28.02.2014 Views

New trends in physics teaching, v.4; The ... - unesdoc - Unesco

New trends in physics teaching, v.4; The ... - unesdoc - Unesco

New trends in physics teaching, v.4; The ... - unesdoc - Unesco

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

<strong>New</strong> Trends <strong>in</strong> Physics Teach<strong>in</strong>g IV<br />

do arguments about changes <strong>in</strong> numbers of microstates appear, nor is dS = kd (In a) to be found.<br />

Is he then ‘do<strong>in</strong>g statistical mechanics’? Yes, <strong>in</strong> the sense that the Boltzmann factor conta<strong>in</strong>s the<br />

statistics. No, <strong>in</strong> the sense that it is not clear how it does so.<br />

Feynman can f<strong>in</strong>esse the statistics, because of the form common to all the applications he<br />

discusses, <strong>in</strong>deed to any. Briefly, any process that borrows energy E from a heat bath <strong>in</strong>volves<br />

an entropy decrease of the heat bath of AS = - E/T (leav<strong>in</strong>g aside qualifications about constant<br />

volume etc.). Writ<strong>in</strong>g this as A In 52 = - E/kT, and A In 52 as In ( after/a before) we always get<br />

<strong>The</strong> ratio of the values of 52 after and 52 before borrow<strong>in</strong>g energy E gives the probability that it<br />

can be borrowed, and so the numbers of particles able to borrow that much energy. Thus the<br />

Boltzmann factor works because it is a ratio of numbers of microstates.<br />

<strong>The</strong> f<strong>in</strong>esse shows up <strong>in</strong> what looks like a trivial po<strong>in</strong>t <strong>in</strong> Feynman’s argument. <strong>The</strong> gas must<br />

be isothermal, so he <strong>in</strong>troduces a ‘copper bar’ to keep it so (figure 2). Now of course the Boltzmann<br />

factor does not just apply to gases under gravity (the examples Feynman uses are to show this).<br />

So the argument is not really <strong>in</strong>volved with the system be<strong>in</strong>g a gas, even though it appears to be<br />

all about that. <strong>The</strong> ‘copper bar’ is what does the trick. <strong>The</strong> energy rngh for a molecule to get up a<br />

height h cannot come from the gas if the gas must not become cooler <strong>in</strong> any part. It must come<br />

from the ‘copper bar’ heat s<strong>in</strong>k; <strong>in</strong>deed, this is just what a canonical ensemble argument relies<br />

on. <strong>The</strong>re are fewer molecules at great heights because the s<strong>in</strong>k is unlikely to yield up that much<br />

energy by a chance fluctuation. <strong>The</strong>re are fewer molecules at great heights because the ones<br />

below must hold them up, too, but the two ‘becauses’ are very different. <strong>The</strong> first is an essentially<br />

statistical mechanical argument; the second is particular and ad hoc. Feynman uses the second<br />

and not the first, and so is unable to show why the result he gets is general, not special.<br />

-<br />

Mechanism<br />

for equaliz<strong>in</strong>g d<br />

temperatures.<br />

h<br />

Feynman’s ‘copper bar’ to ma<strong>in</strong>ta<strong>in</strong> gas<br />

isothermal state<br />

Figure 2. Feynman’s ‘copper bar’ to ma<strong>in</strong>ta<strong>in</strong> gas <strong>in</strong> isothermal state.<br />

Feynman and heat<br />

In speak<strong>in</strong>g of the second law, Feynman says th<strong>in</strong>gs like, ‘. . . it ought to be possible to lift<br />

weights. . . and thus to do work with heat’; ‘If the whole world were at the same temperature,<br />

160

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!