# Insights The Pantheon of Derivatives - Part II - Comments

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1. Mar 17, 2017

### Staff: Mentor

2. Mar 17, 2017

### zwierz

the space $C^\infty_{\mathbb{R}}([0,1])$ equipped with the uniform norm is not complete. That is not the point but above it is written that all the spaces are Banach

The definition of a vector field on a manifold is very bad. It looks like one implicitly assumes that the manifold is embedded into $\mathbb{R}^m$.

the gradient of a function is not a vector field, it is an 1-form at least until the manifold is not equipped with a metric tensor

the operation rotor as well as cross product give pseudo vectors

Last edited: Mar 17, 2017
3. Mar 17, 2017

### Staff: Mentor

... which (is not claimed and) emphasizes my point to watch out the definitions made in the relating contexts.
True. I removed $\mathbb{R}^n$.
It is a line bundle, and the set $\{(x,\nabla_xf)\,\vert \,x \in M\}$ is a vector field.
$\vec{a} \times \vec{b}$ is a vector in $\mathbb{R}^3$.

4. Mar 17, 2017

### zwierz

In principle I can dig textbooks up and show you corresponding paragraphs but what for?
It would be great if you ask a specialist in differential geometry whom you trust in , to read your text carefully.

5. Mar 17, 2017

### Staff: Mentor

So can I. But what for? Your goal is apparently one which I'm not interested in.

6. Mar 19, 2017

### Stephen Tashi

Is the "material derivative" of fluid mechanics a special case of a Lie derivative? Or is it yet another kind of derivative?

7. Mar 19, 2017

### Staff: Mentor

I'll have it in the fourth part (where I also included an example from Wiki): The material derivative is a special case of a derivative in order to describe the flow of fluids or gases. It is more of a special tool for these currents rather than a special concept of a differentiation process.

\begin{equation*}
D_v\Phi = \frac{d_v}{dt} \Phi = \frac{\partial \Phi}{\partial t}+ (v \cdot \nabla)(\Phi) = \left(\,\frac{\partial}{\partial t}+v_x\,\frac{\partial}{\partial x}+v_x\,\frac{\partial}{\partial x}+v_x\,\frac{\partial}{\partial x}\,\right)\,(\Phi)
\end{equation*}
where $v$ represents the velocity of the flow at point $x$ and time $t$. The first summand is the local behavior in time at a fixed point, the second is the convective change of a particle in the flow.

I would rather call it by its name Euler operator (cp. http://www.math.nyu.edu/faculty/childres/fluidsbook.pdf ; p.8 f.) because the two summands are treated differently: the time dependent part keeps a particle location fixed, whereas the second is to analyze the velocity (flow, integral curve) of a fluid. So strictly speaking I'd say no, since I cannot imagine how to combine this in a single view of a vector field (spacetime aside), will say it's an operator that combines two Lie derivatives. It is also called a total derivative. As I understand it, are Lie derivatives directional derivatives.

Last edited: Mar 19, 2017
8. Mar 19, 2017

### zwierz

Good question.

For simplicity sake let's consider a stationary flow of a fluid with velocity field $v=v(x),\quad v=(v^1,\ldots,v^3)(x).$
Let $g_v^t(x)$ stand for corresponding one-parametric group:
$$\frac{d}{dt}g_v^t(x)=v(g_v^t(x)),\quad g_v^0(x) =x.$$
Then take a smooth manifold $S\subset \mathbb{R}^3,\quad \mathrm{dim}\,S=k$. The manifold $S$ can be a curve (k=1), a surface (k=2), or a domain of $\mathbb{R}^3$ (k=3) and let $\omega$ stand for a $k-$form in $\mathbb{R}^3$.
Theorem. The following formula holds
$$\frac{d}{dt}\Big|_{t=0}\int_{g^t_v(S)}\omega =\int_SL_v\omega,$$ here $L_v$ is the Lie derivative.

This very simple theorem remains true in all dimensions, not only in $\mathbb{R}^3$.
For example, let $\rho(x)$ be a density of the fluid and let $\Omega =\sqrt gdx^1\wedge dx^2\wedge dx^3,\quad g=\mathrm {det}(g_{ij}(x))$ be the volume form; $g_{ij}$ is a metric tensor. Then the mass conservation law is written as follows
$$\int_{g^t_v(D)}\rho\Omega =const,$$ here $D$ is an arbitrary volume and the constant does not depend on time but it surely depends on $D$.
Applying the above theorem we get
$$\frac{d}{dt}\Big|_{t=0}\int_{g^t_v(D)}\rho\Omega=\int_DL_v(\rho\Omega)=0.\qquad (*)$$
It is not hard to show that $L_v(\rho\Omega)=\mathrm{div}\,(\rho v) \Omega$ and sinse $D$ is an arbitrary volume, formula (*) gives the standard continuity equation: $\mathrm{div}\,(\rho v)=0.$
In the same way Helmholtz's theorems https://en.wikipedia.org/wiki/Helmholtz's_theorems and many other useful things from Hamiltonian dynamics follow.

(I have dropped math details such as smoothness of $v$, finite measure of $S,D$ etc)

Last edited: Mar 19, 2017