Science Squabbling

no. thats not how they work

Two kinds of argument can be made. One is thermodynamic - the gibbs free energy of the gamma/fcc/austenite phase drops below that of the alpha/bcc/ferrite phase at some temperature. That's pretty abstract but is what a metallurgist would tell you.

The other argument is a more intuitive one but needs some quick calculations. In short, the iron atomic radius can increase more in the FCC structure without reducing the density too much. This is because the FCC structure is actually a close packed structure in disguise - think stacking cannon balls - and these structures have the best possible packing fraction (~74%). If you calculate the iron atomic radius in the alpha/BCC phase it is 0.124 nm, compared with 0.126 nm in the gamma/FCC.
Ref

Edit: that being said,
@snottyotter might be right too. The BCC alpha ferrite is ferromagnetic and that will lower its gibbs free energy. I guess ideally a true metal would be FCC/CCP (cubic close packed) to increase density but a guy with more expertise than me points out here that transition metals aren't actually true/perfect metals and (presumably) have some covalent character to their bonds. That may prefer a slightly longer bond length which negates the density increase available from an FCC structure, especially when you include the energy benefit of allowing ferromagnetism.

Well, the ferrite is ferromagnetic and the austenite is only paramagnetic...

thanks,

so to paraphrase the top idea in laymanic:

The energy of the gamma phase iron is sufficient to maintain the shorter therefore higher energy bonds in FCC structure however as the temperature drops the gibbs free energy of the iron drops and it cannot maintain the FCC structure and goes to the BCC structure?

Apologies if I've missed the plot entirely.

for the second one, why does the iron radius change?

Apologies in advance - feel free to PM me instead...

Laymanic:
Consider the diagram below. Above T_eq, the solid phase has a lower (gibbs free) energy, so the system will minimise this and exist as a liquid. Above T_eq, the liquid phase has a lower energy so that becomes the preferred phase. These two lines are different because there are energy pros and cons to each phase. Liquid = high entropy = awesome. Solid = regular strong bonds = lots of energy released forming these = awesome.

Now considering our system (ie the iron). It can lower its free energy by changing phase above 912 degC. This is because the gamma/FCC phase has a more space per iron atom, so it can wiggle around more (increased temperature = increased wiggling = larger radius); that increases its entropy. However, it moves the atoms further apart, which costs energy. It's only able to afford this energy cost when the temperature is high enough to compensate through the increase in entropy.

In a bit more detail:
The gibbs free energy is a combination of the enthalpy (the net energy released in the reaction) and the energy required to adjust the level of entropy. It's why endothermic reactions (ones that get colder, like instant ice-packs) can take place, because they massively increase entropy - enough to compensate for their taking in energy. As an equation, Gibbs free energy (G) = Enthalpy (H) - Temperature (T,in K) x Entropy (S). G can be evaluated for any chemical, liquid, crystal etc. That 'times T' is why entropy increases are the stronger driving force for transitions at higher T. Note that the diagram above didn't show straight lines (as you would expect from G=H-TS). This is because H and S are both dependent on T as well, but they vary more slowly so you still see this downward trend with increasing T.

TLDR;
Magnets.

I got as far as the Gibbs free energy bit, but I'm not sure why the structure goes back to BCC above 1394 °C. I'm guessing the curves of G against T for the two structures have different curvatures so that there are two crossing points, but I can't find a diagram which shows this. For obvious practical reasons, nobody takes much interest in pure iron.

This phase diagram suggests that there is no flip back to BCC on the way to liquid above about 10kbar, so the surfaces defined by G vs T vs P for BCC and FCC must have some weird intersection which I'm having trouble visualising :-)

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There are probably things I haven't got, but I do know that alpha-iron is ferromagnetic, which will assist in lowering the free energy. The Curie temperature lies around 1100 degree, so delta-iron is paramagnetic and has no residual magnetisation. This may have some effect.

IANAM

Excellent discussion.

Does FCC really allow more wiggle room per atom? I'd have thought the lower packing efficiency of BCC would do that?

In that case, the wiggle room argument might help to explain the reappearance of BCC at high temperature, while we have to go looking for other effects at low temperature.

I've also seen it said that BCC has smaller spaces for interstitial atoms, which doesn't seem to line up with the lower density either. Is it that BCC has both more space and more spaces (points maximum distance from atoms, counted per atom), so the individual spaces come out smaller?

You can have as much wiggle room as you want, if you make the lattice paramter large enough. According to the source I found (here) the BCC/alpha-iron lattice parameter is 0.286 nm, whereas FCC/gamma-iron is 0.356 nm. My *guess * is that the FCC will minimise the bond stretching (due to increased packing fraction) but maximise the unit cell volume (to allow thermal wiggle room - technical term, of course). Your point about the reversion to BCC is a good one, but I don't really know and can't make a decent educated comment.

As for intersticial space, I assumed we were talking pure iron and so it was irrelevant. As soon as you allow carbon (as you should in all realistic scenarios), the phase diagram looks quite different (ref). As you say, the carbon diffuses through interstitial space, as its atomic radius is comparatively small. The BCC has more interstitial spaces, but they are all fairly evenly sized. The FCC/gamma-iron has fewer, larger, spaces and so carbon diffusion is easier and the allowable carbon content is (much) higher for austenite/gamma-iron (ref).

Thanks, I think I half get most of that. In the second and third sentence is confusing me though. did you mean higher:

Above T_eq, the solid phase has a higher (gibbs free) energy, so the system will minimise this and exist as a liquid. Above T_eq, the liquid phase has a lower energy so that becomes the preferred phase.

Sorry, I was already well past tipsy when I wrote that last post (sun! BBQ!)...
Your suggestion is factually correct but ends up as a tautology -it should read as below.

Below T_eq, the solid phase has a lower (gibbs free) energy, so the
system will minimise this and exist as a solid. Above T_eq, the
liquid phase has a lower energy so that becomes the preferred phase.

Ah, thought as much. Thanks.

Pleasure! I may be able to ask a tame metallurgist in my department (Imperial Materials) if there's anything I've missed. I'll try that tomorrow.

pleasure is mine.
It is likely to fly over my head, I wouldn't want you to inconvenience yourself...

This is coming up again. Was really good last year.
https://pintofscience.com

I can't quite tell if you're taking account of FCC unit cells having more atoms per cell than BCC. (8 8ths + 6 halfs = 4 for FCC vs. 8 8ths + 1 = 2 for BCC.) So FCC is denser despite the larger lattice parameter. I did the maths on the nearest neighbor distances from your link above, and got 0.258 nm for FCC and 0.248 nm for BCC. (The BCC number is room temperature, but the 'Steels' book doesn't say what temp the FCC number relates to.)

Still thinking what to make of this all. I fear we'll need to dig deep into how to calculate the phases' free energy to get much further, and that might need both more quantum mechanics and computing time than i have access to.

I too calculated the nearest neighbour for the atomic radius. I think that's probably a more important measurement than the unit cell volume/theoretical density. As for why it prefers a BCC phase a room temp, I have no idea. As you say, a calculation of the free energy is the real answer, but even then you have to know what to account for! I'll try and follow up with those metallurgists to see if they have an indication of what the real driving force is (bond energy/length? configurational entropy? magnetisation?).

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So far i like most the 'empty d orbitals' argument from one of the links in your first post. I'm less keen on the 'magnets' explanation because the Curie temperature doesn't line up with the transition to FCC, so what's holding it BCC between the Curie point and the phase transition?

Maybe it's something like this: at low temperature iron's d orbitals favor the BCC structure, but its metalness objects to the poor packing efficiency and the atoms get squashed in closer to each other than they'd really like to be. As it gets hotter the atoms are jiggling so much that the d orbitals aren't lining up enough of the time to claim the energy benefit from it, so close packing wins. Hotter still and the d-orbitals conspire with entropy (which favors things being spread out) to switch back to BCC before it melts.

Haha, the 'magnets' response wasn't meant to be definitive - as you say, the transition temperatures don't line up with the curie temperature, but it will have a non-negligible contribution. Also, it was a nice chance for 'magnets' to be a vaguely interesting and non-sarcastic response for once!

I found a paper that points out how easy a BCC-FCC transition is, a deformation in a {100} direction so c/a=sqrt2 will give an FCC lattice. There's an interesting paper here too that I haven't had a chance to read in full detail that indicates that the magnetism may play a reasonable part in stabilising the BCC. I'll get back to you once I've perused it in more detail.

Do orthohombic crystals (thinking of martensite specifically) have an affect on formation/disruption of slip systems other than interfeirance of stress and dislocation feilds?

http://www.informationisbeautiful.net/visualizations/snake-oil-supplements/

Slip systems are inherent to the symmetry of a crystal system. I expect, because of the reduced symmetry of an orthorhombic crystal, that it will have fewer slip systems and so the critical resolved shear stress of a random direction will be higher.

Hadn't seen this.
Yeah. So like a simillar affect to varying grain pattern would have.

Back to the ongoing discussion: I had asked my teacher about this and he gave me a roundabout his explanation (dumbed down a lot I can imagine):

At the different energies the electrons are more likely to be present with differing spacing (smaller spacing at higher energies). If the electrons can overlap with the metal ions they'd be electrostatically more stable. Hence at low energy a wider spacing is favoured (lower packing fraction hence BCC) and when the energy increases smaller gaps are favoured hence FCC. That may explain why it returns to the BCC at an even higher temperature.

Does this sound plausible? I know I must have missed/brushed over complications but as an idea...

I have a feeling that til you ask n metallurgists you'll end up either n+1 slightly different explanations... You can argue in several different general directions and they're all right in their own way. Whether one phase is stable or not depends on the relationship between lots of different properties and half of these will be competing against the other half. It's all a balancing act!
Your best bet to finding a satisfactory answer (as with a lot of empirical science like this) is to hear as many answers as possible and understand that they probably all contribute to some degree.

RE your specific hypothesis, what is your meaning of spacing? Inter-electron spacing, or electron-nucleus spacing? Why would it decrease with increasing (thermal) energy?

In other news, autocorrect was not designed for this kind of conversation...