Quoting directly from my textbook:
We define average coefficent of linear expansion in the temperature range $\Delta T$ as:
$$ \bar \alpha = \frac 1L \frac{\Delta L}{\Delta T} $$
The coefficient of linear expansion at temperature $T$ is the limit of average coefficient as $\Delta T \to 0$, i.e.,
$$ \alpha = \lim_{\Delta T \to 0} \frac 1L \frac{\Delta L}{\Delta T}=\frac 1L \frac{\mathrm{d}L}{\mathrm{d}T}$$
Suppose the length of a rod at $0^\circ \mathrm{C}$ is $L_0$, and at temperature $\theta$ (measured in Celsius) is $L_{\theta}$. If $\alpha$ is small and constant over given temperature interval,
$$ \alpha = \frac{L_{\theta} – L_0}{L_0 \theta} $$
How did they get the last part? Because while integrating, we must keep $L$ variable, and not as a constant, and should thus yield:
$$\alpha \cdot \theta = \ln{\frac{L_{\theta}}{L_0}} $$
Even considering an approximate form, how do we get to that formula? I do not see any other way except treating $L$ as a constant.
Best Answer
You obtain your formula assuming that $\alpha$ is constant. Now assume it is small. Then $\exp(\alpha\theta)\cong1+\alpha\theta$ and \begin{equation} 1+\alpha\theta=\frac{L_\theta}{L_0}\iff\alpha=\frac{L_\theta-L_0}{L_0\theta} \end{equation}