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98

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 & = \...


15

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 ...


14

I will try to answer these questions from different views. Macroscopic view The "quantitative" rather than qualitative difference in a liquid-gas phase transition is due to the fact that the molecules arrangement does not change so much (there is no qualitative difference) but the value of the compressibility changes a lot (quantitative difference). This ...


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 ...


10

Temperature is related to kinetic energy in the rest frame of the fluid/gas. In non-relatvistic kinetic theory the distribution function is $$ f(p) \sim \exp\left(-\frac{(\vec{p}-m\vec{u})^2}{2mT}\right) $$ where $\vec{u}$ is the local fluid velocity. The velocity can be found by demanding that the mean momentum in the local rest frame is zero. Then $\vec{u}...


10

A system is in a heat bath of temperature T so we work with the canonical ensemble. We consider $N$ degrees of freedom $x_1, x_2, ..., x_N$ and $x$ is the vector $(x_1~ x_2 ~ ... ~ x_N)^T$. The potential energy is quadratic so it can be expressed as a function of its second derivatives: $ U=\sum_{i,j} x_i ~H_{i,j} ~x_j = x^T H x ~~ $ with $H_{i,j}=\frac{\...


10

You have to be careful to distinguish between microstates and macrostates. Thermodynamic equilibrium is a macrostate which consists of a mixture of all possible microstates of energy $E$ weighted by a Boltzmann weight $e^{- \beta E} / Z$. A state in macroscopic thermal equilibrium can be thought of as "moving through phase space" ergodically (i.e. the ...


9

Preliminaries: How do we define 'localized?' For a single particle, or for multiple non-entangled particles, it is easy to tell from the expressions for the wavefunctions whether they are localized or delocalized. For example, you might say that if the wavefunction is falling off exponentially or faster for large $x$, that is with a form like $\psi(x)\sim e^...


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 ...


9

For a pure substance that can exist in the solid, liquid, and vapor states (i.e., wood is not in this category), let's assume that a closed container is half full of liquid and half full of vapor. As the temperature rises, the liquid expands and the liquid density falls. Also, as the temperature rises, the pressure in the container rises due to the vapor ...


8

There are two definitions of entropy, which physicists believe to be the same (modulo the dimensional Boltzman scaling constant) and a postulate of their sameness has so far yielded agreement between what is theoretically foretold and what is experimentally observed. There are theoretical grounds, namely most of the subject of statistical mechanics, for our ...


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 ...


7

[June 19,2016: thoroughly revised, giving a more detailed, comparative presentation and better references] General case. In relativistic thermodynamics, inverse temperature $\beta^\mu$ is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the ...


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 I'...


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 d}{\...


5

The basic idea is that the term with the highest weight in the exponential series is exactly the desired term. Further, all of the weight is in terms with closeby particle number, and small fluctuations in this number do not matter in many cases. (Note: I will omit the $k=0$ subscript and just write $b^\dagger$ in the following for convenience.) First, we ...


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) \tag{...



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