Poisson relationship between area and length of electric wire

Question

I was given the following formula to relate the change of area against the change of length of an electric wire with a Poisson ratio:

$ {\Delta A \over A} = -2 \nu {\Delta L \over L} $

where $ \Delta A \over A $ represents the change in cross-sectional area of the wire due to the transverse strain as the wire gets pulled longitudinally stretching length $ L $ to $ L+\Delta L$.

I don't get how this equation is derived. The Poisson ratio is defined by $ \nu = -{\epsilon_{lateral} \over {\epsilon_{longitudinal}} } = - {{\Delta d / d} \over {\Delta L / L}} $ where $ d $ is the diameter of the cross section. Then the ratio of the area:

$ {\Delta A \over A} = {{0.25\pi(d+\Delta d)^2 - 0.25\pi d^2} \over {0.25\pi d^2}} = {{2d \Delta d} \over d^2} + {{\Delta d^2} \over {d^2}} = -2 \nu { \Delta L \over L} + \big( \nu {\Delta L \over L} \big)^2 \neq -2 \nu {\Delta L \over L} $

Answer 1

Since you are essentially using infinitesimal changes, then higher order differences can be neglected.

I.e. following from your equation $${\Delta A \over A} = {{0.25\pi(d+\Delta d)^2 - 0.25\pi d^2} \over {0.25\pi d^2}} = {{2d \Delta d} \over d^2} + {{\Delta d^2} \over {d^2}} = -2 \nu { \Delta L \over L} + \big( \nu {\Delta L \over L} \big)^2 $$

because $\left(\frac{\Delta L}{L}\right)^2$ is a second order difference, you can assume that $\left( \nu {\Delta L \over L} \right)^2\approx 0$.

Therefore: $$ -2 \nu { \Delta L \over L} + \underbrace{\left( \nu {\Delta L \over L} \right)^2}_{\approx 0}\approx -2 \nu { \Delta L \over L} $$