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Rearranging letters in a book destroys semantic information, the information the text gives a reader who can understand it. But it does not change the information in the Shannon or thermodynamic sense, as distinguishable states.
When I read a book I am exploiting pre-existing correlations between my brain and the text, so that reading a certain set of ...
5
In principle, if you observed the isolated gas-filled container for long enough, you would occasionally see a state in which all gas molecules were congregated on the right side of the box. But you would likely have to wait for an exceedingly long time, well in excess of the expected life-time of the visible universe, before it happened for the first time. ...
5
Consider a thermal ensemble comprised of $N$ degrees of freedom at a temperature $T$. Let us look at this system from the perspective of the microcanonical ensemble. In the microcanonical ensemble, such a system with $N$ degrees of freedom is constructed such that its entropy is $\Delta S = \log{N}$, and such that its energies are spreaded in an interval $E ...
4
Why does minimization of free energy result into an almost uniform distribution of protein foldings?
You are forgetting about the aqueous environment the protein is folding in. Typically proteins fold in such a way that hydrophobic sections of the protein are hidden from the aqueous environment and hydrophilic sections are not. This means that even though the entropy decreases in just the confirmation of the protein, the entropy of the aqueous environment ...
2
I think the validity of thermodynamic requiring large number of degrees of freedom, pointed by Bruce Lee, restricts entropy to be rather a large number. However, if you consider the two-state system (spin $\uparrow, \downarrow$), you may get such a situation. Let the $p$-probability for spin to point upwards, and this state will have energy = $\varepsilon$, ...
2
There are many ways of creating entangled photons, the mainly used methods are:
Spontaneous Parametric Down Conversion
In this case they use a special crystal, and input a single (pump) photon, and the output is a pair (or more) photons, whose total energy equals the input photons'. Momentum is conserved as well, the total input photon momentum and the ...
2
There are way more microstates where the gas particles aren't in the corner than there are microstates where the gas is in the corner. So if you're talking the probability of "in the corner" vs "not in the corner" they latter is going to win out.
2
Any reversible cyclic process will be isentropic (since entropy is a state function), whereas it's not necessary that it's adiabatic as well. A famous example is the Carnot cycle.
2
In conventional thermodynamic language "adiabatic" process means being adiabatic (no heat exchnage) at every instant of the process not just in the sense that the total heat exchange is zero. The process itself can be reversible or irreversible, only the heat exchanged must be zero. The so-called adiabatic legs of the Carnot cycle (isothermal-...
2
If by "a microscopic/physical argument" you're looking for an argument that only relies on the axioms usually assumed for ordinary quantum mechanics, then that's not possible. To see why, let's look at classical statistical mechanics.
We know, in principle, how to exactly compute the microscopic time-evolution of a classical system with fixed ...
2
So, you want to know about entropy. Well, in thermodynamics it is defined as the measure of how much energy or heat is 'spread'. Then there is the second law:
Entropy always increases a.k.a. energy tends to spread out over time. - Second Law of thermodynamics
But, in a modern sense, we don't think of entropy just as a thermodynamic quantity measuring the ...
1
The idea that microcanonical ensemble would be "more fundamental" than other ensembles is strongly related to a mechanics-based approach to statistical mechanics, where the clean starting point would be a Hamiltonian isolated system of N particles. Therefore, a system at constant energy.
However, that this is not the only possible point of view was ...
1
The fundamental mistake here is that the Sun radiation system is not in equilibrium past the surface. As it radiates out, the photon density decreases.
Once the energy density is sufficiently reduced, BB-radiation-curve peaking at visible wavelengths (like the Sun) has lower entropy than BB-radiation-curve peaking at infrared wavelengths (like the Earth).
...
1
If the energy level of the 'bitgas' does not depend on the number of bits in state 0 and state 1, then all microstates have the same energy level, and the system is an example of the microcanonical ensemble.
The thermodynamical equilibrium of this system is the macrostate where all microstates have the same probability. This state has entropy S = N*log(2), ...
1
it seems we can have reversible isentropic processes that are not
adiabatic..
No you cannot.
An isentropic process is by definition a process that is both adiabatic and reversible. So you can't have an isentropic process that is not adiabatic.
However, you can have an adiabatic process that is not isentropic, if it is not a reversible process. An example is ...
1
No.
If the universe is bounded and finite, then via the Poincaré recurrence theorem[1], the Fluctuation theorem[2], and some assumptions (e.g. all previous states accessible), the universe will get arbitrarily close to any previous state an infinite amount of times. Over time parts will spontaneously fluctuate to a lower entropy state, and over arbitrary ...
1
Temperature can be thought of as average energy per microscopic degree of freedom of the system. For example, in a gas, temperature is proportional to the average kinetic energy of the particles.
The entropy can be thought of as a measure of the number of microscopic degrees of freedom. If you imagine a discrete system of $n$ binary bits for instance, the ...
1
When you have increased the temperature and left the system alone
The system will eventually try all the possible microstates. And the ordered arrangement is one of the possible various microstates.
But one of the specific well-behaved order is probably very less likely to happen among large number of microstate.
Entropy is a logarithm of number of ...
1
They are indeed the same relation, just written in different forms and with all the exact constants filled in in the second version. Note that, because of the logarithms, and the fact that the molar specific heat of an ideal gas is $c_{v}^{*}=\frac{3}{2}R$, the first expression can be rewritten as
$$\frac{S}{n}=R\log\left(\frac{VT^{3/2}}{N}\right)+X,$$
...
1
The second law of thermodynamics does not say anything about the rate at which entropy increases.
As a society, we have collectively decided that the existence of our
species is more important than accelerating the production of
entropy in our little corner of the galaxy.
1
Why does minimization of free energy result into an almost uniform distribution of protein foldings?
An important factor here is that protein functioning requires specific conformation. If it misfolds, then, at best, it will be disfunctional, at worst it may result in harmful effects for the cell (indeed, protein misfolding is the cause of some serious illnesses). Note also, that proteins fold into the required configuration only under specific normal ...
1
The one-body entropy of a system at equilibrium is equivalent to the thermodynamic entropy per particle. It is ubiquitous in discussions of cooling ultracold gases to reach low-entropy ordered states (here is one example of many), and is often referred to as the "entropy per particle" ($S/N$) or just the "entropy" of the gas. This ...
1
Despite the question already having an accepted answer, I think I've come up with an alternative way of associating a "temperature" with a deck of cards. Assume that, instead of suits and usual labels, the cards are just numbered $1$ through $n$. To each microstate of the deck (i.e. its ordering), we can assign the energy as the number of ...
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