Conductivity of metals at 0 kelvin?

  1. What is the conductivity of metals at 0 kelvin? i think it will be zero because at 0 k entropy is zero. Every motion is cease.
     
  2. jcsd
  3. Re: conductivity

    plz tell me hypothetically it will zero or not?
     
  4. Re: conductivity

    If you measure conductivity, you apply a voltage that moves electrons. No need for entropy.

    Beyond this, experimental data exists for very low temperature.
     
  5. Drakkith

    Staff: Mentor

    Re: conductivity

    IF we did reach 0k and you applied a voltage to the metal, then you would introduce energy and it would no longer be at 0k anyways. In as soon as you applied a voltage it would probably be a superconductor.
     
  6. DrDu

    DrDu 4,351
    Science Advisor

    It depends on how pure the metal is. If there are scattering centers, the conductivity will remain finite. For very pure metals, the conductivity becomes very high and will ultimately only be limited by scattering from the surfaces.
     
  7. Re: conductivity

    thank's
     
  8. Low-Q

    Low-Q 258
    Gold Member

    Re: conductivity

    If you apply voltage anywhere on a superconductor, the voltage will be measured as a result of loss in the wires from the instrument you're measuring with. Current flow will not cause voltage drop over a superconductor, thus no heat.

    Vidar
     
  9. Drakkith

    Staff: Mentor

    Re: conductivity

    I see your point. I have to ask though, if you are adding energy into the superconductor by applying a voltage, is that not related to the internal energy and thus the temperature somehow? Being a minimum energy state, I would expect that any difference in potential or current flow would make it so that the material is no longer in that minimum state.
     
  10. this is wrong.Entropy is not zero at zero k (don't think about formula).it is because of Heisenberg principal because if there will not be any randomness then you can measure both position and momentum simultaneously perfectly!
     
  11. mfb

    Staff: Mentor

    If everything is in its ground state, you have minimal energy and zero entropy. The uncertainty relation does not matter, it just tells you that minimal energy in quantum mechanics is above the minimal energy in classical mechanics.

    I doubt that you can apply a voltage without changing the entropy.
     
  12. Low-Q

    Low-Q 258
    Gold Member

    The current will not cause voltage drop over the superconductor. Energy is in this case voltage x current. If voltage is zero the product will be zero energy. However, the voltage drop over the wires from the instrument will cause heat, but only in the wires. So you actually dont apply energy into a superconductor - only in the wires you try to tranfer that energy with.
     
  13. I don't think zero entropy is any proper world because uncertainty relation really matters when one deals with subatomic things.
     
  14. mfb

    Staff: Mentor

    This is wrong.
    Entropy is defined via the states of the system - and those states already take the uncertainty relation into account.
     
  15. you probably mean to those phase space element where there is certain volume of phase cell but they don't say anything about zero entropy.
     
  16. mfb

    Staff: Mentor

    $$S=-k_B \sum_i P_i \ln(P_i)$$
    If the ground state is not degenerate, ##P_i=0## everywhere apart from the ground state, where ##P_g=1##. Take the limit to avoid ln(0), and you get ##S=-k_B (1 ln(1)+0)=0##.
    Who needs volumes of anything? This is a general result, you can apply it to all thermodynamical systems - spins, gases, crystals, whatever.

    If the ground state is degenerate, you get some tiny amount of entropy.
     
  17. ZapperZ

    ZapperZ 30,165
    Staff Emeritus
    Science Advisor
    Education Advisor

    This is a very clear example where a seemingly simple, elementary question does not have a simple, elementary answer.

    The answer you get depends very much on how complex and at what level you wish to receive:

    1. High School.
    The conductivity is infinite, meaning the resistivity approaches zero. This is imply based on extrapolating what we know from looking at the dependence of conductivity with temperature.

    2. Undergraduate level.
    The conductivity is expected to be infinite, i.e. resistivity approaches zero. This is because the predominant source of resistivity (lattice vibrations) diminishes to zero (theoretically) at T=0.

    3. Graduate/professional level.
    The answer has two forms: theoretical and experimental. Theoretically, the properties of a "typical" metal can be accurately described by Landau's Fermi Liquid theory. Here, one can employ the Drude model and arrive at a description of the scattering rate of the charge carrier (quasiparticles) in a metal that depends on (i) electron-phonon scattering (ii) electron-electron scattering (ii) electron-impurity/defect scattering. Scattering rate of (i) and (ii) are temperature dependent and can approach zero as T approaches zero. However, scattring rate (iii) does not. It is almost a constant.

    Thus, one needs to look if one is asking about our ordinary, REAL metals, or some idolized, perfect, single-crystal, no impurity/defect metal. Any metal of any considerable size will have impurity and defect (i.e. grain boundaries, etc. even without impurities). Thus, what will happen here is that there will be something called "residual resistivity" at T=0. And this is where the experiment comes in, because such a study has been done a long time ago, showing not only such resistivity at very low temperatures, but also the T^2 dependence of the electron-electron scattering (as predicted by the Fermi liquid theory).

    So there!

    Zz.
     
  18. classical calculation does not apply as I already said.It is pointed out in feynman lectures vol. 1 that uncertainty principle must be invoked for non zero entropy.see some early chapter,it is written there.
     
  19. ZapperZ

    ZapperZ 30,165
    Staff Emeritus
    Science Advisor
    Education Advisor

    What classical calculation? The scattering rates are obtained using QFT!

    Zz.
     
  20. just did not see the second page there(too hurry).I replied to the last post of page 1.Sorry for causing trouble.
     
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