I'm trying to make sense of the link between a (linear) Eulerian (i.e., at a given point) and Lagrangian (following a fluid element) perturbation. I will here express not only where I'm stuck, but also my current understanding of the topic, so I beg that if you find any misconception, please point it out!

>> Background

So far, I (think I) know that in the Eulerian perturbation, if $h(\boldsymbol{r},t)$ is the perturbed quantity, where $\boldsymbol{r}=\boldsymbol{r}_0+\boldsymbol{\delta r}$ is the new position vector and $\boldsymbol{\delta r}$ is a small displacement around $\boldsymbol{r}_0$, then $$h(\boldsymbol{r},t)=h_0(\boldsymbol{r})+h'(\boldsymbol{r},t),\ \ \ \ \ \ \ \ (1)$$ where $h_0(\boldsymbol{r})$ is the quantity at an equilibrium state evaluated at $\boldsymbol{r}$ and $h'(\boldsymbol{r},t)$ is a small perturbation.

On the other hand, the Lagrangian perturbation $\delta h(\boldsymbol{r})$ can be written as $$\delta h(\boldsymbol{r})=h(\boldsymbol{r}_0+\boldsymbol{\delta r})-h_0(\boldsymbol{r}_0)=\left[h(\boldsymbol{r}_0)+\boldsymbol{\delta r}\cdot \nabla h_0(\boldsymbol{r}_0)\right]-h_0(\boldsymbol{r})$$ or, by using the relation found for the Eulerian perturbation, eq. $(1)$, evaluated at $\boldsymbol{r}_0$, $$\delta h(\boldsymbol{r})=h'(\boldsymbol{r}_0)+\boldsymbol{\delta r}\cdot \nabla h_0(\boldsymbol{r}_0).$$

>> The question

What I have trouble relating is $\delta h(\boldsymbol{r})$ and $h'(\boldsymbol{r},t)$. Both are small perturbations but, in this case, there is a explicit dependance on $\boldsymbol{r}_0$ by $\delta h(\boldsymbol{r})$. However, if I fix $\boldsymbol{r}_0$, both represent a perturbation about that point so...why are they different? My intuition (probably messed up by know; thanks QM) tells me that they are equal, but the notation is different in almost every book I've read on the topic. Can anyone shine some light here?


1 Answer 1


The Eulerian description is a map from a point in real space $x$ (indexes suppressed) and the fluid label $\Phi$ that is found in $x$ at the time $t$, namely $(t,x)\rightarrow (t,\Phi(t,x))$. The physical observables are, typically, functions of $\Phi$ and its derivatives $\partial\Phi$(unless there are external sources), such as $h=h(\Phi,\partial\Phi)$. This corresponds to a standard field theory approach. Vice versa,the Lagrangian description is the inverse map, at any time, from a fluid element labeled by $\Phi$ to its position in real space $x(t)$, namely $(t,\Phi)\rightarrow (t,x(t,\Phi))$. Relating the two descriptions is just matter of changing the variables, e.g. $h(\Phi)$ becomes $\tilde{h}(t,x)=h(\Phi(t,x))$ when passing to the Lagrangian formulation (assuming that $h$ is a scalar quantity). With a slightly abuse of notation, it's customary not to include the \tilde above.

Back to your original example in perturbation theory: when you say $x_0+\delta x$ in the lagrangian formulation you are actually referring to some fluid element $\Phi_0$ which is not found in $x_0$ anymore but rather in $x(t,\Phi_0)=x_0(t,\Phi_0)+\delta x(t,\Phi_0)$. I can also read it saying that in the point $x_0$ I find a new fluid element, that is $\Phi_0+\delta\Phi$ (Eulerian description). By consistency of these two descriptions, I.e. $(\Phi_0+\delta\Phi)(t,x_0+\delta x)=\Phi_0(t,x_0)$, we get $$\delta\Phi=-\left.\frac{\partial \Phi^i}{\partial x^j}\right|_{x_0}\delta x^j\,.$$ Now, imagine you are interested in some physical quantity $\tilde h(t,x)=h(t,\Phi(t,x))$. The perturbation thus reads $$ \delta \tilde{h}=-\left.\frac{\partial h}{\partial \Phi^i}\right|_{\Phi_0}\left.\frac{\partial \Phi^i}{\partial x^j}\right|_{x_0}\delta x^j\,. $$

  • $\begingroup$ Can you explain, physically, the difference? I see the mathematical one; is the physical one I don't get. $\endgroup$
    – Néstor
    Commented Nov 5, 2012 at 20:39
  • $\begingroup$ @argopulos Please register your account, you will be able to edit and comment on your posts without hassle. $\endgroup$
    – user68
    Commented Nov 6, 2012 at 0:27
  • $\begingroup$ @nestor: I've tried to make more clear what the perturbations are by re-editing my answer. There is no physical difference, it's just about what coordinates you like more. $\endgroup$
    – argopulos
    Commented Nov 6, 2012 at 11:25

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