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*From*: John Denker <jsd@av8n.com>*Date*: Thu, 14 Jan 2010 08:51:24 -0700

On 01/13/2010 07:35 PM, I wrote:

For example, stirring a gas is a dissipative process.

Sudden movements of a piston stir the gas; the

equations of fluid dynamics, even in their simplest

form, leave no doubt about this.

Here is another way of coming to the same conclusion.

I know there are about 50% high school teachers on

this list, and the following explanation is not one

I would present directly to typical HS students ...

but I'm not talking directly to students today. The

folks on this list may find this interesting, or may

find it sufficient to just file away the qualitative

fact that we are on reeeeally solid grounds when we

say that sudden movements of the piston stir the gas.

Let's forget about all the complexities of the

aforementioned fluid dynamics approach. Instead

let's take the quantum mechanical approach. Let's

simplify the gas down to a single particle, the

familiar particle in a box, and see what happens.

As usual, we assume the box is rigid and thermally

isolated / insulated / whatever. No entropy flows

across the boundary of the box. Also, no energy

flows across the boundary except for the work done

by the piston.

Since we are interested in entropy, it will not

suffice to talk about "the" quantum state of the

particle. The entropy of any particular quantum

state (microstate) is zero. We can however

represent the thermodynamic state (macrostate)

using a density matrix ρ.

For some background on density matrices in the

context of thermodynamics, see

http://www.av8n.com/physics/thermo-laws.htm#sec-rho

The entropy is

S := -Tr ρ log ρ [1]

which is _the_ gold-standard most-general definition

of entropy; in the classical limit it reduces to

the familiar workhorse expression S = - ∑ P log P.

Reference: op. cit.

For simplicity we consider the case where the initial

state is a pure state, i.e. a single microstate.

That means the initial entropy is zero, as you can

easily verify. Hint: equation [1] is particularly

easy to evaluate in a basis where ρ is diagonal.

Next we perturb our particle-in-a-box by moving one

wall of the box inward. We temporarily assume this

is done in such a way that the particle ends up in

the corresponding microstate, i.e. identical to the

original quantum state except for the shorter

wavelength as required to fit into the smaller box.

It is a straightforward yet useful exercise to show

that this does P dV "work" on the particle. The KE

of the new state is higher than the KE of the old

state.

Now the fun begins. We retract the previous assumption

about the final state; instead we calculate the final

macrostate using perturbation theory. In accordance with

Fermi's golden rule we calculate the _overlap integral_

between the original quantum state (original wavelength)

and each of the possible final quantum states (slightly

shorter wavelength).

In the original basis set, each basis function is

orthogonal to each of the other original basis

functions. In the final basis set, each of the basis

wave functions is orthogonal to each of the other final

basis wave functions, but it is only approximately

orthogonal to the original unperturbed state function.

So the previous assumption that the particle would

wind up in the corresponding state is provably not

quite true; when we do the overlap integrals there

is always some probability of transition to nearby

states.

It is straightforward to show that if the perturbation

is slow and gradual, the corresponding state gets the

lion's share of the probability. Conversely, if the

perturbation is large and sudden, there will be lots

of state transitions. The final state will not be a

pure quantum state. It will be a mixture. The entropy

will be nonzero, i.e. greater than the initial entropy.

The repeated assertion that the compression of the gas

is the same no "matter whether done slowly or quickly"

is completely untenable.

To summarize:

-- slow and gradual ==> isentropic, non dissipative

-- sudden ==> dissipative

So, like I said, we are on reeeally solid grounds

when we say that thermally insulated + gradual is

isentropic, while a sudden motion of the piston is

dissipative. Saying that the system is adiabatic in

the sense of thermally insulated does *not* suffice

to make it adiabatic in the sense of isentropic.

Note that in the quantum mechanics literature the slow

and gradual case is conventionally called the "adiabatic"

approximation in contrast to the "sudden" approximation.

These terms are quite firmly established ... even though

it conflicts with the also well-established convention

in other branches of physics where "adiabatic" means

thermally insulated; see next message.

========================

I file this example of gaseous dissipation under the

same category as Rumford's cannon boring. Actually

I find it to be a better pedagogical example, because

we understand the microscopics of dissipation in a

gas better than we understand the microscopics of

metal-on-metal sliding friction.

However, we can discuss sliding friction using many

of the same ideas. Let's model friction in terms

of _asperities_ on each metal surface. Each of

the asperities sticks and lets go, sticks and

lets go. When it lets go it wiggles and radiates

ultrasound into the bulk of the metal. This

produces in the short term a nonequilibrium state

due to the sound waves, but before long the sound

field dissipates, depositing energy and creating

entropy in the metal.

Again, if you think in terms only of the

(average force) dot (average dx)

you will never understand friction or dissipation.

You need to model many little contributions of the

form (short term force) dot (short term dx) and

then add up all the contributions. This is where

you see the work being done against the radiation

field.

Also: there is a nice introduction to the idea

of "radiation resistance" in Feynman volume I

chapter 32.

**Follow-Ups**:**Re: [Phys-l] physics of dissipation (was: T dS versus dQ)***From:*Bernard Cleyet <bernardcleyet@redshift.com>

**Re: [Phys-l] physics of dissipation (was: T dS versus dQ)***From:*chuck britton <cvbritton@mac.com>

**References**:**[Phys-l] Thermodynamics question***From:*"LaMontagne, Bob" <RLAMONT@providence.edu>

**[Phys-l] T dS versus dQ***From:*John Denker <jsd@av8n.com>

**Re: [Phys-l] T dS versus dQ***From:*Carl Mungan <mungan@usna.edu>

**Re: [Phys-l] T dS versus dQ***From:*John Denker <jsd@av8n.com>

**Re: [Phys-l] T dS versus dQ***From:*Carl Mungan <mungan@usna.edu>

**Re: [Phys-l] T dS versus dQ***From:*"LaMontagne, Bob" <RLAMONT@providence.edu>

**Re: [Phys-l] T dS versus dQ***From:*"Jeffrey Schnick" <JSchnick@Anselm.Edu>

**Re: [Phys-l] T dS versus dQ***From:*John Mallinckrodt <ajm@csupomona.edu>

**Re: [Phys-l] T dS versus dQ***From:*"LaMontagne, Bob" <RLAMONT@providence.edu>

**Re: [Phys-l] T dS versus dQ***From:*John Denker <jsd@av8n.com>

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