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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339 comments to Advanced concepts in black hole cosmology

  • Erik

    Wladimir:

    I think the decay of 11Be beautifully illustrates the greatness of the established models. As pointed out by JR the probability of a certain decay depends on the overlap between the states – including excited states. Even from the shell model we know which these states are and which transitions in the excited states are forbidden. Is there any way you can do something similar with your semi-classical stuff? Or will the proton just drop down into the ground state?

  • Erik

    Wladimir:

    Regarding everyting you would like to know about the isotopes of Li, pls refer to JR’s excellent references. To be honest with you I find the agreement between the calculated and experimental values rather good!

    What you may perhaps have missed is that the spins of the nucleons do not simply add and combine to the magnetic moment of the nucleus. States will also have a contribution from angular momentum and protons and neutrons need to be treated separately. Using the simple nuclear shell model you can get a good idea about how the ground state and some excited states of a certain nucleus look like. But since this model does not include interactions between nucleons, energies, magnetic moments and quadrupole moments of the various states will not be reliable. If you are interested in these quantities you will have to introduce an Hamiltonian similar to the one used in JR’s reference (Eq. 1).

  • Wladimir Guglinski

    JR wrote in July 15th, 2013 at 11:48 AM

    Dear Wladimir,
    Finally, the coulomb interaction is something like a factor of 100 times weaker than the strong force (or significantly more, depending on how you decide to compare things). So even though the distant neutron is more loosely bound, the binding from the strong force (‘weak’ relative to other nuclei) is still large compared to the coulomb effect.

    COMMENT
    Your explanation is UNACCEPTABLE.

    The strong force does not actuate in the distance 7fm:

    From wikipedia:
    ————————————–
    In the context of binding protons and neutrons together to form atoms, the strong interaction is called the nuclear force (or residual strong force).

    Unlike the strong force itself, the nuclear force, or residual strong force, does diminish in strength, and in fact diminishes rapidly with distance. The decrease is approximately as a negative exponential power of distance.
    http://en.wikipedia.org/wiki/Strong_interaction
    —————————————

    Besides, your explanation is UNACCEPTABLE because:

    1- There is not Coulomb repulsion on the halo neutron in the 4Be11

    2- If your argument should be correct, the halo neutron would not exist in the 4Be11, because the neutron would go back to the nucleus 4Be11.

    3- As you claim that a halo proton (under Coulomb repulsion) in the new-born 5B11 goes back to the nucleus 5B11, then why a halo neutron in 4Be11 does not succeed to go back to the nucleus 4Be11 ???? (since there is not any Coulomb repulsion on it).

    4- From your argument the 4Be11 cannot have halo neutron

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in July 15th, 2013 at 9:24 AM

    Dear Wladimir and orsobubu,

    Calculations of 7Li magnetic moment in conventional theories (some new, some 15 years old, all giving results very close to the measurement):
    http://arxiv.org/pdf/0810.0547.pdf
    http://cds.cern.ch/record/351635/files/9804014.pdf
    http://arxiv.org/html/nucl-th/9802073

    ===========================================
    COMMENT 1:
    In the link bellow the authors use the shell-model:
    http://cds.cern.ch/record/351635/files/9804014.pdf

    Look at the Calculation of magnetic moment for 4Be7:

    THEORETICAL : µ = +0,802
    EXPERIMENTAL: µ = -1,398 (see nuclear table)

    If you believe that they are close, I dont understand what you consider very close.
    ================================================

    ================================================
    COMMENT 2

    In the link
    http://arxiv.org/html/nucl-th/9802073
    the authours use other method.

    They say:

    Many researchers have investigated Coulomb form factors of lithium nuclei using the resonating group method (RGM).
    However, the RGM calculations [2] do not reproduce the experimentally observed second maximum at 3 – 4 Fm-1. ”

    So, the authors propose a new method with many ARBITRARY assumptions, so that to calculate the magnetic moment of 3Li7.

    Look what they say:
    ———————————————
    For example, the form factor of 4He can be approximated by the expression

    F = ( 1 – ( a q2) n ) exp( – b q2) (3)

    where a = 0.09986 Fm2, b = 0.46376 Fm2, and n = 6. For the form factors of 3H and 3He, we can also use formula (3) with the parameters a = 0.0785 Fm2, b = 0.4075 Fm2, n = 5.46 and a = 0.09275 Fm2, b = 0.50875 Fm2, n = 8.1, respectively. We parametrize the deuteron form factor by the expression

    F = exp( – a q2) + b q2 exp(- c q2) (4)

    with the parameters a = 0.49029 Fm2, b = 0.01615 Fm2, and c = 0.16075 Fm2.
    ————————————————

    Well, dear JR,
    by such an ARBITRARY method we can get everything we wish. We need only to choose the suitable parameters.
    ==============================================

    ==============================================
    COMMENT 3
    In the link
    http://arxiv.org/pdf/0810.0547.pdf

    the authors use other two different methods, GMC and GFMC, and in both them, again, several ARBITRARY assumptions are taken in consideration.

    In the end of the article they say:

    Acknowledgments
    The many-body calculations were performed on the parallel computers of the Laboratory
    Computing Resource Center, Argonne National Laboratory

    Therefore, such a method of calculation is actually an attempt so that to fit the theory to the experimental data, since they even use a computer so that find the suitable numbers to be applied, to be changed, to be aplied again… and so one, and after so many attempts they finally get some values near to experimental data.

    Then I ask you, dear JR:

    as each of the three methods use DIFFERENT ARBITRARY ASSUMPTIONS, what is the correct arbitrary assumption to be taken as the correct ??????????

    regards
    wlad

  • JR

    Wladimir,

    I sent you three links to papers calculating 7Li. While I sent preprint or online pdf versions (so that anyone could look at them without having subscriptions) two of the three were published in Physical Review C.

    I’m disappointed that less than 8 hours after sending you three links showing exactly what you asked for, you’re demanding links to the same thing of someone else. If the difference is requiring publication in reputable peer reviewed journals, then I’m just amused (and would like to see links to your arguments in a reputable peer-reviewed journal)

  • Wladimir Guglinski

    Erik wrote in July 15th, 2013 at 10:19 AM
    Wladimir:

    1-
    Using the following link you will find a diagram illustrating the decay of 11Be. As you can see, the proton formed may end up in various excited states which will then decay into the ground state (or into 7Li by alpha decay). There is nothing strange about this from the “established” nuclear physics point of view.

    http://www.tunl.duke.edu/nucldata/GroundStatedecays/11Be.shtml

    COMMENT
    The diagram does not show how the halo proton far away 7fm from the rest of the nucleus goes back to the nucleus 5B11

    The diagram only explains WHAT HAPPENS

    The diagram does not explain HOW IT HAPPENS

    2-
    I think I finally also figured out how you (wrongly) assume the nuclear spins are calculated. In the nuclear shell model (you may want to google this or check the page on wikipedia) the angular momentum combines with the spins of the fermions. Spin 3/2 in 7Li does not mean you have a couple of nucleons aligned but the unpaired proton is in the 1p3/2 state.

    COMMENT
    Show us any paper published in peer review journal where the magnetic moment of 3Li7 is calculated

    regards
    wlad

  • JR

    Dear Wladimir:

    Your request: “please explain to orsobubu what quantum mechanics,, nuclear theory, etc…. actually say about these observations.

    My reply:
    http://arxiv.org/pdf/1201.0724.pdf

    For future inquiries:
    http://bit.ly/u8YVBf

  • JR

    Dear Wladimir,

    If you need to be disproven in yet another way, I’ll go ahead and give you a more formal answer. I strongly suspect you’ll just consider it more bla-bla-bla, because it requires even more actual knowledge of physics than the previous ones.

    If the 11Be nucleus is bound with respect to 10B + proton, then the neutron can’t just decay into a proton and go flying off, as the system would have to gain energy to do so. So the only time the decay can happen is if the neutron decays into a proton and then goes into the nucleus. So no matter how rare this decay might be, it’s going to be the dominant decay if it’s the only allowed decay. (It isn’t the only one, as Erik showed, but what you consider to be the main prediction of conventional theory just is not possible).

    The probability that a decaying neutron would end up close enough to form 11B depends on the overlap of the wavefunction of the 11Be ground state and the 11B ground state (and any excited states, since it can decay into an excited state) – the neutron does spend some time nearer the others, it’s just further away on average. Because the sizes are different, this overlap is small, and thus the probability to decay is suppressed, weakening the decay and yielding a longer lifetime. But since it can’t just ‘fall apart’ the way you want without violating energy conservation, this suppressed decay is still the most favorable. By the way, the fact that 11B is more tightly bound (i.e. is a lower energy state) makes it a more favorable state for 11Be to decay into, so that tends to increase the decay rate, and offset some of the suppression from the size difference.

    Finally, the coulomb interaction is something like a factor of 100 times weaker than the strong force (or significantly more, depending on how you decide to compare things). So even though the distant neutron is more loosely bound, the binding from the strong force (‘weak’ relative to other nuclei) is still large compared to the coulomb effect. The strong interaction is much, much, much more important than the coulomb interaction in light nuclei, but it’s only much, much more important for a weakly bound system like this. But that doesn’t mean that the coulomb interaction has a very large effect.

    So now, between this and may last few answers to this question, I’ve explained why it can happen in a simple picture, given a more formal explanation in terms of quantum mechanical principles, and explained why your basically every piece of your argument that it can’t happen is wrong. I concede that you have managed to prove that a really stupid picture of what’s going on is incorrect, but since this picture disagrees with most principles of modern physics, it doesn’t prove anything at all about current nuclear theory.

  • Wladimir Guglinski

    Dear JR,
    please explain to orsobubu other question, now concerning 3Li8.

    Proton has µ =+2,793 and neutron has µ =-1,913

    3Li8 :
    It has i= 2 and µ= +1,65

    Suppose the following structure for 3Li8:
    a) 2 protons with contrary spins, so they yield i=0 and µ =0
    b) 2 neutrons with contrary spins, so they yield i=0 and µ =0
    c) one proton and three neutrons with aligned spins, in order to perform i= 2.
    So they yield µ = +2,793 +(-1,913 -1,913 -1,913) = -2,94

    Therefore, while the experiments show that 3Li8 has a POSITIVE µ= +1,65 , unlike from the current nuclear models 3Li8 would have to have a NEGATIVE µ near to -2,94.

    Dear JR,
    please explain to orsobubu what quantum mechanics,, nuclear theory, etc…. actually say about these observations.

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in July 14th, 2013 at 8:09 PM
    Dear Wladimir,

    1-
    Other minor errors on 11Be: I never said that the ‘distant’ neutron had to be the one to beta decay,

    COMMNENT
    You dont need to say it.
    The other neutrons are strongly connected to the nucleus 4Be11, and therefore there is no reason why they would decay, since the halo neutron is weakly tied to the nucleus (as you yourself had claimed.

    Besides, even if a neutron strongly tied to the 4Be7 should suffer decay, you have answer the following questions:

    1- The halo neutron is weakly tied to the nucleus. Then why would the halo neutron go back to the nucleus in the decay 4Be11 -> 5B11 ?????????

    2- The halo neutron is under the action of a centripetal force 3 times stronger than that existing over the other neutrons of 5B11. Then by what reason the halo neutron in the 5B11 would go back to the center of the 5B11, since the halo neutron of 4Be11 did not go back to the center of the 4Be11 ???????

    2-
    the Coulomb force is not very strong (or even mildly strong) compared to nuclear forces relevant to this problem,

    COMMENT
    Why not ??????
    Since you yourself had said that the halo neutron (or proton) is weakly bound ??????

    Have I remember you that the strong force had been incorporated in Nuclear Physics because the strong force is need to neutralize the Coulomb forces between protons ????????????
    But perhaps you had forgoten it… ha-ha-ha-ha

    3-
    there’s no reason why a weakly bound nucleon can’t end up as part of a more tightly bound nucleus, this decay happening 97% of the decay’s doesn’t tell you how strong the decay is only how strong it is relative to other decay channels. The long half-life tells you that it’s a relatively weak decay, so even thought I agree that the decay would be suppressed by the difference in nuclear sizes, that doesn’t seem to contradict the data.

    COMMENT
    again bla-bla-bla…
    and not any satisfactory explanation, based on the current principles of nuclear physics.

    You avoid to speak about the fundamental point: to explain how the halo neutron(or proton) comes back to the 5B11.

    I am tired of your bla-bla-bla
    And I think the readers of Rossi`s JNP are either

    regards
    wlad

  • Erik

    Wladimir:

    Using the following link you will find a diagram illustrating the decay of 11Be. As you can see, the proton formed may end up in various excited states which will then decay into the ground state (or into 7Li by alpha decay). There is nothing strange about this from the “established” nuclear physics point of view.

    http://www.tunl.duke.edu/nucldata/GroundStatedecays/11Be.shtml

    I think I finally also figured out how you (wrongly) assume the nuclear spins are calculated. In the nuclear shell model (you may want to google this or check the page on wikipedia) the angular momentum combines with the spins of the fermions. Spin 3/2 in 7Li does not mean you have a couple of nucleons aligned but the unpaired proton is in the 1p3/2 state.

  • JR

    Dear Wladimir and orsobubu,

    Wladimir has asked that I explain what conventional theory says about the 11Be –> 11B decay. As I’ve said repeatedly, I see nothing about it that isn’t consistent with conventional theory. However, I am not enough of an expert to try and make a first-principles calculation of the decay, so all I can say is that the various arguments that Wladimir makes about why it is impossible are wrong (see, for example, my comment from July 14, 8:09pm).

    As for his question about 3Li7, I’d say that the result he gets suggest that is simple suppositions are incorrect. However, since he isn’t coming close to doing a real calculation, the fact that his quick and dirty estimate based on very simple assumptions doesn’t work isn’t exactly something to worry about.

    Finally, it’s pointless to try and make an argument by insisting over and over and over and over that the person you’re debating explain one thing after another after another. (Actually, it’s not pointless, it’s just a way to avoid actually addressing the points raised that disagree with what you claim). Since I’m not Wlad’s personal tutor, I feel no need to solve every homework problem he comes up with. If he wants to know what nuclear theory says about something, he can try learning some nuclear theory. Alternatively, he can try searching the web. For example (I only looked at the first 10 results from a google search):

    Calculations of 7Li magnetic moment in conventional theories (some new, some 15 years old, all giving results very close to the measurement):
    http://arxiv.org/pdf/0810.0547.pdf
    http://cds.cern.ch/record/351635/files/9804014.pdf
    http://arxiv.org/html/nucl-th/9802073

    This next one doesn’t provide the calculation, but apparently it was assigned as a homework problem, so I’d have to assume that it doesn’t actually demonstrate that all of conventional nuclear theory is incorrect.
    http://www.physicsforums.com/showthread.php?t=445803

  • Andrea Rossi

    Bill Nichols:
    1- I think the E-Cat is not the right system to absorb efficiently heat, not directly. Of cpourse you can make conditioned air using the E-Cat as a heat source, but is not this that you mean. Our effect is not endothermic.
    2- See above
    Thank you anyway for the interesting report of your experience in the USAF in the nineties.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Dear JR,
    please explain to orsobubu other question, now concerning 3Li6 and 3Li7.

    Proton has µ =+2,793 and neutron has µ =-1,913

    3Li6 :
    It has i= 1 and µ= +0,822

    Suppose the following structure for 3Li6:
    a) 2 protons with contrary spins, so they yield i=0 and µ =0
    b) 2 neutrons with contrary spins, so they yield i=0 and µ =0
    c) one proton and one neutron with aligned spins, in order to perform 1=1.
    So they yield µ = +2,793 -1,913 = +0,88

    The difference D = 0,822 -0,88 = -0,058 can be credited to the magnetic moment due to the spin of the nucleus 3Li6.
    So, we can explain the µ of 3Li6 by considering the current nuclear models very well.


    3Li7 :

    It has i= 3/2 and µ= +3,25

    Suppose the following structure for 3Li7:
    a) 2 protons with contrary spins, so they yield i=0 and µ =0
    b) 2 neutrons with contrary spins, so they yield i=0 and µ =0
    c) one proton and two neutrons with aligned spins, in order to perform i= 3/2.
    So they yield µ = +2,793 +(-1,913 -1,913) = -1,033

    Therefore, while the experiments show that 3Li7 has a POSITIVE µ= +3,25, unlike from the current nuclear models 3Li7 would have to have a NEGATIVE µ near to -1,033.

    Dear JR,
    please explain to orsobubu what quantum mechanics,, nuclear theory, etc…. actually say about these observations.

    regards
    wlad

  • Erik

    Wladimir:

    Regarding 4He, 8Be, 12C etc.

    Obviously I can not answer for JR, but I believe it can have something to do with spin zero. For some reason they have only included nuclei and states of non-zero spin. I guess you will find the zero-spin nuclei in volume 2 of this table. Or perhaps it is a conspiracy or discrimination against zero-spin states? Or do you suggest they do not even exist because they are not in the table?

    Best regards

    Erik

  • Wladimir Guglinski

    JR wrote in July 14th, 2013 at 3:54 PM

    Dear orsobubu,

    The other aspect of Wladimir’s approach to these arguments is to claim that conventional nuclear physics cannot explain certain observations, based on either

    (3) misrepresenting what quantum mechanics, nuclear theory, etc…. actually say about those observations.

    Dear JR,
    please explain to orsobubu what nuclear theory says about the following observations:

    1- The strong force actuates in the range of maximum 3fm

    2- In 4Be11 the halo neutron is far away 7fm from the rest of the nucleus.

    3- You claim that there’s nothing to explain, other than the fact that 11Be must be very weakly bound

    4- But in 97% of the decays the 4Be11 transmutes to 5B11

    5- 5B11 is stable, and it has no halo neutron.

    6- Therefore the halo neutron decays in a proton, and so, according to the current nuclear theory the proton would have to be expelled, because of the strong Coulomb repulsion on it.
    So, the decay 4Be11 -> 5B11 cannot occur.

    7- Beyond the Coulomb repulsion, there is on the halo proton a strong centripetal force, due to the nuclear spin of the new-born 5B11. Such centripetal force is 3 times stronger in the halo proton than the force on the nucleons in the rest of the nucleus, since the orbit of the halo proton has radius 3 times larger.

    8- Therefore, according to nuclear theory, the decay 4Be11 -> 5B11 is IMPOSSIBLE to occur.

    Dear JR,
    please explain to orsobublu what quantum mechanics,, nuclear theory, etc…. actually say about these observations.

    regards
    wlad

  • Bill Nichols

    Andrea Rossi…

    A couple of questions…

    It has been described (by yourself and others) the E-CAT is basically a exothermic reaction (heat…as we understand it…is “released”?).

    1.) What is your view about the potential for eventually creating an endothermic reaction (heat…as we understand it…is “absorbed”) based E-CAT?

    2.) Is there any evidence from your tests so far of this possibility? If so, can you at least generalize on this question?

    Working with thermodynamics…radiant…chemical…nuclear and mechanical energies for now nearly 40 years and nature has always been 2 way. Why shouldn’t it be here?

    As I’ve mentioned quite a while back to you, saw LENR phenomena in early 1990s while in USAF military.

    Am increasingly questioning if we understand what heat, temperature and other thermodynamic concepts really are. Inclined to think it is routed in an incomplete defining of mass (plus all thermodynamic concepts and the other 4 branches or disciplines of energy mentioned above) that is rooted on gravity which is the weakest of the 4 forces (other 3 forces use what we vaguely grasp as “charge”. Never seen anything in nature that wasn’t “2-way”…like natural exothermic and endothermic process we see everyday. Have seen a lot of really peculiar phenomena that can’t be explained using convectional science in decades of R&D. Gravity is not even a constant on the earth…much less understood fully. This variance should not be dismissed.

    If you have evidence of endothermic processes with the E-CAT, maybe when you have something on the market and have the resources, you could do some out of the box tests on this potential phenomena.

    Visualize the economic benefits commercially of a endothermic system that could supplement a exothermic system. By definition this would provide a more efficient and powerful energy system that is “2-way”.

    FWIW…would be very surprised if you don’t eventually run into an endothermic reaction that can be commercially harnessed.

    Recall…energy…in derived units we call “Joules” and equal to the energy expended both as a Newton-meter, and also a Watt-sec. This point or bridge of gravity and charge w/r/t energy may be helpful in future tests to consider.

    Please consider the thoughts provided above. Offer nature may provide you and your colleagues some insight to future E-CAT tests. Especially radiant energy interactions which are much more dynamic than we understand…including impacts on earth phenomena. This is being proven more an more each day…especially in my field we describe in the short term as weather, and longer term as climate. Isn’t the E-CAT a micro scaled version, or at least a cousin?

    Apologize in advance if you’ve already addressed the 2 above listed questions. I may have missed it.

    Thanks.

    Kind Regards,

    Bill Nichols
    Atmospheric Scientist

  • Here are the 4 of the segments of my interview with Andrea Rossi:

    Andrea Rossi talks about Fleischmann and Pons
    https://vimeo.com/70294469

    Skepticism:
    https://vimeo.com/69864491

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

    Andrea Rossi talks about working with Dr. Sergio Focardi
    https://vimeo.com/68983909

  • Here is another segment of my interview with Andrea Rossi.

    In this week’s segment, Andrea speaks about the work of Fleischmann and Pons, and discusses the importance of their work in relation to LENR.

    https://vimeo.com/70294469

  • JR

    Dear Wladimir,

    I did not suggest that there was a conspiracy, that was your doing and I interpreted it to mean that you agreed that such measurements had not been made or at least not admitted. As for you other questions:

    1) The decay of 11Be: My answer is correct. The fact that it sounds like ‘bla-bla-bla’ to you is not my problem. In fact, that’s my whole point: you don’t understand conventional theory, and you don’t want to, so you dismiss attempts at a real physics conversation as ‘bla-bla-bla’.

    2) You assert that when a measurements is made and a value zero is found, it is left out of the table. I assert that you are wrong. If you felt like finding a reference that says that one of these states of yours has been measured in accordance with your assertion, that would be great and would support your interpretation. I showed you two stating that it hadn’t been measured (at least for one of the states).

    I suspect that some nuclei are left out because standard theory says that they must have mu=0, i=0 (spin-0 non-excited even-even N=Z nuclei). But even if they are left out because the answer is assumed to be known, that doesn’t tell you whether or not such measurements were performed (and I pretty sure that at some-to-many of the techniques don’t even apply to spin-0 nuclei, if they require measuring samples with aligned spins). It also doesn’t support your assertion that nuclei that ARE included use a blank space to indicate zero, without any information at all about how that zero was measured, what the uncertainty was, etc….

    I also note that SAYING SOMETHING IN BOLD FONT AND ALL CAPS IS NOT A VALID PHYSICS ARGUMENT. If you disagree with that statement, then note that it was in bold caps, so it must be true.

    Other minor errors on 11Be: I never said that the ‘distant’ neutron had to be the one to beta decay, the Coulomb force is not very strong (or even mildly strong) compared to nuclear forces relevant to this problem, there’s no reason why a weakly bound nucleon can’t end up as part of a more tightly bound nucleus, this decay happening 97% of the decay’s doesn’t tell you how strong the decay is only how strong it is relative to other decay channels. The long half-life tells you that it’s a relatively weak decay, so even thought I agree that the decay would be suppressed by the difference in nuclear sizes, that doesn’t seem to contradict the data.

  • Wladimir Guglinski

    Dear JR,

    please explain us the reason why the following non-excited nuclei are NOT quoted in the nuclear tables:

    2He4
    4Be8
    6C12
    8O16
    10Ne20
    12Mg24
    etc
    etc
    etc.

    ALL THOSE NUCLEI ARE NOT QUOTED IN THE NUCLEAR TABLES

    Please explain us: WHY ????????????


    Please do not use bla-bla-bla

    regards
    wlad

  • Wladimir Guglinski

    Dear Jr

    you did not give us yet a satisfactory for the halo neutron of 4Be11 and its decay 4Be11 -> 5B11

    Theb I will refresh your memory, and I will be wainting your explanation.

    Please do not give us bla bla bla again

    YOUR ARGUMENT:

    ============================================
    There is no reason that the size of a bound state must be less than the ‘approximate range’ of the relevant force. You can solve a system with square well potential with a well defined maximum range, and the particles can be further from the center than this range. In fact, even a point interaction can yield a bound state with finite size. So there’s nothing to explain, other than the fact that 11Be must be very weakly bound.
    ============================================

    Pay attention to your argument:
    ——————————————–
    So there’s nothing to explain, other than the fact that 11Be must be very weakly bound.
    ——————————————–

    And my reply:

    =================================================
    I would like to emphasize that your explanation is 100% INCORRECT and UNACCEPTABLE

    Because:

    In 97,1% of the decays of 11Be, it transmutes to 11B.

    Therefore, the halo neutron has a decay and it transmutes to a proton.

    Well, earlier the decay of 11Be the halo neutron was in a distance of 7fm from the rest of the nucleus.
    You claim that the halo neutron is very weakly bound to the rest of the nucleus.

    However, when the neutron transmutes to a proton, a very strong Coulomb force tries to expel the proton from the rest of the nucleus.
    Remember that the proton is far away 7fm from the nucleus, and so it is very weakly bound.

    Therefore the proton cannot go back to the nucleus, so that to form the 11B, since the proton is quickly expelled because of the strong Coulomb repulsion.

    CONCLUSION:
    from the principles of current Nuclear Physics, the decay 11Be -> 11B is IMPOSSIBLE !!!!
    but experiments show that it occurs in 97,1% of 11Be decays.
    =================================================

    Dear JR
    as you know, 5B11 is stable, and it has NO halo neutron.

    Therefore my interpretation is the unique way to explain what occurs in the 4Be11 decay:

    1- the neutron decays to a proton
    2- the proton goes back to the nucleus (which is IMPOSSIBLE by considering the current nuclear theories, since the proton has a big repulsion with the rest of the nucleus 5B11, and so it cannot go back to the nucleus, since it is very weakly bound (as YOU yourself had claimed).


    Therefore, dear JR
    we would like to hear from you the explanation:

    ==============================================
    how is explained by the current nuclear models the decay 4Be11 -> 5B11 ????????
    ==============================================

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in July 13th, 2013 at 12:16 PM

    1-
    So your theory is that physicists avoided measuring these particular quantities for these nuclei (or measured them but covered up the results) because they worried that they would destroy conventional nuclear physics.

    COMMENT
    Dear JR,
    I dont know, among the two possibilities bellow, what is the real cause of your lack of understanding:

    1- You really did not understand the point

    2- You deceive us trying to convince us that you did not understand.

    Dear JR
    The physicists did not avoid measuring these particular quantities for these nuclei because they worried that they would destroy conventional nuclear physics, as you are claiming

    What happens is the following:

    When the measurement does not detect a value different of zero, it is not quoted in the nuclear table

    For instance:
    The values of i, µ, and Q(b) of the non-excited nuclei with Z=N= pair ARE NOT QUOTED IN THE NUCLEAR TABLES, just because the experiments did not find values different of zero.

    Therefore, the non-excited nuclei 2He4, 4Be8, 6C12, 8O16, 10Ne20, etc, etc, etc, ARE NOT QUOTED IN NUCLEAR TABLES because the experiments did not detect values different of zero.

    The same happens with the excited 6C12 and non-excited 4Be7.
    The µ of 6C12 and Q(b) of 4Be7 ARE NOT QUOTED BECAUSE:

    =======================================================
    THE EXPERIMENTS DID NOT DETECT VALUES DIFFERENT OF ZERO

    for 6C12 and 4Be7
    =======================================================

    But of course the readers already had understood you strategy, dear JR.

    As you know that it`s IMPOSSIBLE to explain the nuclear properties of 6C12 and 4Be7 by considering the current nuclear theories, you use a dishonest strategy, by continuing your bla, bla, bla, bla, trying to deceive the people who are reading such our discussion.

    regards
    wlad

  • JR

    Dear orsobubu,

    Just to add my 2 cents to you comments, I should point out that Wladimir’s representation of current theory as being nothing more than mathematical constructs with no underlying picture is just plain wrong. Some aspects, e.g. quantum mechanics, involve a lot of mathematics and very non-intuitive concepts. But this, and most of the things that go into the nuclear models are far from being purely mathematical constructs.

    The other aspect of Wladimir’s approach to these arguments is to claim that conventional nuclear physics cannot explain certain observations, based on either (1) making up observations that don’t exist, (2) misinterpreting measurements that do exist, and (3) misrepresenting what quantum mechanics, nuclear theory, etc…. actually say about those observations.

    So while I too want focus as much as possible on real, underlying physical pictures of what’s going on, I find Wladimir’s approach to the discussion to be entirely dishonest, and it’s entirely clear from his reactions to points others raise that he has no interest in better understanding the issues or correcting his mistakes as they appear to be the cornerstones of his arguments.

  • KD

    From J-O-N-P

    Frank Acland

    Dear Andrea,
    You mention that the E-Cat is undergoing changes — which is to be expected in product development. Are these changes going to lead to:
    1. Redesigned E-Cat reactors?
    2. Redesigned plants?
    2. Cheaper products?
    3. Higher COP?
    4. Smaller products?
    5. Larger products?
    Many thanks and best wishes,
    Frank Acland

    Andrea Rossi

    Frank Acland:
    All the points you touched are encompassed in what I said recently, which is: the E-Cat technology is undergoing rigorous testing and the results, whatever they are, will provide further guiodance about its potential.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Frank Acland:
    All the points you touched are encompassed in what I said recently, which is: the E-Cat technology is undergoing rigorous testing and the results, whatever they are, will provide further guiodance about its potential.
    Warm Regards,
    A.R.

  • orsobubu

    Wlad,

    I enjoyed the debate between you, JR and Joe and although lacking the tools that allow me to say my opinion onto specific issues, I find more convincing your method of investigation. Being a materialist, I judge particularly idealistic some aspects of frontier theoretical physics, which seems to try and derive concrete reality from an abstract mathematical model, rather than the opposite.

    The situation is similar to that which occurs in political economy, where hundreds of scientists, even Nobel laureates, have developed many theories all contradictory to each other and especially with the rality of production social relations, which they are not able to control. The reason is similar to what you can see in your field of research, namely the particular material interests of a party. In the broader case of the economy, this is the interest of the owners of the means of production, the capitalists and financiers, to create an ideology that would convince the lower classes that their free (or socialist, state protected) market model is the only viable in society, avoiding or deferring any revolutionary attempt.

    Precisely in this regard, allow me to move a little criticism to your sentence, although it has nothing to do with physics. When you say that “the fragility of life makes us to worry about following this fact: the future of humanity is in the hand of one man”. You are right to argue that “all of us fell ourselves very afraid,” but surely not because this is someway related to LENR. The future is not jeopardized by the energy issue, in the same way it is not true that wars are made for oil. The capitalist, imperialist wars have to export their capitals, expand their sphere of influence and gain market share from competitors, to avoid the falling rate of profit. And this is the Marxist revolutionary theory that the bourgeoisie wants to hide to the working classes.

    In the developed world, even now, too much cheap energy is a worrying issue, some commodities prices are artificially kept high on purpose and already some “green frontier” industries are starting to fail. Paradoxically, if after a “war for oil” would be discovered that oil is no longer needed because – thanks to LENR – there is no shortage of energy, it would be much worse, because the fighting parties should resume with even more violence the conflict to destroy other enemy’s assets and secure themselves resources of different types. And as the fossil fuel industry is among those from which the majority of the shares of world surplus quotas are extracted by workers’ exploitation, replacing it with something that takes enormously less amounts of fixed and variable capital would be really disrupting for the economic and military geostrategic balance, and the fight would jump to multiple different unknown fronts to compensate the missing profits.

    The future of humanity is therefore dependent not by technology but by social-economic relationships and struggle among the classes and the problem is much more complex than what can involve a single man; then I think Rossi is right to shirk this responsibility that many attribute to him and concentrate on his business.

  • Frank Acland

    Dear Andrea,

    You mention that the E-Cat is undergoing changes — which is to be expected in product development. Are these changes going to lead to:

    1. Redesigned E-Cat reactors?
    2. Redesigned plants?
    2. Cheaper products?
    3. Higher COP?
    4. Smaller products?
    5. Larger products?

    Many thanks and best wishes,

    Frank Acland
    6.

  • Andrea Rossi

    Martin:
    Exactly the same. So far, because 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 lonf as needed, as I already said, until such tine as the team believes the technology is able to fulfill its promise in commercial settings.
    Warm Regards,
    A.R.

  • RiccardoDG

    To Wlad, Hope my English is understandable, to answer at “Why” someone could hidden or conspire against common sense or the truth, i could submit you the results of scientific researches about an hidden component of society and mankind, called “psychological predators” which are about the 2-4% of ALL mankind, (http://www.psychologytoday.com/articles/199401/charming-psychopath)including Scientist (As per Psycologist Christie back in ’70 who found that most of the people voted to control others usually choose the legal careers, the psychiatric and scientist career !!!!!, As most of them are NOT violent (only verbally) and they seems to be integrated with society (few of them are just killers, most are your neightbor). That’s why to some inexplicable hystory happenings and the decay of the societies. They could be very bright or be high in their career, nevertheless they are fighitng against society, so twist the truth and hidden the discoveries, usually with ironic accuse…

  • Martin

    Dear Andrea,

    I was thinking about the ecats produced by your partner. Did they make the ecats
    Exactly the same or is it a different design?

    Best regards,

    Martin

  • JR

    Dear Wladimir,

    So your theory is that physicists avoided measuring these particular quantities for these nuclei (or measured them but covered up the results) because they worried that they would destroy conventional nuclear physics.

    1) Finally, you admit that there are no results available indicating that mu=0 for the excited state of carbon or that Q=0 for the 7Be ground state. I’m glad to hear that you’re finally acknowledging that you’re just making up these “measurements”, so that we can now completely ignore all demands that we must prove they are possible in conventional nuclear physics. This will save time. (Unless you’ve made such measurements and kept them hidden to avoid challenging traditional dogma – if so, I recommend that you publish).

    2) You clearly don’t understand scientists, who like nothing more than making measurements that show some unusual and unexpected effect and confront conventional wisdom.

  • Erik

    Wladimir:

    I do not think there is a boycott against any isotopes or states. I think they have simply not been measured yet for various reasons.

    Regards,

    Erik

  • Erik

    Wladimir:

    I never claimed the magnetic moment of the first excited state be zero. That is your claim. I am convinced it is not equal to zero due to the combination of the 1p1/2 and 1p3/2 states. Give me any piece of evidence that the magnetic moment is zero and I will reveiw my opinion.

    Best regards,

    Erik

  • Wladimir Guglinski

    Erik wrote in July 13th, 2013 at 5:19 AM

    I also find it hard to understand how you find three possible cases for the excited state and try to calculate the resulting magnetic moments. To my understanding one of the fermions just goes from the 1p3/2 to 1p1/2 state. But please prove me wrong.

    COMMENT
    Dear Erik,
    then try the best you think is the correct way, and show us that it is possible to have the excited 6C12 with i=2 and µ=0.

    Good luck

    regards
    wlad

  • Wladimir Guglinski

    CONSPIRACY IN THE SCIENCE

    To the readers of Rossi`s blog:

    The nuclear tables quote values of magnetic moments µ and quadrupole electric moment Q(b) for all the nuclei, from the neutron 0n1 up to 99Es255, for all nuclei which have µ and Q(b) different of zero.

    So, all the values of µ and Q(b) different of zero are quoted in the tables.

    However, there is a conspiracy of the physicists against the excited 6C12 and the non-excited 4Be7.

    The reason is because according to the prevailing current nuclear models the two nuclei have to have the following properties:

    1- Excited 6C12 has to have µ very biggest than zero

    2- Non-excited 4Be7 has to have Q(b) very biggest than zero.

    However, due to some unknown mysterious reason, the physicists who make the nuclear tables decided to boycott the 6C12 and the 4Be7.

    Indeed, while all the nuclei with µ and Q(b) different of zero are quoted in the nuclear tables, however 6C12 and 4Be7 are not quoted.

    Such conspiracy against 6C12 and 4Be7 is very intriguing, because as they have µ and Q(b) different of zero (as we expect from the current nuclear models), then there is no reason why to eliminate them from the nuclear tables.

    But probably I am very stupid, and that`s why I cannot understand such a conspiracy.
    Probably there is a good reason for such boycott against 6C12 and 4Be7.

    So, I would like somebody smarter than me could help me to understand such conspiracy.

    I am sure there is a good reason for such boycott, and I will be very thankfull if somebody explain it to me.

    Dears Joe ,Steven N. Karels, and Erik,

    may you explain it to me ?

    regards
    wlad

  • Erik

    Wladimir:

    If you have a closer look on (professor) Stone’s table you will find that each line corresponds to a reference in which either the magnetic moment or the quadrupole moment has been measured (provided we use the same table). The only reference for 12C refers to its quadrupole moment so it is not surprising that there is no magnetic moment given.

    I also find it hard to understand how you find three possible cases for the excited state and try to calculate the resulting magnetic moments. To my understanding one of the fermions just goes from the 1p3/2 to 1p1/2 state. But please prove me wrong.

    Best regards,

    Erik

  • Wladimir Guglinski

    JR wrote in July 12th, 2013 at 6:46 PM

    Dear Wladimir,
    We disagree on how to read the tables, so picking a second example where you treat a blank space as a zero doesn’t help. If I am right about how the tables present the data, then your second example is just as wrong as your first. If you are right, then you should be able to confirm it by finding actual measurements of the quantities that you are claiming are zero.

    COMMENT
    Dear JR
    dont be so ingenuous

    It seems you think that those scientists who make the tables are stupid guys.

    It seems you think that they did not understand the meaning of the data collected, because they are silly.

    But you are wrong.
    When they had measured the magnetic moment of excited 6C12, of course they had understood that the value zero defies the current theories.

    Than of course that they had repeated the measurement one, two, three, and more times, so that to get assurance on that value zero obtained.
    And after so many measurements they finally become convinced that excited 6C12 really has magnetic moment zero, in spite of it is impossible to explain it from the current theories.

    But perhaps you are right, dear JR, and the scientists who made the experiments are stupid as you believe.

    In this case I suggest you to repeat the experiment in the Argonne National Laboratory, so that to eliminate the controversy.

    But I am sure you are not interested to confirm the magnetic moment zero of excited 6C12, because you actually prefer to belive that there is nothing wrong with the current nuclear models.

    The Nobel Laureate G. t`Hooft did the same regarding to Borghi experiment.
    As he knows that Borghi experiment imply in the failure of the current Nuclear Physics, he prefers to see the Borghi expeiment the most far away possible from him.

    regards
    wlad

  • JR

    Dear Wladimir,

    I’m glad that I’m able to both impress and amuse you so much.

    You keep saying that “the measurements detected nothing”, but there have to actually be measurements before you conclude that seeing nothing was meaningful. You should try looking for some, you might find them interesting.

    A few minutes with the google provides the following for 7Be.

    “In Ref. [9] we have shown that the zero energy cross section of the 7Be(p,gamma)8B reaction scales linearly with the, unfortunately yet unknown, quadrupole moment of 7Be.” Page 2 of http://arxiv.org/pdf/nucl-th/9802003.pdf

    Of course that’s from 1998, the tables you pointed me to are bit more recent. Let’s jump to 2010: “The quadrupole moment of 7Be was not measured yet. The values listed in the table is a theoretical prediction.” Table 2.1, page 7 of
    http://www-alt.gsi.de/documents/DOC-2010-Oct-77-1.pdf

    So rather than throwing out the tables, maybe you should suggest that Dr. Stone fill every blank space with “This space intentionally left blank, Wladimir” in very, very small writing.

  • JR

    Dear Wladimir,

    We disagree on how to read the tables, so picking a second example where you treat a blank space as a zero doesn’t help. If I am right about how the tables present the data, then your second example is just as wrong as your first. If you are right, then you should be able to confirm it by finding actual measurements of the quantities that you are claiming are zero.

  • Wladimir Guglinski

    JR wrote in July 12th, 2013 at 8:38 AM

    1-
    Let me get this straight. Your assertion is that the lack of any published value for some quantity means that it has been measured to be zero with extremely high precision. It is known to be zero at such a high precision that it’s not even worth quoting uncertainties or even publishing these ultra-high precision results anywhere? Now that is laughable.

    COMMENT
    Oh my God !!!!!
    your lack of comprehension is very impresive, dear JR !!!!

    Look:
    The excited 6C12 has i=2.

    Then let us analyse what can be the POSSIBLE values of magnetic moments of the excited 6C12, by considering the current models:

    a)
    4 protons yielding i=2
    2 protons yielding i=0
    6 neutrons yielding i=0
    Such a combination performs µ= +4×2,793 = +11,18

    b)
    4 neutrons yielding i=2
    2 neutrons yielding i=0
    6 protons yielding i=0
    Such a combination performs µ= – 2×1,913= -3,82

    c)
    2 protons yielding i=1
    2 neutrons yielding i=1
    4 protons yielding i=0
    4 neutrons yielding i=0
    Such a combination performs µ= +2×2,793 – 2×1,913= +1,76

    So, the possible values of magnetic moment for excited 6C12 are:
    +11,18
    -3,82
    +1,76

    Those 3 values are VERY FAR AWAY of zero !!!!

    But the experiments had detected NOTHING !!!!

    It is not the case that it has been measured zero with extremely high precision, as you wrongly claim !!

    It is the case that it would have to be measured a value VERY, VERY, VERY… FAR AWAY of ZERO !!!!!

    But the experiments had detected NOTHING
    … or, if you preffer:
    they had detected a value VERY, VERY, VERY, VERY, VERY… CLOSE TO ZERO.

    Oh, come on, dear JR, you are joking…

    Then let me laugh: ha ha ha ha ha

    2-
    If you’re hanging the utility of your approach based on your assertion that traditional nuclear theory is unable to explain a missing entry in a table, then your position is even weaker than I realized.

    COMMENT
    Well, then I had supposed wrongly that nuclear table are made with the aim to help us to verify if the current nuclear models are correct, or not.

    But from the viewpoint defended by you I realize that I was wrong.
    So, I have to change my mind.
    Now I understand, thanks to your help, that nuclear tables are made with the folowing objective:

    1- For NOTHING. The nuclear tables are a joke. We dont need trust them.

    2- TO BE REJECTED. When the data of the nuclear table disprove the nuclear models, they must be rejected.

    3- TO DECEIVE US. When the data are disagree to the theoretical values obtained from nuclear models, we have to trust the models, and never in the experiments.

    Dear JR,
    I have to confess that it`s a very interesting method of investigation followed by you.

    I will think very seriously to follow such a method of you, dear JR.

    However, I will suggest to Dr. N.J.Stone, the publisher of nuclear tables, to throw his tables to the trash…

    ha ha ha ha ha

    regards
    wlad

  • Erik

    Wladimir:

    I am afraid I share JR’s view that absence of numbers does not indicate that the quantity under consideration equals zero. If you happen to have any other information (or experimental data from your lab) on this isotope I would very much appreciate if you could post it (or a suitable reference) here.

    Best regards,

    Erik

  • JR

    Dear Wladimir,

    Let me get this straight. Your assertion is that the lack of any published value for some quantity means that it has been measured to be zero with extremely high precision. It is known to be zero at such a high precision that it’s not even worth quoting uncertainties or even publishing these ultra-high precision results anywhere? Now that is laughable.

    If you’re hanging the utility of your approach based on your assertion that traditional nuclear theory is unable to explain a missing entry in a table, then your position is even weaker than I realized.

  • Wladimir Guglinski

    Dear JR

    as you has not a good understanding on how the nuclear table exhibit the nuclear data, then I will challenge you so that to give explanation for other nuclear phenomenon.

    The non-excited 4Be7 exhibits, according to the nuclear table, the followig data:

    nuclear spin 1=3/2
    magnetic moment µ= = -1,398
    quad elect mom Q(b) =0

    I challenge you to quote any theory , (or you yourself can solve the problem), so that to show a combination of the 4 protons and 3 neutrons of the 4Be7, so that to get 1=3/2, µ= -1,398 , and Q(b)=0.

    As you know, in spite of nuclei with Z=N=pair have non-sherical shape, however the experiments do not detect their Q(b), because as all they have i=0 and µ= 0, there is no way to align them along an external magnetic field during the experiments, and therefore, in spite of they do NOT have Q(b)=0, however they behave as if they should have Q(b)=0.

    It`s not the case of 4Be7, because it has spin 1=3/2, µ= -1,398.

    Therefore, from the current nuclear models, it`s IMPOSSIBLE for the 4Be7 to have Q(b)=0, , because a nucleus with odd number A= 7 of nucleons cannot have a spherial shape, and so it cannot have Q(b)=0, since in the experiments it can be aligned along an external magnetic field,

    I hope you will not claim, again, that there is a conspiracy of the experimental physicists against the 4Be7, so that do not measure its Q(b), in order to justivy why the nuclear tables do not quote its quadrupole electric moment.

    regards
    wlad

  • Wladimir Guglinski

    To the readers of the Journal of Nuclear Physics:

    As I already told herein, in 2011 I had signed an agreement with the publisher Victor Riecansky of the Cambridge International Science Publishing, and according to the agreement my book The Missed U-Turn had to be published in the middle of 2012.

    The book had not been published, and I suspect that the publisher Victor Riecansky had been convinced do not publish the book by academic physicists, who probably told him that I have not a good understanding of the fundamental problems of Physics, and my book is full of misunderstandings.

    If so, sure that Riecansky had been convinced by physicists like our friend JR.
    As I have shown here, JR has (as the most academic physicists) a poor comprehension on many fundamental questions which disprove the current theories of Modern Physics.

    Like JR, the academic physicists do not analyse with accuracy the experiments which disprove the current theories of Physics. When any experiment defies the current theories in which they believe, they have not interest to think seriously about the experiment, as we have seen herein in the case of the interpretation of JR with respect to the experiment which detected the halo neutron of the 4Be11, far away 7fm of the rest of the nucleus.
    JR did not analyse the experiment, and its consequences for the Physics.
    He only had a superficial understanding of the experiment, and in order to keep his belief that the experiment does not defy the current principles of Nuclear Physics, he with single carelessly he use to claim:
    So there’s nothing to explain, other than the fact that 11Be must be very weakly bound.

    However, as I have shown here, such a claim by JR does is disagree to the decay 4Be11 -> 5B11.

    So, the academic physicits adopt the procedure of do not think in deep about experiments which disprove the current theories.
    And when we ask them to explain some experiment which defy the current theories, he uses to give an explanation unacceptable from two viewpoints:

    1- his explanation is not able to explain the experiments by considering the current theories.

    2- his explanation shows that he did no understand in deep the meaning of the experiment.

    The most academic physicists have such misunderstanding.
    For instance, in Oct 2012 the physicist John Arrington sent me an email where he says:
    —————————————————–
    Date: Wed, 3 Oct 2012 06:34:07 -0700
    From: johna_6@yahoo.com
    Subject: Re: plagiarism in the journal Nature
    To: wladimirguglinski@hotmail.com

    Dear Wladimir,

    What I said is correct.

    1) I am not aware of any deficiencies in the current models, and in particular, not in the context of our recent measurement.
    ——————————————————-

    So, I suspect that the publisher Victor Riecansky had been convinced by such sort of physicist like JR, who told him that I have not a good understanding on the fundamental questions in Physics.

    If this is the case, Riecansky had been victim of academic physicists who theyselves do not have a good comprehension of the questions that defy the current Modern Physics.

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in July 11th, 2013 at 12:09 PM

    1- I’ll have to stick with my original answer: As far as I know, no such state exists.

    COMMENT
    ha ha ha ha
    Dear JR, you are pathetic.

    The nuclear tables give the value of magnetic moments FOR ALL THE EXCITED NUCLEI WITH Z=N=PAIR, because all they are non-null.

    Only the magnetic moment of excited 6C12 is not quoted in the nuclear tables.

    Probably, according to you, the authors of the nuclear tables are boycotting the poor 6C12… ha ha ha ha ha…

    Dear JR, you are very funny.

    2- FYI, the table is not THAT difficult to understand. Each line is one measurement, in almost all cases, a given measurement extracts EITHER mu or Q(b). The method used for each measurement is also quoted, and the only measurement for the 12C excited state used the “CER” (Coulomb Excitation Reorientation) technique which only provides Q(b) values for all cases I could find.

    COMMENT
    Of course the only measurement for the 12C excited state is only CER.
    As the methods RIV/D,R and TF (for detection of magnetic moments) did not detect any magnetic moment for the excited 6C12 (just because excited 6C12 HAS NULL MAGNETIC MOMENT), the methods are not quoted in the table.

    Dou you expect that a nuclear table would have to quote a method which had been detecting nothing ??????

    If you expect it, you are wrong.
    The nuclear tables do not quote methods when do not detect values different of zero.

    For instance, the non-excited nuclei with Z=N=pair have, all they, null magnetic moment and null spin.
    They are NOT quoted in the nuclear tables, NEITHER the method which had been used.

    regards
    wlad

  • Wladimir Guglinski

    Eric Ashworth wrote in July 11th, 2013 at 1:49 PM

    Your mention of ‘the sound of aether’ intrigues me because you mention a velocity faster than light.
    […]
    It could in theory travel at the same velocity as a photon.

    The faster then light propagation has been detected by experiments.
    1- Koryu Ishii T. and Giakos G. C. (1982), Transmit Radio Messages Faster than Light, Microwaves & RF

    http://130.203.133.150/showciting;jsessionid=3801AF91FCFBDAE58DEB7CB90CC1FDED?cid=11402884

    regards
    wlad

  • Joe

    Wladimir,

    One way in which classical physics can explain a proton being sent to the core of an obviously repulsive nucleus is by invoking the Pauli Exclusion Principle. If a nuclear shell is available for the proton, it will go to it regardless of Coulomb forces. (This situation is similar to the electron stationed in its own shell about the nucleus, disregarding any Coulomb forces acting on it by nearby electrons.)

    One potential advantage that QRT might have is in explaining the great velocity of the emitted electron. In the decay of a free neutron, the electron has the time to slow down, change from a high-speed classical trajectory to a slower helical trajectory before it leaves the proton. In the decay of a bound neutron, the electron would be leaving the proton at the stage of the classical trajectory due to a critical change in magnitude of the gravitational flux n(o) as its radius expanded. This flux (that of 2He4) would be countering the neutron’s own gravitational flux n(o) in such a way as to upset its magnetic moment and hasten its demise.

    All the best,
    Joe

  • Andrea Rossi

    Dear Wladimir Guglinski:
    If you refer to our technology, it is shared now and does not anymore depend only on me. The E-Cat technology is undergoing rigorous testing and the results- positive, negative, or inconclusive- will provide further guidance about its potential.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Dear Andrea Rossi

    the Sergio Focardi death makes us to think about how fragile is the life.

    And the fragility of life makes us to worry about this following fact: the future of humanity is in the hand of one man.

    All of us fell ourselves very afraid

    regards
    wlad

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