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I am doing a problem set about dimensional analysis and water waves. Before the last question which I have trouble with, I found the following two relations: $$\begin{equation*} v\propto\sqrt{g\lambda}\\ v\propto\sqrt{\frac{\gamma}{\rho\lambda}} \end{equation*}$$ Where $\gamma$ is the surface tension, $\lambda$ is the wave length. However the very last question asks us to find under what circumstances the wave motion is dominated by surface tension rather than gravity using dimensional analysis. In which case, I don't know how to do it, although I know relation between depth of the water and the phase velocity is: $$\begin{equation*} v\propto\sqrt{gh} \end{equation*}$$ and the actual equation between the phase velocity and the wavelength is: $$\begin{equation*} v=\sqrt{g\lambda}\tanh{\sqrt{h/\lambda}} \end{equation*}$$ And despite I also know the answer of this question is when $h\ll\lambda$, but I don't know how to derive this from dimensional analysis.

It will very much helpful if anyone can tell me if it is possible to find.

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3 Answers 3

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For large wavelengths the wave motion is dominated by gravity, and for short wavelengths it is dominated by surface tension. You already found these approximate formulas for the wave speed $v$ depending on wavelength $\lambda$: $$v \propto\begin{cases} \sqrt{g\lambda} & \text{, if }\lambda\text{ very large} \\ \sqrt{\frac{\gamma}{\rho\lambda}} & \text{, if }\lambda\text{ very small} \end{cases}$$

So your $v$-$\lambda$ graph will look like this. enter image description here

But you don't know yet what "very large" and "very short" wavelength exactly means. For this you need to find where the two asymptotes in the graph intersect. This is roughly at $$\sqrt{g\lambda} \approx \sqrt{\frac{\gamma}{\rho\lambda}}$$ or $$\lambda \approx \sqrt{\frac{\gamma}{\rho g}}$$

So this is the characteristic wavelength which separates the long waves (dominated by gravity) from the short waves (dominated by surface-tension). By the way, this is the only way how to combine $\gamma$ (dimension N/m), $\rho$ (dimension kg/m$^3$) and $g$ (dimension N/kg) to get a length. So we could have guessed this characteristic $\lambda$ already by dimensional analysis only.

Using this you can write your formulas from the beginning more exactly: $$v \propto\begin{cases} \sqrt{g\lambda} & \text{, if }\lambda \gg \sqrt{\frac{\gamma}{\rho g}} \\ \sqrt{\frac{\gamma}{\rho\lambda}} & \text{, if }\lambda \ll \sqrt{\frac{\gamma}{\rho g}} \end{cases}$$

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Assuming you understand what is meant by "dimensional analysis:" yes, it is possible. You should ask yourself how to compare surface tension and gravity. These are not dimensionally equivalent. What could you multiply or divide them by to make them equivalent, and could these factors illuminate the "circumstances" in which one or the other is dominant? Alternatively, how would you compare the effects of surface tension and gravity.

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Let's suppose that there is a physically meaningful equation involving the variables $(v,g,\lambda,\rho,h,\gamma)$. Since these six variables have three dimensions $[L, T, M]$, you can form $6-3 =3$ independent dimensionless quantities with them, e.g. $ ({v^2 / g\lambda},\ {h / \lambda},\ {v^2 \rho\lambda/ \gamma}) $ or $ ({v^2 / g\lambda},\ {h / \lambda},\ {\gamma/ \rho g \lambda^2})$. The Buckingam $\pi$-theorem tells us that the equation we are looking for must be a relation between these three dimensionless variables: $$ F\left({v^2 \over g\lambda},\ {h \over \lambda},\ {v^2 \rho\lambda \over \gamma} \right) = 0 \qquad or \quad F\left({v^2 \over g\lambda},\ {h \over \lambda},\ {\gamma \over \rho g \lambda^2} \right) = 0 $$

If we neglect the surface tension with the variables $ (v,g,\lambda,\rho,h) $ we can built $5-3=2$ independent dimensionless quantities $ ({v^2 / g\lambda},\ {h / \lambda}) $ and the $\pi$-theorem now suggests the simple pre-formatted relation: $$ F\left({v^2 \over g\lambda},\ {h \over \lambda}\right) = 0 \qquad\qquad alias: \qquad v = \sqrt{g\lambda}\cdot f\!\left(h\over\lambda\right) $$

It's worth noting that the velocity $v$ does not depend from the density $\rho$. The explicit form of the function $ f () $ cannot be obtained by dimensional analysis alone.

In the case of very deep waters, we can assume that the parameter $h$ may be irrelevant and that the meaningful quantities are $(v,g,\lambda,\rho,\gamma)$. The two dimensionless variables are $ ({v^2 / g\lambda},\ {\gamma/ \rho g \lambda^2})$ and so we have $$ F\left({v^2 \over g\lambda},\ {\gamma \over g\rho\lambda^2}\right) = 0 \qquad\qquad alias: \qquad v = \sqrt{g\lambda}\cdot f\!\left(\gamma \over g\rho\lambda^2 \right) $$

Airy's wave theory gives the following explicit expression for the velocity of the waves:

$$ v = \sqrt{\left({g\lambda \over 2\pi} + {2\pi\gamma \over \rho\lambda}\right) \tanh\left(2\pi h \over \lambda\right) } \qquad\qquad alias: \qquad {1 \over 2\pi} {g\lambda \over v^2} + 2\pi {\gamma \over v^2\rho\lambda } = 1 / \tanh\left(2\pi {h \over \lambda} \right) $$

As you can see, Airy's equation actually has the form required by the Buckingam $\pi$-theorem.

If we neglect the surface tension ($\gamma=0$) we have: $ v = \sqrt{{g\lambda \over 2\pi} \tanh\left(2\pi h \over \lambda\right) } $

If $h\ll \lambda \ \tanh \to 2\pi h/\lambda)$ we have $ v \simeq \sqrt{gh\left(1 + {4\pi^2\gamma \over g\rho\lambda^2} \right)} \simeq \sqrt{gh} \qquad\qquad\qquad$ (shallow waters)

If $h\gg \lambda \ (\tanh \to 1)$ we have $ v \simeq \sqrt{{g\lambda \over 2\pi} + {2\pi\gamma \over \rho\lambda} } = \sqrt{ {g\lambda \over 2\pi} \left(1+ {4\pi^2\gamma \over g\rho\lambda^2}\right) } \simeq \sqrt{g\lambda \over 2\pi} \qquad $ (deep waters)

Of course such results are outside of the dimensional analysis.

However I don't quite understand your ''last question''.

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  • $\begingroup$ Great name! did you get this name from Candide?the philosophy teacher? Anyway, I can't understand my last question either. In order to make things clear, I have displayed this question below: $\endgroup$ Sep 9, 2020 at 16:06
  • $\begingroup$ Consider a wave of wavelength $\lambda$ propagating at the surface of an infinitely deep lake. Based on dimensional analysis, under what circumstances would you expect the wave motion to be dominated by surface tension effects rather than gravity? (Hint: as always, think about dimensionless quantities!) $\endgroup$ Sep 9, 2020 at 16:08
  • $\begingroup$ I don't know if I understand it correctly. But as far as I can tell, I think this question implies that we can use dimensional analysis to derive the fact that it asks for. What do you think? $\endgroup$ Sep 9, 2020 at 16:09
  • $\begingroup$ @Pangloss, The point is not to do someone's homework for them; the poster asked: "if it is possible to find" a solution by dimensional analysis. Point them on the right path, and give some insights into how to think about such things. $\endgroup$ Sep 9, 2020 at 20:01
  • $\begingroup$ @Raghu Parthasarathy If we don't know Airy's equation, the fact that $\gamma/g\rho\lambda^2$ is a dimensionless quantity does not allow us to establish if the wave motion is ''dominated'' by surface tension $\gamma$ rather than gravity $g$ using dimensional analysis alone. $\endgroup$
    – Pangloss
    Sep 10, 2020 at 8:04

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