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...