The Pantheon of Derivatives - Part II - Comments

[fresh_42]

fresh_42 submitted a new PF Insights post

The Pantheon of Derivatives - Part II

Continue reading the Original PF Insights Post.

 

[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

 

[fresh_42]

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

... which (is not claimed and) emphasizes my point to watch out the definitions made in the relating contexts.

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

True. I removed $\mathbb{R}^n$.

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

It is a line bundle, and the set $\{(x,\nabla_xf)\,\vert \,x \in M\}$ is a vector field.

the operation rotor as well as cross product give pseudo vectors

$\vec{a} \times \vec{b}$ is a vector in $\mathbb{R}^3$.

 

[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.

 

[fresh_42]

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.

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

 

[Stephen Tashi]

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

 

[fresh_42]

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

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.

 

[zwierz]

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

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)