Impact of superconductivity on mechanical properties?

[fluidistic]

It is often said that the mechanical properties of metals are mostly due to their free electrons. I'm talking about malleability, ductility and hardness.
So, when they suffer a superconductive transition, I expect a drastic change in their mechanical properties, but I haven't found anything on the Internet about such consequences.

My questions are: What can we expect, regarding the mechanical properties of metals, when they pass from a non superconducting state to a superconducting state?

What about type II superconductors?

 

[Entr0pic]

Not all electrons are involved within the superconducting phase - only those within a gap about the Fermi surface.

The thermal and electronic properties of the material would change, but why do you think the mechanical properties would change drastically at all?

 

[fluidistic]

Not all electrons are involved within the superconducting phase - only those within a gap about the Fermi surface.

The thermal and electronic properties of the material would change, but why do you think the mechanical properties would change drastically at all?

For the exact same reason their thermal and electronic properties would change.

 

[ZapperZ]

It is often said that the mechanical properties of metals are mostly due to their free electrons

Who said that?

Zz.

 

[fluidistic]

Who said that?

Zz.

Starting with Ron Maimon:

• Why are metals strong?

Because you are cutting delocalized electrons when you slice through a metal. The whole metal is like one gigantic chemical bond binding all the atoms together, because the electrons are delocalized over the whole material. It isn't like salt, where if you cleave, it's easy for the electrons to pick sides.

• Why are metals malleable?

Because the electrons holding a metal together are delocalized, not holding nearest neighbors together. So any local change in shape of the atoms is not so bad, the imporatant thing is that the whole Fermi-energy not change. So you can break the weak nearest-neighbor bonds and shuffle the atoms, without breaking the metal. This is also why brittle materials are insulators.

Following by socratic.org:

Why are metallic bonds ductile?
Because the delocalised electrons are free to move.

Metallic bonds are formed by the electrostatic attraction between the positively charged metal ions, which form regular layers, and the negatively charged delocalised electrons. These are the electrons which used to be in the outer shell of the metal atoms. These delocalised electrons are free to move throughout the giant metallic lattice, so as one layer of metal ions slides over another, the electrons can move too keeping the whole structure bonded together.

This is the opposite of what happens in a giant ionic lattice, where both the positive ions and the negative ions are locked in place. If the crystal is stressed and one layer moves with respect to another, the positive ions can end up lined up with each other, and the negative ions lined up with each other. This causes repulsion, so the crystal fractures.

Lastly, Wikipedia:

Metallic bonding is a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons (in the form of an electron cloud of delocalized electrons) and positively charged metal ions. It may be described as the sharing of free electrons among a structure of positively charged ions (cations). Metallic bonding accounts for many physical properties of metals, such as strength, ductility, thermal and electrical resistivity and conductivity, opacity, and luster.[1][2][3][4]

So I would expect that if a "condensation" of the electrons near the Fermi surface occur, like in a superconductivity transition, the mechanical properties I've mentioned in my first post would drastically change.

 

[king vitamin]

The solution is basically what entr0pic said: only a small fraction of electrons undergo Cooper pairing. In particular, only the electrons within about a Debye energy of the Fermi surface pair up. A typical Fermi energy of a metal is on the order of 1-10 eV, while the typical Debye energy is .01 eV, so typically less than 1 out of every 100 electrons "localizes" into a Cooper pair (though Cooper pairs are fairly delocalized themselves relative to the typical microscopic length scales).

 

[fluidistic]

The solution is basically what entr0pic said: only a small fraction of electrons undergo Cooper pairing. In particular, only the electrons within about a Debye energy of the Fermi surface pair up. A typical Fermi energy of a metal is on the order of 1-10 eV, while the typical Debye energy is .01 eV, so typically less than 1 out of every 100 electrons "localizes" into a Cooper pair (though Cooper pairs are fairly delocalized themselves relative to the typical microscopic length scales).

I am sorry, I am still not undestanding!
If "only" about 1 free electron in 100 is able to condense into a Cooper pair, that is still quite significant to me. For example, the electrons responsible for electrical conductions are of the order of $v_\text{drift} / v_\text{Fermi}$ times the number of free electrons. That is insanely small, i.e. about $10^{-10}$. Less than one free electron in a billion actually contributes to the electrical current, when the current is a small perturbation (as it usually is). Yet the effect is quite measurable.
Another example is the electrons that contribute to the specific heat are the ones having an energy within $k_BT$ of the Fermi energy, i.e. about 1 free electron in 100 near room temperature. That contribution is small near room temperature, but becomes dominant at very low temperatures (in particular near superconducting temperatures for pure metals). Again that's measurable and is something that is not neglected at low temperatures.

So, as is, it is impossible for me to buy the argument that it is because only 1 electron out of 100 free electrons would condense into a bosonic pair, that the mechanical properties are left (almost?) intact.

I need deeper explanations.

 

[f95toli]

Well, they are. For the reasons that were explained above.
It also depends on what you mean by "mechanical properties". Some properties like velocity of sounds and attenuation of ultrasound do change as you go through Tc. However, if you are talking about ductility and hardness nothing at all changes; at least not at a level where it makes any practical difference.
It is perhaps worth remembering that superconductivity is generally a low-temperature phenomena meaning real metals/alloys (with impurities, GBs etc) that exhibit SC can also exhibit some non-trivial behavior (some phase transitions etc) but this is just " normal" behavior, not anything that has to do with SC.

 

[king vitamin]

So, as is, it is impossible for me to buy the argument that it is because only 1 electron out of 100 free electrons would condense into a bosonic pair, that the mechanical properties are left (almost?) intact.

Your point isn't a bad one, and in order to resolve it, we need to go back to the derivation of these quantities (electrical conductivity, specific heat, and the pressure) in a Fermi gas, and see what portions of the electron gas contribute.

For example, for the specific heat, one needs to know how the internal energy changes as a function of temperature. But if you look at the way that the Fermi-Dirac distribution changes as a function of temperature when temperature is very low, you see that the leading effect is simply that there is a small change in probabilities close to the Fermi energy. In particular, there is a larger probability of electrons occupying states a little larger than $\epsilon_F$, and a smaller probability of electrons occupying states a little less than $\epsilon_F$. States with energies much larger or smaller than $\epsilon_F$ are not affected at all by a small increase in temperature, so only states close to the Fermi energy contribute to the specific heat. This is also expressed in the general expression for the specific heat at low temperatures:
$$
C_v = \frac{\pi^2}{2} k_B^2 T D(\epsilon_F)
$$
Here, $D(\epsilon)$ is the density of states, so we see explicitly that the density near the Fermi energy is all that matters. (Then clearly, for a gapped superconductor, see get a drastically different answer!)

Similar considerations apply to electrical conductivity. If one applies a small external electric field, this effectively displaces the Fermi surface by some small amount. The electrons deep inside the Fermi surface do not feel this displacement, while the ones close to the edge either leave the surface or enter it depending on where they are relative to it. Then one gets an expression for the conductivity that depends on the density of states at the Fermi level rather than the absolute magnitude of it.

Finally, we get to the pressure. This is determined by the change in the internal energy with respect to the volume, which, unlike temperature and electric field, is an extensive rather than intensive quantity. More heuristically, because all of the electrons are delocalized in real space (even those far below the Fermi energy), they all can feel even a small deformation of the volume. You can also look at the explicit equation for the pressure of a Fermi gas:
$$
P = \frac{(3 \pi^2)^{2/3} \hbar^2}{5m} \left( \frac{N}{V} \right)^{5/3}
$$
Here, we can explicitly see why the point I made my point above: the pressure depends on the total number of electrons in the Fermi gas. If I take one out of every 100 electrons and remove them from my system, the pressure only changes by $P' = (1-1/100^{5/3})P = 0.9995P$. And of course, the Cooper pairs will have some nonzero contribution to the pressure as well. So the pressure will not change much.

 

[fluidistic]

Thank you very much king vitamin, now that was an extremely convincing argument!