Why does the assumption of waves amplitude being small, compared to the wavelength, (as used for surface gravity waves in water) lead to the equivalent condition: $$ {\partial v \over \partial t } >> { (v \nabla )v}$$

This seems to be one of the basic assumptions of certain classes of fluid mechanic problems.

I spend two hours in the library today reading the applicable chapters of about nine books and not in one the explanation was more detailed that to claim that $ (v \nabla )v$ must be on the order of $v^2<<1$.


1 Answer 1



The nonlinear steepening term, $\mathbf{v} \cdot \nabla \mathbf{v}$, is proportional to (unit-wise) a speed squared divided by a scale length, $L$. It is more important for larger values of $v$ and/or smaller values of $L$.

To illustrate this, let's assume a linear approximation. This allows us to use the following: $$ \partial_{t} \rightarrow -i \ \omega \\ \nabla \rightarrow +i \ \mathbf{k} $$ where $\omega$ is the angular frequency, $\mathbf{k}$ is the wavenumber, and we have assumed all quantities can be written as $Q \approx Q_{o} + \delta Q$, where: $$ Q_{o} \equiv \text{ constant} \\ \delta Q \propto e^{i \left( \mathbf{k} \cdot \mathbf{x} - \omega \ t \right)} $$ In other words, we assume that $\partial_{t} Q_{o} = \nabla Q_{o} = 0$.

Using these assumptions, we can rewrite the total derivative as: $$ \frac{ d }{ dt } = \partial_{t} + \mathbf{v} \cdot \nabla \\ \approx -i \ \omega + i \ \mathbf{v} \cdot \mathbf{k} $$ which shows that we have: $$ \frac{ d \mathbf{v} }{ dt } \rightarrow \frac{ d \delta \mathbf{v} }{ dt } \approx -i \ \omega \ \delta \mathbf{v} + i \ \left(\delta \mathbf{v} \cdot \mathbf{k}\right) \delta \mathbf{v} $$ where in the one-dimensional limit with $\delta \mathbf{v}$ entirely parallel to $\mathbf{k}$ reduces to: $$ \frac{ d \delta v }{ dt } \sim -i \ k \left[ \frac{ \omega }{ k } \ \delta v + \left( \delta v \right)^{2} \right] $$


Why does the assumption of waves amplitude being small, compared to the wavelength, (as used for surface gravity waves in water) lead to the equivalent condition: $$ \partial_{t} \mathbf{v} \gg \mathbf{v} \cdot \nabla \mathbf{v} $$

The idea is that the term $\delta v$ is the wave amplitude, namely that there is a velocity fluctuation about some mean value. If the wavelength is large, that means $k$ is small which corresponds to a large phase velocity, $\omega/k$, assuming a constant frequency.

Thus, the small amplitude, $\delta v$, means that we have the following: $$ \frac{ \omega }{ k } \ \delta v \gg \left( \delta v \right)^{2} \\ \text{or, in another form:} \\ \frac{ \omega }{ k } \gg \delta v $$

It is just another way of stating that the velocity fluctuations are not large compared to the quasi-static terms. One makes assumptions like this to avoid things like nonlinear wave steepening, dispersion, strong damping, etc.

Side Note
You will find this linearization method used in nearly all branches of physics for numerous topics for any system that can be decomposed into superpositions of multiple terms. It is incredibly handy and often annoyingly accurate.

  • $\begingroup$ thanks for the answer. It is missing one crucial point for me though: why is $\delta v$ the amplitude? If it is the relation follows easily form a linear approximation but this fact is the essential point that does not seem to follow from anything. Why do we say this and how can we see it in general and not by example? Or is that not possible? $\endgroup$
    – Kuhlambo
    Dec 20, 2015 at 11:42
  • $\begingroup$ Were you expecting the perturbed quantity to be something other than velocity? In some cases, one can assume the fluctuating terms are, for instance, density. It depends upon the specific situation, which is why I did not elaborate on that part. In surface gravity waves in the small amplitude limit, the fluctuating term is the velocity field. $\endgroup$ Dec 21, 2015 at 14:03
  • $\begingroup$ sorry but maybe you're strill interested in answering this. What i don't understand is why we assume in general the amplitude should be the $\delta v$ ? $\endgroup$
    – Kuhlambo
    Jan 12, 2016 at 13:40
  • $\begingroup$ I am asking this because from the premise namely the amplitude is small the main conclusion ${\partial v \over \partial t } >> { (v \nabla )v}$ follows trivially. But only if we know that the amplitude will always be determined by the velocity. So the crux really is: Why is the amplitude determined by the velocity. $\endgroup$
    – Kuhlambo
    Jan 12, 2016 at 13:45
  • $\begingroup$ @pindakaas - It's not a general rule, it just happens to be the fluctuating term in this example. In a sound wave, one can look at the density as the fluctuating term. In an Alfvén wave, the magnetic field fluctuates. It just depends upon the wave mode. $\endgroup$ Jan 12, 2016 at 18:19

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