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6

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 ...


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, ...


4

It's true quite generally (for seperable Hilbert spaces at least). The original proof is in "Proof of the Strong Subadditivity of Quantum Mechanichal Entropy", by Lieb and Ruskai (J. Math. Phys. 14, 1938–1941 (1973)). The main idea is to prove the finite-dimensional case, and then extend it by taking a limit of the inequality on finite-dimensional subspaces ...


3

Regarding your new formulated question "my question boils down to whether a device similar to a voltmeter or thermometer exists for entropy": Well in that case then the answer is no, you can only measure entropy by studying the change of its dependencies on extensive/intensive variables that influence it (as they are different in different systems). To make ...


3

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 ...


3

You can set the entropy of your system under zero temperature to zero in compliance with the statistical definition $S=k_B\ln\Omega$. Then the S under other temperature should be $S=\int_0^T{\frac{dQ}{T}}$.


3

Unitarity of quantum mechanics prohibits information destruction. On the other hand, the second law of thermodynamics claims entropy to be increasing. If entropy is to be thought of as a measure of information content, how can these two principles be compatible? I don't think there's anything inherently quantum-mechanical about this paradox. The same ...


3

Based on some "google research" I get the impression that the popularity of the perfume thought experiment stems from a 1975 Scientific American article written by David Layzer called The Arrow of Time. The article featured this figure visualizing the thought experiment: Of course, the notion that the second law of thermodynamics implies an asymmetry ...


2

One cannot "derive" any non-equilibrium rate law from thermodynamics, simply because they are beyond the scope of the theory. Thermodynamics simply does not deal with such phenomena and hence cannot tell you how such processes occur (in this case heat conduction). All that thermodynamics does is relate mean values of certain properties of systems amongst ...


2

The Second Law of Thermodynamics states that the entropy of the universe always increases or stays constant. This means that we can reduce the entropy of the gas in the box (gas compressed from the whole box to half the box at constant temperature), only if we increase the entropy somewhere else. For example, we can compress the gas, doing work on it, and ...


2

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 ...


2

The current entropy in the Universe is all stored in photons. The first reference by Qmechanic gives you the precise value. Since the photons of the CMBR do not at present interact with anything, the entropy of the Universe is very close to being a constant. What evolution there is, is all due to non-reversible processes in baryonic matter, but it amounts to ...


2

As a general rule, physics gets easier when the mathematics gets harder. For example, algebra-based physics comprises a bunch of seemingly unrelated formulae, each and every one of which needs to be memorized separately. Add calculus and wow! Many of those supposedly disparate topics collapse into one. Add mathematics beyond the introductory calculus level ...


2

John Baez has a nice article on his website that goes into some detail on this question. Broadly, your intuition is right that at face value, it looks like structured systems are born out of nearly featureless initial conditions. However, as (for instance) a gas cloud collapses into a galaxy, it heats up (see the virial theorem, a theorem any ...


2

First, you have to understand that Rudolf Clausius put together his ideas on entropy in order to account for the losses of energy that was apparent in the practical application of the steam engine. At the time he had no real ability to explain or calculate entropy other than to show how it changed. This is why we are stuck with a lot of theory where we ...


2

In classical thermodynamics only the change of entropy matters, $\Delta S = \int \frac{dQ}{T} $. At what temperature it is put zero is arbitrary. You have the similar situation with potential energy. One has to arbitrarily fix some point where the potential energy is put zero. This is because only differences of potential energy matters in mechanical ...


1

I would strongly disagree with your statement that entropy has always been quite a mysterious quantity. Quite the contrary. What makes water (without something else interfering) always flow down the hill? Gravity. One does not have to know anything further about gravity to make good use of this statement, and people have used this fact for thousands of ...


1

Short answer is no but even then it depends on what you mean by "the 2nd law of thermodynamics". In conventional treatments of so-called equilibrium thermodynamics Fourier's law of heat conduction is completely independent of the rest. In what is called "rational thermodynamic" where the 2nd law is formulated as the "Clausius-Duhem inequality" it is in fact ...


1

What it means is that the number of bits required to specify the exact physical state the system is in, increases by 100/log(2) bits after the gas is heated. I think measuring the temperature in energy units is a step in the right direction, but what is even better is to do without any units. I.e. while units may be introduced for convenience, the formalism ...


1

I think the assumption that radiation is required for a collapse in general is mistaken. Think about a cloud of gas. If it is going to gravitationally collapse it must have a negative total energy; if it doesn't parts of the gas will fly off. If it has a negative total energy then there is some finite maximum size for the gas cloud, where it only has ...


1

A free dark matter cloud (without the presence of ordinary matter) will simply not "collapse" the same way a radiating gas cloud does. In both cases total momentum, angular momentum and energy are conserved, but in the case of a gas cloud the photons can carry away some of the angular momentum and most of the energy, in case of a dark matter cloud they ...


1

A higher entropy equilibrium state can be reached from the lower entropy state by an irreversible but purely adiabatic process. The reverse is not true, a lower entropy state can never be reached adiabatically from a higher entropy state. On a purely phenomenological level the entropy difference between two equilibrium states, therefore, tells you how "far" ...


1

Could anyone explain how this process complies with the second law of thermodynamics? The Stefan-Boltzmann law. I'll start with an ideal black body. Black bodies absorb all incoming radiation. They also emit radiation as a function of temperature. The peak frequency and the intensity increase as temperature increases. The emitted power is given by the ...


1

Strictly speaking, you're using the relation $a\ln b = \ln(b^a)$ outside of its domain of validity. When $a$ is a quantity with units (and $b$ is dimensionless), it's perfectly valid to write $a\ln b$, but it is not equal to $\ln(b^a)$, because $b^a$ is undefined and so is its logarithm. If you want, for notational convenience you could specify that the ...


1

If we assume our universe as an isolated system, then its entropy can only increase. It cannot decrease because of the second law of thermodynamics. It cannot stay unchanged because the universe is undergoing all kinds of irreversible processes.


1

Are not we simply saying that things more likely to occur, occur more times? Isn't it then, that the second law is simply an inmense tautology? No, this argument doesn't suffice to prove the second law. This argument only proves that thermal fluctuations away from equilibrum should be rare and short-lived. That's a statement that doesn't have anything ...


1

Now we have the second law of thermodynamics, that says that entropy always increases. Second law does not say exactly that. It has more formulations, some of which use the concept of entropy. One such formulation is When thermally insulated system changes its state from one equilibrium state to another, its entropy cannot decrease. This statement ...


1

Henri Poincaré, in discovering limit cycles, used a thought experiment containing a box with a partition. One side had a gas, and the other didn't. When the partition was removed, the gas would diffuse through the opening and occupy both sides of the container. He first published works describing limit cycles somewhere in 1881-1882. I am unsure if he ...


1

Can the law of conduction be derived from the assumptions of classical thermodynamics? The answer should be no, because classical thermodynamics does not deal with description of irreversible processes in time; it only deals with equilibrium states. Second law of thermodynamics does not assert that entropy decreases in time, only that after irreversible ...


1

A certain volume of space with a uniform distribution of particles has maximum entropy. That is correct for non-interacting particles, but wrong for particles with the gravitational interaction. When gravity condenses these particles, it increases the entropy of the system, not decreases it, at least when the Jeans instability condition is satisfied. ...



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