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Mathematical Derivation

Is there any way to derive a formal (mathematical) criterion for the condition of thermal equilibrium, starting from a hypothetical knowledge of  $\Omega_1$  as a function of  U₁  and  $\Omega_2$  as a function of   U₂ = U - U₁?  Of course! Why else would I be doing this? The thing about a maximum of a function (or a minimum, for that matter; either type of extremum obeys the same rule) is that the slope of the function must be zero at the extremum. [Otherwise it would still have further up or down to go!] Since the slope is given by the derivative, this reads

$$\hbox{\rm Criterion for an \underline{extremum}: } \qquad {\partial \Omega \over \partial U_1} \; = \; 0$$ (15.6)

In this case, since   $\Omega = \Omega_1 \cdot \Omega_2$,  the PRODUCT RULE for derivatives gives

$${\partial \Omega \over \partial U_1} \; = \; {\partial \Ome . . . . . . ega_1 \cdot {\partial \Omega_2 \over \partial U_1} \; = \; 0$$ (15.7)

Now,  $\Omega_2$  is a function of  U₂,  not  U₁;  but we can get around that by using the CHAIN RULE,

$${\partial \Omega_2 \over \partial U_1} \; = \; {\partial \O . . . . . . \partial U_2} \cdot {\partial U_2 \over \partial U_1} \cdot$$

where   U₂ = U - U₁  and  U  is a constant, so

$${\partial U_2 \over \partial U_1} \; = \; - 1$$

We can therefore substitute   ${\displaystyle -{\partial \Omega_2 \over \partial U_2} }$  for   ${\displaystyle {\partial \Omega_2 \over \partial U_1} }$  in Eq. (7):

$${\partial \Omega_1 \over \partial U_1} \cdot \Omega_2 \; - \; . . . . . . ega_1 \cdot {\partial \Omega_2 \over \partial U_2} \; = \; 0$$

or

$${\partial \Omega_1 \over \partial U_1} \cdot \Omega_2 \; = \; \Omega_1 \cdot {\partial \Omega_2 \over \partial U_2}$$

If we now divide both sides by the product   $\Omega_1 \cdot \Omega_2$,  we get

$${1 \over \Omega_1} \cdot {\partial \Omega_1 \over \partial U_ . . . . . . \over \Omega_2} \cdot {\partial \Omega_2 \over \partial U_2} .$$ (15.8)

Now we need to recall the property of the natural logarithm that was so endearing when we first encountered it:  $\ln(x)$  is the function whose derivative is the inverse,

$${d \over dx} \ln(x) = {1 \over x}$$

and, by the CHAIN RULE,

$${d \over dx} \ln(y) = {1 \over y} \cdot {dy \over dx}$$

In this case "y" is  $\Omega$  and "x" is  U,  so we have

$${\partial \over \partial U} \ln(\Omega) = {1 \over \Omega} \cdot {\partial \Omega \over \partial U}$$

which means that Eq. (8) can be written

$${\partial \over \partial U_1} \ln (\Omega_1) \; = \; {\partial \over \partial U_2} \ln (\Omega_2)$$

But the logarithm of the MULTIPLICITY FUNCTION  $\Omega$  is the definition of the ENTROPY  $\sigma$, so the equation can be simplified further to read

$${\partial \sigma_1 \over \partial U_1} \; = \; {\partial \sigma_2 \over \partial U_2}$$ (15.9)

where of course we are assuming that all the other parameters (like  N₁  and  N₂) are held constant.

Note that we have recovered, by strict mathematical methods, the same criterion dictated by common sense earlier. The only advantage of the formal derivation is that it is rigourous, general and involves no questionable assumptions.^15.14