theory The Chemistry of God Metals

[skaa]

On 12/13/2016 at 7:11 AM, Pagerunner said:

I know people have proposed similar things (like atium is Ruin’s electrum) based on allomantic behavior, and I didn’t like them at the time (the ideas and logic behind them, not the people), but the way Brandon spoke of Harmonium makes me think that approach was on the right track. However, it’s on a more fundamental level: the very protons and electrons are either of Ruin or Preservation, and when you combine the two, you get a very reactive metal.

That would be me. To be fair, I did not have any way of knowing harmonium's chemical properties at the time.

On 12/14/2016 at 3:48 AM, Pagerunner said:

But there are some good points to praseodymium and neodymium, considering both aren't found in their pure form in earth's crust and the former makes a green oxide when exposed to air (my big hang-up with cobalt, since I can't find an oxide that matches lerasium's greenish coating).

The idea that lerasium is green is something I've been trying to verify ever since @Phantom Monstrosity asserted it three years ago. I've yet to find evidence of it. I've re-read the Well scene several times. Nothing green was mentioned there. Apparently a bunch of people believe it's green without knowing why. Perhaps you could share your source?

[skaa]

2 hours ago, FiveLate said:

Well we have have electromagnetic force here.    Gravitational force on Roshar.   Now we just need to figure out how the Strong and Weak Shardic (nuclear) forces work to develope a Unified Theory of Sanderson.

You might be interested in this (warning: Stormlight Archives spoilers!).

Back on topic, I've just had a thought. @Pagerunner has already mentioned how the Realmatic effects of two opposing Shardic powers sharing a single atom could make harmonium more reactive. This reminded me of how relativistic effects can make certain heavier elements deviate from some trends in the Periodic Table (e.g. this is why cesium has a golden hue while the other alkali metals are silvery white). And it got me wondering: what if the atomic number of harmonium isn't even that of an alkali metal, but it acts like an alkali metal only because of Realmatic effects?

Pagerunner, you're the chemical engineer, so tell me what you think of this idea: Suppose harmonium is a non-alkali metal, would it make sense to say that the repulsive forces between the Ruin particles and the Preservation particles could somehow be strong enough to give a harmonium atom only one valence electron?

I'm asking because we know that certain Scadrians are planning to create a bomb that can destroy a city, and I'm starting to think a magical uranium-235 (with 117 Preservation nucleons and 118 Ruin nucleons) could be handy when they figure out E=mc^2.

Interestingly enough, "finely divided uranium" reacts with cold water (according to Wikipedia), so maybe giving it a Realmatic push towards more reactivity is enough to make it a "super cesium".

Pardon me if anything I said above sounded stupid. I am not a chemist, just someone relying on Wikipedia.

Edited December 15, 2016 by skaa

[SpeakoftheDeval]

Though this tickles my happy science magic spot, i cant help but feel that brandon has something bigger planned with the extra ruin. The way hes been teasing us... i dont know. It could be red herring but i want it to be more. I still really love this. Upvotes!!!

[Pagerunner]

10 hours ago, skaa said:

The idea that lerasium is green is something I've been trying to verify ever since @Phantom Monstrosity asserted it three years ago. I've yet to find evidence of it. I've re-read the Well scene several times. Nothing green was mentioned there. Apparently a bunch of people believe it's green without knowing why. Perhaps you could share your source?

I, uh... have no idea where I got that color information from. I may have picked it up from that thread, originally; I just don't remember. I looked through the Well scene, the deleted Well ending, glanced at the opening couple of chapters of HoA with Vin and Elend, and the last dozen-or-so epigraphs of HoA, and I didn't see anything. If it wasn't described as such in the well scene, maybe it's in a later mention in Hero of Ages, where they're thinking back on it or recounting it to someone. I'll have to do some more looking. But it certainly would make things easier of lerasium wasn't green. (Although gold wouldn't fit cobalt any better.)

7 hours ago, skaa said:

Back on topic, I've just had a thought. @Pagerunner has already mentioned how the Realmatic effects of two opposing Shardic powers sharing a single atom could make harmonium more reactive. This reminded me of how relativistic effects can make certain heavier elements deviate from some trends in the Periodic Table (e.g. this is why cesium has a golden hue while the other alkali metals are silvery white). And it got me wondering: what if the atomic number of harmonium isn't even that of an alkali metal, but it acts like an alkali metal only because of Realmatic effects?

Pagerunner, you're the chemical engineer, so tell me what you think of this idea: Suppose harmonium is a non-alkali metal, would it make sense to say that the repulsive forces between the Ruin particles and the Preservation particles could somehow be strong enough to give a harmonium atom only one valence electron?

I'm asking because we know that certain Scadrians are planning to create a bomb that can destroy a city, and I'm starting to think a magical uranium-235 (with 117 Preservation nucleons and 118 Ruin nucleons) could be handy when they figure out E=mc^2.

Interestingly enough, "finely divided uranium" reacts with cold water (according to Wikipedia), so maybe giving it a Realmatic push towards more reactivity is enough to make it a "super cesium".

Pardon me if anything I said above sounded stupid. I am not a chemist, just someone relying on Wikipedia.

The specific avenue you suggest... I'm gonna say it involves a lot more screwing with physics to actually change the energy levels. Electron shielding gives the periodic table its shape, so if the new electron shielding is too powerful, then d-block electrons (a.k.a. all the transition metals' valence electrons) might behave very differently, and might make them look more like post-transition metals or metalloids. I'm hesitant to go too far in this direction, since my education is even a step removed from the chemistry of it, much less the physics. The underlying idea, though, that anything made of a mixture of the two condensed opposing Investitures would be more reactive, is valid nonetheless; harmonious uranium would be more reactive than regular uranium. But, again, Brandon specifically told me that harmonium's behavior was not a nuclear property, that it's an 'electron thing,' so that approach might be going about it backwards. Let's get a nucleus that fits with the observed reactivity, not see how we can fit the observed reactivity into a desired nucleus.

You can gauge how violently different metals react with water by comparing tabulated ionization energies, on Wikipedia. Lower ionization energies correspond to faster reactions (that's not the whole story, but a big part of it, and if we're just comparing reactions with water then that's most of what we need for this specific discussion). It's not linear, it has to do with changing the bounds of integration for the Maxwell-Boltzmann distribution graph for that temperature, and technically I don't think water molecules follow that energy distribution plot, but whatever. Less energy required to remove means it happens more quickly, and I expect rate will be roughly exponential with respect to ionization energy. Cesium has the lowest first ionization energy, at 375 kJ/mol. It's 1/16 less (d'oh!) than the next alkali metal up, rubidium, at around 400 kJ/mol. Uranium is at 600 kJ/mol, (actually, pretty comparable to calcium), so it would need some very significant effects on its electron cloud to react quicker than cesium. If I were Brandon, I'd be worried about breaking physics if I were adding a new force with this magnitude into the atoms. So, that's why I lean with it being an actual alkali metal whose reactivity has been slightly enhanced, rather than another metal that has been made extremely more reactive.
 

9 hours ago, FiveLate said:

Well we have have electromagnetic force here.    Gravitational force on Roshar.   Now we just need to figure out how the Strong and Weak Shardic (nuclear) forces work to develope a Unified Theory of Sanderson.

I don't think this is an enhancement of the electromagnetic force; I think it's a new force that exists in tandem with the electromagnetic force, derived from the opposition between Ruin and Preservation. Gravity is technically still affecting the electrons around a nucleus in an atom; it's just that the forces are so small compared to the electromagnetic forces, they're negligible. I think this Realmatic force operates like the electromagnetic force, on a similar scale, and causes similar effects. Close, but not quite identical.

I wonder though, now that you mention it, if there is a 'gravitic' Investiture attraction as well, a force without dipoles that causes Investiture to congregate in Perpendicularities.

Edited December 15, 2016 by Pagerunner

[+robardin]

This is a really cool line of thought. On a much simpler one though, also prompted by "ettmetal + water = boom" and what that paralleled in our so-called "real" world ;), the obvious question is, "what is left after the reaction?"

Using the Cesium example, the chemical equation (that I looked up) is:

    2Cs(s) + 2H₂O(l) → H₂(g) + 2Cs⁺(aq) + 2OH⁻(aq)

So, you mix solid cesium and liquid water, get a big boom, and are left with some hydrogen gas, Cesium ions, and hydroxide ions.

Is it possible on Scadrial that while harmonium may not be (easily) separated into lerasium / atium, the "ionic leftovers" of such a reaction might be? (Too simple, of course, for the positive and negative "ions" from adding water to harmonium were themselves lerasium and atium, or the Southern Continent would be replete with unstoppable full Mistborn warriors.)

Since the title of the next Mistborn book is projected to be "The Lost Metal", we may soon find out.

 

 

[YungDankBlast]

On 12/12/2016 at 3:11 PM, Pagerunner said:

...(although we do know that at this point, if Sazed died, he would drop a single powerful Shard.)...

Do we know this from Brandon?  If so, would someone please link me?

[Pagerunner]

6 minutes ago, CayJoBla said:

Do we know this from Brandon?  If so, would someone please link me?

http://www.theoryland.com/intvmain.php?i=979#174

[skaa]

14 hours ago, Pagerunner said:

But, again, Brandon specifically told me that harmonium's behavior was not a nuclear property, that it's an 'electron thing,' so that approach might be going about it backwards. Let's get a nucleus that fits with the observed reactivity, not see how we can fit the observed reactivity into a desired nucleus.

Well, yes, it's clear that the explosive effect of adding water to harmonium is due to the mechanism that you explained in the OP. That's why when the Set's researchers tried building a bomb out of it, they only managed to create something as powerful as dynamite. Because the explosion wasn't nuclear.

But if the reference to city-destroying bombs ("they needed something that could end cities") was a hint from Brandon that harnessing the nuclear energy of harmonium will be key to future Scadrian technology, then it would make sense for Brandon to use uranium (or perhaps plutonium) as the base element for harmonium, because that's one kind of fissile material his readers would be familiar with.

I admit it's also possible that Brandon wants a magical cesium that is fissile. So in the end we go back to the test you offered: Which one breaks physics less? A variant of fissile uranium that acts as an alkali metal? Or a variant of cesium that can be used as fissile material?

Edit: That line about cesium's first ionization energy being ~1/16th less than that of rubidium is very nice. Upvoted you for that.

Edited December 16, 2016 by skaa

[HK-42]

26 minutes ago, FiveLate said:

Would you definitely get hydrogen gas, or would it perfect to stay as hydronium? My real-world chemistry is a bit weak on this since most of my chemistry was technical chemistry having to deal with double buffer solutions to maintain a specific basic pH in pressurized water reactors...

When a Group I metal (cesium, sodium, etc) reacts with water you definitely get a hydroxide salt and hydrogen gas. This creates a basic (alkaline) environment. Hence the name alkali metal. 

26 minutes ago, FiveLate said:

Second....I was under the impression that the ettmetal was used up in the process....burned away....so there has to be more going on bc otherwise you could boil off the water and have harmonioum hydroxide which should be able to be returned to pure harmonious if I recall correctly....

Well, they could mean "burned up" as in the explosive reaction resulting in harmonium hydroxide. To someone who isn't familiar with chemistry, they might not be able to distinguish this from water. Depending on the abundance of harmonium, it probably wouldn't be energy intensive to try and return ettmetal back to its elemental state. I've never heard of anyone regenerating cesium from CsOH. If I remember correctly, you can react NaOH with molten magnesium to yield sodium metal, but it definitely is not worth the effort in terms of cost. 

It is a lot easier to return some metal hydroxides to elemental forms though, like copper hydroxide to elemental copper. Very simple process, but still not energy intensive unless metal is very rare. 

Edited December 16, 2016 by NavySealsGuy

[Pagerunner]

Alas, ninja'd in my own thread! Whatever, I'll post it anyways.

16 minutes ago, FiveLate said:

Would you definitely get hydrogen gas, or would it perfect to stay as hydronium? My real-world chemistry is a bit weak on this since most of my chemistry was technical chemistry having to deal with double buffer solutions to maintain a specific basic pH in pressurized water reactors...

 

Second....I was under the impression that the ettmetal was used up in the process....burned away....so there has to be more going on bc otherwise you could boil off the water and have harmonioum hydroxide which should be able to be returned to pure harmonious if I recall correctly....

Hydronium is from protons, H+, hydrogen ions. This reaction will transfer the positive charge from protons in the solution (pure water has 1E-7 M hydronium) to the harmonium, leaving behind hydrogen radicals which will combine to form H2 gas.

Unless matter is created or destroyed in this process (one possibility that I only briefly touched on, that harmonium might not actually form oxides, and when it loses an electron the whole atom destabilizes and releases energy and Investiture), then the harmonium atoms aren't going anywhere, and they will be bound as harmonium hydroxide after the reaction. Recovery, however, would require as much energy in as you get out. It's kind of like cracking methane (CH4) with steam (H2O) to get carbon dioxide (CO2) and Hydrogen (H2), something my company does. That reaction takes place in a furnace that gets above 1000 degF. So, any practical way of cracking harmonium hydroxide after separating it from the environment (remember, it explodes) would require a ton of energy itself, and not really be feasible. You could get various harmonium salts through ion exchange, but any way of getting electrons back to the metal would be more effort than it's probably worth.

[skaa]

3 hours ago, FiveLate said:

If anything......I think it would have to be Uranium 256......simply bc that is 16 preservation x 16 ruin.....

Well, I was trying to incorporate @Pagerunner's idea of one extra Ruin nucleon in harmonium, but I guess at this point your idea is as valid as any, though note that uranium-256 is not one of the known isotopes of uranium (like, that's a lot of extra neutrons you're adding there...).

Also, @FiveLate, you should edit your post if you want to add stuff instead of creating a new post. Double-posting is frowned upon by the mods here.

Edited December 16, 2016 by skaa

[elezraita]

Hi, I'm new here, but I thought I'd weigh in. This is going to be long. You may not agree with what I'm about to say, but here goes...Also, disclaimer: I'm tired and I don't want to proofread this right now. It's probably barely coherent. Sorry. I'll proofread it tomorrow when after someone has blasted me for pointing this out.

So, if I'm understanding the theory correctly, we are suggesting that there are two different types of electrons in a Harmonium atom because the repulsive force exerted between the Ruin electrons and the Preservation electrons is not uniform across all subatomic particles. This implies that either the fundamental constants regulating the strength of the electrostatic force differs between the two god subatomic particles, or that the electrostatic force is uniform, but there is an extra force associated with each type of electron when it comes in contact with the other type. It doesn't matter which is the case; either way, you've got two different and (more importantly) distinguishable types of negatively charged Fermions in the system. I'm not entirely sure that the physics involved with with having essentially two different types of negatively charged Fermions in an atom would necessarily be those that you're obviously expecting. The Hamiltonian (total energy operator) of that system would be very different from that of a system containing only one type of Fermion because the potential energy portion would be completely different. This would lead to a completely different set of eigenfunctions (in this case, solutions to the Schrodinger equation that describe the atomic orbitals) for the system than you would get for an atom with a set of non-distinguishable (interchangeable electrons).

The time independent Schrodinger equation can is H*Y=E*Y where I'll pretend that Y is the Greek letter, Psi, called the wavefunction and E is the energy of the quantum state described by the wavefunction. H is the Hamiltonian, or total energy operator and H = T + V. T is the kinetic energy part and V is potential energy part. The kinetic energy part is just the sum of all the kinetic energies for all of the particles, and this will not be affected by differences in repulsive/attractive forces of the electrons, but it would be affected if the two types had different masses. Note that I am using the Born-Oppenheimer Approximation, that the motions of the electrons are so fast compared to the motion of the nucleus, that we can decouple the kinetic energy of the nucle(us/i) that of the electrons when solving the Schrodinger equation for an atom or molecule. Essentially, we just pretend that the nucleus is fixed and the electrons are the only things that move. This is true in most cases (but not always). Anyway, that was kind of a tangent. Sorry. The potential energy term, V, is a little more complicated and because of this, we can't solve the Schrodinger equation exactly for any atom but hydrogen.

In the case of a regular atom (one with only one type of electron) the potential energy is the negative sum of the energy of the electrostatic attractions of every electron with its nucleus minus the sum of all of the electrostatic attractions to any other nuclei around (in the case of molecules) plus the coulomb repulsion of all the other electrons around. This would be easier to show if I could write down the equation, but just to remind you, the coulomb potential (energy) is the product of some constants (which isn't important to the argument) multiplied by the product of the two charges then divided by the distance between the charges, r_ij. You can imagine that each electron feels the electrostatic interaction with all the other charged particles around in inverse proportion to the its distance, r, from the other particles. When one particle changes position, all the rest want to as well, because they want to be as close to the nucleus as possible and as far away from the other electrons as possible (basically, they want to minimize the total energy of the system). Because the magnitude of the charge is the same for every electron, the only variables in this scenario are the distances between particles. We can factor out everything else and make one sum of electrostatic repulsions and one sum of attractions. When we solve the Schrodinger equation with this Hamiltonian, we get eigenfunctions (wavefunctions) that describe the s, p, d, and f orbitals you learned about in gen chem.

Now, what would happen if there were two sets of electrons, each with a different magnitude of force it exerts on the other type? There would be more sums in the potential energy term than the ones we got in the first case. There would be (presumably) two sums of the the electrostatic attraction between the nucleus, which is made up of a mixture of ruin protons and preservation protons. One would be for the preservation electrons and one would be for the ruin electrons. I imagine that the ruin electrons would be slightly more attracted to the nucleus than the preservation electrons because there are more ruin protons. Furthermore, instead of there being a single sum of repulsions, there would now be three different sums: one for the ruin/ruin replusion, one for the preservation/preservation repulsions, and one for the ruin/preservation repulsions. This difference in the Hamiltonian from one set of indistinguishable fermions to two of them, would likely change the eigenfunctions of the Hamiltonian in the Schrodinger equation to such a degree that there is no guarantee that the atomic orbitals would even be hydrogen-like, and the element's physical and chemical properties would defy all rational attempts to place it in the periodic table at all. At the very least, the energy levels would change substantially, which would definitely affect it's reactivity. I don't want to go through trying to solve a differential equation like that. Actually, the real world version is already impossible to solve analytically, except for the case of the Hydrogen atom. Solving the Harmonium version of the equation with approximate solutions wouldn't be fun either. I certainly wouldn't be able to do it. Note that all this is assuming that the potential of the god repulsions is goes as 1/r like coulomb (electrostatic) repulsions and gravity. What if the god repulsion potential goes as r^-2 or r^-3...or r^-(3/2)? If that is the case, the Hamiltonian will have even more terms, and will yield wavefunctions that are even less Hydrogen-like, thus throwing all this wonderful chemical intuition that has been demonstrated here out the window, so to speak.

Okay, I get that that was super long and probably not well explained. I'm lazy, I guess. My apologies.

One last thing. The Pauli exclusion principle states that two identical Fermions cannot occupy the same set of quantum numbers (n,l, m_l, m_s). This fact leads to the Aufbau principle, which is that in the electronic ground state, electrons occupy successively higher energy atomic orbitals in pairs. By this, I mean that each electron shares the principle quantum number, n; the angular momentum quantum number, l; and the magnetic quantum number, m_l. Only the spin (projection) quantum number, m_s differs between the two. One is m_s=-1/2 (spin down), and one is m_s=1/2 (spin up). The Pauli exclusion principle again dictates how bonding works. Atomic orbitals combine in such a way as to form molecular orbitals. I could go more in depth on this, but it is sufficient to understand that bonding (and anti-bonding) molecular orbitals are filled the same way as atomic orbitals: in pairs. How would the filling of energy levels work if there two types of electrons? This sounds like a graduate level quantum mechanics homework question. The Pauli exclusion principle only refers to identical Fermions. If there were two types of fermions, would this mean that each orbital (assuming that the orbitals are hydrogen-like) would be able to contain four electrons instead of two? This would certainly be interesting. We are assuming 55 electrons. Let's see:

1s4,2s4,2p12,3s4,3p12,4s4,3d15

which would put it at...Zinc, but it would be paramagnetic, like Manganese.

That's weird. Anyway, sorry for this being so long.

tl;dr version: Having two different, distinguishable types of electrons would break Physics, and therefore Chemistry to the point where our Chemical understanding would not really apply.

If Harmonium is an alloy, I'm all for that. I just don't think it makes physical sense for Harmonium to be an atom.

I just noticed NavySealsGuy's post. Yes, very probably, for the reasons I talked about relating to the eigenfunctions of the Hamiltonian.

 

 

 

Edited December 29, 2016 by elezraita
I forgot to finish one sentence.

[Pagerunner]

9 hours ago, elezraita said:

Hi, I'm new here, but I thought I'd weigh in. This is going to be long. You may not agree with what I'm about to say, but here goes...Also, disclaimer: I'm tired and I don't want to proofread this right now. It's probably barely coherent. Sorry. I'll proofread it tomorrow when after someone has blasted me for pointing this out.

tl;dr version: Having two different, distinguishable types of electrons would break Physics, and therefore Chemistry to the point where our Chemical understanding would not really apply.

Hey, no need to apologize; this is exactly the kind of discussion I was looking for, and it was a pretty good explanation of the Schrodinger equation. I'm pretty weak on my PChem, but I can always dig out my old textbook if I need to. (And it sounds like I will have to, since I don't remember the derivation of atomic orbitals.) I hadn't considered that additional forces would change the actual orbitals; would it just change the energy levels, or would it actually change the 2-6-10-14 progression of electrons in orbitals? (i.e. Positronium has a similar energy level structure to hydrogen, but with different spectral lines.) You'd have additional potential energy terms, but as long as:

• Investiture forces are the same order with respect to distance as electromagnetic (which, as you point out, would definitely be necessary)
• The atom is sufficiently large enough that the Preservation and Ruin forces coming from the nucleus are similar (say, less than 5% difference, like 22 vs 23 protons), the difference in additional repulsion from the nucleus will be negligible (since that's only a portion of the potential energy, anyways)
• Ruin and Preservation electrons are evenly distributed throughout the atom (so that a pair of electrons which share the first 3 quantum numbers will have a Ruin Electron and a Preservation electron, their combined potential energy with any given electron will be the two electromagnetic potentials plus the Investiture potential, means that for either of those two electrons, the average total potential is their electromagnetic potential plus half of the Investiture potential)

Then shouldn't the Investiture potential energy terms combine with the Electromagnetic potential energy terms (all a function of the same r), leaving us with, basically, the same potential energy form with a more complex constant in place of the electromagnetic constant? (They really do need an equation editor on this board, just for this thread!) And a changing a constant won't change the form of the eigenfunctions, correct?

For reconciling the Pauli exclusion principle, I was debating between two different possibilities, neither of which I've mentioned yet because I didn't find it relevant. The first was piggybacking it on the spin quantum number, to avoid four electrons per orbital, but then it would have to be limited in this effect for just harmonium, since atium and lerasium obviously have both spins. The other one, which I liked a little bit better, was that Investiture just straight-up didn't play into the Pauli exclusion principle, and that the specific mixing of electrons in harmonium was set by Harmony when he created this metal, but it sounds like the extra forces have a 'feedback' effect into the original physics of orbitals, so that might not be feasible. Maybe it's a combination of the two; when Harmony created his metal, he recognized that it would be all skiwampus if he didn't pair up the electrons one-to-one, so he has to be intentional when manifesting it.

But, all-in-all, great thoughts, and I'll need to pull out my old PChem textbook and do some reading. Unfortunately, I'm a chemist, so when we start to look at one atom at a time, that gets a little outside my comfort zone. That's why I'm glad we've got some physicists to chime in!

[elezraita]

Pagerunner: I'm going to have to think about what you said a bit before I respond. I think that your points are worth considering, but I'll have to actually consider them. I really just want to clarify that I'm neither a real chemist, nor a real physicist. I'm a physical chemist. My research in graduate school (before I got screwed out of my PhD and had to take a masters) was in reaction dynamics. Quantum Mechanics is not exactly my area of expertise, though I did TA P-Chem II one semester (and three semesters of gen chem) while I was a graduate student. My point is that I'm not infallible, and even I still have to check things to make sure I'm not providing incorrect information. However, this stuff is fairly ubiquitous knowledge among p-chemists, though not necessarily among other divisions. I still struggle with inorganic chemistry because it requires such a breath of knowledge, so I will always approach a discussion like this in terms of the underlying physical principles, while I really appreciate the people who have memorized a lot of reaction and reaction classes, for example.

[Oversleep]

On 15.12.2016 at 2:08 AM, aeromancer said:

EDIT: The atium/lerasium alloy turns you into a Seer if you burn it, by the way.

Ehh... Don't be so sure.

See, lerasium-gold alloy would turn people into Augurs and atium-gold alloy would let you see other's people alternate selves, rIight? Now, let's alloy the lerasium-gold with atium and atium-gold with lerasium.

What do we have here now? Is there a difference between lerasium alloy of malatium and atium alloy of lerasium-gold? I think there is (I mused on those things here).

[elezraita]

22 minutes ago, Pagerunner said:

Hey, no need to apologize; this is exactly the kind of discussion I was looking for, and it was a pretty good explanation of the Schrodinger equation. I'm pretty weak on my PChem, but I can always dig out my old textbook if I need to. (And it sounds like I will have to, since I don't remember the derivation of atomic orbitals.) I hadn't considered that additional forces would change the actual orbitals; would it just change the energy levels, or would it actually change the 2-6-10-14 progression of electrons in orbitals? (i.e. Positronium has a similar energy level structure to hydrogen, but with different spectral lines.) You'd have additional potential energy terms, but as long as:

• Investiture forces are the same order with respect to distance as electromagnetic (which, as you point out, would definitely be necessary)
• The atom is sufficiently large enough that the Preservation and Ruin forces coming from the nucleus are similar (say, less than 5% difference, like 22 vs 23 protons), the difference in additional repulsion from the nucleus will be negligible (since that's only a portion of the potential energy, anyways)
• Ruin and Preservation electrons are evenly distributed throughout the atom (so that a pair of electrons which share the first 3 quantum numbers will have a Ruin Electron and a Preservation electron, their combined potential energy with any given electron will be the two electromagnetic potentials plus the Investiture potential, means that for either of those two electrons, the average total potential is their electromagnetic potential plus half of the Investiture potential)

Then shouldn't the Investiture potential energy terms combine with the Electromagnetic potential energy terms (all a function of the same r), leaving us with, basically, the same potential energy form with a more complex constant in place of the electromagnetic constant? (They really do need an equation editor on this board, just for this thread!) And a changing a constant won't change the form of the eigenfunctions, correct?

For reconciling the Pauli exclusion principle, I was debating between two different possibilities, neither of which I've mentioned yet because I didn't find it relevant. The first was piggybacking it on the spin quantum number, to avoid four electrons per orbital, but then it would have to be limited in this effect for just harmonium, since atium and lerasium obviously have both spins. The other one, which I liked a little bit better, was that Investiture just straight-up didn't play into the Pauli exclusion principle, and that the specific mixing of electrons in harmonium was set by Harmony when he created this metal, but it sounds like the extra forces have a 'feedback' effect into the original physics of orbitals, so that might not be feasible. Maybe it's a combination of the two; when Harmony created his metal, he recognized that it would be all skiwampus if he didn't pair up the electrons one-to-one, so he has to be intentional when manifesting it.

You are right that changing the constant won't change the form of the solutions to the differential equation (eigenfunctions), but the r is not the same r. In fact, r is indexed such that every r in the sum is a different variable. That is why the Schrodinger equation is impossible to solve without approximate methods for multi-electron atoms: there are the number of electrons coupled variables and one equation. Anyway, I'm still trying to convince myself that having a different constant for each sum wouldn't cause a different form for the solutions because we wouldn't be able to factor out the same constant from all the terms to get all of them in the same sum. I'm going to try to work this out. Imagine if you had three particles all coupled to together such that V = a/r_12 + b/r_13+c/r_23 where a/=b/=c are constants. No matter what you do, you can't factor out the numerator with affecting each denominator in a different way. If we tried, we would get abc(1/(bc*r_12)+1/(ac*r_13)+1/(ab*r_23)). But then we could just say that x_0=r_12/bc, x_1=r_13/ac and x_2=r_23/ab and rewrite this as abc*SUM(1/x_n) from n=0 to 2. I'm guessing we'd have an annoying time with the change of variables for the kinetic energy term, but maybe you are right. Maybe the orbitals would remain hydrogen-like as long as the investiture potential goes as 1/r. It seems like they would be now. huh. It's not the first time I've thought about something more and tentatively changed my mind.

Whenever I have to think about things like this, I always feel like I'm missing something in the math. Are there any math people here? That is what we really need. My differential equations skills are rusty, at best. All of this is rusty, as I don't think about it much anymore.

The thing about the Pauli exclusion principle is that it only applies to identical fermions, as I said before. Harmony would have to have somehow, as you say, added a fabricated constraint to his metal to force it to behave as if it were following the Pauli exclusion principle. If that is the case, I wonder what happens when an x-ray photon is absorbed by Harmonium. An x-ray photon will knock out a core rather than a valence electron. This, as you can imagine, is a highly energetic electronic configuration, so at that point, another electron from a higher energy orbital will relax and fill that vacancy. The difference in energy between those states is either emitted as a photon or the energy is used to eject another (what is called Auger) electron from the atom. This is called the Auger effect. In real life, the Auger effect is much less likely than the ejection of a photon, but I wonder if balance would cause this not be the case depending on whether is was a ruin or preservation core electron initially ejected. It's a pointless question, but it might give some insight into how precisely balance must be satisfied.

 

 

[Pagerunner]

18 hours ago, elezraita said:

You are right that changing the constant won't change the form of the solutions to the differential equation (eigenfunctions), but the r is not the same r. In fact, r is indexed such that every r in the sum is a different variable. That is why the Schrodinger equation is impossible to solve without approximate methods for multi-electron atoms: there are the number of electrons coupled variables and one equation. Anyway, I'm still trying to convince myself that having a different constant for each sum wouldn't cause a different form for the solutions because we wouldn't be able to factor out the same constant from all the terms to get all of them in the same sum. I'm going to try to work this out. Imagine if you had three particles all coupled to together such that V = a/r_12 + b/r_13+c/r_23 where a/=b/=c are constants. No matter what you do, you can't factor out the numerator with affecting each denominator in a different way. If we tried, we would get abc(1/(bc*r_12)+1/(ac*r_13)+1/(ab*r_23)). But then we could just say that x_0=r_12/bc, x_1=r_13/ac and x_2=r_23/ab and rewrite this as abc*SUM(1/x_n) from n=0 to 2. I'm guessing we'd have an annoying time with the change of variables for the kinetic energy term, but maybe you are right. Maybe the orbitals would remain hydrogen-like as long as the investiture potential goes as 1/r. It seems like they would be now. huh. It's not the first time I've thought about something more and tentatively changed my mind.

Whenever I have to think about things like this, I always feel like I'm missing something in the math. Are there any math people here? That is what we really need. My differential equations skills are rusty, at best. All of this is rusty, as I don't think about it much anymore.

The thing about the Pauli exclusion principle is that it only applies to identical fermions, as I said before. Harmony would have to have somehow, as you say, added a fabricated constraint to his metal to force it to behave as if it were following the Pauli exclusion principle. If that is the case, I wonder what happens when an x-ray photon is absorbed by Harmonium. An x-ray photon will knock out a core rather than a valence electron. This, as you can imagine, is a highly energetic electronic configuration, so at that point, another electron from a higher energy orbital will relax and fill that vacancy. The difference in energy between those states is either emitted as a photon or the energy is used to eject another (what is called Auger) electron from the atom. This is called the Auger effect. In real life, the Auger effect is much less likely than the ejection of a photon, but I wonder if balance would cause this not be the case depending on whether is was a ruin or preservation core electron initially ejected. It's a pointless question, but it might give some insight into how precisely balance must be satisfied.

 

 

It should be the same constant, almost all the way across the board:

[elezraita]

 

I hate to admit this, but I'm confused. Why should sharing and orbital make any difference? The potential in the Schrodinger equation makes no a priori claim of orbitals. In fact, orbitals don't really exist; they are a construct to help us visualize the solutions to the Schrodinger equation. They arise because of the Schrodinger equation and are not taken into account when setting it up and solving it. 

Any given electron in an atom has a finite probability to be found anywhere in space except for radial and spherical nodes, the number of which depends on the principle and angular momentum quantum numbers, respectively. But quantum numbers don't exist either really, at least not until we solve the Schrodinger equation. They arise from certain boundary conditions that must be satisfied in order for the solutions of the Schrodinger equation to make physical sense. The orbital shapes are simply a description of the probability of finding an electron at a given point in space. And they don't look like planetary orbits either, as some people may imagine. They are more like probability density clouds. For example, s orbitals are spherical; they have n-1 radial nodes, and 0 spherical nodes because l = 0. All of this means that there is a probability of finding a one s electron sitting right next to a 4p electron in space. It isn't likely, but there is nothing forbidding it. The other 1s electron could be right at the nucleus at that moment. This is more likely.  In this case, the r between the 4p electron the first 1s electron would be very different to that of the r between the 4p electron and second 1s electron. Even if two 1s electrons were close together at that instant, it is likely that the distance between each one and the 4p electron would be different.

What I'm trying to say is that you cannot make the assumption that r_ik = r_jk, because the likelihood of it being true for even an instant is minuscule beyond belief. That is one reason for which this is so complicated. 

As for the nuclear potential, are we assuming that there is the same number of Preservation and Ruin protons? Also, do the neutrons exert an investiture force on the electrons? I didn't catch that part.

Finally, it may help to remember that electrons act like waves more than particles, in a lot on cases. We treat them mathematically like point charges, but they really don't act like particles when they are confined to spaces on the order of their deBroglie wavelength (lambda=h/p, where h is Planck's constant and p is the momentum). Notice how we talk about nodes in the orbitals? In classical mechanics, nodes are not used when discussing particles, only when discussing waves.

I do tend to ramble. 

[Pagerunner]

28 minutes ago, elezraita said:

 

I hate to admit this, but I'm confused. Why should sharing and orbital make any difference? The potential in the Schrodinger equation makes no a priori claim of orbitals. In fact, orbitals don't really exist; they are a construct to help us visualize the solutions to the Schrodinger equation. They arise because of the Schrodinger equation and are not taken into account when setting it up and solving it. 

Any given electron in an atom has a finite probability to be found anywhere in space except for radial and spherical nodes, the number of which depends on the principle and angular momentum quantum numbers, respectively. But quantum numbers don't exist either really, at least not until we solve the Schrodinger equation. They arise from certain boundary conditions that must be satisfied in order for the solutions of the Schrodinger equation to make physical sense. The orbital shapes are simply a description of the probability of finding an electron at a given point in space. And they don't look like planetary orbits either, as some people may imagine. They are more like probability density clouds. For example, s orbitals are spherical; they have n-1 radial nodes, and 0 spherical nodes because l = 0. All of this means that there is a probability of finding a one s electron sitting right next to a 4p electron in space. It isn't likely, but there is nothing forbidding it. The other 1s electron could be right at the nucleus at that moment. This is more likely.  In this case, the r between the 4p electron the first 1s electron would be very different to that of the r between the 4p electron and second 1s electron. Even if two 1s electrons were close together at that instant, it is likely that the distance between each one and the 4p electron would be different.

What I'm trying to say is that you cannot make the assumption that r_ik = r_jk, because the likelihood of it being true for even an instant is minuscule beyond belief. That is one reason for which this is so complicated. 

As for the nuclear potential, are we assuming that there is the same number of Preservation and Ruin protons? Also, do the neutrons exert an investiture force on the electrons? I didn't catch that part.

Finally, it may help to remember that electrons act like waves more than particles, in a lot on cases. We treat them mathematically like point charges, but they really don't act like particles when they are confined to spaces on the order of their deBroglie wavelength (lambda=h/p, where h is Planck's constant and p is the momentum). Notice how we talk about nodes in the orbitals? In classical mechanics, nodes are not used when discussing particles, only when discussing waves.

I do tend to ramble. 

It might be putting the cart before the horse, but if they share the same orbital, they occupy the same space (r isn't a distinct value for any electron, it's a probability distribution obtained by applying a different operator to the wavefunction, correct? And if they share the same distribution for r, then they can be used interchangeably in the equation), so they'll exert the same average electromagnetic force. Yes, it assumes that electrons will order themselves into orbitals in the end, but as long as they do that, I don't see why there'd be a problem with the math.

I haven't included neutrons in the potential energy calculation, but I did assume that the functional electromagnetic force from both kinds of protons is similar (23 Ruin, 22 Preservation, it's within 5%, lets me fudge the math and say they're identical).

[elezraita]

I said nothing about your math. It looks fine. The fact is I don't know if your approximations are valid or not. I suppose it would depend upon how close you want to be to the correct answer in the end. People make approximations all the time and then spend ages talking about why the answers they get differ so much from experiment and from other answers obtained with different approximations. Your assumptions are probably fine, as long as you understand what you lose (and gain) by making them, as you have just demonstrated that you do. I didn't realize that you understood your assumptions, and I didn't understand them satisfactorily because you did not justify them to my satisfaction the first time. You just used what looked to me like math based on a misunderstanding of physical principles. Call my reaction a throwback to being a TA. You had skipped steps in your explanation, so I assumed you were missing fundamental understanding. This is often what happens with students. My apologies for treating you as such.

It was obvious that you weren't including the neutrons. I asked those questions to get you to think about your assumptions. Again, a throwback to teaching. Please forgive me for my presumptuousness.

[The One Who Connects]

I feel like I could earn a Chemistry Degree just from the research required to properly discuss what is going on in this thread. That said, what I do have to add will be more simple terminology, as I do not have the aforementioned degree in Chemistry. (Or Physics, since several lines end off in "this goes into the physicists territory..") So without further ado, what I have gathered and understood in this thread, in simple terms for easy understanding and quick correction.
Note: I gave them Atomic Symbols for easy abbreviation

---

Harmonium (Hm) atoms may be made of atoms of Lerasium (Le) and atoms of Atium (A). At the very least, the subatomic particles; Protons (+), Neutrons (0), and Electrons (-) are made up of Investiture from Ruin (R) and Preservation (P)

Alkali metals (Cesium, Potassium, etc..) are very reactive because their outer shell only has 1 Electron. It stands to reason that Harmonium, being a "super-cesium," will also follow this rule. In the original post of this thread, it is theorized that the outer electron is a Ruin Electron, meaning that the Nucleus has 1 more Ruin Proton than it has Preservation Protons, because of Harmony wanting to offset his extra Ruin Investiture.

Electrons, being negative, are naturally pulled to Protons, as they are positive. Negatives are pushed away from other Negatives (other Electrons in this case). Ruin electrons also push away from Preservation electrons, which makes the outer ring on electrons even less stable. This also follows for Protons made of R&P.
|  R- is pulled towards R+  |  P- is pulled towards P+  |  R- is pushed away from P+ and from P-  |  R+ is pushed away from P+  |
I have made the assumption that a Ruin Proton(R+) will only attract a Ruin Electron(R-), and the same for Preservation. Given that normal elements want to be in a state of balance (keeping the # of Protons equal to the # of Electrons) it makes sense that it would want the # of R+ to equal the # of R-, and the same for Preservation.

---

@Pagerunner, as the OP, has theorized that Harmonium is quite literally "super-cesium", atomic number and all. Cesium has 55 Protons, so if it is a combination of Lerasium and Atium, they have to have a 1 Proton difference, to keep with the uneven/valence electron balance. Meaning that Le is magic Cobalt, and A is a magical form of Nickel. Cobalt (27 Protons) + Nickel (28 Protons) add together to Cesium (55 Protons).
There is an issue with the Atomic Mass however. The stable masses of Co (58.9) and Ni (58.7) add together to 117.6. Stable Cesium has a mass of 132.9. This is a difference of 15.3 units of mass (mostly neutrons). I do not feel that the reduced "gravitational" force of the nucleus could overpower the "anti-gravitational" force of R+ and P+ pushing away from each other. This is my personal opinion with what I remember from high school, and is probably incorrect in some way.

@skaa has brought up the comment of "city destroying bombs" (which drove the in-story reactivity testing IIRC) and theorized that in reference to that, Brandon is implying the metal is one with nuclear potential. They have narrowed it down to 2 possibilities. 1) The extra reactivity of R&P Investiture has made cesium capable of going nuclear. 2) The increased reactivity of R&P Investiture has made a non-alkali element appear to be alkali when reacting with water.

I am proposing a different idea, which.. somewhat works around the problems in the prior solutions. (It likely has it's own problems, so feel free to point them out)
Cesium-135. It has low decay energy, no gamma radiation, and a 2.3 million year half life, which makes it much less hazardous than other Cesium isotopes. Xenon-135 (where Cs-135 comes from) has a half life of 9.2 hours, which removes most issues about the required time to form our "Harmonium." Xe-135 also has as much as a 90% chance to form stable Xe-136, which could account for Harmonium still being a semi-rare metal to find. Also, natural fission occurs on Earth, so in pre-TLR Scadrial, it is entirely possible that such a thing occurred undiscovered until now.
For the purpose of balance, you can keep the uneven split of R&P protons/electrons through Harmony's interference if you wish.

---

I lack the knowledge, and google lacks the easy explanations thus far , for me to comment on the mathematical arguments of Pagerunner and Elezraita, so I won't.

Questions, Comments, Concerns anybody?

 

[Pagerunner]

4 hours ago, elezraita said:

I said nothing about your math. It looks fine. The fact is I don't know if your approximations are valid or not. I suppose it would depend upon how close you want to be to the correct answer in the end. People make approximations all the time and then spend ages talking about why the answers they get differ so much from experiment and from other answers obtained with different approximations. Your assumptions are probably fine, as long as you understand what you lose (and gain) by making them, as you have just demonstrated that you do. I didn't realize that you understood your assumptions, and I didn't understand them satisfactorily because you did not justify them to my satisfaction the first time. You just used what looked to me like math based on a misunderstanding of physical principles. Call my reaction a throwback to being a TA. You had skipped steps in your explanation, so I assumed you were missing fundamental understanding. This is often what happens with students. My apologies for treating you as such.

It was obvious that you weren't including the neutrons. I asked those questions to get you to think about your assumptions. Again, a throwback to teaching. Please forgive me for my presumptuousness.

Like I said before, no apologies needed - the onus is on me to validate my assumptions, and I did brush over a couple of them. By all means, keep on ragging on me until I've answered to your satisfaction. (Although, if I had known how far down the rabbit hole we would have gone, I'd probably have taken my time and posted a more rigorous explanation in the first place! I'm rushing a little bit, since I've got family over and don't have a ton of time to spend drafting and revising a forum post, however much fun it is.)

You've taken the conversation to another level, which is good, because it highlighted an assumption that I had made for thematic reasons (each electron pair has a Ruin and a Preservation) and showed why it is necessary (we don't want to fundamentally change the Schroedinger equation). As long as the they're paired up like that, we can broadly assume (since we're both just rusty enough at the math to not be 100% confident) that the additional Investiture force will effectively 'alter' the electromagnetic constant, but won't change how the electrons order themselves.

So, yeah, it's been a great conversation, and I think I would have enjoyed having you as a TA. Please feel free to continue asking about details about this specific theory (although it might not always sound like it, I am having fun), and I hope to see you around in some other threads.

@The One Who Connects, good summary, a few minor points:

• I wasn't quite clear on the way you presented the various attractions/repulsions, if those are all the interactions are just the Investiture-related ones or includes the normal electromagnetic attractions. For example, I don't think there's any additional attraction between P+ and P- than a normal proton and electron (which is why I think that lerasium and atium can behave exactly like real metals, since there are no extra forces). Also, R+ might not actually repel P-; the Investiture-related force and electromagnetic force are opposing, but the Investiture force might not be greater than the electromagnetic force, so there might still be a net attraction. So, a definitive assertion might need to be a little more... squishy.
• Because god metals aren't naturally occurring, they won't have decimal points in their atomic masses; that's due to the percentages of isotopes occurring in nature. God metals would be based on a particular stable isotope of real metals, so it would be a whole number.
• Can you explain what benefit there is of bringing Xe-136 into the conversation? Harmonium isn't naturally occurring; it's something that is manifested by a Shard in a specific way (like the geodes at the Pits of Hathsin). I wouldn't say that every atom of cesium is harmonium; it's more like a solid-gold Gameboy, which is the same as a regular Gameboy but made out of gold. But the gold one wasn't made at the factory like the rest, it was made out of something different, somewhere different, for a special purpose. This analogy is getting really weird, so I'm going to stop. But, are you suggesting that Harmony manifests as a vapor form as Xenon, which can undergo radioactive decay to 'condense' into Harmonium?

 

[elezraita]

2 hours ago, The One Who Connects said:

I feel like I could earn a Chemistry Degree just from the research required to properly discuss what is going on in this thread. That said, what I do have to add will be more simple terminology, as I do not have the aforementioned degree in Chemistry. (Or Physics, since several lines end off in "this goes into the physicists territory..") So without further ado, what I have gathered and understood in this thread, in simple terms for easy understanding and quick correction.
Note: I gave them Atomic Symbols for easy abbreviation

---

Harmonium (Hm) atoms may be made of atoms of Lerasium (Le) and atoms of Atium (A). At the very least, the subatomic particles; Protons (+), Neutrons (0), and Electrons (-) are made up of Investiture from Ruin (R) and Preservation (P)

Alkali metals (Cesium, Potassium, etc..) are very reactive because their outer shell only has 1 Electron. It stands to reason that Harmonium, being a "super-cesium," will also follow this rule. In the original post of this thread, it is theorized that the outer electron is a Ruin Electron, meaning that the Nucleus has 1 more Ruin Proton than it has Preservation Protons, because of Harmony wanting to offset his extra Ruin Investiture.

Electrons, being negative, are naturally pulled to Protons, as they are positive. Negatives are pushed away from other Negatives (other Electrons in this case). Ruin electrons also push away from Preservation electrons, which makes the outer ring on electrons even less stable. This also follows for Protons made of R&P.
|  R- is pulled towards R+  |  P- is pulled towards P+  |  R- is pushed away from P+ and from P-  |  R+ is pushed away from P+  |
I have made the assumption that a Ruin Proton(R+) will only attract a Ruin Electron(R-), and the same for Preservation. Given that normal elements want to be in a state of balance (keeping the # of Protons equal to the # of Electrons) it makes sense that it would want the # of R+ to equal the # of R-, and the same for Preservation.

---

@Pagerunner, as the OP, has theorized that Harmonium is quite literally "super-cesium", atomic number and all. Cesium has 55 Protons, so if it is a combination of Lerasium and Atium, they have to have a 1 Proton difference, to keep with the uneven/valence electron balance. Meaning that Le is magic Cobalt, and A is a magical form of Nickel. Cobalt (27 Protons) + Nickel (28 Protons) add together to Cesium (55 Protons).
There is an issue with the Atomic Mass however. The stable masses of Co (58.9) and Ni (58.7) add together to 117.6. Stable Cesium has a mass of 132.9. This is a difference of 15.3 units of mass (mostly neutrons). I do not feel that the reduced "gravitational" force of the nucleus could overpower the "anti-gravitational" force of R+ and P+ pushing away from each other. This is my personal opinion with what I remember from high school, and is probably incorrect in some way.

@skaa has brought up the comment of "city destroying bombs" (which drove the in-story reactivity testing IIRC) and theorized that in reference to that, Brandon is implying the metal is one with nuclear potential. They have narrowed it down to 2 possibilities. 1) The extra reactivity of R&P Investiture has made cesium capable of going nuclear. 2) The increased reactivity of R&P Investiture has made a non-alkali element appear to be alkali when reacting with water.

I am proposing a different idea, which.. somewhat works around the problems in the prior solutions. (It likely has it's own problems, so feel free to point them out)
Cesium-135. It has low decay energy, no gamma radiation, and a 2.3 million year half life, which makes it much less hazardous than other Cesium isotopes. Xenon-135 (where Cs-135 comes from) has a half life of 9.2 hours, which removes most issues about the required time to form our "Harmonium." Xe-135 also has as much as a 90% chance to form stable Xe-136, which could account for Harmonium still being a semi-rare metal to find. Also, natural fission occurs on Earth, so in pre-TLR Scadrial, it is entirely possible that such a thing occurred undiscovered until now.
For the purpose of balance, you can keep the uneven split of R&P protons/electrons through Harmony's interference if you wish.

---

I lack the knowledge, and google lacks the easy explanations thus far , for me to comment on the mathematical arguments of Pagerunner and Elezraita, so I won't.

Questions, Comments, Concerns anybody?

 

Much like Pagerunner, I'm a bit confused about what you are proposing in terms of Xenon. Xenon is not a metal; it is a noble gas. This means that it is extremely difficult to get it to react with anything, because it has a full valence shell. I will note, however, that oxides of Xenon that are highly explosive under certain circumstances (but not upon contact with water) can be synthesized. Furthermore, elemental Xe's melting point is around -112 C so it is unlikely to be solid anywhere on Scadrial. Xe just doesn't fit the physical description of Ettmetal. Also, I think that OP said that Brandon said something about super-Cesium, so I'm thinking that Pagerunner is right about the Hm group even though I'm still a little bothered by the Pauli exclusion thing. I want to ask Brandon about it. This is actually a good thought though. You are thinking outside the box.

I think Pagerunner did a good job of explaining that the extra investiture force in Hm is not necessarily great enough to overcome the natural electrostatic forces in the atom, so I'll say nothing further about that.

I just want to clarify that we are talking about electrostatic forces, not gravitation. They are very different forces, of very different strengths. Gravity has to do with mass/mass interaction and electrostatic interaction has to do with charge interaction. Investiture force...who knows at this point?  

 

[The One Who Connects]

@Pagerunner @elezraita

I forgot about the "not naturally occurring" part, and foolishly brought up how the non-super cesium occurs to show that it could easily have occurred within that 350 some odd year timeframe since harmony ascended. Feel free to ignore that.

Pagerunner:

1) The Investiture related reactions. Didn't mean to imply that P+ to P- was any different than normal + to -  I meant it as Ruin Protons would only pull in Ruin Electrons, and not electrons in general. Also, I hadn't considered that the natural force might be stronger than the Investiture one, which would make R&P repelling less of a stability issue than I assumed.

2) The periodic table I used as reference stated that "mass numbers in parentheses are of the most stable isotope" so tired me forgot that the rest are average mass numbers. If I rounded my numbers to the nearest whole number, adding Co and Ni is still 13 short of Cesium, but that is besides the point.

3) See top about "naturally occurring."

3a) No, I wasn't implying that gaseous Harmonium can decay into solid form. If we had any precedent for the vapor form of Investiture condensing into the solid form, then by all means we can take that as what I meant. I don't think it does, or else Lerasium wouldn't be as rare as it was.

Elezraita:

1) Yea, I screwed up on that one and assumed part of our physics could be applied to god metals

1a) Xenon oxides can be explosive? Fascinating.

2) My "grav" and "anti-grav" comments were... Attraction and Repulsion escaped me yesterday afternoon and I used terms that I thought people would understand in the context.

Upon reread, I may have actually meant gravity when I said it, as less neutrons would mean a reduced nucleus mass, and thus less gravitational pull. Physics overwriting Chem, my bad. I only meant "a repelling force" when I used "anti-grav" so I am fairly sure that I didn't truly mean anti-gravity.

3ish) Before I make another wild assumption, can multiple interactions be directly involved in somethings creation at the atomic/subatomic level? Not necessarily gravity and electrostatic, perhaps some other interactions that haven't been brought in the thread yet. I may have an idea on how Investiture interaction might work, but I figured it's better to ask that question first

Edited January 1, 2017 by The One Who Connects
Spellcheck, twice

[elezraita]

@The One Who Connects

I figured that you didn't really mean gravity, but as Pagerunner has seen, I like to be as rigorous as possible. The force involved in holding nuclei together is, interestingly enough, called the "Strong interaction". It is, according to hyperphysics, 1.67*10^38 times stronger than the force of gravity, though its range is on the femtometer (10^-15 m) scale. Because the mass of subatomic particles is so tiny, their gravitational pull...well, calling it neglible would be the understatement of the history of the world. The thing is, I'm not really a physicist; I'm a physical chemist, so my knowledge of the the strong force and most of particle physics is spotty at best. I find it fascinating, but I've not studied it the way I've studied more chemistry related physics topics.

I do know that the last fundamental force we haven't named yet on this thread is called the "weak interaction" or the "electroweak" interaction. It is the force responsible for certain types of radioactive decay among other things.

The four fundamental interactions are the strong interaction, electromagnetic interactions (we've been talking about electrostatic forces, but magnetism is related to it because they both originate only from charged particles), the weak interaction, and gravity.

Again, the strong and weak interactions are not my area of expertise, but I'm pretty sure I haven't said anything wrong. I hope this answers your question.