8 ms·

Why are folks trying to prove that this material is a superconductor in roundabout ways, like levitation, dimagnetism, etc? What is the reason they don't just t

Why are folks trying to prove that this material is a superconductor in roundabout ways, like levitation, dimagnetism, etc? What is the reason they don't just test to see if electrical current flows without resistance? Surely there is something I am missing here.

rendang 3y ago
Purely speculating, but might it be easy to doubt resistance measurements by claiming that equipment was faulty or used incorrectly?
jacquesm 3y ago
They do both. But measuring the resistance of something that is that small is super hard because it will be close to zero anyway even if it isn't superconducting. So a larger sample would give much more conclusive results.
caconym_ 3y ago
I have heard people much more knowledgeable than me say that measuring true zero resistance is actually quite difficult and takes some degree of specialized equipment, especially with such small samples. That may be part of it.
- jacquesm 3y ago
  Just getting the probes to connect reliably is tricky. depending on how large the superconducting features are it may be anywhere from just difficult to next to impossible to do accurately (for instance, if the size of the superconducting features is smaller than the probe size).
- balaji1 3y ago
  so none of the previous more esoteric superconductors were ever measure for zero resistance?
  - whimsicalism 3y ago
    They were, but they had more time to resolve the difficulties
  - scarmig 3y ago
    They were. E.g. see here for a major, good paper announcing a novel class of superconductors: https://sci-hub.ru/https://link.springer.com/article/10.1007/bf01303701 https://sci-hub.ru/https://link.springer.com/article/10.1007... The difference now is that we're seeing a premature preprint being replicated in real time. Even in that paper, the authors note: "The way the samples have been prepared seems to be of crucial importance: Michel et al. [21] obtained a single-phase perovskite by mixing the oxides of La and Cu and BaCOa in an appropriate ratio and subsequent annealing at 1,000 ~ in air. We also applied this annealing condition to one of our samples, obtained by the decomposition of the corresponding oxalates, and found no superconductivity." And you can see that in their resistivity/temperature graph of samples prepared using different protocols.
    - pyrale 3y ago
      > a premature preprint Considering how that preprint has sparked interest in other research institutions and multiplied the resources allocated to the problem, I would say this publication was not premature, it's most other research results that are late.
      - squeaky-clean 3y ago
        The paper was leaked early. The team wanted more time to get more attempts at producing it and improving their yield. The paper also wasn't generally up to their writing standards. It's essentially an early draft that was leaked.
        gus_massa 3y ago
        It looks like this worked in this material for years, perhaps the would have taken a few more years.
- frisco 3y ago
  Right, if our best regular conductors (used in your ohmmeter) are ~10^-8 and superconductivity is (by convention) less than 10^-11, one can see right away the simple regular methods won’t work and some cleverness is needed.
  - someplaceguy 3y ago
    > superconductivity is (by convention) less than 10^-11, Ah, so you're saying that superconductivity is not actual zero resistance, but something close to it, and in fact only a factor of 1000x less resistive than the best conductor? If that is so, this is something that I had previously thought would make a lot more sense to me. But in that case it's not intuitive to me how SMES is possible with a 0% discharge rate. Shouldn't a significant fraction of the electrons looping around the coils be lost after many loops? (I know very little about electricity, as you can probably tell, never mind superconductors).
    - mort96 3y ago
      No, I believe it's literally zero but we don't have a measurement apparatus with infinite precision so we need some cut-off.
      - someplaceguy 3y ago
        Thanks, that makes sense.
      - deleted 3y ago
        [deleted]
      - fooker 3y ago
        It's only literal zero for superconductors close to 0K temperature. For high temperature superconductors (50-70+K), it's not literal zero for superconducting mechanisms discovered so far.
  - crote 3y ago
    The conductors of your ohmmeter are not that important, though. You can work around that by using four-terminal sensing, and you can of course also calibrate your probes by directly touching them together. Even if your ohmmeter conductors have a resistance of several ohm, you could still get an accurate measurement if your tool has a high enough resolution. A bigger issue is going to be sample size. A 1mm-diameter 1mm-long rod of silver has a resistance of about 20 μΩ (or 2e-5) at room temperature. That's already getting tricky to measure with lab-grade equipment without pushing insane currents through it, let alone anything even smaller. If you want to measure a 1m-diameter 1m-long silver rod (which would be 0.02μΩ or 2e-8) you could just push a few thousand amps through it and reliably measure that using a household multimeter in the mV range - but do that with a small sample and it'll evaporate.
Daneel_ 3y ago
It’s difficult to rule out external factors - a badly attached probe could also result in zero resistance for example. Showing diamagnetism is one of the least error-prone ways to demonstrate the superconductor effect. That’s my understanding anyway.
- jacquesm 3y ago
  > a badly attached probe could also result in zero resistance for example No, a badly attached probe would usually show a larger resistance, not a smaller one. That's actually the easiest error to make, making improper contact with the sample. The resistance is measured indirectly using a reference current. So you'd measure a higher resistance or a break rather than zero if a probe were not attached correctly (unless the two voltage probes are touching but that would normally speaking be spotted). The diamagnetism is simply easier to verify using an impure or small sample.
  - Daneel_ 3y ago
    Glad to be corrected, thank you.
  - voidmain 3y ago
    It's apparently hard to measure zero vs extremely low resistance with the two probe setup you are imagining (I guess because the probes and wires aren't superconducting, so most of the resistance in the circuit is not in the sample). The graphs I have seen are all made with a four probe setup [1], where a constant current is run through the outer probe pair and the voltage across the inner pair is measured. If the inner probes have contact issues, the voltage (and inferred resistance) drops, potentially to zero. [1] https://www.ni.com/docs/en-US/bundle/ni-daqmx/page/measfunds/4wireres.html https://www.ni.com/docs/en-US/bundle/ni-daqmx/page/measfunds...
    - applied_heat 3y ago
      That’s what he said. You inject a known current and measure the voltage drop across the item you want to measure the resistance of, and then use ohms law
bhaney 3y ago
It's difficult to produce a sample large enough and pure enough to have its resistance measured.
ProjectArcturis 3y ago
Not an expert, but I would guess a direct measurement couldn't distinguish between a true superconductor and something with very very low resistance.
- Zardoz84 3y ago
  No. Any device to measuring resistence ALWAYS have a minimal value that can measure. They can never measure exactly zero ohms.
  - ProjectArcturis 3y ago
    Isn't that exactly what I said?
httpz 3y ago
With the current fabrication process, they're only getting a chunk with LK-99 particles sprinkled in. Since nobody yet knows how to fabricate a pure chunk of LK-99, it'll be hard to measure the true resistance of LK-99.
- appplication 3y ago
  Thank you, this is the best explanation I’ve heard for why it’s so hard of measure.
- hgrbrm 3y ago
  How do you explain the almost three orders of magnitude drop in resistance in Fig. 6d in one of the original articles ( https://arxiv.org/pdf/2307.12037.pdf https://arxiv.org/pdf/2307.12037.pdf ) with a few LK-99 particles sprinkled here an there? There must be a current path along which more than 99.8% of the material is in the supposed superconducting state. So the particles almost touch but not quite yet? I think the more likely explanation is that the particles do touch each other but the interface is not superconductive. In other words, it is a polycrystalline material, and most of it is LK-99, but the grain boundaries are not a very good conductor. In conventional superconductors grain boundaries don't disrupt superconductivity because they are 3D superconductors, but in this allegedly 1D superconductor the superconducting channels in most cases don't meet at the grain boundaries, so the current has to overcome the resistance of some material that is almost an insulator. If that is the case it will be difficult to produce a material that is macroscopically superconducting. But I hope researchers will be able to make single crystals that are large enough for resistance measurements so that finally it can be determined if this material is a superconductor or not. For practical uses the best result that can be achieved with this material may be a metal-LK-99 composite where the LK-99 particles lower the resistivity of the metal by 50-90%.
- adrian_b 3y ago
  I believe that with the synthesis method proposed by the Korean team, i.e. by the chemical reaction between a certain kind of lead sulfate with copper phosphide, the chances of progress are slim. The Korean team appears to have been stuck for several years by the lack of reproducibility of this synthesis method. While it was a great discovery that has shown that this material must have some very interesting properties, perhaps even superconductivity at ambient temperature and pressure, in order to be able to measure its properties and be able to evaluate the possible practical applications, a much more precise method for enforcing the desired crystal structure is required, than mixing powders and baking them into a ceramic. Perhaps such a method for producing samples with deterministic properties would be to develop first a method to grow monocrystals of the special kind of lead phosphate that forms the base crystal structure, maybe by drawing the crystals from melt. Once monocrystals of this kind of lead phosphate are available, they could be doped with copper, e.g. by ion implantation. By controlling and varying the parameters of the process, e.g. the angle of incidence and the velocity of the ions and the thermal profile used for annealing, it is likely that reproducible samples can be produced, where the copper ions substitute lead in the useful places and not in the others. By this method it would be possible to produce only thin layers of LK-99, but that should be enough to enable the characterization of the material. Moreover, because LK-99 is very fragile, it is unlikely that it could be used to make cables or coils. Practical uses where LK-99 would be deposited as thin films are much more likely. As an alternative to ion implantation, which might be able to produce thicker layers, perhaps once monocrystals of the base lead phosphate are available it may be possible to develop some method of chemical vapor deposition, to grow epitaxially a layer of LK-99 over the base crystal, but with such a method it is less obvious if there is any way to control which lead atoms are substituted, though this may depend on the orientation of the base crystal.
dontwearitout 3y ago
My understanding is that synthesis of the exactly right crystal structure is very hard (for every 10 lead atoms, 1 - the exact right one - needs to be substituted with copper), so the samples are small and inhomogeneous. As a result measuring resistivity won't be illuminating until large amounts of perfect material can be produced.
- octacat 3y ago
  kinda like with Graphene, cool material, hard to produce.
postalrat 3y ago
How would you prove there isn't any water in a cup?
- bagels 3y ago
  I didn't see any references to cups or water in these papers.
  - postalrat 3y ago
    If you can't answer the question then you won't ever understand how to measure zero resistance.
    - svnt 3y ago
      It's a limitation of the most basic approach to the problem (measuring resistance), but not a limitation of every approach (such as measuring current). You can inject a current into a superconducting coil and take measurements of the resultant magnetic field as the current circles for an indefinite period of time. I'm not able to see how this approach analogizes to water in a cup.
      - davrosthedalek 3y ago
        I don't think that works as advertised: Any measurement of the current will likely produce a magnetic field itself -- for example from the current in a Hall probe. This current will induct a small, opposite current into the superconducting coil. So the current will go down. And even if not, what you would need is to measure the change of the field over time. This has finite resolution, so you can't distinguish no resistance from very very small resistance.
        svnt 3y ago
        There are several tests that have been ongoing over decades doing exactly this. The small loss of current during measurement occurs, but you only engage with the field intermittently, and you can calculate approximately what the loss should be. Yes it could be some tiny resistance, but the same issue occurs with the resolution/accuracy of the voltage or current measurement you would make.
Zardoz84 3y ago
Because (like someone explained very well in another HW notice) you can't measure zero ohms. Any device to measure resistance ALWAYS would have a minimal value that can measure. It's far more easy to detect superconductivity using the weird things that does all superconductors in presence of magnetic fields (ie, levitation).
hgomersall 3y ago
Without being intentionally snarky, the same logic applies to measuring current and voltage. What is observed is an effect of zero resistance, not zero resistance itself (whatever that would mean).
m3kw9 3y ago
The korean guy say this is a 1d superconductor (as opposed to 2d sheet or 3d, which means you would need a single line of superconducting molecules from end to end. Its likely why it doesn't float all the way as only parts of it has these and scattered all over the chunk of solid. However, if they perfect 1d production, they can layer in a bunch of them to create a quasi 2d or 3d superconductor.

rendang 3y ago

Purely speculating, but might it be easy to doubt resistance measurements by claiming that equipment was faulty or used incorrectly?

jacquesm 3y ago

They do both. But measuring the resistance of something that is that small is super hard because it will be close to zero anyway even if it isn't superconducting. So a larger sample would give much more conclusive results.

caconym_ 3y ago

I have heard people much more knowledgeable than me say that measuring true zero resistance is actually quite difficult and takes some degree of specialized equipment, especially with such small samples. That may be part of it.

jacquesm 3y ago
Just getting the probes to connect reliably is tricky. depending on how large the superconducting features are it may be anywhere from just difficult to next to impossible to do accurately (for instance, if the size of the superconducting features is smaller than the probe size).
balaji1 3y ago
so none of the previous more esoteric superconductors were ever measure for zero resistance?
- whimsicalism 3y ago
  They were, but they had more time to resolve the difficulties
- scarmig 3y ago
  They were. E.g. see here for a major, good paper announcing a novel class of superconductors: https://sci-hub.ru/https://link.springer.com/article/10.1007/bf01303701 https://sci-hub.ru/https://link.springer.com/article/10.1007... The difference now is that we're seeing a premature preprint being replicated in real time. Even in that paper, the authors note: "The way the samples have been prepared seems to be of crucial importance: Michel et al. [21] obtained a single-phase perovskite by mixing the oxides of La and Cu and BaCOa in an appropriate ratio and subsequent annealing at 1,000 ~ in air. We also applied this annealing condition to one of our samples, obtained by the decomposition of the corresponding oxalates, and found no superconductivity." And you can see that in their resistivity/temperature graph of samples prepared using different protocols.
  - pyrale 3y ago
    > a premature preprint Considering how that preprint has sparked interest in other research institutions and multiplied the resources allocated to the problem, I would say this publication was not premature, it's most other research results that are late.
    - squeaky-clean 3y ago
      The paper was leaked early. The team wanted more time to get more attempts at producing it and improving their yield. The paper also wasn't generally up to their writing standards. It's essentially an early draft that was leaked.
      - gus_massa 3y ago
        It looks like this worked in this material for years, perhaps the would have taken a few more years.
frisco 3y ago
Right, if our best regular conductors (used in your ohmmeter) are ~10^-8 and superconductivity is (by convention) less than 10^-11, one can see right away the simple regular methods won’t work and some cleverness is needed.
- someplaceguy 3y ago
  > superconductivity is (by convention) less than 10^-11, Ah, so you're saying that superconductivity is not actual zero resistance, but something close to it, and in fact only a factor of 1000x less resistive than the best conductor? If that is so, this is something that I had previously thought would make a lot more sense to me. But in that case it's not intuitive to me how SMES is possible with a 0% discharge rate. Shouldn't a significant fraction of the electrons looping around the coils be lost after many loops? (I know very little about electricity, as you can probably tell, never mind superconductors).
  - mort96 3y ago
    No, I believe it's literally zero but we don't have a measurement apparatus with infinite precision so we need some cut-off.
    - someplaceguy 3y ago
      Thanks, that makes sense.
    - deleted 3y ago
      [deleted]
    - fooker 3y ago
      It's only literal zero for superconductors close to 0K temperature. For high temperature superconductors (50-70+K), it's not literal zero for superconducting mechanisms discovered so far.
- crote 3y ago
  The conductors of your ohmmeter are not that important, though. You can work around that by using four-terminal sensing, and you can of course also calibrate your probes by directly touching them together. Even if your ohmmeter conductors have a resistance of several ohm, you could still get an accurate measurement if your tool has a high enough resolution. A bigger issue is going to be sample size. A 1mm-diameter 1mm-long rod of silver has a resistance of about 20 μΩ (or 2e-5) at room temperature. That's already getting tricky to measure with lab-grade equipment without pushing insane currents through it, let alone anything even smaller. If you want to measure a 1m-diameter 1m-long silver rod (which would be 0.02μΩ or 2e-8) you could just push a few thousand amps through it and reliably measure that using a household multimeter in the mV range - but do that with a small sample and it'll evaporate.

Daneel_ 3y ago

It’s difficult to rule out external factors - a badly attached probe could also result in zero resistance for example. Showing diamagnetism is one of the least error-prone ways to demonstrate the superconductor effect. That’s my understanding anyway.

jacquesm 3y ago
> a badly attached probe could also result in zero resistance for example No, a badly attached probe would usually show a larger resistance, not a smaller one. That's actually the easiest error to make, making improper contact with the sample. The resistance is measured indirectly using a reference current. So you'd measure a higher resistance or a break rather than zero if a probe were not attached correctly (unless the two voltage probes are touching but that would normally speaking be spotted). The diamagnetism is simply easier to verify using an impure or small sample.
- Daneel_ 3y ago
  Glad to be corrected, thank you.
- voidmain 3y ago
  It's apparently hard to measure zero vs extremely low resistance with the two probe setup you are imagining (I guess because the probes and wires aren't superconducting, so most of the resistance in the circuit is not in the sample). The graphs I have seen are all made with a four probe setup [1], where a constant current is run through the outer probe pair and the voltage across the inner pair is measured. If the inner probes have contact issues, the voltage (and inferred resistance) drops, potentially to zero. [1] https://www.ni.com/docs/en-US/bundle/ni-daqmx/page/measfunds/4wireres.html https://www.ni.com/docs/en-US/bundle/ni-daqmx/page/measfunds...
  - applied_heat 3y ago
    That’s what he said. You inject a known current and measure the voltage drop across the item you want to measure the resistance of, and then use ohms law

bhaney 3y ago

It's difficult to produce a sample large enough and pure enough to have its resistance measured.

ProjectArcturis 3y ago

Not an expert, but I would guess a direct measurement couldn't distinguish between a true superconductor and something with very very low resistance.

Zardoz84 3y ago
No. Any device to measuring resistence ALWAYS have a minimal value that can measure. They can never measure exactly zero ohms.
- ProjectArcturis 3y ago
  Isn't that exactly what I said?

httpz 3y ago

With the current fabrication process, they're only getting a chunk with LK-99 particles sprinkled in. Since nobody yet knows how to fabricate a pure chunk of LK-99, it'll be hard to measure the true resistance of LK-99.

appplication 3y ago
Thank you, this is the best explanation I’ve heard for why it’s so hard of measure.
hgrbrm 3y ago
How do you explain the almost three orders of magnitude drop in resistance in Fig. 6d in one of the original articles ( https://arxiv.org/pdf/2307.12037.pdf https://arxiv.org/pdf/2307.12037.pdf ) with a few LK-99 particles sprinkled here an there? There must be a current path along which more than 99.8% of the material is in the supposed superconducting state. So the particles almost touch but not quite yet? I think the more likely explanation is that the particles do touch each other but the interface is not superconductive. In other words, it is a polycrystalline material, and most of it is LK-99, but the grain boundaries are not a very good conductor. In conventional superconductors grain boundaries don't disrupt superconductivity because they are 3D superconductors, but in this allegedly 1D superconductor the superconducting channels in most cases don't meet at the grain boundaries, so the current has to overcome the resistance of some material that is almost an insulator. If that is the case it will be difficult to produce a material that is macroscopically superconducting. But I hope researchers will be able to make single crystals that are large enough for resistance measurements so that finally it can be determined if this material is a superconductor or not. For practical uses the best result that can be achieved with this material may be a metal-LK-99 composite where the LK-99 particles lower the resistivity of the metal by 50-90%.
adrian_b 3y ago
I believe that with the synthesis method proposed by the Korean team, i.e. by the chemical reaction between a certain kind of lead sulfate with copper phosphide, the chances of progress are slim. The Korean team appears to have been stuck for several years by the lack of reproducibility of this synthesis method. While it was a great discovery that has shown that this material must have some very interesting properties, perhaps even superconductivity at ambient temperature and pressure, in order to be able to measure its properties and be able to evaluate the possible practical applications, a much more precise method for enforcing the desired crystal structure is required, than mixing powders and baking them into a ceramic. Perhaps such a method for producing samples with deterministic properties would be to develop first a method to grow monocrystals of the special kind of lead phosphate that forms the base crystal structure, maybe by drawing the crystals from melt. Once monocrystals of this kind of lead phosphate are available, they could be doped with copper, e.g. by ion implantation. By controlling and varying the parameters of the process, e.g. the angle of incidence and the velocity of the ions and the thermal profile used for annealing, it is likely that reproducible samples can be produced, where the copper ions substitute lead in the useful places and not in the others. By this method it would be possible to produce only thin layers of LK-99, but that should be enough to enable the characterization of the material. Moreover, because LK-99 is very fragile, it is unlikely that it could be used to make cables or coils. Practical uses where LK-99 would be deposited as thin films are much more likely. As an alternative to ion implantation, which might be able to produce thicker layers, perhaps once monocrystals of the base lead phosphate are available it may be possible to develop some method of chemical vapor deposition, to grow epitaxially a layer of LK-99 over the base crystal, but with such a method it is less obvious if there is any way to control which lead atoms are substituted, though this may depend on the orientation of the base crystal.

dontwearitout 3y ago

My understanding is that synthesis of the exactly right crystal structure is very hard (for every 10 lead atoms, 1 - the exact right one - needs to be substituted with copper), so the samples are small and inhomogeneous. As a result measuring resistivity won't be illuminating until large amounts of perfect material can be produced.

octacat 3y ago
kinda like with Graphene, cool material, hard to produce.

postalrat 3y ago

How would you prove there isn't any water in a cup?

bagels 3y ago
I didn't see any references to cups or water in these papers.
- postalrat 3y ago
  If you can't answer the question then you won't ever understand how to measure zero resistance.
  - svnt 3y ago
    It's a limitation of the most basic approach to the problem (measuring resistance), but not a limitation of every approach (such as measuring current). You can inject a current into a superconducting coil and take measurements of the resultant magnetic field as the current circles for an indefinite period of time. I'm not able to see how this approach analogizes to water in a cup.
    - davrosthedalek 3y ago
      I don't think that works as advertised: Any measurement of the current will likely produce a magnetic field itself -- for example from the current in a Hall probe. This current will induct a small, opposite current into the superconducting coil. So the current will go down. And even if not, what you would need is to measure the change of the field over time. This has finite resolution, so you can't distinguish no resistance from very very small resistance.
      - svnt 3y ago
        There are several tests that have been ongoing over decades doing exactly this. The small loss of current during measurement occurs, but you only engage with the field intermittently, and you can calculate approximately what the loss should be. Yes it could be some tiny resistance, but the same issue occurs with the resolution/accuracy of the voltage or current measurement you would make.

Zardoz84 3y ago

Because (like someone explained very well in another HW notice) you can't measure zero ohms. Any device to measure resistance ALWAYS would have a minimal value that can measure. It's far more easy to detect superconductivity using the weird things that does all superconductors in presence of magnetic fields (ie, levitation).

hgomersall 3y ago

Without being intentionally snarky, the same logic applies to measuring current and voltage. What is observed is an effect of zero resistance, not zero resistance itself (whatever that would mean).

m3kw9 3y ago

The korean guy say this is a 1d superconductor (as opposed to 2d sheet or 3d, which means you would need a single line of superconducting molecules from end to end. Its likely why it doesn't float all the way as only parts of it has these and scattered all over the chunk of solid. However, if they perfect 1d production, they can layer in a bunch of them to create a quasi 2d or 3d superconductor.