# In the equation $\delta L/L=\alpha \delta T$ why is $\alpha$ called ''coefficient'' of linear expansion? [closed]

Also what else can we say about a material if we know its coefficient of linear expansion?

• It is a coefficient as it is called. This is measured in test and needs some curve fitting to get the value. If we know it, we can select material for design. For example, if we want to make a thermometer, we prefer to use a material which has a large coefficient. And if we make a high precision tool, we prefer a material with less coefficient. – user115350 Apr 12 '18 at 22:26

Naturally you expect the length to be a function of temperature, i.e.: $$L = f(T),$$ assuming the function $f(T)$ is sufficiently well-behaved it can be taylor expande around $T_0$ as: $$L = f(T_0) + f'(T_0)(T-T_0) + f''(T_0)\frac{(T-T_0)^2}{2} + \cdots,$$ assuming that $T-T_0$ is sufficiently small we can neglect the quadratic and higher terms finding that: $$L = f(T_0) + f'(T_0)\Delta(T),$$ where $f'(T_0) = \alpha$ is the linear coefficient of the expansion, and $f(T_0) = L_0$ the rest length at $T_0$.