it’s a challenging problem because it’s very close to being bound. However, all kinds of heavy even-even nuclei are not bound and that’s perfectly well understood.

    5 – This argument is still based on misreading the tables of nuclear excited states, as previously discussed.

    6 – This one I don’t know the details of off the top of my head, so all I can say at this point is that there may be 33 N-N potentials available, but many of them are old, based on fits to limited data sets, and are known to be wrong. So the fact that these results differ is not surprising and not very meaningful. There are probably about half a dozen modern N-N potentials that do a good job of fitting p-p, n-p, and n-n scattering data, and they generally agree pretty well in their predictions for he deuteron structure. So the question isn’t if you get different results when you compare modern potentials to old ones that are known to be incorrect. The question is if the modern results give consistent and correct results for the deuteron structure. If they do give correct and consistent results, then the problem IS solved. In the listing of potentials given in the snippet you quote from the paper, I don’t see any of the modern N-N potentials, which makes me think that this is not a meaningful test of how well we can describe the deuteron based on the N-N potential extracted from scattering data.

    Finally, you end by discussing the range of models used to describe nuclei. As I attempted to clarify in my recent post, there are many approaches to performing the calculations, but they are for the most part based on the same underlying physics. They differ in the approximations made and numerical techniques applied in trying to solve the extremely complex and challenging problem of calculating nuclear structure. So it’s really not correct to say that there are a whole bunch of fundamentally different theories that we pick and choose from in order to reproduce measurements. Different approaches work for different nuclei and we are only able to use those approaches which simply the problem enough for it to be solved without missing important physics.

    So I see nothing in your examples to support the conclusion that we will never be able to find a satisfactory nuclear theory based on our current understanding. We certainly cannot make precise calculations for all nuclei at this point, but that is due to limitations in the techniques to calculate these properties. There may be missing pieces of the puzzle or even some fundamental flaw in our assumptions that we’re not aware of. But nothing you described suggests that to be the case.

  • Here is the entire set (so far) of interview segments with Andrea Rossi, conducted by Thomas Florek:

    Epoché
    https://vimeo.com/70753081

    Fleischmann and Pons
    https://vimeo.com/70294469

    Skepticism
    https://vimeo.com/69864491

    Dr. Sergio Focardi
    https://vimeo.com/68983909

    The End of the Middle Ages
    https://vimeo.com/68776012

  • Here is another segment in my series of interviews with Andrea Rossi:

    Andrea Rossi: Epoché
    https://vimeo.com/70753081

  • Wladimir Guglinski

    To Steven, Joe, and the readers of the JNP:

    If the nuclei were bound via strong nuclear force

    If the nuclei were bound by strong force as the nuclear theorists suppose, they would have to exhibit DIFFERENT NUCLEAR PROPERTIES of those detected in the experiments

    For instance:

    1-
    Light Nuclei with Z=N=pair like 8O16 and 10Ne20, would have to have spherical shape.
    But the experiments published in 2012 have shown that they have non-spherical form:
    http://www.nature.com/nature/journal/v487/n7407/full/nature11246.html

    2-
    Heavy nuclei with Z=N=pair like 92U238 when non excited have spherical form. When they are excited they had to continue having spherical form. But the experiments show that they are ellipsoidal.

    3-
    Heavy nuclei with Z=N=pair as the 86Rn220 and 88Ra224 could not be pear-shaped, they had to be spherical
    https://plus.google.com/113624764391012151708/posts/4HvW9VUrK3e

    4-
    4Be8 would have to be stable, as all the other light nuclei with Z=N=pair, as 2He4, 6C12, 8O16, 10Ne20, etc.

    5- 4Be7 could not have quadrupole electric moment Q(b) near to zero, as the experiments have detected, because it has A= 7, and therefore it cannot have a spherical shape. As it has spin and magnetic moments different of zero, 4Be7 could not have Q(b) near to zero if the nucleons would be bound via strong force within the nuclei.

    6- Along more than 50 years 34 attempts had been made so that to calculate the quadrupole moment of the deuteron. The last attempt had been published in the Turkish Journal of Physics in 2005, where the authors say:
    ——————————————————-
    So, in our investigation of deuteron tensor polarization we employ thirty-three local potential models of the nucleon-nucleon force. The thirty-three local potentials are denoted by the following notation: GK1, GK3 and GK8 of Glendenning and Kramer [5]; PARIS of Lacomb et al. [6]; RHC, RSC, RSCA of Reid [7]; TSB and TSC of de Tourreil and Sprung [8]; HJ of Hamada and Johnston [9]; TRS of de Tourreil et al. [10]; L1, L2, 2, 4, ::: , 6 of Mustafa [11]; r1, r3, ::: , r7 of Mustafa et al. [12]; MHKZ of Mustafa et al. [13]; and a, b, c, ::: , i of Mustafa [14]. These thirty-three potential
    models have different deuteron properties, such as deuteron quadrupole moment QD, D-state probability PD, asymptotic D-state amplitude AD and asymptotic ratio .
    The values of these properties are not equal, but have wide range of values in all potential models.
    ——————————————————-
    http://journals.tubitak.gov.tr/physics/issues/fiz-05-29-3/fiz-29-3-1-0408-3.pdf

    Then obviously the following questions arises:

    6.1 – Deuteron 1H2 is the most simple nucleon formed by two nucleons, one proton and one neutron

    6.2 – Suppose the proton and the neutron were bound in the deuteron via strong force as claim the theorists. However it is hard to understand why along more than 50 years of attempts, and after the publication of 34 papers, the puzzle is not satisfactorily solved yet.

    ———————————
    CONCLUSIONS
    ———————————

    The nuclear theorists try to explain such puzzles by adopting several arbitrary assumptions introduced in some nuclear models, because:

    A) as in their models the nucleons are bound via strong force within the nuclei,

    B) but as the nuclei actually are not bound the strong force

    C) then of course it is impossible to conciliate the current models with the nuclear properties detected in the experiments

    As they use different models, as the Monte Carlo model, the shell model, the models with clusters, the core-model, etc., then obviously with suitable mathematical improvements it is possible to suit the results of calculations to some experiments.
    But a fundamental problem with those several models adopted arises because one arbitrary assumption adopted for the explanation of a nucleus A does not work for a nucleus B. Then they have to use, for instance the shell model so that to make calculations concerning the nucleus A, and they have to use core model so that to make calculations concerning the nucleus B.

    Sure that they will never succeed to find a satisfactory nuclear theory so that to calculate the nuclear properties of the whole nuclei.

    Regards
    wlad

  • JR

    Dear Wladimir,

    You once again point to the neutron halo of 11Be as something that cannot be explained by conventional nuclear physics. I’ve tried to explain why that argument is incorrect, but you dismissed that argument, so let me give a more direct demonstration that there is no problem in this case.

    You can look at the original measurement here: http://arxiv.org/pdf/0809.2607.pdf

    What they actually measure is the charge radius, which means that they’re basically only measuring where the protons are, not where the neutrons are. The 7fm number you quote is extracted from the measurement using a simple (but not unreasonable) model, but it’s not actually what they measure. You can see from figure 3 in their paper that their data are very well described by conventional models. In fact, they conclude the paper by stating “Comparison with elaborate nuclear structure caluclations (GFMC, FMD, and LBSM) shows in all cases good agreement with our measurements”. So this is quite clearly not a case where conventional theory doesn’t have any problem explaining the data.

    As a side note, the 7fm number comes from a model assuming a static 10Be nucleus plus one extra neutron, and the offset is determined by the difference between the 10Be charge radius and the 11Be charge radius (assuming that the increase in size is entirely related to the neutron orbiting the 10Be core. But looking at the figures, I’d say that the difference between the radii is only about 4 times the size of the uncertainty on the difference, so the 7fm number has a large error, something like 2fm. On top of that, if you just go from A=10 to A=11, the naive estimate is that the volume and mass goes up by a factor of 11/10, so the radius would go like the cube root of this, or a factor of 1.032 (so a 3.2% increase in the radius). Their data shows a 4.5% increase. This suggests that a good two-thirds of the size increase is what you expect if you have NO neutron halo, and just make the nucleus larger. So by assuming that ALL of the effect comes from the neutron halo, they could be significantly overestimating the size of the halo with the 7fm number.

  • Wladimir Guglinski

    Steven N. Karels wrote in July 20th, 2013 at 7:38 PM

    Wlad – Although I am sure it seems there is a conspiracy of Physicists against you, the ones I know couldn’t organize a protest let along a conspiracy. Maybe they honestly disagree with you?

    COMMENT
    Dear Steven

    I think it is hard for the physicists to accept some proposals of my theory.
    For instance, in my QRT the nucleons are not bound via strong nuclear force within the nuclei.

    In my QRT it is proposed that there is not the classic Coulomb repulsion within the nuclei.
    The classic Coulomb repulsion is due to the secondary field of particles (proton and electron).

    A nucleus has a big secondary field involving it (which is composed by the overlap of the secondary fields of protons and neutrons within the nucleus). Such field repels any charge particle (as the proton) being outside of the nucleus, via the classic Coulomb repulsion.

    But when the proton enters the nucleus, while the body of the proton perforates the secondary field, the secondary field of the proton is additioned to the secondary field of the nucleus.
    Therefore, within the nuclei there is not classical repulsion between protons.

    However it is very hard for the physicists to accept my congectures.
    I think this is the reason why it is very hard to accept my nuclear model.

    But there are some evidences suggesting that strong nuclear force is not responsible for the nucleons aggregation (as for instance the fact that two neutrons do not form the dineutron).
    The experiment which had detected the halo neutron of 11Be 7fm far away of the nucleus is reinforcing my argument.

    Unfortunatelly the nuclear theorist did not realize yet that it is impossible to get a satisfactory nuclear theory via aggregation by the strong force, and they are keeping their of goal of trying to solve the nuclear puzzles by keeping their model via strong force interactions.

    I hope new upcoming experiments will finally disappoint them, and they will finally understand that it is impossible to develop a satisfactory nuclear theory via strong nuclear interactions.

    regards
    wlad

  • Wladimir Guglinski

    Joe wrote in July 20th, 2013 at 6:55 PM

    1. When you say that a vacuum does not exist, do you mean that

    a) no amount of space is ever free of aether particles (100% density)

    COMMENT
    There is no way to get a space empty of aether particles

    or
    b) space itself does not even exist (illusion)?

    COMMENT
    of course space exists, since it is filled by aether

    2. What is the source for the intrinsic spin of the electron?

    COMMENT
    When the electron is created a flux n(o) is formed, and the body of the electron (its quarks) has spin regarding to the fluw n(o).

    Like in the case of the Faraday laws of electromagnetism, there are laws ruling the behavior of the particle and its flux n(o). One of the laws is concerning the electron having spin regading the flux n(o).

    3. Do you believe that an electron can be stationary in the aether if there is a balance of forces acting on it, as there would be within a charged sphere for example?

    COMMENT
    I do not believe

    regards
    wlad

  • Andrea Rossi

    Robert Curto:
    Please don’t talk of base ball to me today: Yankees -Red Sox 5-2 ( sigh!)
    Warm Regards,
    A.R.

  • Robert Curto

    Dr. Rossi, just read your excellent Baseball Post.
    You hit a home run.
    1.First Base
    2.Second Base
    3.Third Base
    4.Home Plate

    Robert Curto
    Ft. Lauderdale, Florida
    USA

  • Andrea Rossi

    I am spamming all the comments that continue to repeat the same questions. Anyway, I want to repeat the answer to the FAQ:
    1- The E-Cat technology is undergoing rigorous testing and all the results- positive, negative or inconclusive- will provide further guidance about its potential
    2- We have great hopes for the E-Cat and what it can accomplish and we are pleased about the findings of the other scientists who have participated in evaluating it so far. As this technology is still in the development stage and undergoing rigorous review, we want to allow the continued process of testing that technology to determine its potential and uses. We are pleased with our progress to date and will share more as our work continues.
    3- Technological development can require a long process, involving many changes as a technology moves forward. E-Cat is undergoing that process now. This process will continue as long as needed, until such time as the team believes the technology is able to fulfill its promise in commercial settings.
    4- E-Cat is still also in a phase of R&D, as we continue this work more findings will be released and additional technical information will be provided once practicable. As I focus on continuing my research, I will not be able to respond to each specific question.
    Warm Regards,
    Andrea Rossi

  • Andrea Rossi

    Koen Vandewalle:
    The name will be always E-Cat.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Steven N Karels:
    He,he,he…this time you outsmarted the vile robot!
    Warm Regards,
    A.R.

  • Steven N. Karels

    Dear Andrea Rossi,

    I have been posting responses but your robot keeps diverting my responses. I suspect there is a major conspiracy by your robot, and others of the same manufacture, to suppress my information. I would call your robot “dumb” but he appears to be smarter than me so I will continue to insult him until he realizes the error of his ways and relents.

    Wlad and JR,

    I have enjoyed your postings and responses, although I find the anger accompanying the postings to divert my reception of the information. I am old – my last nuclear physics course was in 1968 – I recall the hot topic was neutrinos. So I comprehend little of your arguments and context other than you both seem passionate regarding the subject.

    JR – I find your postings clear, well written and coherent although the anger content diminishes what otherwise would be an excellent response. To paraphrase Klingon – “Vengeance is a dish best served cold”. Lose the anger and you will be more effective.

    Wlad – Although I am sure it seems there is a conspiracy of Physicists against you, the ones I know couldn’t organize a protest let along a conspiracy. Maybe they honestly disagree with you?

    Well, maybe, this posting will get through AR’s nasty robot? Robot – Your robot mother was a trash compactor and your robot father a blender – Take That! There, I feel better. Steve

  • Joe

    Wladimir,

    1. When you say that a vacuum does not exist, do you mean that

    a) no amount of space is ever free of aether particles (100% density)
    or
    b) space itself does not even exist (illusion)?

    2. What is the source for the intrinsic spin of the electron?

    3. Do you believe that an electron can be stationary in the aether if there is a balance of forces acting on it, as there would be within a charged sphere for example?

    All the best,
    Joe

  • Koen Vandewalle

    Dear Andrea,

    will the final product be given a name that links to you or to e-cat?
    Do you have agreements on that?

    I find that important.

    Kind regards,
    Koen

  • Andrea Rossi

    Steven N. Karels:
    Your comment of today has been erroneously put in the spam by the robot. Can you please send it again?
    Thank you,
    JoNP

  • JR

    Dear Wladimir,

    You said “If he were honest, he would have recognized, among other things, that current nuclear models cannot explain the instability of 8Be.”

    8Be is certainly a challenging case, because it is unbound by a only a very small amount. Therefore, only very precise calculations can hope to make a meaningful prediction and making simple calculations with significant approximations is not sufficient.

    However, a quick google search provides a couple nice discussions:
    http://www.scholarpedia.org/article/Clusters_in_

    http://books.google.com/books?id=7bI4S_07Yn0C&pg=PA67&lpg=PA67&dq=%228be+binding+energy%22&source=bl&ots=2C5SqQ2eS-&sig=i-Ib8w4f4AgLekItWmPd2DJaGXk&hl=en&sa=X&ei=OoXqUZf9IMyoqQGZ-4DIBg&ved=0CCsQ6AEwAA#v=onepage&q=%228be%20binding%20energy%22&f=false

    which then lead to at least one modern calculation of high precision:
    http://arxiv.org/pdf/nucl-th/0002022.pdf

    Because knowing whether or not 8Be is bound requires extremely high precision on the calculation, it is probably not possible at the moment to make a definitive prediction on whether 8Be is bound according to our current understanding due to the uncertainties in the calculation. But just because we can’t evaluate the result with enough precision to be 100% certain doesn’t mean that the models cannot possibly yield an unstable 8Be nucleus, which is your assertion. For example, the calculation mentioned above DOES yield an unstable nucleus, and it’s not the only one,

    Finally, let me clarify what I’m trying to prove, since you’re still misrepresenting me. I’m not saying that current theory can explain everything – there are certainly things it can’t yet calculate with sufficient precision, and therefore we don’t know in those cases if our physics understanding is complete, and there certainly could be something wrong or missing in our understanding.

    However, you are stating that the current understanding cannot possibly explain several things, and I’m simply pointing out that those assertions (at least all the ones that you’ve brought up in our discussion so far) are incorrect. I explain why I think they are mistaken (whether it’s misreading the data tables, using extremely naive methods to ‘calculate’ what conventional theory states, or just misstating what quantum mechanics or nuclear theory say about various measurements). If you think my arguments are incorrect, I’m happy to discuss them further. I’m also happy if you stick with your decision not to discuss this any further, but I couldn’t let it stand with a misrepresentation of what I’ve claimed and an entirely unsupported (and incorrect) statement about the inability of conventional theory to explain 8Be.

  • Wladimir Guglinski

    Joe wrote in July 20th, 2013 at 1:13 AM
    Wladimir,

    1. Does an electron have a helical trajectory when it is traveling through

    i) the aether?

    COMMENT
    Of course.
    The helical trajectory is caused by the interaction with the aether

    ii) a vacuum (no aether)?

    COMMENT
    There no exist vaccum

    2. Does an electron have an intrinsic spin when it is

    i) traveling through the aether?

    COMMENT
    Yes.
    The helical trajectory is due to the intrinsic spin. If the electron should had no intrinsic spin it would not have helical trajectory.

    ii) stationary in the aether?

    COMMENT
    A fundamental particle can never be stationary in the aether.
    There is a limit of minimum speed of the elementary particles in the aether, as there is the limit of maximum speed c=300.000km/h.
    When the velocity of the particle tends to the minimum speed, the radius of its helical trajectory tends to infinite.
    When the velocity of the particle tends to velocity c of light, the radius of its helical trajectory tends to zero.

    iii) traveling through a vacuum (no aether)?

    COMMENT
    There no exist vaccum

    iv) stationary in a vacuum (no aether)?

    COMMENT
    There no exist vaccum

    regards
    wlad

  • Andrea Rossi

    Giuliano Bettini:
    Your question answers to itself.
    Thank you for your attention,
    Warm Regards,
    A.R.

  • Giuliano Bettini

    Dear Andrea,
    we know that the E-Cat technology is undergoing extremely rigorous tests.
    We trust that from this, surely, important results will come.
    Without causing any disturbance, however we would ask:
    your feeling of “Big Surprises arriving soon” is still valid?
    Big Regards,
    Giuliano

  • Joe

    Wladimir,

    1. Does an electron have a helical trajectory when it is traveling through

    i) the aether?

    ii) a vacuum (no aether)?

    2. Does an electron have an intrinsic spin when it is

    i) traveling through the aether?

    ii) stationary in the aether?

    iii) traveling through a vacuum (no aether)?

    iv) stationary in a vacuum (no aether)?

    All the best,
    Joe

  • Wladimir Guglinski

    Steven N. Karels wrote in July 19th, 2013 at 8:39 AM
    To All,

    I realize that difficult questions (and answers) can bring emotions out but we need to retain respect to each other. That is one of the things that makes this forum great. Even when I suggest a “dumb” idea to AR, he responds with respect. I ask that each of us do the same and maintain this forum to the standard that AR has demonstrated.

    Dear Steven

    When you suggest a ‘dumb’ idea to AR, you are not trying to deceive the readers of this blog.

    It`s not the case of Mr. JR.
    He uses ‘dumb’ arguments so that to deceive everybody here, so that to make them to believe that current theories are able to explain all the nuclear phenomena.

    I have no patience anymore with physicists who try to deceive the lay man with dishonest arguments.

    It is obvious that JR is trying to convince everybody that current Nuclear Physics is able to explain all the nuclear phenomena.

    However I have shown that JR arguments make no sense.

    If he were honest, he would have recognized, among other things, that current nuclear models cannot explain the instability of 8Be.

    I would had respect to him if he had recognized that current theories cannot explain why 8Be is unstable.

    Unfortunatelly he prefers to keep his dishonest via of argumentation, by using the stupid inversion of causality in his way of reasoning.

    I dont have patience to hear stupid arguments. That`s why I decided to stop the discussion, telling him that it`s a waste of time to continue the discussion, and I said goodbye to him.

    regards
    wlad

  • Andrea Rossi

    Gianfranco:
    This is the blog of the Journal Of Nuclear Physics, and all the comments regarding the papers and the Authors published on the JoNP are welcome. There is no reason to make separate sections, the Readers read what they prefer and send the comments they want. All the comments are published, so far they do not contain insults or stupidities, or statements that are false at first sight, or questions to me that I cannot or want not to answer to.
    Warm Regards,
    Andrea Rossi, on behalf of JoNP

  • gianfranco

    Caro ing.Rossi,
    l’enorme interesse suscitato dalla pubblicazione di Wladimir Guglinski merita una trattazione distinta da quella sulle Sue attività E-Cat.
    Suggerisco che i commenti sul lavoro di Guglinski vengano trattati in coda al lavoro stesso e che non appaiano uniti a quelli specifici sull’E-Cat,rispettando l’ambito delle diversità degli argomenti trattati.
    Lasciamo a Guglinski ciò che è di Guglinski ed ad A. Rossi ciò che è di Andrea Rossi.

    Cordiali saluti

    Dear ing.Rossi,
    the enormous interest aroused by the publication of Wladimir Guglinski deserves separate discussion from your activities on E-Cat.
    I suggest that comments on the work of Guglinski are treated in the queue to his work and which do not appear together with those specific on the E-Cat, respecting the scope of the diversity of the topics covered.
    Give to Guglinski what is of Guglinski and to A. Rossi what is of Andrea Rossi.

  • JR

    Dear Wladimir,

    You’re imagining things again. I never said one word about packing fraction or packing loss; I don’t even know what you’re mean by that (nor do I really desire to, at this point). I also did not call Bethe, you, or anyone else stupid, and I did not change my story, I simply provided the additional details you requested.

    I have stated that you appear to have little understanding of modern nuclear theory and tend to make assertions about facts/measurements/etc… without any actual support for those fact, and that you favor insults and mocking to physics. These statements I stand by, though I’m always open to being proved wrong.

    I agree with Steven that insults get us nowhere, and am happy to town down my sarcasm, which I have let get the best of me on occasion, assuming that you can bring yourself to address the questions at hand rather than fabricating quotes, dismissing my (and others’) arguments as “bla bla bla”, and dismissing anyone who disagrees with you as a “betrayer of the scientific method”.

  • JR

    Dear orsobubu,

    The article you linked to on neutrinoless double beta decay is very interesting, and this is one of a few experiments trying to address this very interesting question: Are neutrinos their own antiparticles?

    If so, then a decay which normally releases two neutrinos may be able to proceed in a way where the neutrinos annihilate (if a neutrino is also an anti-neutrino), leaving a decay with no neutrinos. It’s hard to measure, but a very exciting question.

    It doesn’t really ‘challenge’ the standard model, as such, because the standard model doesn’t know if a neutrino is it’s own antiparticle or not. So this is one of the big open questions in the standard model.

    I suppose that the standard model must, by default, assume one option or the other. But I would think that it assumes that neutrinos are NOT their own antiparticles, in which case this supports the standard model. But as I say, it’s really an open question, and so seeing the neutrinoless double beta decay or excluding it at the level that we know that neutrinos can’t annihilate will certainly tell us which is correct and become a part of the standard model.

    So many interesting things going on!

  • Steven N. Karels

    To All,

    I realize that difficult questions (and answers) can bring emotions out but we need to retain respect to each other. That is one of the things that makes this forum great. Even when I suggest a “dumb” idea to AR, he responds with respect. I ask that each of us do the same and maintain this forum to the standard that AR has demonstrated.

  • Wladimir Guglinski

    JR wrote in July 18th, 2013 at 10:19 PM

    1-
    ———————————————–
    Why isn’t 8Be bound as tightly as 4He? Because two protons and two neutrons can go into the lowest orbitals (one with spin up, one with spin down). Addition nucleons have to go into higher energy states and so aren’t bound as tightly.
    ————————————————

    COMMENT

    WOW !!!
    Now you have changed your argument !!!!. It is not anymore the packing loss the responsible for the instability of 8Be… ha-ha-ha

    However, you continue claiming that the Nobel Laureate Hans Bethe is an idiot, since he did never succeed to explain why 8Be is not stable, while you had solved the problem with only few words.

    Dear JR,
    I promise you: I will recommend your name for the Nobel Commission, so that to give you the next Nobel Prize

    2-
    ———————————————–
    Of course, I assume at this point your entire goal is to mock and insult, since your total lack of meaningful arguments is getting to be pretty clear.
    ———————————————–

    COMMENT

    No, dear JR,
    you had insulted yourself showing to everybody that you dont know to make distinction between CAUSE and EFFECT

    It is clear to everybody why you believe that current Nuclear Physics is able to explain all the nuclear phenomena.

    After all, everybody now know that you use the inversion of causality as a method of reasoning. You explain a phenomenon by considering that it is caused by the effect produced by the own phenomenon.

    From such method of reasoning it is obvious that current Nuclear Physics is able to explain everything.

    I will not waste my time with a person with such method of reasoning.
    It`s a lost of time

    Goodbye

    wlad

  • JR

    I also promised you some clarification on the issues with these ‘conventional nuclear physics’ calculations. One thing that seems to be causing trouble is confusion between two related but different questions: 1) is the physics model/understanding complete enough to explain the relevant measurements, and 2) whether we can perform sufficiently precise calculations based on this underlying physics picture.

    For example, we can describe the interaction between a nucleus and its electrons with high precision, and we can describe the interactions between the electrons in different atoms very well. But that doesn’t mean that we make precise calculations of the strength of a steel girder based on a first principles calculation of what’s going on atom by atom. The problem is just to big and complex, even if we fully understand the interactions between individual atoms.

    Because interactions in the nuclear are stronger and more complicates, they become very difficult to calculate as you go to heavier nuclei. Several techniques can calculate the structure in 3He and 4He precisely, without having to make approximations to simplify the calculations. But such first principles are difficult to extend to heavy nuclei, and I don’t know of any such calculations that go beyond 12C

    For heavier nuclei, there are simplifications and approximations that can be made for nuclei with A=Z= one of the ‘magic numbers’, e.g. 4He, 16O,…. Because of things like the symmetries of the problem, these calculations can be simplified numerically, and are thus easier to calculate reliably. For nuclei which differ from these by just one or two protons and/or neutrons, calculations can be performed by starting with one of these magic nuclei as the ‘core’, and then modeling the impact of a couple additional nucleons.

    Other techniques and simplifications allow for calculations of even more complex systems. In general, the underlying physics picture is the same, but the approach to actually calculating the structure varies in these different techniques: VMC (Variational Monte Carlo), GFMC (Green’s Function Monte Carlo), shell model, no-core shell model, coupled cluster calculations, DFT (Density Functional Theory), etc…. While they are often discussed as different ‘models’, they aren’t using different explanations for the underlying physics; they’re using different approaches to evaluate the structure, different approximations which allow different nuclei to be studied while trying to include all of the important physics.

    For experts in the field, you can make predictions about which approximations are likely to be sufficient to fully capture what’s going on in a given nucleus. For someone who isn’t familiar with all of the technical details, it’s hard to tell which approaches are expected to be precise enough for a given calculation. So it’s no surprise that some calculations will give the wrong result for some specific excited state of some nucleus. But that doesn’t mean that the underlying physics picture is wrong, it just means that the particular approach to calculating it gives an incorrect result. It could be a problem with the underlying physics, or it could be a problem with the approximations used in making the calculation. So it’s not enough to look at one ‘wrong’ calculation, you have to see if any of the appropriate calculations can do a good job. It may look like random picking and choosing if you don’t know which approaches are better/worse for different cases, but that’s not the case.

    You just gave an example on 8Be. It may well be that the calculating the expected mass of 8Be was difficult to do when the paper was written (over 40 years ago). But now the calculations tools are better, as is the detailed understanding of the nucleon-nucleon interaction. So now it can be calculated and it’s been shown that it’s unbound, with what is largely if not entirely the exact same underlying physics that could not be used to make a solid prediction in 1971.

    So when you say that modern theory can’t calculate something or the other, your examples often involve showing that it can’t be described in a couple lines of math using only the simplest principles of quantum mechanics and nuclear theory. But that doesn’t mean that a real calculation, including all of the things we know, can’t explain it well. This is why I point you to references for calculations that have directly explained the observation rather than arguing over the details of the very naive estimates/arguments that you provide. Just because a simple calculation with lots of approximations doesn’t work is not reason to say that the entire underlying physics framework is entirely wrong. But I think that this is what your examples often boil down to, at least when you’re comparing to real measurements of known nuclear states.

  • JR

    You asked why 8Be is not bound, and I said that it is because it is not bound as tightly as two 4He nuclei. It’s not complete, but it’s not confusing cause and effect.

    Why isn’t 8Be bound as tightly as 4He? Because two protons and two neutrons can go into the lowest orbitals (one with spin up, one with spin down). Addition nucleons have to go into higher energy states and so aren’t bound as tightly.

    If you want an actual calculation, you can look up calculations which show that the energy of the 8Be state is higher then twice that of 4He. This shows that it can be well described by conventional nuclear physics.

    I didn’t call you or anyone else stupid for asking the question, I gave a simple and straightforward answer to the question you asked. You meant something different, and so I’m providing a more detailed and direct answer to your questions now. If you have a specific thing in mind when you are asking a question, try phrasing it so that it is clear.

    Of course, I assume at this point your entire goal is to mock and insult, since your total lack of meaningful arguments is getting to be pretty clear.

  • Wladimir Guglinski

    To the readers of the JoNP:

    Let me explain you why our friend JR is unable to understand some fundamental questions in Physics

    Suppose that I do the following question to Erik:

    – Erik, why the bodies fall down?

    And Erik gives us the following reply:

    – The bodies fall down because of the force of gravity

    So, his answer is correct, because:
    1- The gravity is the cause
    2- The falling down of bodies is the consequence of the gravity actuation.

    Now suppose I do the following question to JR:

    – JR, why do the gravity exists in our planet ?

    And JR replies:

    – The gravity exists because the bodies fall down

    So, his explanation is incorrect, because:
    1- The falling down of the bodies is not the cause of the existence of the gravity
    2- The falling down is the consequence of the gravity existence
    2- The cause of the existence of the gravity is the property of matter which gives gravitational attraction between the bodies

    So, JR uses an inversion in the causality. He uses it so that to explain a phenomenon by stating that the effect caused by the phenomenon is the cause which yields the phenomenon.

    It is hard to believe that JR can really understand the fundamental questions in Physics, since he even is unable to understand that a phenomenon cannot be caused (and explained) by the effect produced by the phenomenon.

    regards
    wlad

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