# Tag Info

32

We can only approach absolute zero asymptotically because we can't suck heat out of a system. The only way we can get heat out is to place our system in contact with something cooler and let the heat flow from hot to cold as it usually does. Since there is nothing colder than absolute zero, we can never get all the heat to flow out of a system. We can ...

31

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) ...

28

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. ...

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 ...

17

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 accounts for the volume of the particles. Also if they have internal structure, that structure can have ...

13

DavePhD's answer explains the specifics. The separation decreases the enthalpy of the oil-water mixture. But there's one more step: When the enthalphy of the dressing decreases by $\Delta H$, it causes the entropy of the dressing and its surrounding environment to increase by $\Delta H / T$. The reason is: The decrease in enthalpy releases heat, which ...

10

Frustrated total internal reflection is an optical phenomenon. It's such a close analogue to quantum tunneling that I sometimes even explain it to people as "quantum tunneling for photons". But you can calculate everything about it using classical Maxwell's equations.

9

It's intuitive if you are careful in remembering what is held constant. $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V + \mu \mathrm{d}N$$ and $$\mu = \left(\frac{\partial U}{\partial N}\right)_{S,V}$$ means that the entropy and the volume are kept constant. Remember that entropy is a measure of how many microstates you have available to realize the ...

8

Here's an intentionally more conceptual answer: Entropy is the smoothness of the energy distribution over some given region of space. To make that more precise, you must define the region, the type of energy (or mass-energy) considered sufficiently fluid within that region to be relevant, and the Fourier spectrum and phases of those energy types over that ...

7

There is quite a big controversy these days about the correct definition of the entropy in the microcanonical ensemble (the debate between the Gibbs and Boltzmann entropy), which is closely related to the question. Everyone agrees that the correct definition of the density matrix is given by $$\rho(E)=\frac{\delta(E-H)}{\omega(E)},$$ where $H$ is the ...

6

The entropy of a system is the amount of information needed to specify the exact physical state of a system given its incomplete macroscopic specification. So, if a system can be in $\Omega$ possible states with equal probability then the number of bits needed to specify in exactly which one of these $\Omega$ states the system really is in would be ...

6

Standard Monte Carlo samples the canonical (NVT) ensemble. So it maintains constant temperature but the potential energy is free to fluctuate - both up and down. This will only seem odd if you incorrectly imagine the equilibrium state of a system to correspond to that with the minimum energy. The equilibrium state is actually determined by the minimum free ...

6

You will almost never encounter a calculation that is intended to account for every detail of a phenomenon with perfect accuracy. That isn't possible, and in fact many times adding more detail to a calculation only takes away from the insight it grants. Why make a complicated calculation when a simple one tells you everything you want to know? Gamow is ...

5

The distribution you use depends on the ''ensemble'' you are working in. There are three kinds of ensembles: Mircocanonical ensemble (or practical: an isolated system): the number of particles $N_s$ and the energy $E_s$ is fixed. This is ONE SPECIFIC REALISATION of the system. You have only one possible value for the energy and number of particles, so the ...

5

The term canonical gives it away. The canonical ensemble density matrix $\rho$ is defined as follows in terms of the Hamiltonian $H$ and inverse temperature $\beta = 1/kT$: \begin{align} \rho(\beta) = \frac{1}{Z(\beta)}e^{-\beta H}, \qquad Z(\beta) = \mathrm{tr}(e^{-\beta H}) \end{align} Then the canonical ensemble average of any observable $O$ is given ...

5

Negative temperatures are only defined for systems where there are a limited number of energy states. Consider rising the temperature of such a system, then as the temperature starts rising, particles begin to move into higher energy states, and as the temperature continues rising, the number of particles in the lower energy states and in the higher energy ...

5

I believe it was Boltzmann who first made the connection between entropy and micro states. chapter 12 of "Classical and Statistical Thermodynamics" by Ashley H. Carter discusses Boltzmann's arguments. To summarize from that book: Entropy ($S$) corresponds to a particular configuration of an ensemble of particles called a macro state. A macro state can be ...

5

No. You wouldn't say that pair of beams has a temperature. Temperature is defined by the zeroth law of thermodynamics, which states that if $A$ is in thermal equilibrium with $B$ and $B$ is in thermal equilibrium with $C$ then $A$ is in thermal equilibrium with $C$ and $A$, $B$ and $C$ are said to have the same temperature. Temperature is fundamentally a ...

5

What is entropy, more than disorder. Mathematically, entropy is just a measure of spread of a probability distribution: The lower the entropy, the more spiked the distribution. In statistical mechanics, a state is generally only partially defined via some macroscopic constraints, and entropy is a measure of microscopic indeterminacy. Maximizing entropy ...

5

Here's an elegant way to show that any linear combination of (anti)symmetric states is always (anti)symmetric. We use Dirac notation here for the states, and we assume, for simplicity, that we are dealing with a two-component system so that states of the system are linear combinations of products $|\psi\rangle = |\psi_1\rangle|\psi_2\rangle$. First, we ...

5

Ultimate physical motivation Strictly in the sense of physics, the entropy is less free than it might seem. It always has to provide a measure of energy released from a system not graspable by macroscopic parameters. I.e. it has to be subject to the relation $${\rm d}U = {\rm d}E_{macro} + T {\rm d} S$$ It has to carry all the forms of energy that cannot be ...

5

In terms of the temperature, the entropy can be defined as $$\Delta S=\int \frac{dQ}{T}\tag{1}$$ which, as you note, is really a change of entropy and not the entropy itself. Thus, we can write (1) as $$S(x,T)-S(x,T_0)=\int\frac{dQ(x,T)}{T}\tag{2}$$ But, we are free to set the zero-point of the entropy to anything we want (so as to make it convenient)1, ...

5

I'll attempt an explanation without any equations. In kinetic theory, the pressure exerted by an ideal gas (point-like, non-interacting particles, elastic collisions) depends on the rate at which momentum is exchanged with the walls of any container as the molecules bounce off them. The momentum is of course proportional to the molecule's speed and mass, ...

5

Here's something I believe is a simple proof. Unfortunately it uses a little bit of cohomology. Consider the canonical 2-form in extended phase space $T^*M \times \mathbb{R}$ $$\omega = \sum_{i=1}^N dq_i \wedge dp_i - dH(\vec{q},\vec{p},t) \wedge dt ,$$ where $N = dim(M)$. A function $f: M \to M$ is said to be a canonical transformation iff $f^* \omega = ... 5 The quantum-mechanical proof is actually pretty much identical to the classical one given in the link. You simply replace the integral over the phase space with a trace over the states in the Hilbert space. The equilibrium density operator is $$\rho = \frac{e^{-\beta H}}{Z},$$ where the partition function is$Z = \mathrm{Tr}(e^{-\beta H})$and the inverse ... 5 Classical physics (intuitive) answer: temperature is a property of matter, that is a sort of estimate of the kinetic energy of particles. If you have no particles, there's nothing to measure the temperature of. Mathematical answer: since temperature is changed by multiplying by a real number, the only way to reach 0 would be to have an object already at ... 5 I would argue that this maybe due to the way you calculate your autocorrelation. An autocorrelation like that straight line is the result of a large square signal. The Ising model has a phase transition at the critical temperature. Above it, it's disordered; below it, it becomes ordered, which means that the magnetization stops flipping back and forth. This ... 5 A process is ergodic if you get the same statistical momenta considering a single realization for a sufficiently long time, and by considering a sufficiently big number of realizations at a precise moment. In other word you can extract all the information on a process by looking at a single realization for some time. Having said that is clear that ... 5 A quick google search leads one immediately to the wikipedia page on this particular theorem. The first paragraph of this page states: The Bohr–van Leeuwen theorem is a theorem in the field of statistical mechanics. The theorem states that when statistical mechanics and classical mechanics are applied consistently, the thermal average of the ... 5 A footnote on Wikipedia actually explains this. The$h\$ appearing in the partitioning of the phase space is not Planck's constant, it is simply the size of a phase space cell in which we do not distinguish single states anymore. It's initially arbitrary. Since Planck's constant provides a natural scale below which we should not apply classical thinking, it ...

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