43 ms·

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

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
    - SV_BubbleTime 3y ago
      And what he saying, and you weren’t listening to is it all your testing equipment is introducing resistance. You can mesaure current by electrical fields, but have the same issues as before. Your testing equipment and your very tiny sample sizes.
      - fnordpiglet 3y ago
        I sort of don’t get this either. If your testing equipment is introducing resistance can you not “tare” or calibrate your measurements by measuring the equipments current flow in a direct circuit between probes to determine its base resistance then introduce your sample and measure the difference? The resistance in a serial circuit is additive, no?
        deleted 3y ago
        [deleted]
        proto_lambda 3y ago
        The parasitic resistance will be orders of magnitude larger than your sample resistance, any detector capable of detecting the sample resistance would be completely swamped out by it.
        jacquesm 3y ago
        They are measuring voltage, not resistance (the resistance is computed). It's impossible to measure resistance without current and some voltage across the resistance, and while the parasitic effects are small the resistance we're talking about (and hence the voltage across that resistance) is so small that second order effects introduced by the probe wires can have a real effect, for instance if the probe wires have a higher resistance that makes them more susceptible to electric fields, which in turn would show up as a voltage. So this is anything but trivial.
        jacquesm 3y ago
        If you pick your reference current properly this effect should be extremely small. It will have a little bit of effect but nothing that you should normally have to worry about.
  - jacquesm 3y ago
    I am not imagining a 'two probe' setup (see other comment in this thread).
- fsckboy 3y ago
  >No, a badly attached probe would usually show a larger resistance, not a smaller one. if the area is small, couldn't it be "badly attached" directly to the other probe?
  - jacquesm 3y ago
    That could happen, but you'd have such a perfect zero across the whole temp range that I think any researcher worth their pay would spot the error.
qnleigh 3y ago
That's right. The standard way to measure resistance without being sensitive to contact resistance is called the four-terminal method, see (https://en.m.wikipedia.org/wiki/Four-terminal_sensing https://en.m.wikipedia.org/wiki/Four-terminal_sensing). You drive a current using two outer wires, but then detect the voltage across the sample using a different pair of wires. You'll measure zero voltage if it's superconducting, since V=IR. Or if one of the probe wires became detached. The first LK-99 paper used this method to claim zero resistivity, but people complained that if the inner probes lost contact, that would also be consistent with their data. This criticism doesn't totally make sense to me, since the apparent superconductivity came and went in the expected way as they changed an external magnetic field. I don't understand how a loose terminal could mimic that figure (I think it was in figure 1).
- fnordpiglet 3y ago
  It also doesn’t make sense to me because you would expect they didn’t just say “omg super conductivity!” on their first measurement and presumably measured more than once, verified connections, and generally assumed scrutiny would be high on any claim and (from what I’ve read of the history so far) the reason it’s from 1999 and published now is they didn’t believe it was real to begin with but believed it was a measurement mistake. This doesn’t mean others should be credulous and replication is crucial. But it seems to beggar reality to assume they made such an amateur mistake. I would assume outright fraud first.
- EvgeniyZh 3y ago
  One case in which you can see something similar to SC transition is percolation, where the sample is mix of regular metal and insulator, and you (accidentally) connect current contacts to metal but voltage probes to insulator [1] [1] https://arxiv.org/abs/1808.05871 https://arxiv.org/abs/1808.05871
- stolsvik 3y ago
  Don’t you also measure the current of the “inner probes” (the ones “driving a current”), to verify that you do have contact?!

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
  - SV_BubbleTime 3y ago
    And what he saying, and you weren’t listening to is it all your testing equipment is introducing resistance. You can mesaure current by electrical fields, but have the same issues as before. Your testing equipment and your very tiny sample sizes.
    - fnordpiglet 3y ago
      I sort of don’t get this either. If your testing equipment is introducing resistance can you not “tare” or calibrate your measurements by measuring the equipments current flow in a direct circuit between probes to determine its base resistance then introduce your sample and measure the difference? The resistance in a serial circuit is additive, no?
      - deleted 3y ago
        [deleted]
      - proto_lambda 3y ago
        The parasitic resistance will be orders of magnitude larger than your sample resistance, any detector capable of detecting the sample resistance would be completely swamped out by it.
        jacquesm 3y ago
        They are measuring voltage, not resistance (the resistance is computed). It's impossible to measure resistance without current and some voltage across the resistance, and while the parasitic effects are small the resistance we're talking about (and hence the voltage across that resistance) is so small that second order effects introduced by the probe wires can have a real effect, for instance if the probe wires have a higher resistance that makes them more susceptible to electric fields, which in turn would show up as a voltage. So this is anything but trivial.
      - jacquesm 3y ago
        If you pick your reference current properly this effect should be extremely small. It will have a little bit of effect but nothing that you should normally have to worry about.
- jacquesm 3y ago
  I am not imagining a 'two probe' setup (see other comment in this thread).
fsckboy 3y ago
>No, a badly attached probe would usually show a larger resistance, not a smaller one. if the area is small, couldn't it be "badly attached" directly to the other probe?
- jacquesm 3y ago
  That could happen, but you'd have such a perfect zero across the whole temp range that I think any researcher worth their pay would spot the error.

qnleigh 3y ago

That's right. The standard way to measure resistance without being sensitive to contact resistance is called the four-terminal method, see (https://en.m.wikipedia.org/wiki/Four-terminal_sensing https://en.m.wikipedia.org/wiki/Four-terminal_sensing). You drive a current using two outer wires, but then detect the voltage across the sample using a different pair of wires. You'll measure zero voltage if it's superconducting, since V=IR. Or if one of the probe wires became detached. The first LK-99 paper used this method to claim zero resistivity, but people complained that if the inner probes lost contact, that would also be consistent with their data. This criticism doesn't totally make sense to me, since the apparent superconductivity came and went in the expected way as they changed an external magnetic field. I don't understand how a loose terminal could mimic that figure (I think it was in figure 1).

fnordpiglet 3y ago
It also doesn’t make sense to me because you would expect they didn’t just say “omg super conductivity!” on their first measurement and presumably measured more than once, verified connections, and generally assumed scrutiny would be high on any claim and (from what I’ve read of the history so far) the reason it’s from 1999 and published now is they didn’t believe it was real to begin with but believed it was a measurement mistake. This doesn’t mean others should be credulous and replication is crucial. But it seems to beggar reality to assume they made such an amateur mistake. I would assume outright fraud first.
EvgeniyZh 3y ago
One case in which you can see something similar to SC transition is percolation, where the sample is mix of regular metal and insulator, and you (accidentally) connect current contacts to metal but voltage probes to insulator [1] [1] https://arxiv.org/abs/1808.05871 https://arxiv.org/abs/1808.05871
stolsvik 3y ago
Don’t you also measure the current of the “inner probes” (the ones “driving a current”), to verify that you do have contact?!