Charge Asymmetry of Muons Generated in a Muon Generator from Ultra-Dense Hydrogen D(0) and p(0)

L. Holmlid
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引用次数: 1

Abstract

Laser-induced nuclear reactions in ultra-dense hydrogen H(0) (review in Physica Scripta 2019) create mesons (kaons and pions). These mesons decay mainly to muons. The muons created are useful (patented source) for the muon-induced fusion process. The sign of the muons from the source depends on the initial baryons used. With D(0) (ultra-dense deuterium) the source produces mainly positive muons and with p(0) (ultra-dense protium) the source produces mainly negative muons. Negative muons are required for muon-induced fusion. This charge asymmetry was reported earlier, and has now been confirmed by experiments with a coil current transformer as the beam detector. The current coil detector would give no signal from the muons if charge symmetry existed. The charge asymmetry could indicate unknown processes, for example, caused by the different annihilation processes in D(0) and p(0). The conclusions of a new analysis of the results are presented here. Using D(0) in the muon source, the asymmetry is likely due to the capture of µ- in D atoms and D2 molecules. This leads to emission of excess µ+ from D(0). With p(0) in the muon source, the capture rate of µ- is lower than in D(0). The emitted number of µ+ will be decreased by the reaction between µ+ and the surrounding abundant electrons, forming neutral muonium particles. This effect decreases the amount of emitted µ+ for both p(0) and D(0), and it is proposed to be the main reason for a larger fraction of emitted µ- in the case of p(0). Thus, there is no dominant emission of negative muons which would violate charge conservation.
超致密氢产生μ子的电荷不对称性D(0)和p(0)
激光诱导的超致密氢H(0)核反应(参见《物理脚本》2019)产生介子(介子和介子)。这些介子主要衰变成介子。产生的μ子对于μ子诱导的聚变过程是有用的(专利源)。来自源的介子的符号取决于最初使用的重子。对于D(0)(超致密氘)源主要产生正μ子,对于p(0)(超致密质子)源主要产生负μ子。负介子是介子诱导核聚变所必需的。这种电荷不对称早前被报道过,现在用线圈电流互感器作为光束探测器的实验证实了这一点。如果电荷对称存在,电流线圈探测器将不会从μ子发出信号。电荷不对称可能预示着未知的过程,例如,由D(0)和p(0)中不同的湮灭过程引起的。这里给出了对结果进行新的分析的结论。使用介子源中的D(0),不对称可能是由于μ - in D原子和D2分子的捕获。这导致从D(0)发射多余的µ+。在μ介子源p(0)时,µ-的捕获率比D(0)时低。µ+与周围丰富的电子发生反应,形成中性的介子粒子,从而减少µ+的发射数。这一效应降低了p(0)和D(0)的µ+发射量,这被认为是p(0)的情况下µ-发射比例较大的主要原因。因此,不存在违背电荷守恒的负μ子的主导发射。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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