${\partial\over{\partial x_j}}\left(\partial x_i\over\partial t\right)\ne{\partial\over{\partial t}}\left(\partial x_i\over\partial x_j\right)$?

Question

$ \boldsymbol x = f(\boldsymbol X,t)$ is the position of a particle in an instant of time

$\boldsymbol X$ is the initial position

$t$ time

$\boldsymbol u$ velocity

In my opnion $f$ is continuos...

Considering:

$$u_i={{\partial x_i}\over{\partial t}}$$

Then:

$${\partial \over {\partial x_j}} \left(\partial x_i \over \partial t \right) = {{\partial u_i}\over{\partial x_j}}=\boldsymbol{\nabla}\boldsymbol{u}$$

But we can't Invert the Order of the partial derivative, otherwise we would have:

$${\partial \over {\partial t}} \left(\partial x_i \over \partial x_j \right) \neq 0$$

That is not the same result.

Answer 1

You should really consider using vector and/or index notation more consistently; have in index notation: $$u_i={{\partial x_i}\over{\partial t}}$$ which gives: $${\partial \over {\partial x_j}} \left(\partial x_i \over \partial t \right) = {{\partial u_i}\over{\partial x_j}}=\boldsymbol{\nabla}\boldsymbol{u}$$ The bold symbols are in vector notation. Then: $${\partial \over {\partial t}} \left(\partial x_i \over \partial x_j \right) \neq 0$$

necessarily right?

Answer 2

In the equation $\vec{u}=\frac{\partial \vec{x}}{\partial t}$, it is $\vec{X}$ that is being held constant (i.e., the material particle is being held constant):$$\vec{u}=\left(\frac{\partial \vec{x}}{\partial t}\right)_{\vec{X}}$$In the expression for the velocity gradient tensor $\nabla \vec{u}$, it is time that is being held constant, while the gradient is taken with respect to the current positions of all particles. So the operations are not commutative.