I've suggested (& published in 21 journal papers) a new theory called quantised inertia (or MiHsC) that assumes that inertia is caused by horizons damping quantum fields. It predicts galaxy rotation & lab thrusts without any dark stuff or adjustment. My University webpage is here, I've written a book called Physics from the Edge and I'm on twitter as @memcculloch. Most of my content is at patreon now: here

Tuesday 10 November 2015

How can MiHsC be applied to a hot star?

Peter Reid just asked me this following question: 'For one of those stars near the edge of a galaxy, wouldn't its individual particles still be accelerating quite a bit, since they are part of a seething ball of plasma? It seems like that should make the minimum acceleration not apply to the individual particles, and so not apply to the star as a whole. How do you account for this?'

This is a good question because MiHsC usually only produces anomalies for things at very low accelerations, so how can it predict anomalies for stars with huge thermal accelerations inside? The diagram below explains how. It shows a star (the yellow circle) orbiting the galaxy (the centre of which is to the left), with an orbital speed shown by the arrow pointing up. The schematic shows five hot, highly-accelerated particles inside the star (the red circles) and their acceleration vectors (the black arrows). Each particle has a large acceleration and so a Rindler horizon forms close by in the opposite direction (the black curves) and this horizon affects the particle's own inertial mass due to MiHsC, but since there are a lot of particles and they are moving randomly in the plasma, the black Rindler horizons are distributed randomly around the star and they therefore have no effect on the star as a whole (we'll forget the star's spin for now, which is a small acceleration in comparison). In this schematic there are only five particles, so it may not look like the black horizons quite average out, but in a star there are a very large number so the average will be very good.

Each particle within the star also has a tiny acceleration that it shares with all the other particles due to the orbit of the star in the galaxy. In the diagram this is shown by the light blue arrows, which are all pointing at the galactic centre to the left. The Rindler horizons associated with these smaller accelerations are much further away because the orbital acceleration is ridiculously smaller than the thermal, and these horizons must be far off to the right hand side, something I can't represent well on the diagram (see the blue curves surrounded by the dashed circle). This circled pack of horizons is the composite horizon that MiHsC applies to stars, or any object, as shown in the previous blog. If you consider an object as a whole, you can ignore its particle's individual horizons which average out, and just consider the composite horizon due to its combined acceleration, and figure out how Unruh waves hitting it are sheltered by that.

I've been asked other insightful questions recently, so I'll answer those in following blogs...

4 comments:

Analytic D said...

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.

On the real hand, the mass of the galaxy gravitationally pulls the star inward, creating the star's "Rindler volume".

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.

Mike McCulloch said...

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.

Unknown said...

Wow, I did not expect such a thorough answer. Thanks!

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?


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.

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) .

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.

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.

I imagine somewhere in this chain of thinking I am misunderstanding the theory, but it is not clear to me where.

Mike McCulloch said...

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.