tag:blogger.com,1999:blog-4637778157419388168.post1664370601079871360..comments2024-03-16T04:13:10.154-07:00Comments on Physics with an edge: How can MiHsC be applied to a hot star?Mike McCullochhttp://www.blogger.com/profile/00985573443686082382noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4637778157419388168.post-36017493091697894902015-11-12T13:00:49.362-08:002015-11-12T13:00:49.362-08:00Nice and useful thought experiment. If you draw a ...Nice and useful thought experiment. If you draw a diagram of the 2 particles, showing the horizons, it's clear that for a local mutual acceleration of your particles round each other then the horizons that form are symmetric around their centre of mass, so no matter what Unruh waves are disallowed by MiHsC the effect will be symmetric and will not move their mutual centre of mass, though it may cause their mutual orbit to expand or contract around that centre of mass. So, even the spin of stars should not effect the inertial mass I use in MiHsC to model the orbit of their centres of mass around the galaxy. Thinking solely in terms of horizons like this grows naturally out of MiHsC, and answers some question, like this, that were unanswered before.Mike McCullochhttps://www.blogger.com/profile/00985573443686082382noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-53525137370896994392015-11-11T22:16:36.938-08:002015-11-11T22:16:36.938-08:00Wow, I did not expect such a thorough answer. Than...Wow, I did not expect such a thorough answer. Thanks!<br /><br />If I am understanding your answer right, you are saying that each particle has two Rindler horizons that we should consider for this example -- one for thermal acceleration and one for acceleration toward the galaxy's center. Is that the right interpretation?<br /><br /><br />I am trying to not be a lazy questioner, so I have thought about an idealized situation that I think illustrates where my confusion is coming from. Suppose we have two pairs of particles out towards the edge of the galaxy, where their acceleration towards the galactic center predicted by Newtonian gravity is only 1e-8 m/s². In pair A, the two particles are just floating alongside each other. In pair B, the two particles are orbiting each other (because they're charged, let's say) with a radial acceleration of 1 m/s. Other than that, the two pairs of particles are identical.<br /><br />We should be able to calculate the inertial mass of each of these particles using the formula in http://arxiv.org/pdf/1302.2775v1.pdf : Mi ≈ Mg*(1 - 2c²/aϴ) .<br /><br />If we calculate the inertial mass of pair A's particles, (with a = 1e-8) we would get that the inertial mass is only 93% of the gravitational mass. If we calculate the inertial mass of pair B's particles (with a ≈ 1), we would get the inertial mass to be just about 100% of the gravitational mass.<br /><br />So, pair A would orbit the galactic center fast enough to make Newtonian gravity look wrong, but pair B would make Newtonian gravity look right. And an actual hot star, with lots of particles accelerating around, seems like it would behave more like pair B.<br /><br />I imagine somewhere in this chain of thinking I am misunderstanding the theory, but it is not clear to me where.<br />Anonymoushttps://www.blogger.com/profile/13869713289410909812noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-31348409756366763122015-11-11T01:35:20.538-08:002015-11-11T01:35:20.538-08:00AnalyticD: I like your logical reversal of it. Jus...AnalyticD: I like your logical reversal of it. Just a correction: the Rindler volume isn't pushing the star inwards to the galactic centre, it is pulling it out (ie: damping Unruh waves on the right side of the diagram so the star get pushed outwards) to produce what we normally call inertia (caused in MiHsC by this asymmetric Casimir effect). Then, when accelerations are very low the Unruh waves become too long to be affected by the Rindler volume and this asymmetric Casimir effect fails. This is where MiHsC predicts deviations from the inertial norm that cause a slight unexpected push towards the galactic centre and solves the galaxy rotation problem.Mike McCullochhttps://www.blogger.com/profile/00985573443686082382noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-62478238363373768922015-11-10T13:06:51.185-08:002015-11-10T13:06:51.185-08:00A volumetric Rindler horizon, like that of your or...A volumetric Rindler horizon, like that of your orbiting star, exists as a imaginary object with very interesting properties, especially when you reverse the causality. <br /><br />On the real hand, the mass of the galaxy gravitationally pulls the star inward, creating the star's "Rindler volume".<br /><br />On the other hand, running the math in the other direction, the "Rindler volume" is pushing the star inward, while orbiting the galactic center synchronously with the star. At low centripetal acceleration (c/H0), the "Rindler volume" is projected onto the Hubble shell. At higher accelerations, the "Rindler volume" maintains its form as a "shadow" of its star.Analytic Dhttps://www.blogger.com/profile/14307179997233629815noreply@blogger.com