“…The central portion will not be an exact straight line, and to connect smoothly on the exponential decay part we will need a different coefficient in the exponential of eq 9 and 10. Let us still approximate the central part of the profile by a straight line, calling the Az defined in this way (i.e., as in Figure 2 and in eq 7 and 8) 2, and we replace eq 12 by b0 = /8 (13) As we shall see, in order to have a consistent picture, and 2 will have to behave slightly differently as the critical point is approached. In general, we may expect the actual curvature of the nearly straight-line midsection of the profile to be such that the slope with which it connects to the exponential drop offs is less than it is at z = 0.…”