---
title: "Capillary wave speed by dimensional analysis"
subject: "Fluid Mechanics"
chapter: 10
key: "semester_3/fluid_mechanics:capillary-wave-speed-dimensional"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/capillary-wave-speed-dimensional
site: BCT Engineering
---

# Capillary wave speed by dimensional analysis

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2076 Chaitra, 8 marks
- Video: [Fluid Mechanics: Dimensional Analysis: Buckingham Pi Theorem](https://www.youtube.com/watch?v=pBkfjNMW26Y) — Study from Home with Us
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/capillary-wave-speed-dimensional
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

The speed of propagation C of a capillary wave in deep water is known to be function only of density $\rho$, wavelength $\lambda$, and surface tension $\sigma$. Find the proper functional relationship, completing it with a dimensionless constant. For a given density and wavelength, how does the propagation speed change if surface tension is doubled?

## Short answer

C = K√(σ/(ρλ)). Doubling surface tension at fixed ρ and λ increases speed by a factor √2 ≈ 1.414 (about 41% increase).

## Answer

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

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

| Dimension | Equation |
|---|---|
| M | $0 = a + c$ |
| L | $1 = -3a + b$ |
| T | $-1 = -2c$ |

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%).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are the dimensions of surface tension sigma?
   A. MLT^-2
   B. ML^-1T^-2
   C. MT^-2 (force per length)
   D. ML^2T^-2
   Answer: C. Force per unit length is MLT^-2 / L = MT^-2. (Key point: Dimension of surface tension)

2. Why is there only one pi term?
   A. Because C is dimensionless
   B. Because rho cancels
   C. Because lambda has no dimension
   D. n = 4 variables and m = 3 dimensions, so n - m = 1
   Answer: D. n - m = 4 - 3 = 1. (Key point: Number of pi terms)

3. What is the relationship found?
   A. C = K sqrt(sigma/(rho lambda))
   B. C = K sqrt(rho lambda/sigma)
   C. C = K sigma rho lambda
   D. C = K sqrt(sigma rho/lambda)
   Answer: A. T gives c = 1/2, M gives a = -1/2, L gives b = -1/2. (Key point: Result)

4. What happens to C if sigma is doubled at fixed rho and lambda?
   A. It doubles
   B. It increases by a factor sqrt(2), about 1.41
   C. It increases by a factor 4
   D. It stays the same
   Answer: B. C is proportional to sqrt(sigma), so the ratio is sqrt(2). (Key point: Doubling sigma)

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