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This is more or less Roger Penrose's position: https://web.archive.org/web/20150327124054id_/http://image.s...

He argues that the wavefunction collapses on a timescale of h-bar / E_G, where E_G is the gravitational self-energy of the quantum system.


Didn’t know this but glad I’m in such distinguished company.

I've heard this claim, but I don't think it holds up. The idea is that supposedly the calculation is done in parallel across the different universes which allows you to see an exponential speedup in some calculations. But the whole point of many-worlds is that after a branch event the various branched portions of the wavefunction no longer interact. Even if it were true that quantum computing was really a bunch of parallel classical computations (which it is not), there would be no way for any of the other universes to communicate the result back to the universe you observe it in.

Many worlds doesn't really feature 'hard branches'. In fact the interpretation is pretty specific that quantum interactions becoming de-correlated and stopping to react is the mechanism by which you get the 'many worlds', and this is a soft and messy process not some global hard split.

OTOH, I don't think any of the other interpretations really struggle to explain quantum computing so I agree it's not much of a distinguishing factor.


I think a lot of people have an understanding that the force that binds the nucleus together is quite a bit stronger than the electromagnetic force. (Obligatory xkcd: https://xkcd.com/1162/)

What is not always appreciated is that the forces binding the nucleus together are the strong force analog of the van der Waals force in electromagnetism. So it is not really right to compare the way the electron is bound to the nucleus to the way the nucleus itself is bound together. If you want to compare the electromagnetic force to the strong force, the better comparison is between the force binding the nucleus to a hydrogen bond.

The strong force itself is many orders of magnitude stronger than the force binding the nucleus together.


The outer boundary is the point where a photon has a ~50% chance of escaping without encountering another particle.


Do you have a source? I'd love to read more. :)


Amateur stargazer's understanding:

Depending what kind of reading material you're looking for (e.g. high level details or mathematically/jargon dense papers) look for things discussing "Rosseland optical depth" and "grey atmosphere approximation".

At a high level, the common convention is to define the radius by finding where the optical depth is 2/3 when using Rosselands clever way of calculating a weighted mean of the opacity from that layer in the star to space across different wavelengths of light. 2/3 being a clever derivation from Eddington where, in an idealized model of a star, that's when the actual temperature of the star should equal its blackbody equivalent temperature.

Pedantically, this distance to the point of equality is an ever so slightly different value than the "distance from the center where there is a 50% chance a photon traveling directly outward will escape the star without another interaction" rule of thumb (in the same idealized grey model). Practically, that the difference is so small is why it's a fantastic rule of thumb explanation.


I'd recommend Carroll & Ostlie's Introduction to Modern Astrophysics, though it requires some basic calculus and physics.


Thanks!


We have constraints that primordial black holes must be less than the size of a large asteroid. They would be gravitationally insignificant to the Earth unless they actually passed through it.

I also don't think it would really be detectable in the Sun. Perhaps there would be some transient astroseismology signature.


They did some modeling in a previous paper of how this would impact Ia supernovae and it depends on whether the supernova is driven by Kelvin-Helmholtz instabilities or not. In the case that it is, this would only trigger Ia supernovae for white dwarfs which are already very close to the Chandrasekhar mass. In this case the standard candles would be pretty much the same.

In the case that KH instabilities are weaker they claim that you can get supernovae at masses farther from the Chandrasekhar limit. This could plausibly have more of an impact on Ias as standard candles. (Though perhaps less than you might think becauseit would also likely have an effect on the shape of the light curve, and Ias are normalized based on their light curve shape.)

All that said, even if this process does occur, I doubt the rates would be high enough for it to be a substantial fraction of observed Ias.


To paraphrase the old joke, if you wake up with a $78,000 AWS bill, you have a problem. If you wake up with a $78 million AWS bill, Amazon has a problem.


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