Capillary wave speed by dimensional analysis

Use dimensional analysis with $C = f(\rho, \lambda, \sigma)$. Here $n = 4$ variables and $m = 3$ dimensions (M, L, T), so there is only one $\pi$ term.

Dimensions

$C = LT^{-1}$, $\rho = ML^{-3}$, $\lambda = L$, $\sigma = MT^{-2}$ (force per length).

Form of the relation

$$C = K\,\rho^{a}\lambda^{b}\sigma^{c}$$

From T: $c = \tfrac12$. From M: $a = -\tfrac12$. From L: $b = 1 + 3a = -\tfrac12$.

Result

$$C = K\sqrt{\frac{\sigma}{\rho\lambda}}$$

where $K$ is a dimensionless constant (found by experiment or theory; the theory of capillary waves gives $K = \sqrt{2\pi}$).

Effect of doubling the surface tension

At fixed $\rho$ and $\lambda$, $C \propto \sqrt{\sigma}$:

$$\frac{C_2}{C_1} = \sqrt{\frac{2\sigma}{\sigma}} = \sqrt2 = 1.414$$

Answer: $C = K\sqrt{\sigma/(\rho\lambda)}$; if $\sigma$ is doubled, the speed increases by a factor $\sqrt2 \approx 1.41$ (about 41%).