Showing the thermal expansion coefficient is the sum of the linear expansion coefficients

Question

Given that the thermal expansion coefficient is defined as: $$ \beta=\frac{\frac{\Delta V}{V}}{\Delta T} $$ and the linear thermal expansion coefficient is defined as: $$ \alpha = \frac{\frac{\Delta L}{L}}{\Delta T} $$

Show that $\beta $ is equal to the sum of its linear expansion coefficients in all three directions: $$ \beta = \alpha_x+\alpha_y+\alpha_z $$

I first began as writing: $$ \alpha_i \Delta T= \frac{\Delta L_i}{L_i} $$

Where i denotes a Cartesian coordinate. We then have that:

$$ (\alpha_x+\alpha_y+\alpha_z) \Delta T= \frac{\Delta L_x}{L_x}+\frac{\Delta L_y}{L_y}+\frac{\Delta L_z}{L_z} $$

Simplifying to: $$ (\alpha_x+\alpha_y+\alpha_z) \Delta T=\frac{\Delta L_x*L_y L_z+\Delta L_y*L_x L_z+\Delta L_z*L_y L_x}{L_x L_y L_z} $$

Implying that the numerator is the change in volume of our object. Upon further thinking, I realized that if we had a cube that started off with a one meter edge, and expanded to a cube with a two meter edge, we should have a volume change of 7, but the numerator of my equation gives 3. This leads me to conclude something is wrong with my formulation, but I do not see it.

Answer 1

Formulating this with differentials shows that your reasoning is indeed correct:

If $V = L_xL_yL_z$, then

$$\mathrm{d}V = \mathrm{d}(L_xL_yL_z) = L_yL_z\mathrm{d}L_x + L_xL_z\mathrm{d}L_y + L_xL_y\mathrm{d}L_z$$

and therefore

$$ \beta = \frac{1}{V}\frac{\mathrm{d}V}{\mathrm{d}T} = \frac{1}{L_xL_yL_z} \frac{L_yL_z\mathrm{d}L_x + L_xL_z\mathrm{d}L_y + L_xL_y\mathrm{d}L_z}{\mathrm{d}T} = \alpha_x + \alpha_y + \alpha_z$$

where $\alpha_i = \frac{1}{L_i}\frac{\mathrm{d}L_i}{\mathrm{d}T}$. There is no problem if you define the coefficient properly by considering infinitesimal changes instead of the approximative notation with $\Delta$.