# Conductivity of metals at 0 kelvin?

• spectrum123
In summary: But overall, the entropy is zero because all the states are equally possible and have the same entropy.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##.
spectrum123
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.

please tell me hypothetically it will zero or not?

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

Beyond this, experimental data exists for very low temperature.

spectrum123 said:
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.

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.

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.

thank's

Drakkith said:
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.
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

Low-Q said:
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

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.

i think it will be zero because at 0 k entropy is zero
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!

andrien said:
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!
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.

Drakkith said:
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.
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 don't apply energy into a superconductor - only in the wires you try to tranfer that energy with.

I don't think zero entropy is any proper world because uncertainty relation really matters when one deals with subatomic things.

andrien said:
I don't think zero entropy is any proper world because uncertainty relation really matters when one deals with subatomic things.
This is wrong.
Entropy is defined via the states of the system - and those states already take the uncertainty relation into account.

mfb said:
This is wrong.
Entropy is defined via the states of the system - and those states already take the uncertainty relation into account.
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.

$$S=-k_B \sum_i P_i \ln(P_i)$$
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.

spectrum123 said:
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.

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.

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.

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.

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.

andrien said:
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.

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

Zz.

ZapperZ said:
What classical calculation? The scattering rates are obtained using QFT!
just did not see the second page there(too hurry).I replied to the last post of page 1.Sorry for causing trouble.

I don't see any classical calculation in my post.

does your eqn take into account the uncertainty principle.I don't think so.

The pi take the uncertainty principle into account.
Seriously, I don't understand your problem with my equation. You just repeat "uncertainty principle" every time. It is like I would counter this with "particle mass" all the time. I mean, where is your point?

The problem here is that people think that the HUP is a starting point in QM rather than merely a consequence. It is as if one has to forcibly introduce the HUP into something, rather than realizing that it is naturally built in when one adopts the QM formalism.

Zz.

The problem here is that people think that the HUP is a starting point in QM rather than merely a consequence.
yes,So uncertainty principle must follow from it.I don't see how it is following.Just saying it is already there,does not mean it is really there.
I mean, where is your point?
The point is that you must either provide a reference or something equivalent,if you are saying that uncertianty principle is already taken into account.I believe in what Feynman has said that if uncertainty principle is taken into account then entropy must not be zero at zero temperature.
Also the formula written by you is classical one,it is called Gibbs entropy
S=-kBƩpi lnpi
quoting wiki
For a classical system (i.e., a collection of classical particles) with a discrete set of microstates, if Ei is the energy of microstate i, and pi is its probability that it occurs during the system's fluctuations, then the entropy of the system is
S=-kBƩpi lnpi

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Quantum mechanically, the statistical entropy is defined by von Neumann's formula
##S=-k\mathrm{Tr}\rho \ln \rho ## where ##\rho## is the so called density matrix.
This expression can be evaluated in a basis which diagonalizes the density matrix. The density matrix only has non-negative eigenvalues $p_i$ which can be interpreted as occupation probabilities. Hence von Neumann's expression reduces to Gibbs expression.
Especially for a pure quantum mechanical state, only one of the p_i is 1, the others being 0. Hence the entropy vanishes for a pure state.
Already for a single particle there are pure states with non-vanishing velocity or current.
Hence finite current and vanishing entropy are not contradictory, even when QM effects are taken into account.

andrien said:
The point is that you must either provide a reference or something equivalent,if you are saying that uncertianty principle is already taken into account.

It seems the kind of reference you're after would have to be a general proof of the uncertainty principle, since the 'i's label quantum states, which necessarily satisfy the HUP. I'm sure most QM textbooks will give a derivation. Try Sakurai, Modern Quantum Mechanics second edition for example.

ZapperZ said:
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.
Zz.

It depends how "ideal" our idealised metal is: if the metal is modeled as having a partially filled nearly-free band, and no scattering, then i think you would expect Bloch oscillations giving net zero current, so a small amount of scattering is necessary to get a nonzero conductivity. (I'm basing this on Singleton, Band Theory and Electronic Properties of Solids, chapter 9.1.)

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It seems the kind of reference you're after would have to be a general proof of the uncertainty principle, since the 'i's label quantum states, which necessarily satisfy the HUP. I'm sure most QM textbooks will give a derivation. Try Sakurai, Modern Quantum Mechanics second edition for example.
just an index does not mean any quantum property.Also I have done sakurai some time ago,it does not say anything useful for it.
Hence von Neumann's expression reduces to Gibbs expression.
wiki says'This upholds the correspondence principle, because in the classical limit, i.e. whenever the classical notion of probability applies, this expression is equivalent to the familiar classical definition of entropy'(gibbs one).
in terms of a density matrix this expression is fine and yields the required classical form.It will also yield zero for entropy if pure state is used.But it seems like to build up on classical analogy.It seems to be a matter of definition.One can go with it.But then usual definition of entropy must not apply(randomness)

andrien said:
yes,So uncertainty principle must follow from it.I don't see how it is following.
It is possible to calculate that all physical wavefunctions satisfy the (p,x)-uncertainty principle. This is related to the mathematics of Fourier transformations. Alternatively, it is possible to derive it in a pure algebraic way as well. Read some introduction book about quantum mechanics, or see Wikipedia for the general concepts.

I believe in what Feynman has said that if uncertainty principle is taken into account then entropy must not be zero at zero temperature.
Source?
It can be different from zero (with a degenerate ground-state), but it does not have to.

The dependencies look like that:
[Basics of QM] -> uncertainty principle
[Basics of QM] -> zero entropy at zero temperature (possible)

Therefore, it is meaningless to ask how you can "derive", or "include" the uncertainty principle in (QM) statistical mechanics. It is impossible to avoid it!

mfb said:
It can be different from zero (with a degenerate ground-state), but it does not have to.

I don't think there are actual thermodynamical systems whose true ground state is degenerate. Usually at sufficiently low temperatures some very weak interactions break the degeneracy.

That is true. I just keep track of that special case as it is not forbidden.

It is possible to calculate that all physical wavefunctions satisfy the (p,x)-uncertainty principle. This is related to the mathematics of Fourier transformations. Alternatively, it is possible to derive it in a pure algebraic way as well.
every one knows that.
Source?
It been time.I think it was written in vol. 1 of his and early chapters.

andrien said:
I think it was written in vol. 1 of his and early chapters.

If you could provide a page number, chapter number or something like that that would be helpful, but either way I'm going to check this out when i next visit a suitable library, because i would have expected better of Feynman!

psmt said:
i would have expected better of Feynman!

He was unambiguously referring to zero point energy, not entropy.

## 1. What is the conductivity of metals at 0 Kelvin?

The conductivity of metals at 0 Kelvin, also known as absolute zero, is zero. This is because at this temperature, the atoms in the metal are not vibrating and there is no movement of electrons, which are responsible for conducting electricity.

## 2. Why does the conductivity of metals decrease at 0 Kelvin?

As mentioned earlier, at 0 Kelvin, the atoms in the metal are not vibrating and there is no movement of electrons. This lack of movement makes it impossible for the electrons to flow and conduct electricity, resulting in a decrease in conductivity.

## 3. Can any metal have non-zero conductivity at 0 Kelvin?

No, all metals will have zero conductivity at 0 Kelvin. This is a fundamental property of metals at this temperature and cannot be changed.

## 4. How does the conductivity of metals at 0 Kelvin affect their use in electrical applications?

The lack of conductivity at 0 Kelvin makes metals unsuitable for use in electrical applications at this temperature. However, metals are often used at higher temperatures where their conductivity is not affected by the lack of movement of electrons.

## 5. Is there any practical use for studying the conductivity of metals at 0 Kelvin?

Yes, studying the conductivity of metals at 0 Kelvin helps us understand the fundamental properties of metals and their behavior at extreme temperatures. This knowledge can be applied in various fields, such as materials science and engineering, to develop new materials with specific properties.

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