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When deriving the sound wave equation: $${1 \over c^2} {\partial^2 p' \over \partial t^2 }= \Delta^2 p' $$ by linearizing the Euler equation:

$$\rho {d v \over dt }= - \nabla p $$ and the continuity equation:

$$ {\partial \rho \over \partial t } + \nabla (\rho v)=0$$ using an approach of small deviations $\rho', v', p'$from an equilibrium $\rho_0, v_0, p_0$ with $v_0=0$. So $$p=p_0+p'\\ v=0+v' \\ \rho=\rho_0+\rho'$$

Why can we just neglect the convective term in the euler equation? Meaning why can we use: $$\rho_0 {\partial v' \over \partial t }= \nabla p'$$?

Why can we just assume $ (v'\nabla)v'\approx 0$ ?

Kuhlambo
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  • Second order in the small parameter v' – Thomas Nov 23 '15 at 17:01
  • You can make the assumption that $\mathbf{v}' \cdot \nabla \mathbf{v}' \sim 0$ if there is no nonlinear steepening. It is loosely another way of saying the medium is incompressible because all compressible waves can steepen, by definition. Without energy dissipation, sound waves in Earth's atmosphere would steepen until they formed shock waves. – honeste_vivere Nov 25 '15 at 14:21

2 Answers2

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As Thomas has commented, the trick is that we only assume first order terms and this convective acceleration would be small of the second order.

In fact, that is one of the first assumptions to drop when you consider more general cases. See e.g. Burger's equation for first generalizations and/or Lighthill's equation for source terms arising in the wave equation when not neglecting the convection.

The small amplitude assumption could be given e.g. by experiments: The sound of 94 dB (that's really a shouting) corresponds to $1 \ \mathrm{Pa}$ of RMS sound pressure amplitude $p_A$. Given that atmospheric pressure $p_0 \approx 10^5 \ \mathrm{Pa}$ we obtain:

$$ \frac{p_{A \ shouting}}{p_0} \approx 10^{-5} $$

Of course, there is a nonlinear acoustics of higher amplitudes as well.

Victor Pira
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  • so but why can we assume this? and why is $(v'\nabla)v'\approx 0$ of order $v'^2$? Are we aswell supposing that $ \nabla v'$ is of order $v'$ or how does this work? – Kuhlambo Nov 24 '15 at 15:27
  • For the first question see the edit. For the second one: yes, we usually assume just this. – Victor Pira Nov 25 '15 at 10:22
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The assumption when linearizing is that the deviations/perturbations are very small compared to the reference (averaged) values.

Typically the derivatives of the deviations are of the same order as the deviations themselves. Consider the deviations having this functional form in 1D: $$u'=\Delta u\sin kx\quad \partial_x u'=k\Delta u\cos kx$$ The deviation and its derivative are of order $O\left(\Delta u\right)\ll O\left(1\right)$. The convective terms then are: $$u'\partial_{x}u'=\frac{1}{2}\Delta u^{2}k\sin2kx$$ which is order $O\left(\Delta u^2\right)\ll O\left(\Delta u\right)$ and therefore negligible compared to other terms.

An analysis of the equations without evaluating the derivatives can be done if we start more general with the continuity and Cauchy momentum equation (neglecting viscous stresses): $$\partial_{t}\rho+\boldsymbol{\nabla}\cdot\rho\boldsymbol{v}=0$$

$$\partial_{t}\rho\boldsymbol{v}+\boldsymbol{\nabla}\cdot\left[\rho\boldsymbol{v}\otimes\boldsymbol{v}+p\boldsymbol{I}\right]=0$$

linearizing:

$$\partial_{t}\left(\rho_{0}+\rho'\right)+\boldsymbol{\nabla}\cdot\left(\rho_{0}+\rho'\right)\boldsymbol{v}'=0$$

$$\partial_{t}\left(\rho_{0}+\rho'\right)\boldsymbol{v}'+\boldsymbol{\nabla}\cdot\left[\left(\rho_{0}+\rho'\right)\boldsymbol{v}'\otimes\boldsymbol{v}'+\left(p_{0}+p'\right)\boldsymbol{I}\right]=0$$

i am sure you will agree that $\rho' \boldsymbol{v}' \ll \rho_0 \boldsymbol{v}'$ and $\rho'\boldsymbol{v}'\otimes\boldsymbol{v}'\ll\rho_0\boldsymbol{v}'\otimes\boldsymbol{v}'\ll p'\boldsymbol{I}$, which yields the linearized equations you are looking for:

$$\partial_{t}\rho'+\rho_{0}\boldsymbol{\nabla}\cdot\boldsymbol{v}'=0$$

$$\rho_{0}\partial_{t}\boldsymbol{v}'=-\boldsymbol{\nabla} p'$$

Appendix: Using the identity: $$\boldsymbol{\nabla}\cdot\left(\boldsymbol{A}\otimes\boldsymbol{B}\right)=\boldsymbol{B}\left(\boldsymbol{\nabla}\cdot\boldsymbol{A}\right)+\left(\boldsymbol{A}\cdot\boldsymbol{\nabla}\right)\boldsymbol{B}$$

we can rewrite the momentum equation in simplified form: $$\begin{align}\partial_{t}\rho\boldsymbol{v}+\boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\otimes\boldsymbol{v}\right) &=\boldsymbol{v}\partial_{t}\rho+\rho\partial_{t}\boldsymbol{v}+\boldsymbol{v}\boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\right)+\left(\rho\boldsymbol{v}\cdot\boldsymbol{\nabla}\right)\boldsymbol{v} \\ &=\boldsymbol{v}\left[\partial_{t}\rho+\boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\right)\right]+\rho\left[\partial_{t}\boldsymbol{v}+\left(\boldsymbol{v}\cdot\boldsymbol{\nabla}\right)\boldsymbol{v}\right]\end{align} $$

where the first term on the last line is identically zero due to the continuity equation.

nluigi
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  • Thanks for the answer but I don't see how $\partial_{t}\rho\boldsymbol{v}+\nabla\cdot\left[\rho\boldsymbol{v}\boldsymbol{v}+p\boldsymbol{I}\right]=0$ is the Navier stokes equation. $\nabla\cdot[\rho\boldsymbol{v}\boldsymbol{v}]$ in this expresstion should be $\boldsymbol{v}\cdot \nabla[\rho\boldsymbol{v}]$ or there should be a tensor product right? O – Kuhlambo Nov 24 '15 at 23:44
  • @pindakaas - technically you are right (the best kind of right!) so i have fixed the notation and added a reference to back up the equations. I usually neglect the tensor product as i consider: $$\boldsymbol{v}\boldsymbol{v}=\left[\begin{array}{cc} uu & uv\ vu & vv \end{array}\right] $$ – nluigi Nov 25 '15 at 05:06
  • hey thanks for the attention. The only problem is that i have no idea why $\boldsymbol{\nabla}\cdot[\rho\boldsymbol{v}\otimes\boldsymbol{v}]=\rho (\boldsymbol{v} \cdot \nabla )\boldsymbol{v}$ so this sadly doesn't help me to understand... – Kuhlambo Nov 25 '15 at 09:49
  • @pindakaas - That should be asked as a separate question and also more suitable for Math.SE but i will answer it anyway. Only in steady-state that relation is true, but not in general. See my appendix to see how you go, in general, from one form to another. – nluigi Nov 25 '15 at 10:10
  • thanks for the great answer. Although I have to say i find steady state to be a bit confusing here since this in fluis mechanics normally means $ \partial_t \boldsymbol v =0$ but I venture you mean steady mass state? So $ {d M \over dt } =0$ right ? – Kuhlambo Nov 25 '15 at 10:33
  • @pindakaas - at steady-state any time derivatives vanish, i.e. $\partial_t \rho=0$ and $\partial_t \boldsymbol{v} = 0$. The implication is that density is constant and by the continuity equation $\boldsymbol{\nabla}\cdot\boldsymbol{v}=0$. The term $$\boldsymbol{\nabla} \cdot \left[ \rho \boldsymbol{v} \otimes \boldsymbol{v} \right]=\rho \boldsymbol{\nabla} \cdot \left[\boldsymbol{v} \otimes \boldsymbol{v}\right]=\rho \boldsymbol{v}\left(\boldsymbol{\nabla}\cdot\boldsymbol{v}\right)+\rho \left(\boldsymbol{v}\cdot\boldsymbol{\nabla}\right)\boldsymbol{v} $$ using the identity shown in my answer. – nluigi Nov 25 '15 at 10:55
  • I don't think I am making myself clear. Assuming mass conservation is enaugh. Assuming $\partial_t \rho =0$ and assuming incompressibility is simply not necessary. – Kuhlambo Nov 25 '15 at 18:11
  • Let us continue this discussion in chat. – Kuhlambo Nov 25 '15 at 19:12
  • after revisiting this I just realized that this is still not really a good argument. the derivative of something small just doesn't have to be zero because it is itself small... – Kuhlambo Dec 19 '15 at 15:24
  • @pindakaas: i'm confused about which argument you are talking about; my argument was that for pertubations the derivatives of the pertubations are of the same order as the pertubations themselves. Their product is one order smaller and therefore negligible; you can effectively set these to zero... – nluigi Dec 19 '15 at 19:57
  • for example $\nabla \rho' v' \approx 0$ but this is because $\rho' v' \approx 0$ because of the nature of the pertubation ansatz. BUT this is not true in general the derivative of a funciton that is very close to zeor can easyly be big in value. Do you see what i mean? In the same way the $\nabla (v \otimes v) $ term only vanished because we look at $\nabla (v \otimes v + p'I) $ and we know $v \otimes v + p'I \approx p' I$ but I read everywhere that the convective part should vanish by itself... But I guess we can assume what you stated in the first answer order of deriv. = order of original – Kuhlambo Dec 20 '15 at 11:58
  • @pindakaas: I agree, that why i don't use that argumentation anywhere in my answer (as far as i can see, plz correct me if i am wrong). My only arguments are on products of pertubations and average values not on derivatives, i.e.: $$\rho'\boldsymbol{v}'\otimes \boldsymbol{v}'\ll\rho_0 \boldsymbol{v}'\otimes \boldsymbol{v}'\ll p'\boldsymbol{I}$$ So i'm still confused about why you say: 'that this is still not really a good argument'. What are you refering to when saying 'this'? – nluigi Dec 21 '15 at 12:21
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    I guess you're right. Sorry you do the comparison before deriving so that is indeed completely reasonable. Thats why i accepted the answer again. – Kuhlambo Dec 21 '15 at 12:43