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94

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 ...


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 ...


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 ...


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

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 & = ...


14

Your description of critical temperature isn't quite right. If you increase the temperature of a liquid beyond the critical point, the atoms are moving so quickly that persistent structure fails to form and so you have something that behaves a lot like a very dense gas. Similarly, if you increase the pressure of a gas beyond the critical point, it becomes ...


11

This is really a statistical effect, as pretty much all of thermodynamics. You have two free hydrogen atoms. They tend to move around the space they have, and when conditions are favourable (there's enough energy, the atoms come "close enough" together), they might interact - chemically or otherwise. Now, "enough energy" is the important bit here. When a ...


9

I think there is a misunderstanding. You are perfectly right when you write that the total micro canonical entropy of a combined system will be \begin{equation} S_\textrm{combined}(2E) = k_B\ln \sum_x \Omega(x)\Omega(2E-x) \end{equation} The micro canonical entropy ought to be a function of only the total energy, total amount of matter and total volume of ...


9

I'm going to take a slightly different approach and say it's because we defined energy to make it so. In other words, systems "try" to find the lowest energy state because energy is a concept humans invented in order to describe what we observe. This is the reason that for any given set of constraints, you might need a different "energy" to describe the ...


7

OP's question (v1) is essentially asking Does the operator identity $$ e^{\frac{it}{\hbar}[\hat{H},~\cdot~]}\hat{A}~ =~ e^{i\hat{H}t/\hbar}\hat{A}e^{-i\hat{H}t/\hbar} \tag{1} $$ have an analog using functions/symbols $H$ and $A$ rather than operators $\hat{H}$ and $\hat{A}$, respectively? The answer is: Yes, in terms of the Groenewold-Moyal star ...


7

Both are relevant, and "the misconception that Langevin equation is the universal stochastic differential equation for all kinds of noisy systems is responsible for the difficulties mentioned"* in your post. Take the SDE from Thomas' answer, $$\frac{dy}{dt} = A(y) + C(y)L(t)$$ where $L(t)$ is the noise term. Suppose we can turn the noise off, so we'd only ...


6

In some sense yes. The temperature is defined as an imaginary time in Matsubara Green's functions or some path integrals. Thus, a negative inverse imaginary temperature can be considered as a time. Here is a quotation from Alexander Altland, Ben Simons "Condensed Matter Field Theory": "Thus, real time dynamics and quantum statistical mechanics can be ...


6

This phenomenon has been studied by Vella and Mahadevan and written up in the American Journal of Physics (http://scitation.aip.org/content/aapt/journal/ajp/73/9/10.1119/1.1898523). It's called the Cheerios effect. If the cereal pieces clump together away from the edges of the bowl, they gravitate toward a slight concavity in the surface caused by milk ...


6

It depends on the parameter you consider. The Maxwell-Boltzmann distribution (as the name is applied in the Wikipedia article you link) is the distribution of the particle's speed (absolute value of the velocity) in a gas and this is a chi distribution. You probably were thinking of the distribution in terms of the vectorial velocity and then indeed it ...


6

Let's simplify things down to the barest minimum: one dimension, one particle, and a wall. O | The particle moves to the right, hits the wall, and rebounds, perfectly elastically. If the wall is fixed in place, the particle will leave the collision with exactly the same kinetic energy as it came in with. But what if the wall is moving to ...


6

Major edit: In @gatsu's answer, it is pointed out that only the amount of energy should matter, which is correct, as there's no such thing as distinguishable microstates with only rearranged energy (think stars-and-bars-type entropy calculations). So, I've edited out that part of the first paragraph and equations (in the first draft, I dropped that part of ...


6

A typical velocity dispersion in a globular cluster is 10 km/s. For a typical 1 solar mass subgiant in an old globular, then equating the kinetic energy to $3kT/2$, we get $T = 5\times 10^{60}$ K. Doesn't seem that helpful really... The concept of temperature is only ever applied in a relative sense - i.e. some component is hotter than another. Can't say ...


5

It's true. Special equipment and a long time is required to mix helium and nitrogen. According to one study, a mixture of 2.7% He, 93.3% N at 800 p.s.i.g. required a special cradle to repeatedly upend the cylinder, and 20.5 hours to reach equilibrated gas, which then remained mixed: http://pubs.acs.org/doi/abs/10.1021/je60005a002. The helium repeatedly ...


5

wsc's answer (i.e., Onsager's computation of the free energy) provides one alternative road to a proof of a phase transition in the Ising model. It implies the existence of a phase transition in dimension 2 (for the nearest-neighbor model). Combined with correlation inequalities, this implies existence of a phase transition in any dimension d≥2, and ...


5

Yes, neutrinos should obey Fermi-Dirac statistics and yes, the Pauli Exclusion Principle should operate for neutrinos. But let's examine how dense the neutrino population has to be for this to be important. The Fermi momentum is given by $$ p_F = \left( \frac{3}{8\pi}\right) h n_{\nu}^{1/3} $$ where $n_{\nu}$ is the neutrino number density. In order to be ...


5

So, the short answer is that you're quite correct: if the dynamics of a system is subject to Liouville's theorem, then phase space volume is conserved, so the entropy associated to a given probability distribution remains constant as it evolves under those dynamics. This is actually just one instance of a much more general puzzle: how do we reconcile the ...


5

Non-equilibrium systems are most often considered in the approximation where local equilibrium is valid, yielding a hydrodynamic or elasticity description. Local equilibrium means that equilibrium is assumed to hold on a scale large compared to the microscopic scale but small compared with the scale where observations are made. In this case, one considers a ...


5

$$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$ This means that if we have a function of $t, p, q$ namely $\rho(t,\vec p,\vec q)$ and we have a trajectory that is a curve in $(p,q)$ space, namely $q_i(t), p_i(t), i=1\ldots N,$ then: $$ \frac{\mathrm ...


5

Both Ito and Stratonovich stochastic PDEs can be used to derive a Fokker-Planck equation. Indeed, for simple one-dimensional processes the Ito process $$ dx = a\, dt + b\, dW(t) $$ is equivalent to the Stratonovich process $$ dx =\left( a\,-\frac{1}{2}b\partial_x b\right) dt + b\, dW(t) $$ The answer is then that both are physically reasonable, for a given ...


5

The entropy $S$ is extensive as long as you're consistent about what you mean by entropy. In your case you've mixed up two different definitions. One definition of the entropy has the system at fixed energy $E$ -- the other, a fixed temperature $T$. Fixed-E entropy For a system with fixed energy $E$ the entropy is defined to be $$ S = \log\Omega(E) ...


5

The entropy of a single atom does not make sense per se, unless you specify the preparation. The entropy of a single isolated atom, fixed at a point, is indeed not defined – the entropy is, after all, a property of an ensemble not of a system. The entropy of an ensemble of isolated atoms prepared at a specific energy, on the other hand, is well defined (this ...


4

A microstate is just a particular microscopic configuration of the system, where the state of each particle is fixed. For example, take a three-level system with four particles. Treating the particles as indistinguishable, one particular microstate corresponds to two particles in the lowest state, which has energy $-\epsilon$, one particle in the second ...


4

You can get a nice expression for the leading-order correction to the flat-manifold result via the use of Riemann normal coordinates. Basically, imagine expanding the metric in a power series at the point $x_0$: $$ g_{\mu \nu}(x) = g_{\mu \nu} (x_0) + \partial_\rho g_{\mu \nu} (x^\rho - x_0^\rho) + \frac{1}{2} \partial_\rho \partial_\sigma g_{\mu \nu} ...


4

The equipartition theorem is a mathematical consequence of very specific kind of Hamiltonians. It states that any 'squared' term of deegree of freedom in the Hamiltonian gets $\frac{1}{2}k_bT$ of energy (it is a statement about the energy distribution for this kind of Hamiltonians). For example - classical ideal gas Hamiltonian - ...



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