Tag Info

New answers tagged

0

The problem you are trying to solve is different from the standard textbook situation where $A$ and $B$ are two heat reservoirs. When considering equilibration of temperatures, you have to account for the finite heat capacity of the two objects. This allows you to consider a series of quasi-static reversible heat exchanges with a series of heat reservoirs, ...


2

Boltzmann's version of entropy require a finite number of states, and Planck had asserted that probability has no meaning without a “finite number of equally likely configurations.”{1} That is, in order to be able to use Boltzmann's equation and obtain finite results, he needed to use a discrete number of states, but light was supposed to be continuous. ...


1

My two cents to add to the great response from wetsavanna: a short explanation would be the following. Technically, information never increases or decreases at the microscopic level. Entropy is a measure of information about how much do we know about the microscopic levels when making a macroscopic observation. But the relationship between entropy and ...


5

Your question is an excellent one and I think your IT approach is spot on. The contradiction you rightly point out arises partly from the subtle differences between the different Conditional Entropies (Conditional Informations) at play in this discussion. Leaving aside Universes for the moment, let's instead think about a truly thermodynamically isolated ...


1

If it is adiabatic then $ \Delta Q_i $ will be always zero. The fact that it is irreversible doesnt matter. Any path that thakes you from A to B will result in the same change of entropy, as both initial and final states are in equilibrium. If you choose what is called a quasistatic path, which is idealized as a tranformation that occurrs slow enough so that ...


1

I would not write $dS=0$ for a cyclic process but rather its closed contour integral $\oint dS = 0$, then you also have $\oint \frac {\delta Q}{T} \le 0$ and from $T>0$ you will get that in a cyclic process you must have somewhere, sometime, at some stage of the cycle $\delta Q < 0$, that is some amount of heat must be rejected to complete the cycle. ...


0

For a reversible path between two states (1 and 2), entropy change of a system is NOT zero. It is $$\Delta S = \int_a^b \frac{dQ}{T}$$ For reversible path between two states, entropy of the universe (Or any isolated system) is zero. $$\Delta S + \Delta S_\text{surroundings} = 0$$ So You cannot just take any system and say that entropy change between two ...


0

What is important here is that the entropy is also very low today. So, since the Big Bang the entropy has been increasing but we're still quite far removed from the heat death state. So, the reason why at the macroscopic level time reversal inavariance appears to be broken is simply because the entropy today is low compared to the maximum entropy, and the ...


1

Can this scenario be considered as erasure of information as referred to in Landauer's principle? No. The reason is that the whole process is reversible and Landauer's principle specifically needs an irreversible step - otherwise no "erasure" can have taken place. Where's the problem? It's here: The state $$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$ ...


0

The second law implies that entropy of an isolated system should be expected to go to a maximum given a sufficiently large length of time, but in the short term, for certain isolated systems it would be possible to have the probability that entropy will decrease over a certain time interval be larger than the probability it will increase over the same ...


1

After reaching thermodinamic equilibrium, if you wait enough time, any system will reaach, by random chance, any state of lower entropy. Of couse, the time needed to for such a fluctuation to occur increase exponentially with the amount of decrease in entropy (search poincare recurrence theorem). My guess is that this is not the kind of answer you were ...


0

I understand that there was zero entropy or at least low entropy that became self agitated to bring about the big bang. The closest example to zero or very low entropy is a solution of Dropleons with quasiparicles structures in a liquid state that are self propagating. This is based on the initial conditions of the solution containing super conducting and ...


1

In the standard homogeneous cosmological models the total energy in an expanding volume is zero. This is true for positive, negative or zero curvature and it must take into account the gravitational energy (which is negative), dark energy, matter and heat. Since the gravitational energy is negative the heat can be positive and increasing as you go back ...


1

In Zero-Energy Model, negative energy associated with Gravity counterbalances positive energy associated with matter, photons, etc. So, No, Big Bang wasn't cold. You are just looking at partial picture (you just ignored Gravity). This is what Zero-Energy Model says: With traditional Big Bang model (which doesn't contain Inflation), the universe started out ...


0

One way to derive the existence and expression for entropy is to consider a system that undergoes arbitrary cyclic process $\gamma$ in which its temperature is changing and is the same as temperature of a heat reservoir it is connected to. The cyclic process can be decomposed into a set of infinitesimal Carnot cycles, where the change of state $s$ during ...


0

Properties of the Von Neumann entropy 1. Purity. A pure state ρ | φ 〉〈 φ | has S ( ρ ) = 0. 2. Invariance. The entropy is unchanged by a unitary change of basis S ( U ρ U † ) = S ( ρ ) , because the entropy depends only on the eigenvalues of the density matrix. 3. Maximum. If ρ has D non-vanishing eigen- values, then S ( ρ ) ≤ log D, with equality when all ...


0

But... all quantum states are really pure states right? There exists the density matrix formulation for the many body system quantum mechanically. What happens in a many body problem, order of 10^23 molecules per mole that are the appropriate numbers for a thermodynamic formulation, is that the off diagonal elements are so small that they are ...


2

The issue is that when a wavefunction collapses it has an inherent randomness to it. Since entropy is fundamentally related to information, I'll start with information to explain why this is significant. Information, Randomness, and Entropy If you think of the information of a state as what is needed to completely define a system something interesting ...


2

the entropy of the universe is always increasing True. Let's call this the total entropy. (Well, almost true, since the entropy of the universe remains constant for a reversible process). When a hot stone is dropped in cold water, it's entropy decreases (it gets colder), but the water increases it's entropy at the same time (it receives heat and gets ...


1

Nor entropy, neither enthalpy alone describe the spontaneity of a reaction (and temperature is also important): it is described by Gibb's free energy, which states $ \Delta\ G = \Delta\ H - T *\Delta\ S $ If $ \Delta\ G $ is positive, the process is endoergonic so not spontaneous; if $ \Delta\ G $ is negative, the process is esooergonic so spontaneous. ...


2

It is a basic assumption of statistical physics that the probability distribution ${p_k}$ for microstates $k$ best to use for a system with internal energy $U$ is that which obeys the constraint $$ \sum_k p_k E_k = U $$ where $E_k$ is energy of microstate $k$. There is another constraint $$ \sum_k p_k = 1. $$ There are many different distributions obeying ...


8

I think it is a mistake, in this case, to think of entropy as "a description of our ignorance." Rather, I would suggest that you think of entropy as a well-defined, objective property provided that you specify which degrees of freedom in the universe are inside and outside of your system. The content of this statement isn't really different, but it ...


6

There is a difference between the 'fine grained' and the 'coarse grained' entropy. If we start with a pure state (zero entropy), and we time-evolve it, the entropy indeed stays zero by unitary of time evolution. The fine grained entropy did not change. The coarse grained entropy is what we usually call the thermal entropy, and is the thing that always ...


4

The aging of the body has nothing to do with entropy. As has been pointed out, the body is not a closed system. It takes in energy all during its life and the overall thermodynamic state of 2 bodies of different ages but identical everything else (such as fat content, state of hydration, and so on) are equivalent. Aging is caused by many factors such as ...


0

Will provide an answer to this question based on the paper: "Topological Torsion and Thermodynamic Irreversibility" by R. Kiehn. 2001 The existence of magnetic monopoles is asumed to violate one of Maxwell's equations, specificaly (Gauss' law): $$div{\vec{B}} = 0$$ which implies the relation: $$\vec{B} = curl{\vec{A}}$$ for some vector potential ...


0

Basicaly in generaly we explain entropy as "the disorderness of the system" we measure it on the large scale for macroscopic objects not for microscopic objects.


1

The reasoning in the question is correct. If you have a box with gas particles placed in half of a box but otherwise uniformly random and with random velocities then it is overwhelmingly likely that it entropy will increase with time, but if reverse the velocities, you will still have randomly distributed velocities and the same argument will apply. By time ...


0

If you consider a small volume $v_0$ then the gas is no longer ideal by definition. If you modify the Sackur-Tetride to take this into account (with a hard sphere model for example as suggested by FenderLesPaul) you obtain a Van der Waals equation of state.


2

If you want the gas particles to have extended size then the first approximation would be to use the hard sphere gas model. Assume you have $N$ non-interacting hard spheres each of volume $v_0$ in a box of volume $V$. Then the entropy can be calculated as $S = Nk_B \ln [\frac{e}{N}(V - \frac{Nv_0}{2})(\frac{4\pi mEe}{3Nh^2})^{3/2}]$. C.f. Kardar ...


2

It is true that classical thermodynamic equations emerge from statistical mechanics. And that the increase in entropy depends on the increase in the number of microstates. Decays also increase the number of microstates. They are irreversible because decay releases energy and the thermodynamic system cannot deliver enough energy and combination of ...


-2

You can get at this without even worrying about nuclear physics. An atom in an excited electronic state is unstable, and it decays by emitting light (usually visible light). We commonly observe waves radiating outward spherically, but not converging inward spherically and being absorbed. This is a consequence of the second law, although some people like Ritz ...


0

Entropy change is not better nor worse than any other quantity obtained from a Legendre transformation, but its applicability when is convenient is different. If the system is isolated (internal energy is constant) and some internal constraints are relaxed then entropy is maximized when equilibrium is reestablished. On the other hand, if the interaction is ...


2

The change in internal energy is not a relevant quantity for spontaneous evolution of a system. Consider an isolated system made of two blocks of the same material at two different temperatures such that $T_1>T_2$. Heat will flow from $1$ to $2$ but the total change in internal energy is $\Delta U_{1+2}=0$. This information is therefore not useful. On ...


-1

because G=H_TS, and entropy of system at constant pressure must increase if its enthalpy remains constant and know that we've a spontaneous reaction when ¤s>0,


1

It seems to me you are conflating micro and macrostates. Entropy is not a property defined for a specific microstate but for an ensemble. When we describe a system in terms of pressure, entropy, etc instead of the momentum and position of each particle we are giving up the possibility of discussing specific microstates. The 'states of high order' you are ...


0

A very simple example of a configuration would be 0 heads and 100 tails. I. E. Coin 1 is tails, coin 2 is tails.... Coin N is tails. A configuration is the orientation (heads or tails) of each coin in the system and specifies a microstate. The total number of each state specifies a macrostate.


-1

I believe in this scenario, "configuration" would be synonymous with microstate. In statistical mechanics/thermodynamics, many of these terms are often synonymous in certain circumstances while still having specific meaning. "Configuration" does not have the exact meaning that microstate does, but in this case is used to mean microstate as you are more ...


2

The problem with the Boltzmann definition is, as you have neatly shown, that its usefulness depends on the assumption that your system is in equilibrium with its surroundings. Without first assuming equilibrium and subsequently setting the temperatures as equal, one cannot show that the Boltzmann entropy satisfies the First Law and hence meaningfully define ...


0

A short answer for a specific system might be helpful here. Imagine a system of two containers linked by a valve. One container is full of gas - the other is empty - it contains no gas - it has a vacuum. Answer to question 1 If the valve is opened the gas will fill both containers - and the entropy will increase, but temperature remains constant as no ...



Top 50 recent answers are included