Editing Poisson's ratio (section)

==Poisson's ratio from geometry changes==
=== 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>

=== Volumetric change ===

The relative change of volume {{math|{{sfrac|Δ''V''|''V''}}}} of a cube due to the stretch of the material can now be calculated. Since {{math|''V'' {{=}} ''L''<sup>3</sup>}} and
:<math>V + \Delta V = (L + \Delta L)\left(L + \Delta L'\right)^2</math>

one can derive
:<math>\frac{\Delta V}{V} = \left(1 + \frac{\Delta L}{L} \right)\left(1 + \frac{\Delta L'}{L} \right)^2 - 1</math>

Using the above derived relationship between {{math|Δ''L''}} and {{math|Δ''L''′}}:
:<math>\frac {\Delta V} {V} = \left(1 + \frac{\Delta L}{L} \right)^{1 - 2\nu} - 1</math>

and for very small values of {{math|Δ''L''}} and {{math|Δ''L''′}}, the first-order approximation yields:
:<math>\frac {\Delta V} {V} \approx (1-2\nu)\frac{\Delta L}{L}</math>

For isotropic materials we can use [[Lamé parameters|Lamé's relation]]<ref>{{Cite journal|last1=Mott |first1=P. H. |last2=Roland |first2=C. M. |title=Limits to Poisson's ratio in isotropic materials—general result for arbitrary deformation |journal=Physica Scripta |publisher=Chemistry Division, Naval Research Laboratory |date=3 April 2012|volume=87 |issue=5 |page=055404 |doi=10.1088/0031-8949/87/05/055404 |arxiv=1204.3859 |s2cid=55920779}}</ref>
:<math>\nu \approx \frac{1}{2} - \frac{E}{6K}</math>
where {{mvar|K}} is [[bulk modulus]] and {{mvar|E}} is [[Young's modulus]].

=== Width change ===
[[Image:Rod diameter change poisson.svg|350px|thumb|right|Figure 2: The blue slope represents a simplified formula (the top one in the legend) that works well for modest deformations, {{math|∆''L''}}, up to about ±3. The green curve represents a formula better suited for larger deformations.]] 
If a rod with diameter (or width, or thickness) {{mvar|d}} and length {{mvar|L}} is subject to tension so that its length will change by {{math|Δ''L''}} then its diameter {{mvar|d}} will change by:
:<math>\frac{\Delta d}{d} = -\nu \frac{\Delta L} L</math>

The above formula is true only in the case of small deformations; if deformations are large then the following (more precise) formula can be used:
:<math>\Delta d = -d \left( 1 - {\left( 1 + \frac{\Delta L} L \right)}^{-\nu} \right)</math>
where
*{{mvar|d}} is original diameter
*{{math|Δ''d''}} is rod diameter change
*{{mvar|ν}} is Poisson's ratio
*{{mvar|L}} is original length, before stretch
*{{math|Δ''L''}} is the change of length.

The value is negative because it decreases with increase of length