CraigD Posted January 10, 2013 Report Share Posted January 10, 2013 The inclusion of gravity in quantum mechanics is much sought after. While quantum mechanics appears, by all evidence, to accurately describe the small-scale interaction of everything when the small-scale-negligible effect of gravity is ignored, which General Relativity appears, by all evidence, to accurately describe the large scale movement of light and massive bodies under ordinary conditions (eg: not in the first moments after the big bang, or in or near black holes), several generations of physicists have attempted, and failed, to incorporate these two theories into a single, experimentally testable, “theory of everything”. Most approaches to a ToE attempt to explain gravity by introducing a new particle, the graviton, to the Standard Model, implicitly assuming that the SM’s current force-carrying particles (bosons) are not adequate to explain gravity. What sort of theory can we get if we reverse this assumption, and assume that they are? Succinctly, then, here’s my alternate hypothesis and theory: gravity is due to the attractive electrostatic force being slightly greater than the repulsive. Formally, in the classical approximation:[math]\begin{cases}|\boldsymbol{F}|= k_A k_e \frac{|q_1q_2|}{r^2}, & \mbox{if } q_1q_2 < 0 \\|\boldsymbol{F}|= k_R k_e \frac {|q_1q_2|}{r^2}, & \mbox{if } q_1q_2 > 0\end{cases}[/math] [math]k_A > k_R[/math] Where [imath]k_e[/imath], [imath]q_1[/imath] [imath]q_2[/imath], and [imath]r[/imath] are Coulomb’s constant, the charges of a pair of elementary particles (eg: electrons or quarks, not composite particles such as protons or large ensembles of them such as planets), and the distance between them. [imath]k_A[/imath] and [imath]k_R[/imath] are dimensionless constants nearly equal to 1. There is no need, therefore, for a new particle. Gravity emerges from an asymmetry in the electromagnetic interaction, which according the quantum electrodynamics, is carried between charged particles by virtual photons. An obvious experiment to test this hypothesis, which for lack of a better term, I’ll call CraigD’s electromagnetic theory of gravity (DEMG), is to measure the gravitational force on (weigh) a body that has a net electric charge. The weight should change more than can be accounted for by the addition or subtraction of electrons producing the net charge and the classical gravity formula. Another is weighing objects of different atomic compositions. Because the U, D, and D quarks in a neutron have charge +2/3, -1/3, -1/3, compared to the U, U, and D quarks in a proton, which, with an electron, have charge +2/3, +2/3, -1/3, -1, bodies made of lighter elements should weigh more than can be accounted for by a count of and the atomic mass of their atoms. For example, relative to one of iron, a silicon body should be about 2.4% too heavy. Another is to measure the trajectory of a beam of antimatter (eg: antihydrogen). It should curve in the opposite direction than expected, “falling up”. Such an experiment is actually being done now by the AEGIS experiment group at CERN. While at first glance the second of these experiments seems easy to do and to clearly invalidate DEMG – atomic mass and weight appear to exactly equate – my working assumption in not abandoning it is that, since we ordinarily measure bodies by weighing them, not by counting atoms or massing them using spring scales and centrifuges, the experiment has never actually been done. This approach is so obvious, I can’t imagine that people with better physics skills than I haven’t considered it, but I can’t remember encountering any discussion of it in various physics-for-laypeople books, magazines, webpages, or even science forums. I’d appreciate help from my fellow hypographers, old or new, in tracking down such discussion and/or generating some new. A first step (which I meant to do before posting this, but grew too impatient to share my idea to wait) could be to calculate an approximate value for [math]k_A \over k_R[/math], assuming ordinary materials and bodies such as the Earth and Moon. cal, Moontanman and Buffy 3 Quote Link to comment Share on other sites More sharing options...

Aethelwulf Posted January 20, 2013 Report Share Posted January 20, 2013 Formally, in the classical approximation:[math]\begin{cases}|\boldsymbol{F}|= k_A k_e \frac{|q_1q_2|}{r^2}, & \mbox{if } q_1q_2 < 0 \\|\boldsymbol{F}|= k_R k_e \frac {|q_1q_2|}{r^2}, & \mbox{if } q_1q_2 > 0\end{cases}[/math] [math]k_A > k_R[/math] Where [imath]k_e[/imath], [imath]q_1[/imath] [imath]q_2[/imath], and [imath]r[/imath] are Coulomb’s constant, the charges of a pair of elementary particles (eg: electrons or quarks, not composite particles such as protons or large ensembles of them such as planets), and the distance between them. [imath]k_A[/imath] and [imath]k_R[/imath] are dimensionless constants nearly equal to 1. I have wondered if gravity is just an electrostatic force. I have no evidence for this, only that electromagnetism is very like the gravitational. I like your idea. Quote Link to comment Share on other sites More sharing options...

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