Biological Physics: Energy, Information, Life

6. Track 2[[Student version, January 17, 2003]] 207

waylet’s now evaluate ∫
+∞
−∞

dx 1 ···dxn+1e−x

2

in two ways. On one hand, it’s justYn+1,but then again it’s

∫∞

0 (Anr

ndr)e−r^2 ,where nowAnis

the surface area of ann-dimensional sphere, our quarry. Let’s call the integral in this expression
Hn.UsingY=π^1 /^2 weconclude thatAn=π(n+1)/^2 /Hn.
Wealready know thatH 1 =1/2. Next consider that

−

d

dβ

∣∣

∣∣

β=1

∫∞

0

drrne−βr

2

=

∫∞

0

drrn+2e−r

2

(a trick recycled from Equation 3.14). The right side is justHn+2,while on the left we change
variables tor′=

√

βr,finding−ddβ

∣∣

β=1

[

β−(n+1)/^2 ×Hn

]

.SoH 3 =H 1 ,H 5 =2H 3 ,and in general
for any odd numbernwe haveHn=^12 ×

(n− 1

2

)

!. Substituting into the above formula for the sphere
areaAnand takingn=3N− 1 gives the formula quoted in Equation 6.6 above. (Think for yourself
about the case wherenis even.)

4. Why did we need the Planck constant
   in Equation 6.6? No such constant appears in Equa-
   tion 6.5. Actually, we might have expected that constants with dimensions would appear when
   wepass from purely mathematical constructions to physical ones. One way to explain the ap-
   pearance of
   is to note that in classical physics, position and momentum are continuous variables
   with dimensions. Thus the “number of allowed states” is really avolumein the space of positions
   and momenta, and so has dimensions. But you can’t take the logarithm of a number with dimen-
   sions! Thus we needed to divide our expression by enough powers of a number with the dimensions
   L·MLT−^1 .The Planck constant is such a number.
   Quantum mechanics gives a deeper answer to this question. In quantum mechanics, the allowed
   states of a system really are discrete, and we can count them in the na ̈ıve way. The number of states
   corresponding to a volume of position/momentum space involves
   via the Uncertainty Principle,
   and so
   enters the entropy.


5. We must be careful when formulating the Statistical Postulate, because the form of a probability
   distribution function will depend on the choice of variables used to specify states. To formulate
   the Postulate precisely, we must therefore specify that equilibrium corresponds to a probability
   distribution that is uniform (constant),whenexpressed in terms of a particular choice of variables.
   Tofind the right choice of variables, recall that equilibrium is supposed to be a situation where
   the probability distribution doesn’t change with time. Next, we use a beautiful result from me-
   chanics (Liouville’s theorem): A small region d^3 Npd^3 Nrevolves in time to a new region with the
   same volume inr-pspace. (Other choices of coordinates on this space, like (vi,ri), do not have
   this property.) Thus if a probability distribution is a constant times d^3 Npd^3 Nrat one moment, it
   will have the same form at a later time; such a distribution is suitable to describe equilibrium.

6.3.2′

1. What’s the definition ofT,Equation 6.9, got to do with older ideas of temperature? One could
   define temperature by making a mercury thermometer, marking the places where water boils and
   freezes, and subdividing into 100 equal divisions. That’s not very fundamental. Moreover, if we
   did the same thing with an alcohol thermometer, the individual markings wouldn’t agree: The
   expansion of liquids is slightly nonlinear. Using the expansion of an ideal gas would be better, but
   the fact is that each of these standards depends on the properties of some specific material—they’re