The equation for linear thermal expansion is $$\alpha_L = \frac 1L \frac{dL}{dT}.$$

According to this page, linear expansion can be approximated as $$\frac{\Delta L}{L} \approx \alpha_L \Delta T$$ for approximately constant $\alpha_L$ in the range of $\Delta T$ and $\Delta L/L \ll 1$.

How can this approximation be verified using "formal" mathematics (e.g. Taylor expansion)?

  • 1
    $\begingroup$ You say how can if be verified using a Taylor expansion - have you actually tried doing this? $\endgroup$
    – jacob1729
    Nov 18, 2018 at 13:02
  • $\begingroup$ There is a mean value theorem in calculus that also says almost the same thing. $\endgroup$ Nov 18, 2018 at 13:04
  • $\begingroup$ @jacob1729 I have tried integrating the equation. Assuming the coefficient to be constant and integrating the equation gives $$\Delta L = L_0 (e^{\alpha_L \Delta T} -1 )$$. But now I don't know how to proceed (Taylor expansion was just a guess). $\endgroup$ Nov 18, 2018 at 13:12
  • $\begingroup$ @EricDavidKramer Could you elaborate? I wouldn't know how to apply it. $\endgroup$ Nov 18, 2018 at 13:15

2 Answers 2


While Thomas Jones' answer is correct I thought it might also be helpful to see that you don't need to integrate to get this result.

The initial ODE is:

$$\alpha = \frac{1}{L} \frac{dL}{dT} $$

The quickest way to see your approximation is to replace: $$dL \to \Delta L , dT \to \Delta T$$ And assuming $\alpha$ to be constant gives the result immediately. The formal justification for this is a first order Taylor expansion (which is equivalent to the mean value theorem alluded to in comments):

$$L = L(T_0) + \left(\frac{dL}{dT}\right)_{T_0}(T-T_0) + \dots$$

Then identifying $T-T_0$ as $\Delta T$ and the derivative as $L(T_0)\alpha$ gives the result.


Integrating that expression up there gives you $\Delta L = L_0 \left( e^{\alpha_L \Delta T} - 1\right) $, like you got.

Now, the Taylor expansion of $e^x$ for small $x$ is $1 + x + O(x^2)$, so neglecting higher order terms, we can substitute this into the expression above to get $\Delta L = L_0 \left( 1 + \alpha_L\Delta T - 1 \right) = L_0 \alpha_L\Delta T$.

Since the change in L is small, $L_0 \approx L$, so we can just divide by that and get $\frac{\Delta L}{L} = \alpha_L \Delta T$.


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