Differential expansion | Thermal expansion | Class 11 Physics

Differential Expansion

Differential expansion is the difference in the amounts by which two objects (rods or layers) of different materials expand when subjected to the same temperature difference $\Delta \theta$.

Derivation

Consider two parallel metallic rods rigidly connected to a common base at one end. Let their initial lengths be $l_1$ and $l_1'$ with corresponding coefficients of linear expansion $\alpha$ and $\alpha'$. Let the initial temperature of the setup be $\theta_1^\circ\text{C}$, where the initial difference between their free ends is $d_1 = l_1 - l_1'$.

When the temperature increases uniformly to $\theta_2^\circ\text{C}$, the rods expand to new lengths $l_2$ and $l_2'$. The new separation distance between their free ends becomes $d_2 = l_2 - l_2'$. The individual final lengths are given by:

$$l_2 = l_1(1 + \alpha\Delta\theta)$$ $$l_2' = l_1'(1 + \alpha'\Delta\theta)$$

Substituting these expressions into the final separation distance equation:

$$d_2 = l_1(1 + \alpha\Delta\theta) - l_1'(1 + \alpha'\Delta\theta)$$ $$d_2 = l_1 + l_1\alpha\Delta\theta - l_1' - l_1'\alpha'\Delta\theta$$ $$d_2 = (l_1 - l_1') + (l_1\alpha - l_1'\alpha')\Delta\theta$$

Replacing $(l_1 - l_1')$ with the initial distance $d_1$ yields the relation for the final gap:

$$d_2 = d_1 + (l_1\alpha - l_1'\alpha')\Delta\theta$$ $$d_2 - d_1 = (l_1\alpha - l_1'\alpha')\Delta\theta$$

Here, $d_2 - d_1$ directly quantifies the systemic differential expansion between the two bars.

For the gap distance to remain completely invariant across all temperatures ($d_2 = d_1$, or $d_2 - d_1 = 0$):

$$(l_1\alpha - l_1'\alpha')\Delta\theta = 0$$

Since the temperature change is non-zero ($\Delta\theta \neq 0$):

$$l_1\alpha - l_1'\alpha' = 0$$ $$l_1\alpha = l_1'\alpha'$$ $$\frac{l_1}{l_1'} = \frac{\alpha'}{\alpha}$$

This is the required condition for maintaining zero differential expansion, showing that the initial lengths must be inversely proportional to their linear expansivities.