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I've been told that many systems possess some residual entropy at absolute zero.

This would seem to disagree with the 3rd Law of Thermodynamics? How can this be explained physically speaking?

I am mildly aware that the 3rd law speaks with respect to a perfect system but I've never really understood what is meant by this.

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    $\begingroup$ I have always thought of it in terms of the kinetics of trying to move displaced ions and molecules through a crystal lattice at low temperatures. While displaced they are naturally increasing the entropy, and although they "should" return to their correct positions the energy barrier to this process is fat too high. Further to this impurities which are always present in real systems lead to a residual entropy since they have to sit in interstitial sites or even displace framework species from a lattice. $\endgroup$ – RedPen Dec 4 '14 at 0:04
  • $\begingroup$ So the argument is that only a perfectly crystalline lattice will obey the third law. Could the presence of residual entropy also be connected to the heisnberg uncertainty principle i.e. if at 0K the entropy is 0 J then surely I violate the uncertainty principle by knowing both the momenta and position of my particles/ions/atoms etc. $\endgroup$ – Ari Ben Canaan Dec 4 '14 at 1:31
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I think it's not purely an issue of degeneracy, from what I understand a system can have a residual entropy even with a unique ground state. I'm not sure, but I think the way it works is that every system would end up in its lowest-energy ground state at exactly absolute zero, but in practice if we are interested in the behavior as you approach absolute zero, some systems may have a large number of possible states close to the ground state so that if you lower the temperature and measure energy as a function of temperature, the system may not have enough time to "find" the ground state and spend more time in it than other states as T approaches 0. In this case, if you lowered the temperature over some sufficiently huge span of time it would still be true that the average energy as a function of T would approach the ground state energy as T approached 0, but for a more practical span of time in which you lower the temperature and measure how the energy changes (by measuring the heat capacity, which is the rate of change of internal energy with respect to temperature), the energy may not approach the ground state energy on a graph of U vs. T.

Here's a discussion of residual entropy from An Introduction to Thermal Physics by Daniel Schroeder, p. 94 (note the part I bolded below about needing to 'wait eons' for the system to find the ground state, and the other bolded suggesting there is a unique lowest-energy ground state for the system he's discussing):

In practice, however, there can be several reasons why S(0) is effectively nonzero. Most importantly, in some solid crystals it is possible to change the orientations of the molecules with very little energy. Water molecules, for example, can orient themselves in several possible ways within an ice crystal. Technically, one particular arrangement will always have a lower energy than any other, but in practice the arrangements are often random or nearly random, and you would have to wait eons for the crystal to rearrange itself into the true ground state. We then say that the solid has a frozen-in residual entropy, equal to k times the logarithm of the number of possible arrangements.

Another form of residual entropy comes from the mixing of different nuclear isotopes of an element. Most elements have more than one stable isotope, but in natural systems these isotopes are mixed together randomly, with an associated entropy of mixing. Again, at T=0 there should be a unique lowest-energy state in which the isotopes are unmixed or are distributed in some orderly way, but in practice the atoms are always stuck at their random sites in the crystal lattice.

A third type of "residual" entropy comes from the multiplicity of alignments of nuclear spins. At T=0 this entropy does disappear as the spins align parallel or antiparallel to their nighbors. But this generally doesn't happen until the temperature is less than a tiny fraction of 1 K, far below the range of routine heat capacity measurements.

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In the ideal case, at zero kelvin the system must be in a state with the minimum possible energy, and this statement of the third law holds true if the perfect system has only one minimum energy state, called the ground state. Entropy is related to the number of possible microstates. If the ground state of the system is degenerate system containing a certain collection of particles, it is possible that the grounf state is degenerate (tat is, more than one ground state exists). In such a case the entropy could not be zero, aven at zero K because there will be more than one microstate compatible with the ground macrostate.

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  • $\begingroup$ Hmmm, that's interesting especially since we've just been introduced to the idea of microstates. I would like to pose to you the same question as I have posted in the comments bit above regarding the uncertianty principle (see reply to redpen's comment above yours). $\endgroup$ – Ari Ben Canaan Dec 4 '14 at 1:32
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    $\begingroup$ I do not think the uncertainty principle plays a role here. In quantummechanics zero energy is not equivalent to zero motion. And the trhirs law is related to the number of microstates. You can have a microstate that staisfies the uncertainty relation but is still a single microrstate as far as it had a definite energy, and this energy does not fluctuate with the uncertainty in momentum and position. $\endgroup$ – Wolphram jonny Dec 4 '14 at 1:41

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