Relationship between Linear, Area and Cubic Expansivity

Derivation of The Relationship between Linear Expansivity and Area Expansivity:

Let us consider a sheet of metal that has been heated through a temperature change of θ. Let the initial length and breadth be L₁ and B₁ respectively and the final length and breadth be L₂ and B₂ respectively.

Let the initial area of the sheet, \( \scriptsize A_1 = L_1 \: \times \: B_1 \)

Let the final area of the sheet, \( \scriptsize A_2 = L_2 \: \times \: B_2 \)

From the formula of linear expansivity,

L₂ = L₁( 1 + α Δθ)

B₂ = B₁(1 + α Δθ)

where Δθ = θ₂ – θ₁

Hence \( \scriptsize A_2 = L_2 \: \times \: B_2 \\ = \scriptsize L_1 (1 \: + \: \alpha \Delta \theta) \: \times \: B_1(1 \: + \: \alpha \Delta \theta) \\ \scriptsize = L_1B_1(1 \: + \: \alpha \Delta \theta)^2 \\ \scriptsize = L_1B_1( 1 \: + \: 2 \alpha \Delta \theta \: + \: \alpha^2 \Delta \theta^2) \)

We can neglect α² because α is a very small quantity (e.g. 0.00002). Therefore, α² is approximately zero. I.e. α² = 0.

Hence we have;

\( \scriptsize A_2 = L_1B_1( 1 \: + \: 2 \alpha\Delta \theta) \) …..(1)

From the formula of Area expansivity,

\( \scriptsize A_2 = A_1( 1 \: + \: \beta \Delta \theta) \) …..(2)

Note: \( \scriptsize A_1 = L_1 B_1\)

By comparing equations (1) and (2)

We have \( \scriptsize \beta \Delta \theta = 2 \alpha \Delta \theta \)

Divide both equations by Δθ

\( \frac{ \beta \Delta \theta}{\Delta \theta} = \frac{2 \alpha \Delta \theta}{\Delta \theta} \)

\( \scriptsize \therefore \beta = 2 \alpha \)

Derivation of The Relationship between Linear Expansivity and Cubic Expansivity:

Using similar calculations we can show that the cubic and linear expansivity are related by this equation

\( \scriptsize \gamma = 3 \alpha \)

Let us consider a sheet of metal that has been heated through a temperature change of θ. Let the initial length, breadth, and height be L₁, B₁ and H₁ respectively and the final length, breadth and height be L₂, B₂ and H₂ respectively.

Let the initial volume of the metal, \( \scriptsize V_1 = L_1 \: \times \: B_1 \: \times \:H_1\)

Let the final volume of the metal, \( \scriptsize V_2 = L_2 \: \times \: B_2 \: \times \:H_2\)

From the formula of linear expansivity,

L₂ = L₁( 1 + α Δθ)

B₂ = B₁( 1 + α Δθ)

H₂ = H₁( 1 + α Δθ)

Substitute the values above into the equation \( \scriptsize V_2 = L_2 \: \times \: B_2 \: \times \:H_2\)

V₂ = L₁( 1 + α Δθ) × B₁( 1 + α Δθ) × H₁( 1 + α Δθ)

V₂ = L₁B₁H₁( 1 + α Δθ)( 1 + α Δθ) (1 + α Δθ)

open brackets

V₂ = L₁B₁H₁( 1 + α Δθ)( 1 + α Δθ) (1 + α Δθ)

V₂ = L₁B₁H₁( 1 + α Δθ + αΔθ + α²Δα²Δθ²θ² )(1 + α Δθ)

V₂ = L₁B₁H₁( 1 + 2αΔθ + α²Δθ² )(1 + α Δθ)

Simplify further by opening brackets

V₂ = L₁B₁H₁( 1 + 2αΔθ + α²Δθ² )(1 + α Δθ)

V₂ = L₁B₁H₁( 1 + α Δθ + 2αΔθ + 2α²Δθ² + α²Δθ² + α³Δθ³ )

V₂ = L₁B₁H₁( 1 + 3αΔθ + 2α²Δθ² + α²Δθ² + α³Δθ³ )

But V₁ = L₁B₁H₁

∴ V₂ = V₁( 1 + 3αΔθ + 2α²Δθ² + α²Δθ² + α³Δθ³ )

We can neglect α² and α³ because α is a very small quantity (e.g. 0.00002). Therefore, α² and α³ are approximately zero. I.e. α² = 0, α³ = 0.

∴ \( \scriptsize V_2 = V_1(1 \: + \: 3 \alpha \Delta \theta) \) …(a)

From the formula of cubic expansivity,

\(\scriptsize V_2 = V_1(1 \: + \: \gamma \Delta \theta) \) ….(b)

By comparing equations (a) and (b)

\( \scriptsize \gamma \Delta \theta = 3 \alpha \Delta \theta \)

Divide both equations by Δθ

\( \frac{ \gamma \Delta \theta}{\Delta \theta} = \frac{3 \alpha \Delta \theta}{\Delta \theta} \)

\( \scriptsize \gamma = 3 \alpha \)