# Tag Info

95

A "perfectly efficient" computer can mean many things, but, for the purposes of this answer, let's take it to mean a reversible computer (explained further as we go). The theoretical lower limit to energy needs in computing is the Landauer Limit, which states that the forgetting of one bit of information requires the input of work amounting $k\,T\,\log 2$ ...

74

The anthropomorphic formulation "tries to" is misleading. Under the effect of ambient noise, matter explores the possible configurations around its current state: e.g., two single hydrogen atoms wiggle around and meet. If they happen to bind, this releases energy which goes away, and we say that the energetic state of this new $H_2$ molecule is lower than ...

64

When you say "why aren't things being destroyed", you presumably mean "why aren't the chemical bonds that hold objects together being broken". Now, we can determine the energy it takes to break a bond - that's called the "bond energy". Let's take, for example, a carbon-carbon bond, since it's a common one in our bodies. The bond energy of a carbon-carbon ...

53

The ergodic hypothesis is not part of the foundations of statistical mechanics. In fact, it only becomes relevant when you want to use statistical mechanics to make statements about time averages. Without the ergodic hypothesis statistical mechanics makes statements about ensembles, not about one particular system. To understand this answer you have to ...

35

Let's suppose you have a Rubik's cube that's made of a small number of atoms at a low temperature, so that you can make moves without any frictional dissipation at all, and let's suppose that the cube is initialised to a random one of its $\sim 2^{65}$ possible states. Now if you want to solve this cube you will have to measure its state. In principle you ...

32

How about lifting a litre or kilogram of water? Lifting a kilogram of water $235.3\text{ m}$ (772 feet in sensible units) involves a potential energy change:$$PE=mgh=1\times9.8\times235=2.31 \text{ kJ}$$ Warming a kilogram of water, (specific heat $4.179\text{ kJ/kg/K}$), through a temperature change of $0.556\text{ K}$ (1 degree Farenheit) ...

31

As for references to other approaches to the foundations of Statistical Physics, you can have a look at the classical paper by Jaynes; see also, e.g., this paper (in particular section 2.3) where he discusses the irrelevance of ergodic-type hypotheses as a foundation of equilibrium statistical mechanics. Of course, Jaynes' approach also suffers from a number ...

30

The separation does not violate the 2nd law of thermodynamics, because the oil and water phases being separate is a lower energy state. The water molecules strongly interact with each other, forming hydrogen bonds. The protons of water are shared between two oxygen atoms of two different water molecules, forming a constantly changing network of molecules. ...

29

Arnold Neumaier's comment about statistical mechanics is correct, but here's how you can prove it using just thermodynamics. Let's imagine two bodies at different temperatures in contact with one another. Let's say that body 1 transfers a small amount of heat $Q$ to body 2. Body 1's entropy changes by $-Q/T_1$, and body 2's entropy changes by $Q/T_2$, so ...

25

The resolution to Maxwell's demon paradox is mostly understood to be through Landauer's principle, and it is one of the most compelling applications of information science to physics. Landauer's principle asserts that erasing information from a physical system will always require performing work, and particularly will require at least $$k_B T \ln(2)$$ of ...

24

The question isn't silly. The speed of each molecule in the liquid is much higher than the speed of either the piston or the water shooting out from the nozzle. At room temperature, for water molecules the average is on the order of 500m/s. And yet, the speed of sound in water is three times higher than that, which implies that pressure can propagate in ...

23

If you have only one species of particles then working with $(\mu,p,T)$ ensemble does not make sense, as its thermodynamic potential is $0$. $$U = TS -pV + \mu N,$$ so the Legendre transformation in all of its variables (i.e. $S-T$, $V-(-p)$ and $N-\mu$) $$U[T,p,\mu] = U - TS + pV - \mu N$$ is always zero. The fact is called Gibbs-Duhem relation, i.e. $$0 ... 23 The partition function is strongly related to a very useful tool in probability theory called the moment generating function(al) of the probability distribution. For any probability distribution p of some random variable X, the generating function \mathcal{M}(z) is defined as being: \mathcal{M}(z) \equiv \langle e^{zX}\rangle ... 22 This is a very interesting question which is usually overlooked. First of all, saying that "large scale physics is decoupled from the small-scale" is somewhat misleading, as indeed the renormalization group (RG) [in the Wilsonian sense, the only one I will use] tells us how to relate the small scale to the large scale ! But usually what people mean by that ... 22 Actually, temperature is defined as$$\frac{1}{T} = \frac{\partial S}{\partial E} = \frac{k_B}{\Omega}\frac{\partial\Omega}{\partial E} So in order to have zero temperature, you would need a system with either zero multiplicity, which you can't have by definition, or an infinite derivative $\partial\Omega/\partial E$ even though the multiplicity itself ...

22

Adjacent molecules in a liquid all repel each other because of the electron clouds that surround the nuclei that they contain. In that sense these molecules never even 'touch' each other (at least not in the intuitive sense of the word). When you apply pressure to the liquid you're squeezing them into a (very slightly) smaller volume, thereby increasing the ...

21

From a fundamental (i.e., statistical mechanics) point of view, the physically relevant parameter is coldness = inverse temperature $\beta=1/k_BT$. This changes continuously. If it passes from a positive value through zero to a negative value, the temperature changes from very large positive to infinite (with indefinite sign) to very large negative. ...

21

First, strictly speaking a neutron star is not a nucleus since it is bound together by gravity rather than the strong force. Measuring a surface temperature for any star is deceptively simple. All that is needed is a spectrum, which gives the luminous flux (or similar quantity) as a function of photon wavelength. There will be a broad thermal peak somewhere ...

19

Statistical Mechanics is the theory of the physical behaviour of macroscopic systems starting from a knowledge of the microscopic forces between the constituent particles. The theory of the relations between various macroscopic observables such as temperature, volume, pressure, magnetization and polarization of a system is called thermodynamics. first ...

19

UPDATE: Below I am answering yes to the first question in the post (are the two kinds of entropy the same up to a constant). This has led to some confusion as both Matt and John gave answers saying "the answer is no", however I believe they are referring to the title "Does entropy measure extractable work?". Although the author uses the two interchangeably, ...

18

The realization of non-Abelian statistics in condensed matter systems was first proposed in the following two papers. G. Moore and N. Read, Nucl. Phys. B 360, 362 (1991) X.-G. Wen, Phys. Rev. Lett. 66, 802 (1991) Zhenghan Wang and I wrote a review article to explain FQH state (include non-Abelian FQH state) to mathematicians, which include the explanations ...

18

If the particles are not point-like, they will take up some volume. As the gas is compressed, the collision frequency will rise more quickly, which will make the pressure-volume curve change. The corrections in the Van der Waals model of a real gas account for the volume of the particles. Also if they have internal structure, that structure can have ...

17

$\hbar$ does not need to appear in classical statistical mechanics. You are free to replace it with any quantity with units of angular momentum, say $\hbar_{\mathrm{C}}$. As long as this is choosen smaller than the size you can experimentally probe (i.e., as long as you don't ask questions of the theory that contain structure on this length scale or below) ...

17

This is a consequence of the second law of thermodynamics, which states that In a closed system with fixed internal energy (i.e. an isolated system), entropy is maximized at equilibrium. It can be shown that this statement is equivalent to the following: In a closed system with fixed entropy, the energy is minimized at equilibrium. Callen in his ...

16

It's simple to "roughly prove" the second law in the context of statistical physics. The evolution $A\to B$ of macrostate $A$, containing $\exp(S_A)$ microstates, to macrostate $B$, containing $\exp(S_B)$ microstates, is easily shown by the formula for the probability "summing over final outcomes, averaging over initial states", to be $\exp(S_B-S_A)$ higher ...

16

I wouldn't say the ignorance interpretation is a relic of the early days of statistical mechanics. It was first proposed by Edwin Jaynes in 1957 (see http://bayes.wustl.edu/etj/node1.html, papers 9 and 10, and also number 36 for a more detailed version of the argument) and proved controversial up until fairly recently. (Jaynes argued that the ignorance ...

16

If you were to surround the atmosphere by an adiabatic envelope and allow it to come to equilibrium, it probably would settle into such a state. However, the atmosphere is not a static place. It is actively mixed due to heating of the ground by the sun, and by cooling of the upper atmosphere by radiation into space. This makes the surface air less dense than ...

16

I'll give a very qualitative answer / overview. The classification 'first-order phase transition vs. second-order phase transition' is an old one, now replaced by the classification 'first-order phase transition vs. continuous phase transition'. The difference is that the latter includes divergences in 2nd derivatives of $F$ and above - so to answer your ...

16

If you consider a system $C$ and two subsystems $A,B$ with associated probability distributions $p_C,p_A,p_B$ and want the entropy to add, you must assume the subsystems are independent in the sense that $p_C(X) = p_A(Y)p_B(Z)$ where $Y$ and $Z$ are a partitioning of the variables $X$ belonging to $C$. Then, the total entropy is \begin{align} S_C & = ...

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