Editing Poisson's ratio (section)

=== Length change ===
[[Image:PoissonRatio.svg|thumb|300px|right|'''Figure 1''': A cube with sides of length {{mvar|L}} of an isotropic linearly elastic material subject to tension along the x axis, with a Poisson's ratio of 0.5. The green cube is unstrained, the red is expanded in the {{mvar|x}}-direction by {{math|Δ''L''}} due to tension, and contracted in the {{mvar|y}}- and {{mvar|z}}-directions by {{math|Δ''L''′}}.]]
For a cube stretched in the {{mvar|x}}-direction (see Figure 1) with a length increase of {{math|Δ''L''}} in the {{mvar|x}}-direction, and a length decrease of {{math|Δ''L''′}} in the {{mvar|y}}- and {{mvar|z}}-directions, the infinitesimal diagonal strains are given by
:<math>
d\varepsilon_x = \frac{dx}{x},\qquad d\varepsilon_y = \frac{dy}{y},\qquad d\varepsilon_z = \frac{dz}{z}.
</math>

If Poisson's ratio is constant through deformation, integrating these expressions and using the definition of Poisson's ratio gives
:<math>-\nu \int_L^{L+\Delta L} \frac{dx}{x} = \int_L^{L+\Delta L'} \frac{dy}{y} = \int_L^{L+\Delta L'} \frac{dz}{z}.</math>

Solving and exponentiating, the relationship between {{math|Δ''L''}} and {{math|Δ''L''′}} is then
:<math>
\left(1+\frac{\Delta L}{L}\right)^{-\nu} = 1+\frac{\Delta L'}{L}.</math>

For very small values of {{math|Δ''L''}} and {{math|Δ''L''′}}, the first-order approximation yields:
:<math>\nu \approx - \frac{\Delta L'}{\Delta L}.</math>