New answers tagged

1

The following is from a thermodynamics perspective, rather than statistical mechanics. I've read that equilibrium is the state at which entropy has a maximum. I believe you are referring to the maximum entropy principle. If so, from a thermodynamics perspective, the maximum entropy principle states that for an isolated system (a system for which there is ...


1

Boltezmans entropy equation is what is used for equilibrium situations, thats what the OP has no doubt read in a book, i.e. like this. However, Boltezeman cannot be used in non-equilibrium systems, for that you need to use Shannons entropy equation. That the resolution to the OP's 'paradox'. For more on the connection between entropy definitions, see my ...


2

I have an analogy that is helpul (at least to me). Replace the world energy by income, and energy levels by income ranges. In a country with no restriction for minimum or maximum incomes, and where luck to be born rich and/or gifted is the major factor to get money, there is an income distribution of equilibrium. In that analogy, temperature is related to ...


1

So it should be that the energy is distributed more "evenly" across the molecules, Yes, I that is a good way to describe it. Often entropy is described in terms of “order” and “disorder”, but my experience is that those terms are not understood correctly and lead to all manner of incorrect conclusions about entropy. Instead, if entropy is ...


1

From the paper Entropy and Time by Arieh Ben-Naim: Unfortunately, there is no general definition of equilibrium which applies to all systems. Callen [35*,36**], introduced the existence of the equilibrium state as a postulate. He also emphasized that any definition of an equilibrium state is necessarily circular. I don't know if that view is universal, but ...


2

There are many thermodynamic quantities that can be well-defined for out-of-equilibrium as well as equilibrium states. Internal energy, volume and entropy are a few. We define entropy in general by asserting that it is the sum of the entropies of all the parts of a system. Then you can take the parts small enough that each is in internal equilibrium, yet ...


2

The word “reversible” has (at least) two different meanings in physics. We must distinguish between time reversibility and thermodynamic reversibility. As the orbit and frictionless roller coaster examples in the comments show, falling in vacuum is time reversible. We can travel around a closed loop in phase space (in actual space we return to the same ...


0

It has already been pointed out to you that the first equation is incorrect. You should provide your derivation if you want to find out why. In any case, to calculate the entropy change for any process between two points you can use any convenient reversible path that connects the points (since entropy is a state function) and apply the following definition: ...


1

I guess, there is some error in your first equation, where there should be $C_v$ instead of $C_p$. For a reversible process: $$dQ_{rev}=dU+dW$$ $$dQ_{rev}=nC_vdt+PdV$$ Dividing by $T$, and substituting $P$ from ideal gas equation $PV=nRT$ $$\frac{dQ_{rev}}{T}=nC_v\frac{dT}{T}+nR\frac{dV}{V}$$ Integrating from $T_1$ to $T_2$ and $V_1$ to $V_2$: $$\Delta S_{...


1

I find the following from Enrico Fermi's book to be the most explicit derivation that shows: $$dS \geq \frac{dQ}{T}$$ Looking at a closed loop integral of the ratio of heat absorbed (or surrendered, depending on sign) to the heat bath temperature along each isotherm in a cycle (reversible or not), $$\oint \frac{dQ}{T}=\int_{A}^{B} \frac{dQ}{T}+\int_{B}^{A}\...


3

For the irreversible path between the same two end states, dQ is different than dQ for the reversible path, and in the integral of dQ/T for the irreversible path, you are supposed to use the temperature at the boundary interface between the system and surroundings $T_B$. So for the irreversible path, you should be using $$\int{\frac{dQ_{irrev}}{T_B}}$$So ...


4

Considering the result $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}<S(B)-S(A)$$ for an infinitesimal path, we get $$\mathrm{d}S\ge\frac{\mathrm{d}Q}{T}$$ where the equality holds only for a reversible process (by the definition of entropy). This means that in your expression $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T},$$ $\...


1

You need to clear the difference in entropy of the system and the universe. A phenomenon will happen spontaneously if the entropy of the universe is positive and won't happen if it is negative or zero (again, spontaneously). Now you may object by saying that why does water freezes spontaneously at STP? After all the entropy of ice is less than that of water ...


3

From an atomistic and quantum perspective, the entropy of a system is defined (the "Gibbs definition") in terms of the probability $P_i$ that a randomly selected microsystem from that system is found in the $i$th of the discrete quantum stationary states available for a microsystem to be in, as $$S = -Nk_{\textrm{B}}\sum_{\textrm{All }i}\left(P_i\...


4

The notes should read “reversible work”, which is a useful idealization because it lets us ignore entropy generation. (All real processes generate entropy, often with a rate dependence that overly complicates introductory problems.) Work doesn’t transfer entropy because it raises the energy of all particles in a system equally, with no dispersion. In this ...


7

Entropy is generated in work processes involving viscous dissipation of mechanical energy such as rapid expansion or compression, stirring, etc.


5

Because the entropy in this notes is defined in terms of the heat transfer: $$ dS=\left(\frac{\delta Q}{T}\right)_{int, rev}, $$ which is the increase of the internal energy not associated with work. Remarks As pointed by @ChetMiller in their answer and in the comments, the statement applies to the entropy transfer, but not to the entropy generation, which ...


0

"This is false. Given a low entropy state, the past is either a lower entropy state or a same-entropy state, and there's no time-reversal operator for thermodynamic processes (except those carried out at constant entropy) regardless of boundary conditions." That's a common misunderstanding evident in a number of the answers and comments above, ...


1

Edit: This answer is basically a mathematical supplement to the other two answers provided, which both make sense to me. How do we define entropy? Thermodynamically: studying an ideal gas in a Carnot cycle led to the realization that the ratio of the energy absorbed isothermally at some temperature $T_1$ was equal to the ratio of energy surrendered at some ...


1

"Order" is kind of an imprecise term. What we mean by entropy is "how many microstates could contribute to this macrostate." The way this is expressed mathematically is: $S = k•ln(W)$ $S$ is the entropy of the macrostate of the system, $W$ is the number of microstates that could produce that macrostate $k$ is Boltzmann's constant. ...


3

Probably, the easiest way to understand entropy is to think of it not as of a property of a system, but rather as the property of our knowledge of the state of the system, or better to say, the property of the probability distribution of the system over different internal states. If a system can be in states $1,2,...,n$ with probabilities $p_1,p_2,...,p_n$, ...


1

Suppose you have a single small ball floating in a closed box, bouncing around. If you divide the box in half with an imaginary plane, you can say that one half of the box is "full" if the particle happens to be in that half. The other half is then "empty". As this ball bounces about, the half/empty states will reverse, many times. Add ...


4

The distribution posed can be considered as conditional on the value of $r$, i.e., we have $$S(r)=-\int_0^Rdxp(x|r)\log p(x|r).$$ We then obtain $$p(x|r)=\frac{\theta(r-x)}{r},$$ where $\theta(x)$ is the step-function. Then the joint distribution is $$ p(x,r)=p(x|r)p(r)=\frac{\theta(r-x)}{rR}e^{-\frac{r}{R}}. $$ Marginalizing in respect to $r$ we obtain the ...


0

Let me first point out that entropy may mean different things. As Jaynes points in his article The minimum entropy production principle: By far the most abused word in science is "entropy". Confusion over the different meanings of this word, already serious 35 years ago, reached disaster proportions with teh 1948 advent of Shannon's information ...


3

Your equation for entropy change applies only to a reversible path. It gives the wrong answer for an irreversible path. To get the entropy change for any change in thermodynamic state, you devise (dream up) a reversible path between the same two end states, and evaluate the integral of dQ/T for that reversible path. It doesn't matter whether work is done ...


2

The way I think about is that by introducing the tools of statistical physics you purposefully throw away a lot of information and by doing so you are looking at different dynamics and the time reversibility of the microscopic laws may not be carried over. Take for example a system of $N$ bits $b_1,\dots, b_N$ that can be either in the state $1$ or $0$. A ...


0

In the beginning of the study of thermodynamics, books say that a reversible process is one such that the system can be restored to its initial state without any change in the universe, and then they say that an example of reversible process is the heat flux of two bodies in thermal contact, with a very small difference of temperature. Reader is frustrated ...


7

Yes there absolutely is a well-known apparent (classical) contradiction here. Classical mechanics is symmetric under $t \to - t$, i.e. for any given motion the reverse motion is also possible. As you suggested - we can try to justify it with ode uniqueness theorems that are based on saying the ode's/pde's we started from actually possess some geometric curve ...


2

The remorseless increase in entropy is simply a matter of probability. Consider an idealised pool table set up for a game- one with no friction or air resistance and perfectly elastic balls. You take a cue shot. The moving white ball is now the only energetic object on the table, so the entropy is low. When the white hits the triangle of balls at the end of ...


0

Irreversible change in state occurs over time. Yet time, per se, doesn't determine state in matter - it is the state variable, e.g. pressure, temperature, external field, etc. A volume of gas may have a pressure $p$ at some temperature $T$. But just from that fact alone we know nothing about actual states of particular gas molecules. A number of microstates ...


6

The laws of physics are differential equations, and to solve a differential equation you need two things: the equation itself tells you how the value of some physical variable at each point of a region of time and space is related to the values at its neighbouring points, and you also need the boundary conditions that specify the value of the variable on the ...


11

A long comment. Thermodynamics can be shown mathematically to be an emergent theory from statistical mechanics. Its laws are observational laws, deduced from variables and their measurements, that are needed to get the equations that map and predict the behavior of temperature, pressure etc. Classical mechanics is deterministic, given the equations of motion ...


4

Short answer: mechanics are time reversible on microscopic scale, entropy is never reversible on the macroscopic scale (except in the ideal case where its value does not change): Longer answer: The time-irreversibility of entropy is demonstrated by Clausius' Theorem: $$\oint \frac{dQ}{T} \leq 0 \tag{1}$$ $(1)$ can be derived by analyzing the results of the ...


3

The example you described is not really an illustration of the 2nd Law. Picture rather a helium balloon in a room. Equal pressure inside the balloon and outside. Then the balloon pops. Although at first you have a balloon-shaped cloud of helium in the center of a room full of air (State 1), after some time passes you will have a uniform mixture of helium and ...


26

The arrow of time in thermodynamics is statistical. Suppose you have a deterministic system that maps from states that can have character $X$ or character $Y$, to other states that can have character $X$ or character $Y$. The system is such that, for a randomly selected state $X_n$ or $Y_n$, the probability that the system will map it uniquely and ...


0

In my opinion, you are conceptually "putting the cart before the horse". Physics is the observation of physical phenomena, and the development of a mathematical model that describes the observations. Since practically no mathematical model is unique, it is easily possible to develop several mathematical models that describe the observations to ...


1

I think now my confusion is resolved. I think equation 2 is correct and 1 is wrong This statement is true. But if I use equation 2 I will not get $\rho_{A}=\sum_{i}p_i|\psi^{i}_{A}\rangle\langle \psi^i_{A}| +\sum_{i\neq j}\sqrt{p_ip_j}|\psi^{i}_{A}\rangle\langle \psi^j_{A}|$ The trace of $\rho_{AB}$ in $\mathcal{H}_B$ is: \begin{align*} &=\sum_k(\...


0

Using the definition of the change of entropy in a quasi-static process $\Delta S=\frac{\Delta Q}{T}$ (where $\Delta Q$ is the small amount heat transfered and $T$ is the temperature of the gas in that process) and the relation $\Delta Q=\Delta U + p \Delta V$ between the small amount of heat transferred in a process ($\Delta Q$), the small change of ...


0

As you already provided, Landauer's principle provides a lower limit to the Cost a living entity must pay for computation. However, the possible reward is not bounded a priori. The reason for this is that living agents are not Maxwell demons/Szilard engines trying to eke out a living by seemingly decreasing the entropy of a system or converting information ...


0

In an irreversible process, there are transport and/or chemical reaction processes occurring within the system that proceed at finite rates, all of which generate entropy. These include Heat conduction Viscous friction/dissipation Molecular diffusion Chemical reaction at finite rate These phenomena all occur microscopically (and, in most cases, are ...


2

It needs to be made clear what "reversible" means in this context. A process is thermodynamically reversible if it can be perfectly reversed (so it retraces all past macroscopic states in opposite order) by an arbitrarily small change in system conditions, such as temperature or pressure somewhere. For example, air inside a cylinder with a movable ...


2

In the schrödinger picture, entanglement is usually defined with respect to a fixed factorization of the Hilbert space, say $H_A\otimes H_B$. The factors correspond to complementary subsystems. The key to adapting the concept of entanglement to the heisenberg picture is to define subsystems in terms of observables instead, which is arguably more natural ...


1

When dealing with this kind of magnitudes, the difference between a grain of sand and the observable universe is just 60 orders of magnitude in terms of mass (and hence roughly entropy). That is $10^{-40}$ of $10^{100}$. So the macroscopically indistinguishable states could indeed be entire observable universes, although of course perfectly replicated solar ...


3

You could just as well make an argument for spontaneous clockwise rotation: thermal vibration of the gear bumps the needle and causes it to glide up the gradual ratchet slope much easier than up the sharp ratchet slope. Dropping of the needle behind the sharp ratchet slope prevents reverse (counterclockwise) motion. In reality, both effects are relevant (the ...


1

The information content of a single outcome is just $-\log p$. To make it concrete, if all probabilities are powers of two, then the length of the Huffman code for a signal of probability $p$ is exactly $-\log p$. $-p\log(p)-q\log(q)$ is the expected information content of a trial with two possible outcomes. Only the sum is meaningful, and only if $p+q=1$.


0

The information stored in a random variable $X$ is defined as: $$S=log_b(N) $$ where $N$ is the number of different outcomes we get when we measure $X$. Here the choice of $b$ is not really important and changing $b$ is same as multiplying $S$ with some constant. The common choices for $b$ are: bit for $b=1$ trit for $b=3$ nat for $b=e$ Check here for more ...


0

The important questions is whether zero temperature equals zero entropy. And this is in fact the content of the third law of thermodynamics, which states that the entropy of a system at zero temperature is equal to a well-defined constant. From a microscopical point of view this constant is the log of the number of degenerate ground states. For crystalline ...


0

A classical blackhole does not contain matter. The Schwarzschild metric is a vacuum solution, meaning the blackhole is made from curved spacetime: A stable gravitational soliton completely described by one parameter, $M$. So it's static, contains nothing, and is described entirely by one constant number (with dimensions of mass), for all of eternity. "...


2

They didn't associate entropy with a black hole at all. They had no intuition about it at all, basically because they saw no internal structure to a black hole and evaluating entropy requires counting the number of micro-states. It was the intervention by Bekenstein that raised this question and in fact he was able to write out Hawkings entropy formula ...


1

The multiplicative measure that you seek is perhaps most sensibly understood to simply be $W$ itself - the number of microstates that are compatible with the macrostate. You cannot do the exponentiation $$W^{k_B}$$ because $k_B$ has units. Unitful quantities cannot appear as exponents - just ask yourself, "what is $2^\mathrm{m}$, i.e. '2 to the power of ...


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