Tag Info

Hot answers tagged

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 ...


26

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

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 ...


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 ...


16

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 ...


13

A convenient operational definition of temperature is that it is a measure of the average translational kinetic energy associated with the disordered microscopic motion of atoms and molecules. The underlying framework of all matter is quantum mechanical. This means that the Heisenberg Uncertainty principle holds. Even for a single particle the HUP ...


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

I think gatsu's right: it's because of an entropic attraction resulting from the fact that two spheres whose centers are less than $2a$ apart leave more room for other spheres. To see why this happens, it may help to draw a picture: Here, the blue spheres all have radius $\frac12a$. The gray dashed circle around each sphere has radius $a$, and shows the ...


9

If two particles are close to each other, there is more space for the rest of the particles to move. This gives rise to an effective entropic attraction between the particles because when looking at two particles for different separations while "tracing out" over the degrees of freedom of the rest of the system, the entropy of the rest is higher when the two ...


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

No, it's not a problem. The reason is that, in order for expressions like $$\mu=-T\left(\tfrac{\partial S}{\partial N}\right)_{E,V}.$$ to be meaningful, you have to be using the grand canonical ensemble (or a generalisation thereof), in which particles are able to enter and leave the system. Consequently, $N$ stands not for an integer number of particles, ...


8

I think it will depend the kind of statistical mechanics. For classical statistical mechanics, there is no time, so it is really hard to imagine a nice physical picture of the propagation of something. But nevertheless we still talk of loops as propagating "particles" (we give the "momenta", for instance, which is conserved, etc.). Interestingly, ...


7

I could be wrong, but in my understanding, you're describing and justifying steady state, not detailed balance. In thermal equilibrium, steady state is true always, and detailed balance is true sometimes. Detailed balance means that the rate $X \rightarrow Y$ is always the same as the rate $Y \rightarrow X$. If a system is both in thermal equilibrium and ...


7

Giving a full answer to this one takes quite a bit of information, so I'll first give a few references and then summarise how they all fit in. References Relevant Physics SE Questions Does the scientific community consider the Loschmidt paradox resolved? If so what is the resolution? Theoretical proof forbidding Loschmidt reversal? Perpetual motion ...


7

Mathematical/formal answer: In classical statistical mechanics the answer is "no return to the exact same state". Although the Poincare recurrence theorem indicates that a given isolated system will return to a state that is arbitrarily close to a selected initial state, phase space is continuous and the probability of reaching any specific state is zero.


6

The more natural relationship between the two distributions is the opposite one. The Boltzmann distribution $$\exp(-E/kT)$$ is the more general one (connected with the microscopic definition of the temperature $T$ in any system in physics) and one may simply substitute the kinetic energy $mv^2/2$ for $E$ to get the Maxwell part of the distribution. The ...


6

The third law of thermodynamics states that a quantum system has absolute zero temperature if and only if its entropy is zero: meaning it is not reacting with anything including its environment, which is impossible to achieve. In the QM sense, 0K would be achieved when all the motion of all the particles comprising matter stops and everything comes to a ...


5

How do you prove the second law of thermodynamics from statistical mechanics? You can't. In order to prove the second law, which is time-asymmetric, you need some ingredient that breaks time-reversal symmetry. Statistical mechanics does not have any such ingredient. To remove this symmetry, you need either time-asymmetric boundary conditions or ...


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

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

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

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

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

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

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

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

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 ...


4

Kolmogorov Complexity and Shannon Entropy of an information source have different definitions, which I assume you have read: if not, the Wiki pages on Kolmogorov Complexity, Shannon Entropy and the Shannon's source coding theorem are essential and sound background reading as well as E. T. Jaynes, "Information Theory and Statistical Mechanics" (especially the ...


4

Your second point, which is the most important one I think, is right but is not so problematic I think. You make a point about temperature but the same thing could be said of the density. You can consider a gas (ideal gas to make it simple) in either microcanonical or canonical ensembles and find that if you partition the box into two halves, the 1-particle ...



Only top voted, non community-wiki answers of a minimum length are eligible