Assumption of the conservation of energy to Heat Flow

  • #1

Main Question or Discussion Point

Recently looked at why temperature flows from high Temperatures to Low temperatures.Essentially it was laid on two Fundamental Assumptions:
1.Energy is conserved in the isolated system
2.Entropy in isolated non quasi static systems will always tend to increase.

Lets take a brief look at Noethers theorem:
If a system has a continuous symmetry property, then there are corresponding quantities whose values are conserved in time.
Now without entering the mathematics. conservation of energy is a because of invariance under time translations.That is the langragian is not explicitly depending on time energy is conserved.

Without entering the mathematics, lets assume that we are in a system such that the above assumption is thrown out of the window.Would that mean that temperature does not need to flow from high temperatures to low temperatures?
 

Answers and Replies

  • #2
Orodruin
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First of all, let’s get the nomenclature straightened out. Temperature is an intensive variable and as such it does not flow. What flows (energy) thermally is called heat.

By adding energy to a system you can get heat to flow opposite to the temperature difference. This is the basics of a heat pump, whether used in refridgerators to keep the inside cool or heat pump heating systems (such as the one I have in my house) to keep the inside warn.
 
  • #3
DrDu
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The concept of temperature rests on the transitivity of thermal equilibrium. If two systems A and B are in thermal equilibrium, and B and a third system C are in thermal equilibrium, too, then also A and C are in equilibrium. It can be shown that this is only possible if the thermal equilibria can be parametrized by a parameter ##\theta##, the phenomenological temperature. The direction of the temperature scale is then fixed by the definition that, if two systems which are not in equilibrium are brought into thermal contact, the direction of heat flow is from the system with higher ## \theta ## to the one with lower ##\theta##.
So basically, the relation between heat flow and temperature is fixed by definition.
 
  • #4
First of all, let’s get the nomenclature straightened out. Temperature is an intensive variable and as such it does not flow. What flows (energy) thermally is called heat.

By adding energy to a system you can get heat to flow opposite to the temperature difference. This is the basics of a heat pump, whether used in refridgerators to keep the inside cool or heat pump heating systems (such as the one I have in my house) to keep the inside warn.
My bad I meant heat, just a typo.
 
  • #5
martinbn
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Why is this a quantum physics topic?
 
  • #6
vanhees71
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That's a very deep question! In my opinion the full understanding of Boltzmann's H theorem which is behind the discussion of this thread, indeed can fully understood only in terms of quantum statistics. Heuristically that's easily seen if you think about how to explain statistical physics to beginners in studying statistical physics without quantum theory. I've not been able to find a convincing concept so far, but using quantum theory together with the information-theoretical approach to entropy resolves all these quibbles right away, and you do not need to much quantum theory either. For the ideal gas it's enough to know the description of non-interacting indistinguishable particles and the fact that in 3 dimensions the realization of the permutation symmetry of indistinguishable particles is only through the two most simple representations of the permutation group, namely either the trivial representation (i.e., many-body Fock states and thus all many-body states do not change at all under permutation of indistinguishable particles), leading to the notion of bosons or the representation where each interchange of two particles in a Fock state changes simply the sign of the state, leading to fermions obeying the Pauli exclusion principles, i.e., that there cannot more than 1 particle occupying a single-particle state.

Further considering a single particle in a finite box-sized volume (with periodic boundary conditions to admit the definition of a momentum observable) you also solve the quibble about what's the natural size of phase-space volumes, which plagued Boltzmann from the very beginning in his endeavor to properly formulate statistical mechanics starting from classical mechanics of point particles. At the same time there's no more complicated argument about resolving the issue with Gibbs's paradoxon and the definition of entropy as a truely additive and extensive quantity as it should be definable from the point of view of phenomenological thermodynamics.

Then comes the H theorem (or 2nd Law). This was found by Boltzmann using his famous equation within classical mechanics under the assumption of symmetry of physics under space reflections, leading to detailed balance of the collision term of the Boltzmann equation, which leads to the fact that the total entropy of a closed system cannot decrease (which however doesn't prove the existence of a preferred direction of time, because the direction of time was implicitly put in the derivation of the Boltzmann equation, based on coarse graining in a subtle way through Boltzmann's "molecular-chaos hypothesis", i.e., by throwing away microscopic details about many-body correlations, and factorizing the two-body phase-space distributions into one-body phase-space distributions, thus cutting the BBKY hierarchy at the lowest non-trivial order). Then the equilibrium state follows unambiguously from the Boltzmann equation as the static solution, which is necessarily the state of maximum entropy given the constraints from conservation laws. Knowing that, you can do equilibrium statistical mechanics without the Boltzmann equation at all and just use the maximum-entropy principle to get the equilibrium distribution right away without analyzing the Boltzmann equation (which is not so easy for introductory statistical physics lectures, and it's better to follow the traditional path to first introduce equilibrium statistical physics).

Nowadays, however, we know that Nature doesn't obey any of the "discrete" spacetime symmetries, because the weak interactions breaks all these symmetries (space reflections/parity, time reversal, and charge conjugation symmetries of Minkoswski spacetime and its quantum analogue in connection with local relativistic quantum field theories). Is now the H theorem invalid (at least when weak interactions play a significan role in the physics question at hand)? The answer is clearly no! The reason is that one needs only a rather weak form of the detailed-balance principle, and this is fulfilled for the collision term, when formulated in terms of quantum mechanics via S-matrix elements, defining the transition-probability rates between asymptotic free states. It turns out that the very fundamental quantum mechanical property of the S-matrix being unitary is sufficient to guarantee the validity of the needed weak form of the detailed-balance principle and thus the validity of Boltzmann's H-theorem.

For a very nice detailed explanation, see Landau&Lifshitz volume X.
 

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