AmalgamatedIllusions

joined 1 year ago
[–] [email protected] 6 points 2 months ago (2 children)

Drew mentions this and points out that it's a new OS design and will therefore take a long time. He argues that an OS based on the linux design would be much easier.

[–] [email protected] 2 points 3 months ago

Charge conservation would unambiguously be violated, which is why this decay is not expected. The half-life you quote is an experimental lower-bound.

[–] [email protected] 3 points 3 months ago (1 children)

Charge conservation would indeed be violated, which is why this decay is not expected. Dave is mistaken: the half-life they're referring to is an experimental lower-bound, not a actual expected value.

[–] [email protected] 2 points 3 months ago

They are not expected to decay. The half-life they're thinking of is a lower-bound based on current measurements, not an actual expected half-life.

[–] [email protected] 5 points 5 months ago

Not all radio noise is from the CMB. There's also thermal noise, though this would be minimized too if our hypothetical radio at the end of time is near absolute zero.

[–] [email protected] 8 points 5 months ago* (last edited 5 months ago)

One clarification: electric charge, angular momentum, and color charge are conserved quantities, not symmetries. Time is a continuous symmetry though, and its associated conserved quantity is energy.

Similarly, information isn't a symmetry, but it is a conserved quantity. So I assume you're asking if there's an associated symmetry for it from Noether's theorem. This is an interesting question: while Noether's theorem ensures that any continuous symmetry will have a corresponding conserved quantity, ~~the reverse isn't necessarily true as far as I know.~~ In the case of information conservation, this normally follows naturally from the fact that the laws of physics are deterministic and reversible (Newton's laws or the Schrodinger equation).

If you insist on trying to find such a symmetry, then you can do so by equating conservation of information with the conservation of probability current in quantum mechanics. This then becomes a math problem: is there a transformation of the quantum mechanical wavefunction (psi) that leaves its action invariant? It turns there is: the transformation psi -> exp(i*theta)*psi. So it seems the symmetry of the wavefunction with respect to complex phase necessitates the conservation of probability current (i.e. information).

Edit: Looking into it a bit more, Noether's theorem does work both ways. Also, the Wikipedia page outlines this invariance of the wavefunction with complex phase. In that article, they use it to show conservation of electric current density by multiplying the wavefunction by the particle's charge, but it seems to me the first thing it shows is conservation of probability current density. If you're interested in other conserved quantities and their associated symmetries, there's a nice table on Wikipedia that summarizes them.

[–] [email protected] 3 points 5 months ago

I suspect you may be misunderstanding time dilation. From the perspective of a particle, time always passes by at 1 second per second. If you yourself were to travel at relativistic speeds (relative to, say, Earth) your perspective of time wouldn't change at all. However, observers on Earth would see your "clock" to tick slower. That is, anything you do would progress more slowly from their perspective. In the very early Universe, a given particle would see most other particles moving at relativistic speeds, and so would see their "clocks" tick slower. These sorts of relativistic effects would influence interactions between particles during collisions, decay rates, etc, but are all things we know how to take into account in our models of the early Universe.

[–] [email protected] 5 points 5 months ago* (last edited 5 months ago)

The required temperature depends on the mass of the particles you're considering. You could say photons are always relativistic, so even the photon gas that is the cosmic microwave background is relativistic at 2.7 K. But you're presumably more interested in massive particles.

If you apply the kinetic theory of gases to hydrogen, you'll find that the average kinetic energy will reach relativistic levels (taken to be when it becomes comparable to the rest mass energy) around 10^12^ K. For the free electrons (since we'll be dealing with plasmas at any sort of relativistic temperatures), this temperature is around 10^9^ K due to the smaller mass of the electron. These temperatures are reached at the cores of newly-formed neutron stars (~10^12^ K) [1] and the accretion disks of stellar-mass black holes (~10^9^ K) [2], but not at the cores of typical stars. Regarding time dilation, an individual particle's clock would tick slower from the perspective of an observer in the center-of-mass frame of the relativistic gas, but I don't think this would have any noticeable effect on any of the bulk properties of the gas (except for the decay of any unstable particles). Length contraction would probably affect collision cross-sections, though I haven't done any calculations for this to say anything specific. One important effect would be the fact that the distribution of speeds would follow a Maxwell–Jüttner distribution instead of a Maxwell-Boltzmann distribution, and that collisions between particles could be energetic enough to create particle-antiparticle pairs. This would affect things like the number of particles in the gas, the relationship between temperature and pressure, the specific heat of the gas, etc.

You mention the early history of the Universe in your other comment. You can look through this table on Wikipedia to see the temperature range during each of the epochs of the early Universe, as well as a description of what happened. The temperatures become non-relativistic for electrons at some point during the photon epoch.

[1] https://doi.org/10.1063%2F1.4909560

[2] https://doi.org/10.1016%2Fj.isci.2021.103544

[–] [email protected] 6 points 7 months ago* (last edited 7 months ago) (1 children)

I only have surface level knowledge of String Theory, but my understanding is that strings vibrate in simple harmonic motion and that different frequencies correspond to different particles. Since idealized springs are simple harmonic oscillators, you could perhaps say that, in some sense, the strings in String Theory are springs.

But maybe that's what inspired your question. If you're asking why they can't be springs in a more literal, geometric sense, then I would speculate that it's related to the world sheet that a spring would trace out as it propagates through spacetime. A world line describes a trajectory of a point particle not just through space, but through time as well - thereby describing the history of the particle's motion. In quantum field theory, these world lines are used in Feynman diagrams to describe interactions between particles. However, these diagrams always have sharp interaction vertices. In other words, the interaction occurs at a specific point in spacetime, which is problematic in terms of relativity (different observers should not need to agree on when a spacetime event occurred). For reasons I don't understand, this can give rise to infinities (ultraviolet divergences) when doing certain calculations. These have to removed through renormalization, but apparently this doesn't work when trying to develop a quantum theory of gravity.

In the case of a one-dimensional object like a string, instead of tracing out a world line, it traces out a two-dimensional surface called a world sheet. A consequence of this is that the sharp vertices of Feynman diagrams disappear: while an interaction did occur globally, it did not occur at a specific point in spacetime (different observers will see the event occur at different times, so no relativity issues). This eliminates the ultraviolet divergences and the need for renormalization (again, apparently), allowing for a full quantum theory of gravity. If you were to change the geometry of the strings to something more spring-like, my guess is you would no longer get this nice behavior.

[–] [email protected] 3 points 8 months ago

This is the first I've heard of the effect Mars has on Earth's Milankovitch cycles (unsurprising, given that the paper is recent and the effect is quite small with a very long period). Earth presumably has a similar effect on Mars, but measuring this would be quite difficult. Keep in mind that we're able to do this for Earth by analyzing drill cores (that paper uses data from 293 scientific deep-sea drill holes), which we can't really do for Mars currently. Using other methods, we've been able to measure the effects of axial tilt and precession for Mars, but the effect from orbital interactions with Earth would be much more subtle. I'd be surprised if you could find anything on it in the literature.

I also would not expect the Moon to make much of a difference. The Earth-Moon distance is <1% of the Mars-Earth distance even at closest approach, so the Earth-Moon system is essentially a point mass to first order. Additionally, the mass of the moon is ~1% that of the Earth, so the effect there is quite small as well. As I mentioned, measuring Earth-Mars Milankovitch cycles is already difficult for Earth (we apparently only recently did so) while likely infeasible for Mars (currently), and detecting the effects from the Moon would be harder still.