Dear Sir,

the last part of your question is quite easy to answer (even when not reading this chapter). The dimension of | grad f | /  f  is dim[d/dr] x dim[f] / dim[f] = 1/m x 1, i.e. the dimension of an inverse length.

Sincerely, FM

@ Mukul Bhatnagar - In order to appreciate the cited remark in the book, you will have to read the entire relevant section (covering pp. 534-539). In Eq. (27.17) of the book (p. 538) one has a first-order expansion, in which the gradient of the function f is multiplied by a microscopic vector d whose magnitude is of the order of the linear dimension of the unit cell of the crystal under consideration (for a cubic crystal, this would be the lattice constant a). The assumption at the top of page 536 implies that by assumption the ratio of the second term on the right-hand side of Eq. (27.17) to the first term is at most of the order of a/r0, which in the light of Fig. 27.1 (on p. 535) is considerably smaller than 1. Hence, the second term on the right-hand side of Eq. (27.17) is considerably smaller than the first term; by analogy, on assuming the function f to be arbitrary many times differentiable, the third term, not shown, is considerably smaller than the second term, etc. (the idea explained here forms the basis of the so-called gradient expansion). The validity of Eq. (27.22) (on p. 539) in part depends on the accuracy of the first-order expansion in Eq. (27.17).

I note in passing that the entire analysis by Ashcroft and Mermin is much simplified by expressing the relevant expressions in the Fourier domain (since the relevant integrals are of the convolution type), whereby the condition at the top of page 536 translates into a specific behaviour to be expected of the Fourier transform of the function f in particular in the long-wavelength region.