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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 $v^2<<1$.

Kuhlambo
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  • see pages 4-5 here http://www.physics.umd.edu/grt/taj/374a/waterwaves.pdf –  Dec 19 '15 at 19:25

1 Answers1

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Background

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: $$ \frac{ d \delta v }{ dt } \sim -i \ k \left[ \frac{ \omega }{ k } \ \delta v + \left( \delta v \right)^{2} \right] $$

Answer

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.

honeste_vivere
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  • 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? – Kuhlambo Dec 20 '15 at 11:42
  • 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. – honeste_vivere Dec 21 '15 at 14:03
  • 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 $\delta v$ ? – Kuhlambo Jan 12 '16 at 13:40
  • 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. – Kuhlambo Jan 12 '16 at 13:45
  • @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. – honeste_vivere Jan 12 '16 at 18:19
  • But we are looking not at wave in velocity magnitude but at a wave of water hight right? So why would the waveheight be determined by the velocity (mathematically)? – Kuhlambo Jan 12 '16 at 21:59
  • Try looking at the PDF found at http://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-802-wave-motion-in-the-ocean-and-the-atmosphere-spring-2008/lecture-notes/MIT12_802S08_lec03.pdf on gravity waves. You can also look at papers written by Longhuet-Higgins and/or Stewart. – honeste_vivere Jan 12 '16 at 23:46
  • here the same assertion is made without any base "Also the velocity will have an overall amplitude $v_0$ "(page 5) and then a few lines later this is conflated with the amplitude of the wave being small, as if it was a necasary thing that the two are the same thing. This doesn't make any sense – Kuhlambo Jan 22 '16 at 09:58
  • @pindakaas - Okay, I think I see your issue. So the link to the PDF I sent was intentionally focusing on linear waves, so they need to assume a small amplitude. If the amplitude becomes "large," then the linear approximation (i.e., perturbation approach) often fails for numerous reasons. In this case, the phase speed depends upon the wave amplitude so you can imagine that if the amplitude were large, it would feedback on the phase speed. In this case, the $\mathbf{v} \cdot \nabla \mathbf{v}$ (i.e., steepening term) is no longer negligible... – honeste_vivere Jan 22 '16 at 13:44
  • Another way of saying this is that in the large amplitude limit, the parameters $Q$ can now have forms like $Q \approx Q_{o} \left( \omega, \mathbf{k} \right) e^{i \mathbf{k} \cdot \mathbf{x} - i \omega t + F(k^{2}) + ...}$ and also one can have $\omega = \omega\left( Q_{o}, \mathbf{k} \right)$. As you can see, this will prevent you from assuming things like $\nabla \rightarrow i \mathbf{k}$, which is why one assumes the fluctuations are linear. Otherwise, nearly all the problems one could imagine are no longer analytically tractable. – honeste_vivere Jan 22 '16 at 13:49
  • Am am actually quite sorry to be such a pest but I think my point is still not getting across. I can see that in linear approximation that $\partial_{t} \mathbf{v} \gg \mathbf{v} \cdot \nabla \mathbf{v}$ must hold. This is only equivalent to a small amplitude approximation if amplitude$=a \approx v$ holds. Bur why would we assume $a \approx v$ were does this come from? – Kuhlambo Jan 23 '16 at 09:43
  • @pindakaas - I apologize. I am not sure what seems to be the source of your confusion. – honeste_vivere Jan 23 '16 at 18:53
  • you and all the other sources write "δv is the wave amplitude" or $v_0$ or something like that. But why? We are talking about a velocity field of a fluid here why should this $v$ be the amplitude? – Kuhlambo Jan 25 '16 at 10:54
  • @pindakaas - Ah, okay... It's because in this case the oscillations are in the velocity of the fluid. The perturbations are small fluctuations about some mean in the velocity field. In some specific examples, you can rewrite these expressions as perturbations in some other parameter (e.g., density) because the velocity fluctuations depend upon that other parameter. – honeste_vivere Jan 25 '16 at 13:08