mathematical physics - What makes an abstract physical system describable by a "fluid" equations of motion?

We can describe (some of) the dynamics of many systems using fluid mechanics. Of course these include classical fluids like water, more exotic fluids like photon gases and the universe as a whole and even solid(ish) things over long times, like glasses and ice. Further still we can treat general classical systems in phase space using a phase fluid, quantum systems (e.g. Madelung equations) and with a little bit of hand waving anything that happens on a symplectic manifold (which gives us a Hamiltonian and hence a flow and Louiville's theorem).

So what is it about a system that makes it obey a fluid model? Are there systems that definitely do not fit a fluid model?

(I realise that I haven't said exactly what I mean by "fluid model", this is somewhat deliberate. If you like you can take it to mean "having equations of motion which are (almost) identical in some form to the Euler equations".)

edit: Since admittedly, the original question wasn't quite clear I'll try to clarify a bit. I'm not looking for a high-school answer, or one which describes only actual literal physical fluids, e.g. "a fluid doesn't support a shear stress", "a fluid is something you can wash your hair with". I've given some examples above of situations that fit this description, and without further explanation I think it's non-obvious what a "mean free path" or similar would mean in a generalised fluid. What I'm really looking for (and there may not be any) is some overarching physical or mathematical principle, or failing that an argument as to why there isn't one. I'd even be quite happy to be directed to a book or more appropriate forum. I apologise for not being clearer before and appreciate the answers already given.

general relativity - If the Universe is Flat, has Finite Mass/Energy, and is Simply Connected, Then there MUST be an Edge, Mustn't there?

Assumptions:

The universe is flat (currently supported)

The universe is simply connected (the edges aren't glued together as in a torus)

The universe contains finite mass and energy

Conclusion:

The universe must have an edge.

Yes, there is a similar question here: How can the universe be flat and have no center if universal mass-energy content is finite?

But my question is not answered. In fact, people are neatly dodging the notion of an "edge" by suggesting "unusual topologies"

This is a purely hypothetical question, but since everyone says the universe has no edge and is flat, I am forced to ask the obvious: space or some form of truly empty vacuum might go on forever, but if matter/energy are finite in the universe, then eventually, if we travel far enough past the cosmic horizon, we'll find that there are no more stars, no more galaxies, no more photons... and no more anything. Unless the universe is actually a sphere, in which case eventually we'll end up back where we started.

Is there a flaw in my reasoning? I must have read 100 articles today to get to the bottom of this.

Answer

Yes, if the universe is:

• 
  

  flat (zero spatial curvature)

  
  
  


• 
  

  has finite mass energy (since we know it is uniform this also means it is bounded. If you drop the bounded es because you don't want to admit uniformity or otherwise, i.e., if it is unbounded, then the answer is clearly no)

  
  


• 
  

  is simply connected (has what is called a trivial topology)

  
  

Then it does have to have an edge.

See the zero curvature and other sections of the wiki article on the shape of the universe, it's fairly complete, at https://en.wikipedia.org/wiki/Shape_of_the_universe

The simply connected condition is critical also. If you allow other topologies then both the torus and the Klein bottle topologies are bounded, flat and have no edges.

There are a total of 17 possible different topologies for multiply connected spaces that are flat, in 3D (our spatial dimensions, which is what is referred to when one talks about curvature of the universe) Riemannian space. See fig. 4 in the arXiv paper at https://arxiv.org/abs/0802.2236 for all of them. There are others if the space is not flat.

As far as space being unbounded but mass energy finite, that would violate what we know of the homogeneity and isotropy of the universe. From the CMB we see the (large) scale homogeneity and isotropy. Now, we only see back to 380,000 years after the Big Bang, but no sign of large inhomogeneities. It could theoretically still be true that out inflation bubble is homogeneous, and thus the part of the universe beyond our particle horizon might not be, but there is no theoretical reason to think so. The more prevalent view is that it was as uniform more or less, and the same inflation that created our bubble might have created others. If we ever fully understand our inflation (which at this point looks pretty consistent with observations but those don't rule out various versions, or other unknown mechanisms from an unknown theory of quantum gravity), we might find out better or differently. But presently, a large scale homogeneity with possible bubbles is consistent with all observations.

standard model - Understanding type of force interaction in particle decays

Are there any fundamental rules of thumbs that can be used to identify the type of force interaction (weak, electromagnetic, strong) in a particle decay without drawing the Feynman diagrams at the beginning of the problem. At least to rule out some types of forces from a given decay reaction, before start drawing the Feynman diagrams.

Answer

• 
  

  If a particle changes flavor, it's a charged-current weak decay. Example: $n\to pe\bar\nu$.

  
  


• 
  

  If there's a neutrino in the final state, it's a weak interaction. Decay example: $\pi^+\to\mu^+\nu$. See also neutrino scattering.

  
  


• 
  
  

  If parity isn't conserved, it's a weak interaction. Examples: $K^0 \to 2\pi$ and $K^0 \to 3\pi$. Note that kaon decays and $K\leftrightarrow\bar K$ oscillations also change strangeness, a flavor quantum number.

  
  


• 
  

  If only hadrons are involved, and all the flavor quantum numbers are conserved, it's a strong interaction. Examples: $pp \to p\Delta^{++}\pi^-$, $pp \to p\Lambda^0 K^+$.

  
  


• 
  

  If photons are involved, it's an electromagnetic interaction. Examples: $\pi^0\to\gamma\gamma$, $e^+e^- \to \gamma\gamma$, $\gamma + {}^AZ \to {}^{A-1}Z + n$

  
  

This isn't an ironclad rulebook, because the separation of interactions into strong, weak, and electromagnetic is something we can do artificially since we live in a world where the intrinsic energy associated with each interaction is orders of magnitude different. All the fundamental interactions contribute, at some level, to all of the decays.

For example deuterium formation with cold neutrons ($np\to d\gamma$) is a transition from a strong state (unbound $np$, isospin 1) to a strong state (bound $np$, isospin 0). This transition liberates a magnetic dipole photon because no hadronic degree of freedom exists to carry away the binding energy; the strong matrix elements are presumably the same as in a process like $\pi^0d \to np$.. The photons in $np\to d\gamma$ have a tiny parity-violating asymmetry due to the weak neutral current acting between the neutron and proton.

Why does the Sun's (or other stars') nuclear reaction not use up all its "fuel" immediately?

The temperature and pressure everywhere inside the Sun reach the critical point to start nuclear reactions - there is no reason for it to take such a long time to complete the reaction process.

Just like a nuclear bomb will complete all the reaction within $10^{-6}$seconds.

Why does most of the hydrogen of the Sun still not react even though it reaches the critical point, and why take stars billions of years to run out of fuel?

quantum mechanics - Bose Enhancement Factor

How may one explain the fact that the probability of a boson transferring to a state with an occupation number n is 'enhanced' by a factor of (1+n), compared to the classical case? (In the classical case, the probability is supposed to be independent of the occupation of the final state.)

Does light color change when refracting?

When light refracts from a medium to a second one, its frequency stays the same, and its wavelength changes. If this is true, why we see the refracted light ray's colour is the same as the incident light ray in the second medium? The colors should not be the same. If the wavelength changes, colour should change too.

Answer

The color will not change. What you're not taking into account is the speed of light in the medium. It's not the same $ c $ en vacuo. The frequency stays the same. What changes is that speed of light in the refracting medium and as a result wavelength. This difference for speed is the exact reason we have refractive effects, and I believe was the observation that led to Snell's Law. In symbols $$ \lambda = \frac {c} {\nu} $$ where $\lambda$ is the wavelength and $ \nu $ is the frequency.

The speed of light changes because the photons have to have its energy ( and therefore it's presence ) propagated across very long molecular chains. Electrons have to absorb the incident photons, re-emit, and repeat this in a longitudinal direction. Depending on what the material is made of, this will take variable time in variable media. That notion manifests itself in the different indices of refraction that different objects have.

quantum mechanics - What keeps electrons in an atom from flying away or falling into the nucleus?

In atoms, what force or charge, etc. keeps electrons from flying away or into their nucleus? is there a kind of weak-force at work on the atomic scale?

Note I am aware the electron positions are only abstract variables and can be referred to as the electron field and the like. This is not the question.

What is the reason an electron is bound to that nucleus to the point it can sustain an "orbit" or variable probability path, and not fly away or into it's nucleus?