# Poisson relationship between area and length of electric wire

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:

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

• One man's fish is another man's poisson. – niels nielsen Jan 18 at 0:06

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: $$-2 \nu { \Delta L \over L} + \underbrace{\left( \nu {\Delta L \over L} \right)^2}_{\approx 0}\approx -2 \nu { \Delta L \over L}$$
• I would argue that if $\frac{\Delta L}{L}$ is less than to 0.001 (which is common for materials in the elastic region) then its safe to approximate that to zero. Because in that case $\left(\frac{\Delta L}{L}\right)^2$ would be less than 0.000001. However, I've never seen that written down as a rule of thumb. – NMech Jan 17 at 21:41