forces - How is potential energy actually stored in a steel spring at the atomic level?

Elasticity is one the most intriguing phenomena, wiki gives a summary explanation of what happens in a steel spring:

the atomic lattice changes size and shape when forces are applied (energy is added to the system). When forces are removed, the lattice goes back to the original lower energy state.

Could you explain in detail how potential energy is actually stored in a steel spring and why the material never surrenders to the bending forces taking a new shape?

Update: it would be interesting if someone might add some technical details:

• is the quantity of energy transformed into heat during the compression $H_C$ equal to that lost during the release $H_R$


• is the heat loss during the two phases due to the exactly same causes, process


• how do you minimize heat loss, one factor is surely the mass, what is the max value of the ratio between energy spent in the compression and the energy retrieved ($c, \gamma$)? I read it can be: $E_R/E > 0.99$


• what determines the ratio between energy retrieved and velocity of rebound (CR)


• what is the max CR achieved to date in a spring, and with what material


• choosing the right material, can the speed of relaxation ever increase with the deflection in such a way to allow an elastic collisions

Answer

You are asking two questions really

1)

How is PE actually stored in a steel spring at the atomic level?

The explanation for this lies in quantum mechanics

2)

Could you explain in detail how/where potential energy is actually stored in a steel spring and why the material never surrenders to the bending forces taking a new shape?

Replying to 1) one has to think quantum mechanically. All solids form lattices. These lattices are formed because even though the individual atoms and molecules are neutral, there are electrical spill over fields due to the fact that the orbitals for the electrons are not spherical, except for l=0 angular momentum states, but have shapes. One can think of them as LEGO shapes that fit, positive spill over to negative spill over forces, forming new quantum mechanical solutions at a lower energy level, releasing energy into soft photons in the process ( that is why precipitation releases energy/heat to the environment). Thus the lattice is balanced at a lower quantum mechanical level than the free atoms/molecules themselves.

The state function of the lattice is a collective one of all the molecules/atoms that compose it, and acts as a higher level "particle", responds collectively quantum mechanically as long as the energies it interacts with are within the energy levels of the collective state. You must remember that in quantum mechanical solutions the lowest energy levels are filled but the higher ones exist and are available if energy is supplied.

What happens if an external force is imposed on the lattice? The collective state function, solution of the quantum mechanical state of the lattice, by absorbing soft photons collectively, moves to a higher lattice energy level. When the pressure is removed, it falls back to its lower energy level by releasing soft photons. These are non destructive deformations. If the energy is too high the lattice bonds will be broken and the lattice solutions will no longer be valid.

Elasticity means that some materials have enough energy levels to the lattice to be able to be deformed and fall back to ground state without being destroyed. The energy is stored in the higher energy level attained by the interaction "lattice forrce". So the steel spring lattice at the atomic level is at a higher energy level . Evidently steel lattices can take a lot of deformation so must have many higher than ground state energies.

Now for the second question, the models are continuum models with parameters that do depend on the quantum mechanical underlying framework but are emergent, from the many body problem. This seems to be still a research problem as it has to do with materials useful for industry and life in general. Models are proposed which use the spring modeling of the other answer at the lattice level, avoiding the quantum mechanical complications by approximations.

In a lattice spring model (LSM), the material is discretised into particles linked by springs. However, LSMs always adopt linear springs, which results in a stiff approximation of the corresponding elastic solution. In this work, a high order LSM is proposed to overcome this limitation by introducing additional degrees of freedoms (DOFs) to the particles.

In this case the internal quantum mechanics dynamics are approximated by springs.

Is there a state beyond gas?

If you could boil water in a sealed container until it became vapor and you still kept applying heat to it would something happen? Maybe gas to super-gas? This has been on my mind for a long time I really hope someone can help out.

Answer

Plasma is described as the 4th state of matter, which is what you get if you give so much temperature that the molecules begin to break up and ionize into positively and negatively charged fragments.

Another Claim on the title '4th State of matter' is a 'supercritical fluid'. Sometimes people draw phase diagrams with it to show this '4th state of matter'. (Strictly speaking Supercritical fluids may be a different phase from liquid, solid and gas and more '4th phase of matter' than '4th state of matter' as pointed out in the good comment from Jim. - the state of a supercritical fluid is gas - whether supercritical fluid is a phase or state maybe something that is up for debate at the moment.)

phase diagram of carbon dioxide showing supercritical phase (from wikipedia)

As can be see from the diagram the supercritical state exists above the critical temperature and pressure.

Supercritical fluids have many interesting properties. For example supercritical water dissolves organic molecules (such as organic solvents), which would not normally dissolve in water. It is also very acid and very alkaline at the same time because there is a high concentration in the liquid of both H$^+$ and OH$^-$ ions.

Supercritical carbon dioxide is used to remove caffeine from coffee beans.

Finally to go back to your question, if you took water and heated it in a sealed container it would 'go supercritical' if it did not break the container first - please don't do this at home, because if the container breaks then you would have alot of very hot dangerous steam --- people who research this for science tend to work with very small volumes for safety.

acoustics - Bell in a vacuum, where does the sound energy go?

I'm not educated in physics, but I learned that a bell struck in a small vacuum chamber will not be heard by people around it (in a school science lab). If it had been surrounded by air, there would have been energy propagating outward from the bell in a sound wave. Because this can't happen, does the bell vibrate harder from the impact? Will it break sooner?

Answer

When a bell vibrates in air, it pushes air molecules out of the way which will make the vibrations "decay". If you strike a bell in vacuum, this loss mechanism will not be there so the bell will "ring" for longer (but nobody can hear it). This doesn't mean the initial amplitude is significantly greater - just that it persists longer. Obviously if you rang the bell repeatedly there is a small chance that the moments of impact would amplify the motion of the bell (resonance) but that's not really very likely. I would not expect the bell to break sooner. On the other hand, for an electrical bell the coils used in the mechanism will heat up, and they would normally be air cooled (not terribly efficient but still - it's a cooling mechanism). It is conceivable that an electrical bell operated continuously in a vacuum would burn out because the cooling is less than in air - but that's pure speculation.

Back to the question of damped vibration, a sketch of what this might look like (arbitrary units of time along the horizontal, and amplitude along the vertical axis):

I am assuming there are "some" loss mechanisms for both - but at least one fewer for the bell in vacuum. The initial amplitude after being struck ought to be (very nearly) the same.

astrophysics - Origin of elements heavier than Iron (Fe)

In all the discussions about how the heavy elements in the universe are forged in the guts of stars and especially during a stars death, I usually hear that once the star begins fusing lighter atoms to produce Iron (Fe) that's the end of the star's life and the whole system collapses onto itself and based on how massive the star was initially it has different outcomes like a white dwarf, a neutron star or a black hole.

I have rarely heard a detailed explanation of how the elements heavier than Iron are produced. I would appreciate a convincing explanation of this process.

Answer

Elements heavier than iron are produced mainly by neutron-capture inside stars, although there are other more minor contributors (cosmic ray spallation, radioactive decay). They are not only produced in stars that explode as supernovae. This has now been established fact since the detection of short-lived Technetium in the atmospheres of red giant and AGB stars in the 1950s (e.g. Merrill 1952), and it is tiresome to have to continue correcting this egregious pop-sci claim more than 60 years later.

Neutron capture can occur rapidly (the r-process) and this process occurs mostly inside and during supernova explosions (though other mechanisms such as merging neutron stars have been mooted). The free neutrons are created by electron capture in the final moments of core collapse. At the same time this can lead to the build up of neutron-rich nuclei and the decay products of these lead to many of the chemical elements heavier than iron once they are ejected into the interstellar medium during the supernova explosion. The r-process is almost exclusively responsible for elements heavier than lead and contributes to the abundances of many elements between iron and lead.

There is still ongoing debate about the site of the primary r-process. My judgement from a scan of recent literature is that whilst core-collapse supernovae proponents were in the majority, there is a growing case to be made that neutron star mergers may become more dominant, particularly for the r-process elements with $A>110$ (e.g. Berger et al. 2013; Tsujimoto & Shigeyama 2014). In fact some of the latest research I have found suggests that the pattern of r-process elemental abundances in the solar system could be entirely produced by neutron star mergers (e.g. Wanajo et al. 2004), though models of core-collapse supernovae that incorporate magneto-rotational instabilities also claim to be able to reproduce the solar-system abundance pattern (Nishimura et al. 2017) and may be necessary to explain the enhanced r-process abundances found in some very old halo stars.

The latest word on this debate is the observations of kilonovae and in particular, the spectacular confirmation, in the form of GW170817, that kilonovae can be produced by the merger of two neutron stars. Observations of the presumably neutron-rich ejecta, have confirmed the opacity signature (rapid optical decay, longer IR decay and the appearance of very broad absorption features) that suggest the production of lanthanides and other heavy r-process elements (e.g. Pian et al. 2017; Chornock et al. 2017). Whether neutron star mergers are the dominant source of r-process elements awaits an accurate assessment of how frequently they occur and how much r-process material is produced in each event - both of which are uncertain by factors of a few at least. A paper by Siegel (2019) reviews the merits of neutron star merger vs production of r-process elements in rare types of core collapse supernovae (aka "collapsars"). His conclusion is that collapsars are responsible for the majority of the r-process elements in the Milky Way and that neutron star mergers, whilst probably common enough, do not explain the r-process enhancements seen in some very old halo stars and dwarf galaxies and the falling level of Europium (an r-process element) to Iron with increased iron abundance - (i.e. the Eu behaves like "alpha" elements like Oxygen and Neon that are produced in supernovae).

However, many of the chemical elements heavier than iron are also produced by slow neutron capture; the so-called s-process. The free neutrons for these neutron-capture events come from alpha particle reactions with carbon 13 (inside asymptotic giant branch [AGB] stars with masses of 1-8 solar masses) or neon 22 in giant stars above 10 solar masses. After a neutron capture, a neutron in the new nucleus may then beta decay, thus creating a nucleus with a higher mass number and proton number. A chain of such events can produce a range of heavy nuclei, starting with iron-peak nuclei as seeds. Examples of elements produced mainly in this way include Sr, Y, Rb, Ba, Pb and many others. Proof that this mechanism is effective is seen in the massive overabundances of such elements that are seen in the photospheres of AGB stars. A clincher is the presence of Technetium in the photospheres of some AGB stars, which has a short half life and therefore must have been produced in situ.

According to Pignatari et al. (2010), models suggests that the s-process in high mass stars (that will become supernovae) dominates the s-process production of elements with $A<90$, but for everything else up to and including Lead the s-process elements are mainly produced in modest sized AGB stars that never become supernovae. The processed material is simply expelled into the interstellar medium by mass loss during thermal pulsations during the AGB phase.

As a further addition, just to drive home the point that not all heavy elements are produced by supernovae, here is a plot from the epic review by Wallerstein et al. (1997), which shows the fraction of the heavy elements in the solar system that are produced in the r-process (i.e. an upper limit to what is produced in supernovae explosions). Note that this fraction is very small for some elements (where the s-process dominates), but that the r-process produces everything beyond lead.

A more up-to-date visualisation of what goes on (produced by Jennifer Johnson) and which attempts to identify the sites (as a percentage) for each chemical element is shown below. It should be stressed that the details are still subject to a lot of model-dependent uncertainty.

special relativity - Is it possible that all "spontaneous nuclear decay" is actually "slow neutrino" induced?

This thought was inspired by a comment from the current leading answer, by @Sentry, to the question Where are all the slow neutrinos?

This [slow-neutrino induced nuclear decay] will still be an extremely rare process and the big problem is to distinguish it from normal spontaneous nuclear decay.

Questions which would need to be addressed as corollaries to the main question, I believe, include:

• 
  

  Is such a possibility self-consistent as a theory?

  
  


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  How does the required energy density of slow neutrinos compare to the postulated energy density of dark matter?

  
  

Answer

Consider the average beta decay, which at nucleon level looks like $$ n \longrightarrow p + e^- + \bar{\nu} \,. \tag{1}$$ The distribution of electron energies (as measured in neutron's frame) is controlled by the phase space of the products. We observe an electron energy spectrum consistent with these physics.

What you propose is essentially that this reaction is properly described by $$ n + \nu \longrightarrow p + e^- \,. \tag{2}$$ with a very low energy neutrino. (As an aside, that reaction with high energy neutrinos is seen in accelerator and atmospheric neutrino experiments.)

However, the energy distribution of the electron in the final state of EQN (2) (again, measured in the neuton's rest frame) would be controlled by the incident neutrino's momentum. It could only look like the observed spectrum if the energy-spectrum of the incident neutrinos were like those predicted for the outgoing neutrino in EQN (1). But as those neutrinos have energies of many MeV (depending on the particular decay) they are in no way slow.

Worse, weak universality works using the same effective coupling constant (the Fermi constant) for reactions involving a neutrino in the initial state as for those involving an anti-neutrino in the initial state. (And likewise for (anti-)neutrinos in the final state). So now you need not only a conspiracy to get the right energy spectrum for the neutrinos, but the conspiracy must insure the same number and spectrum for anti-neutrinos as well, despite the nearly factor of two difference in the abundance of quarks for these two kinds to interact with at low energy.

Short answer: No, it's not possible. Not even for weak decays.

potential - Why is the work done on a charge calculated from infinity?

Why is the work done on a charge calculated from infinity to a point? Why not from one particular point to other?

Answer

Consider the form of the potential energy between two point charges in the case that I use a reference distance $r_0$ as the zero (written here in SI units). $$ U_{r_0} = \frac{q_1 \,q_2}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{r_0} \right) \;.$$ This is quite general, but it will get to be very messy to write down and manipulate very quickly indeed. It also means that the sign of the energy depends on the the relative sign of the charges and the relative size of $r$ and $r_0$

Now, the special case of taking $r_0$ as arbitrarily distant, gets us the familiar form \begin{align*} U_\infty &= \lim_\limits{r_0 \to \infty} U_{r_0}\\ &= \frac{q_1 \,q_2}{(4 \pi \epsilon_0) r} \;, \end{align*} which is algebraically simpler and the sign of which can be known at any distance just from the relative sign of the charges.

The conventional form is simply easier to use in the majority of cases.

But it gets better, because the same kind of consideration applies to Newtonian gravitation, and the convention of zero energy at infinite remove means that the total energy bound bodies is negative while that of free bodies is positive (with zero the parabolic edge case).

It really is a natural choice after you've looked at the ways you're going to be using the quantity.

newtonian gravity - What would happen if I were to fall into a massive hollow planet?

Imagine, if you will, there exists somewhere out there a supermassive planet. For some reason, this planet is only a shell of its former existence and all that is left is a crust of substantial size to keep it from collapsing in on itself (let's imagine it is stable at 5000' thick). Now, this planet was HUGE! How large, you ask? Large enough that the remaining shell is still able to maintain a gravity well strong enough to keep you anchored to the surface.

Pretty neat, huh?

Well, since this is a sight to be seen by everyone, you stroll along and, not looking where you are going, you tumble into a chasm and straight down into a hole that permeates the thickness of the shell. As you are falling down and down and down, you think:

"At what point do I stop falling down and I start falling up? And would I be able to retain the velocity, since gravity is a conservative force, and fall back through another hole on the surface on the opposite side? Where would the center of gravity even exist? I mean, aaaaahhhhhhh!"

What say you, browsers and scientists? A stretch of the imagination, sure, but is it physically possible? Could there exists a configuration similar to this where gravity could be concentrated to a point but still allow free travel up to the focus?

Answer

Yes, this is possible. It is perfectly fine for a mass configuration to

• 
  

  produce, for points outside a sphere of radius $R$ centred at $\mathbf r_0$, a gravitational field identical to that of a point mass at $\mathbf r_0$, and still

  
  


• 
  
  

  be completely empty inside a smaller sphere of radius $a$ around $\mathbf r_0$.

  
  

The spherical-shell model you describe is the simplest example of this. (On the other hand, it does not describe a physically realizable model from a materials standpoint, as described in Rod Vance's answer.)

In addition to this, your spherical shell of a planet will have the additional property that once you are inside the shell, the planet's gravitational attraction will vanish. Objects there will perform uniform linear motion with constant velocity, until of course they reach the edge of the shell.

This has been known since the time of Newton and it is one of the standard exercises in electrostatics (which is mathematically identical to newtonian gravity) for a physics undergrad. Schemes like this are fairly common and relatively fun to analyse; you may find the "gravity train" interesting.