# Entropy Zero

The third law was developed by chemist Walther Nernst during the years 1906 to 1912 and is therefore often referred to as Nernst's theorem or Nernst's postulate. The third law of thermodynamics states that the entropy of a system at absolute zero is a well-defined constant. This is because a system at zero temperature exists in its ground state, so that its entropy is determined only by the degeneracy of the ground state.

## Entropy Zero

This version states not only Δ S \displaystyle \Delta S will reach zero at 0 K, but S \displaystyle S itself will also reach zero as long as the crystal has a ground state with only one configuration. Some crystals form defects which cause a residual entropy. This residual entropy disappears when the kinetic barriers to transitioning to one ground state are overcome.[6]

With the development of statistical mechanics, the third law of thermodynamics (like the other laws) changed from a fundamental law (justified by experiments) to a derived law (derived from even more basic laws). The basic law from which it is primarily derived is the statistical-mechanics definition of entropy for a large system:

where S \displaystyle S is entropy, k B \displaystyle k_\mathrm B is the Boltzmann constant, and Ω \displaystyle \Omega is the number of microstates consistent with the macroscopic configuration. The counting of states is from the reference state of absolute zero, which corresponds to the entropy of S 0 \displaystyle S_0 .

In simple terms, the third law states that the entropy of a perfect crystal of a pure substance approaches zero as the temperature approaches zero. The alignment of a perfect crystal leaves no ambiguity as to the location and orientation of each part of the crystal. As the energy of the crystal is reduced, the vibrations of the individual atoms are reduced to nothing, and the crystal becomes the same everywhere.

The entropy of a perfect crystal lattice as defined by Nernst's theorem is zero provided that its ground state is unique, because ln(1) = 0. If the system is composed of one-billion atoms that are all alike and lie within the matrix of a perfect crystal, the number of combinations of one billion identical things taken one billion at a time is Ω = 1. Hence:

The difference is zero; hence the initial entropy S0 can be any selected value so long as all other such calculations include that as the initial entropy. As a result, the initial entropy value of zero is selected S0 = 0 is used for convenience.

An example of a system that does not have a unique ground state is one whose net spin is a half-integer, for which time-reversal symmetry gives two degenerate ground states. For such systems, the entropy at zero temperature is at least kB ln(2) (which is negligible on a macroscopic scale). Some crystalline systems exhibit geometrical frustration, where the structure of the crystal lattice prevents the emergence of a unique ground state. Ground-state helium (unless under pressure) remains liquid.

Glasses and solid solutions retain significant entropy at 0 K, because they are large collections of nearly degenerate states, in which they become trapped out of equilibrium.[citation needed] Another example of a solid with many nearly-degenerate ground states, trapped out of equilibrium, is ice Ih, which has "proton disorder".

For the entropy at absolute zero to be zero, the magnetic moments of a perfectly ordered crystal must themselves be perfectly ordered; from an entropic perspective, this can be considered to be part of the definition of a "perfect crystal". Only ferromagnetic, antiferromagnetic, and diamagnetic materials can satisfy this condition. However, ferromagnetic materials do not, in fact, have zero entropy at zero temperature, because the spins of the unpaired electrons are all aligned and this gives a ground-state spin degeneracy. Materials that remain paramagnetic at 0 K, by contrast, may have many nearly degenerate ground states (for example, in a spin glass), or may retain dynamic disorder (a quantum spin liquid).[citation needed]

The reason that T = 0 cannot be reached according to the third law is explained as follows: Suppose that the temperature of a substance can be reduced in an isentropic process by changing the parameter X from X2 to X1. One can think of a multistage nuclear demagnetization setup where a magnetic field is switched on and off in a controlled way.[9] If there were an entropy difference at absolute zero, T = 0 could be reached in a finite number of steps. However, at T = 0 there is no entropy difference, so an infinite number of steps would be needed.[why?] The process is illustrated in Fig. 1.

A non-quantitative description of his third law that Nernst gave at the very beginning was simply that the specific heat of a material can always be made zero by cooling it down far enough.[10] A modern, quantitative analysis follows.

On the other hand, the molar specific heat at constant volume of a monatomic classical ideal gas, such as helium at room temperature, is given by CV = (3/2)R with R the molar ideal gas constant. But clearly a constant heat capacity does not satisfy Eq. (12). That is, a gas with a constant heat capacity all the way to absolute zero violates the third law of thermodynamics. We can verify this more fundamentally by substituting CV in Eq. (14), which yields

Even within a purely classical setting, the density of a classical ideal gas at fixed particle number becomes arbitrarily high as T goes to zero, so the interparticle spacing goes to zero. The assumption of non-interacting particles presumably breaks down when they are sufficiently close together, so the value of CV gets modified away from its ideal constant value.[citation needed]

In order to get for each object an information about the quality of the classification, I wanted to calculate Shannon's entropy but it does not work when one of the classes has a probability equal to zero (log(0)=-Inf).

According to MacKay's book "Information Theory, Inference, and Learning Algorithms" (chapter 2), the convention is that you can equate to zero the terms including zero probabilities, as the other answers suggest. See an excerpt below:

all packed into a tiny volume with enormous kinetic energies. This hot, dense, expanding, and uniform to within 1 part in 30,000 state would grow into the observable Universe we inhabit today over the next 13.8 billion years. Thinking about what we began with, however, it sure does seem like a disordered, very high-entropy state.

Consider the two systems above, for example. On the left, a box with a divider in the middle has cold gas on one side and hot gas on the other; on the right, the divider is opened and the entire box has gas of the same temperature. Which system has more entropy? The well-mixed one on the right, because there are more ways to arrange (or swap) the quantum states when all the particles have the same properties than when half have one set of properties and half have another, distinct set of properties.

In fact, if we look at the Universe today, even though the overall entropy is enormous, the fact that the volume is so large drives the entropy density to a relatively small number: about 1027 kB/m3 to 1028 kB/m3.

What has occurred and what will occur over the entire history of the Universe is peanuts compared to the biggest entropy growth to ever happen: the end of inflation and the beginning of the hot Big Bang. Yet even during that inflationary state with alarmingly low entropy, we still never saw the entropy of the Universe decrease; it was only the entropy density that went down as the volume of the Universe increased exponentially. In the far future, when the Universe expands to about 10 billion times its present radius, the entropy density will once again be as small as it was during the inflationary epoch.

we know that internal energy of the system is defined in terms of temperature as $(3/2)kT$. so if temperature is zero so internal energy is zero. and that means that particle will not have much kinetic energy. so is the entropy zero, system may not move to different microstates?

At zero temperature, a system must be in its ground state. By the Third Law of Thermodynamics, if there is only one possible non-degenerate ground state (i.e. the object is a "perfect crystal"), then the entropy is zero at zero temperature, because there is only one possible configuration for the system to adopt.

This is manifestly not true in an ideal gas. One of the assumptions of an ideal gas is that there are no interactions whatsoever between individual particles, except for collisions. Therefore, any possible spatial arrangement of particles with zero velocity would have the same internal energy, since there are no interactions to favor one arrangement over another. As such, there are a multitude of possible ground states for the ideal gas, which means that its entropy is nonzero at zero temperature.

Guaschi, John, Llibre, Jaume, and Mackay, Robert S.. "A classification of braid types for periodic orbits of diffeomorphisms of surfaces of genus one with topological entropy zero.." Publicacions Matemàtiques 35.2 (1991): 543-558. .

TY - JOURAU - Guaschi, JohnAU - Llibre, JaumeAU - Mackay, Robert S.TI - A classification of braid types for periodic orbits of diffeomorphisms of surfaces of genus one with topological entropy __zero.JO__ - Publicacions MatemàtiquesPY - 1991VL - 35IS - 2SP - 543EP - 558AB - We classify the braid types that can occur for finite unions of periodic orbits of diffeomorphisms of surfaces of genus one with zero topological __entropy.LA__ - engKW - braid type; surface of genus one; topological entropyUR - ER -

This paper considers the use of autocorrelation properties to design zero reference codes (ZRCs) for optical applications. Based on the properties of the autocorrelation function, the design of an optimum ZRC problem is transformed into a minimization problem with binary variables, and the objective is to minimize the second maximum of the autocorrelation signal sigma. However, the considerable computational complexity for an exhaustive search through all combinations of (nn (1)l) different code patterns is a potential problem especially for large codes, where n and n1 are the length of the ZRC and the number of transparent slits, respectively. To minimize sigma while reducing the computational complexity at the same time, we introduce the Cross-Entropy (CE) method, an effective algorithm that solves various combinatorial optimization problems to obtain a good code. The computer simulation results show that compared with the conventional genetic algorithm (GA), the proposed CE obtains the better sigma with low computational complexity. 041b061a72