5 ms·

At pressures a million times those experienced on earth. Still an advance, of course.

by ttsda 7y ago

saagarjha 7y ago
Also, the temperature (250 K, -23°C) is a bit chilly to be called “room temperature”. Still, this means you don’t need liquid nitrogen for cooling: a good but standard refrigeration system would be enough (and their insane diamond pressure apparatus, of course).
- maxxxxx 7y ago
  Isn’t nitrogen cooling much easier to achieve than this level of pressure?
  - hackerpacker 7y ago
    well the pressure aspect wouldn't require constant energy input. And I assume it is a lot more economical to use a consumer refrigerator than liquid nitrogen. But yah, you can bathe a much more complicated "circuit" in liquid nitrogen a lot easier than compressing it with diamond anvils. Though no commercial superconducting computers have emerged, cmos is just too cost effective currently, and the known examples are still in the 4K range.
    - sfasfsafd 7y ago
      "well the pressure aspect wouldn't require constant energy input." But you have to put energy in to compress it, energy that is lost (since you're compressing it isothermally). Since we're talking about 1E6 atm, any appreciable volume would require insane amounts of E to compress.
      - maxxxxx 7y ago
        Wouldn't that also be pretty much a bomb if the pressure got relieved? I don't know how to do the math but 1E6 atm seems a lot of energy (is it potential energy?) that has to go somewhere.
        bzbarsky 7y ago
        TL;DR: It's pretty much like a bomb, yes. How much energy it is depends on the volume change as you compress. If your material is completely incompressible, there is no volume change, no work done as the pressure rises, and therefore no energy stored. In practice, for a material of bulk modulus K by definition you have -K dV/V = dP. as the basic differential equation relating pressure and volume (if we assume that the bulk modulus is constant through our pressure range, which is quite an assumption in this case). Solving that, we get: V = V_0/exp(P/K) where V_0 is the original volume and V is the volume after the pressure has been applied. Bulk modulus is commonly measured in GPa, so we're going to do that here. 1atm is about 1e5 Pa or 1e-4 GPa. The work done while compressing is the integral of -PdV, which per that first differential equation we can rewrite as the integral of P * V/K * dP (we want to integrate dP because we know what the limits are: it's going from 1atm, or 1e-4 GPa, to 1e6 atm = 1e2 GPa). Plugging in our expression for V we get the integral, from 1e-4 to 100, of P/K * V_0/exp(P/K) dP. The integral there is: K * V_0 * (-P/K - 1) * exp(-P/K) For simplicity, I'm going to approximate that 1e-4 by 0, so we get: K * V_0 * ((-100/K - 1) * exp(-100/K) + 1) when we evaluate at the two endpoints. OK, so let's try an actual material and actual volume. Steel has a bulk modulus of 160 GPa according to <https://en.wikipedia.org/wiki/Bulk_modulus#Selected_values> https://en.wikipedia.org/wiki/Bulk_modulus#Selected_values>. So for steel we have: 160 * V_0 * ((-100/160 - 1) * exp(-100/160) + 1) = 160 * V_0 * 0.13 = 21 * V_0 or so. If we assume we started with a 1 cm^3 volume of steel, that's an energy of: 21e9 N/m^2 * 1e-6 m^3 = 2.1e4 J or about equivalent to 5 grams of TNT per <https://en.wikipedia.org/wiki/TNT_equivalent> https://en.wikipedia.org/wiki/TNT_equivalent>. For comparison, 1 cm^3 of steel is about 7-8 grams of material, so we're right in the "about as much energy as the same weight of TNT" ballpark. Obviously if you start with a larger volume you end up with more stored energy: if we started with 1m^3 we end up equivalent to about 5 tons of TNT. If your material is more compressible, the numbers are higher, as long as it's not too compressible (in which case it just collapses into a tiny volume quickly and not much more work gets done after that). If we take K = 50 (a typical number for glass from that table), you get: 50 * V_0 * ((-100/50 - 1)*exp(-100/50) + 1) = 50 * V_0 * 0.6 = 30 * V_0 If your material is less compressible, you store less energy; for diamond with K = 443, you get something like 9.7*V_0. Again, all this assumes pressure-invariant bulk modulus, which seems moderately unlikely when you are dealing with pressures that are on the order of the size of the bulk modulus or even larger, as here.
    - contravariant 7y ago
      I'm not entirely sure but I think there are no theoretical obstructions to maintaining a ridiculously high pressure indefinitely. Although I suppose there aren't any theoretical obstructions to indefinitely maintaining low temperatures either.
      - imtringued 7y ago
        Even blackholes lose their mass through radiation. Nothing can maintain a high pressure indefinitely.
skunkworker 7y ago
This. lanthanum hydride @ 250K at > 1 million atm. It's still very impressive though.
Zanni 7y ago
Nobody wants "high temperature" superconductors. They want superconductors at roughly standard temperature and pressure. This is only an advance by the extremely artificial metric of considering temperature only. Ceramic superconductors work at liquid nitrogen temperatures and standard pressure, which is much more accessible. I'm not raging at ttsda. It is an advance. Of sorts. But not nearly as much as the headline suggests.
- hoseja 7y ago
  I thought "high temperature", when talking about superconductors, basically means any temperature above the boiling point of nitrogen...
  - chaboud 7y ago
    I think you may mean below. Basically, low temperature superconductors (e.g. niobium titanium) are liquid helium cooled. High temperature superconductors (e.g. YBCO, BSCCO) are everything else, but generally liquid nitrogen cooled.
- jeffwass 7y ago
  The headline is accurate. In the scientific community, High-Tc superconductors typically refer to Type II superconductors, eg cuprates. And the “High-Tc” term means above temperatures relative to the usual elemental superconductor Tc’s, eg about 20K or so. Materials that can superconduct above 77K are especially sought after since they can be cooled with only liquid nitrogen, much cheaper and abundant than liquid helium.
- gaze 7y ago
  You’re vastly underestimating what the significance of what this is part of. High Tc superconductivity was thought to be essentially dead. Extremely few significant new results for decades. Now this and magic angle twisted graphene and some other stuff. It’s really really cool. It’s like we’re watching an entire field thawing and lurching forward.
  - 7y ago
baq 7y ago
this made me think, is the earth core a superconductor?
- effie 7y ago
  Interesting idea. It would help to explain persistence of Earth's magnetic field.
- MichaelApproved 7y ago
  No. The core has pressure but it's way too hot to achieve super conductivity.
taneq 7y ago
Technically correct. The most annoying kind of correct.

At pressures a million times those experienced on earth. Still an advance, of course.

by ttsda 7y ago

At pressures a million times those experienced on earth. Still an advance, of course.

Also, the temperature (250 K, -23°C) is a bit chilly to be called “room temperature”. Still, this means you don’t need liquid nitrogen for cooling: a good but standard refrigeration system would be enough (and their insane diamond pressure apparatus, of course).

maxxxxx 7y ago
Isn’t nitrogen cooling much easier to achieve than this level of pressure?
- hackerpacker 7y ago
  well the pressure aspect wouldn't require constant energy input. And I assume it is a lot more economical to use a consumer refrigerator than liquid nitrogen. But yah, you can bathe a much more complicated "circuit" in liquid nitrogen a lot easier than compressing it with diamond anvils. Though no commercial superconducting computers have emerged, cmos is just too cost effective currently, and the known examples are still in the 4K range.
  - sfasfsafd 7y ago
    "well the pressure aspect wouldn't require constant energy input." But you have to put energy in to compress it, energy that is lost (since you're compressing it isothermally). Since we're talking about 1E6 atm, any appreciable volume would require insane amounts of E to compress.
    - maxxxxx 7y ago
      Wouldn't that also be pretty much a bomb if the pressure got relieved? I don't know how to do the math but 1E6 atm seems a lot of energy (is it potential energy?) that has to go somewhere.
      - bzbarsky 7y ago
        TL;DR: It's pretty much like a bomb, yes. How much energy it is depends on the volume change as you compress. If your material is completely incompressible, there is no volume change, no work done as the pressure rises, and therefore no energy stored. In practice, for a material of bulk modulus K by definition you have -K dV/V = dP. as the basic differential equation relating pressure and volume (if we assume that the bulk modulus is constant through our pressure range, which is quite an assumption in this case). Solving that, we get: V = V_0/exp(P/K) where V_0 is the original volume and V is the volume after the pressure has been applied. Bulk modulus is commonly measured in GPa, so we're going to do that here. 1atm is about 1e5 Pa or 1e-4 GPa. The work done while compressing is the integral of -PdV, which per that first differential equation we can rewrite as the integral of P * V/K * dP (we want to integrate dP because we know what the limits are: it's going from 1atm, or 1e-4 GPa, to 1e6 atm = 1e2 GPa). Plugging in our expression for V we get the integral, from 1e-4 to 100, of P/K * V_0/exp(P/K) dP. The integral there is: K * V_0 * (-P/K - 1) * exp(-P/K) For simplicity, I'm going to approximate that 1e-4 by 0, so we get: K * V_0 * ((-100/K - 1) * exp(-100/K) + 1) when we evaluate at the two endpoints. OK, so let's try an actual material and actual volume. Steel has a bulk modulus of 160 GPa according to <https://en.wikipedia.org/wiki/Bulk_modulus#Selected_values> https://en.wikipedia.org/wiki/Bulk_modulus#Selected_values>. So for steel we have: 160 * V_0 * ((-100/160 - 1) * exp(-100/160) + 1) = 160 * V_0 * 0.13 = 21 * V_0 or so. If we assume we started with a 1 cm^3 volume of steel, that's an energy of: 21e9 N/m^2 * 1e-6 m^3 = 2.1e4 J or about equivalent to 5 grams of TNT per <https://en.wikipedia.org/wiki/TNT_equivalent> https://en.wikipedia.org/wiki/TNT_equivalent>. For comparison, 1 cm^3 of steel is about 7-8 grams of material, so we're right in the "about as much energy as the same weight of TNT" ballpark. Obviously if you start with a larger volume you end up with more stored energy: if we started with 1m^3 we end up equivalent to about 5 tons of TNT. If your material is more compressible, the numbers are higher, as long as it's not too compressible (in which case it just collapses into a tiny volume quickly and not much more work gets done after that). If we take K = 50 (a typical number for glass from that table), you get: 50 * V_0 * ((-100/50 - 1)*exp(-100/50) + 1) = 50 * V_0 * 0.6 = 30 * V_0 If your material is less compressible, you store less energy; for diamond with K = 443, you get something like 9.7*V_0. Again, all this assumes pressure-invariant bulk modulus, which seems moderately unlikely when you are dealing with pressures that are on the order of the size of the bulk modulus or even larger, as here.
  - contravariant 7y ago
    I'm not entirely sure but I think there are no theoretical obstructions to maintaining a ridiculously high pressure indefinitely. Although I suppose there aren't any theoretical obstructions to indefinitely maintaining low temperatures either.
    - imtringued 7y ago
      Even blackholes lose their mass through radiation. Nothing can maintain a high pressure indefinitely.

skunkworker 7y ago

This. lanthanum hydride @ 250K at > 1 million atm. It's still very impressive though.

Zanni 7y ago

Nobody wants "high temperature" superconductors. They want superconductors at roughly standard temperature and pressure. This is only an advance by the extremely artificial metric of considering temperature only. Ceramic superconductors work at liquid nitrogen temperatures and standard pressure, which is much more accessible. I'm not raging at ttsda. It is an advance. Of sorts. But not nearly as much as the headline suggests.

hoseja 7y ago
I thought "high temperature", when talking about superconductors, basically means any temperature above the boiling point of nitrogen...
- chaboud 7y ago
  I think you may mean below. Basically, low temperature superconductors (e.g. niobium titanium) are liquid helium cooled. High temperature superconductors (e.g. YBCO, BSCCO) are everything else, but generally liquid nitrogen cooled.
jeffwass 7y ago
The headline is accurate. In the scientific community, High-Tc superconductors typically refer to Type II superconductors, eg cuprates. And the “High-Tc” term means above temperatures relative to the usual elemental superconductor Tc’s, eg about 20K or so. Materials that can superconduct above 77K are especially sought after since they can be cooled with only liquid nitrogen, much cheaper and abundant than liquid helium.
gaze 7y ago
You’re vastly underestimating what the significance of what this is part of. High Tc superconductivity was thought to be essentially dead. Extremely few significant new results for decades. Now this and magic angle twisted graphene and some other stuff. It’s really really cool. It’s like we’re watching an entire field thawing and lurching forward.
- 7y ago

baq 7y ago

this made me think, is the earth core a superconductor?

effie 7y ago
Interesting idea. It would help to explain persistence of Earth's magnetic field.
MichaelApproved 7y ago
No. The core has pressure but it's way too hot to achieve super conductivity.

taneq 7y ago

Technically correct. The most annoying kind of correct.