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Consider a vector $\vec{\xi}=\vec{X}(\vec{x},t).$ Because $\vec x$ has 3 components x, y,z then we can say that: $\vec \xi$ has 3 components u, v, w and u = u(x, y, z, t) ; v = v(x, y, z, t) ; w = w(x, y, z, t).

Fix t at some value, in a paper I am reading, the author wrote:

$$\vec{\xi}+d\vec{\xi} = \vec{X}(\vec{x}+d\vec{x},t)=\vec{X}(\vec{x},t) + d\vec{x}\cdot\nabla \vec{X}.$$

Here $\nabla$ differentiate with respect to $\vec x$. Hence, $$d\vec{\xi} = d\vec{x}\cdot\nabla X = \sum_{i=1}^3 dx_i \frac{\partial X}{\partial x_i}=dx_i \frac{\partial X}{\partial x_i},$$

where the last form uses the convention that there is an implicit summation on repeated indices, a convention which greatly reduces clutter!


I don't understand the part: $$d\vec{\xi} = d\vec{x}\cdot\nabla X = \sum_{i=1}^3 dx_i \frac{\partial X}{\partial x_i}=dx_i \frac{\partial X}{\partial x_i},$$


Here is what I understand if I have to derive $d\vec{\xi}$:

$d\vec{\xi}$ = du.$\vec i$+ dv.$\vec j$+ dw.$\vec k$ , with:

$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy + \frac{\partial u}{\partial z}dz = \nabla u.d\vec x$ (dt = 0 because t fixed at some value)

$dv =...= \nabla v.d\vec x$

$dw =...= \nabla w.d\vec x$

So we have: $d\vec{\xi}$ = ($\nabla u.d\vec x$).$\vec i$+ ($\nabla v.d\vec x$).$\vec j$+ ($\nabla w.d\vec x$).$\vec k$ (Let me know if I was wrong !!)

I know some about the directional derivative but it still does not help me to see the common of my thought with the author's thought

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  • $\begingroup$ Why do you think that your expression is different from the paper's? $\endgroup$
    – Chappers
    Commented Sep 7, 2017 at 21:42
  • $\begingroup$ @Chappers: Oh I just see it, I can tranform my equation to the author's equation. Now, should I delete this question ?? Sorry, I am new here $\endgroup$
    – Dat
    Commented Sep 8, 2017 at 15:20
  • $\begingroup$ How about you write your own answer to check that you've sorted it out? $\endgroup$
    – Chappers
    Commented Sep 8, 2017 at 15:25
  • $\begingroup$ @Chappers: According to author: $\nabla \vec X$ is vector and it has 3 components which are $\frac{\partial \vec X}{\partial x}$, $\frac{\partial \vec X}{\partial y}$, $\frac{\partial \vec X}{\partial z}$. But what I know is divergence of a vector, i.e $\nabla \vec X$, is a scalar and it equals: $\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}$, how should I understand this point? because of this, I didn't realize the author's idea. $\endgroup$
    – Dat
    Commented Sep 8, 2017 at 16:19
  • $\begingroup$ $\nabla \vec{X}$ is not a vector. Nor is it a scalar. $\nabla \cdot \vec{X}$ is the divergence, which is a scalar. $(\vec{Y} \cdot \nabla) \vec{X}$ is a vector, the directional derivative of $\vec{X}$ in the direction $\vec{Y}$. So $\nabla \vec{X}$ is a tensor (or a matrix if you prefer); it turns into a vector if you dot the $\nabla$ part with a vector. $\endgroup$
    – Chappers
    Commented Sep 8, 2017 at 16:36

1 Answer 1

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@Chapper:

My equation is:

$d\vec{\xi}$ = ($\nabla u.d\vec x$).$\vec i$+ ($\nabla v.d\vec x$).$\vec j$+ ($\nabla w.d\vec x$).$\vec k$

$$= (\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy + \frac{\partial u}{\partial z}dz).\vec i+(\frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy + \frac{\partial v}{\partial z}dz).\vec j+(\frac{\partial w}{\partial x}dx + \frac{\partial w}{\partial y}dy + \frac{\partial w}{\partial z}dz).\vec k $$

$$ = dx.(\frac{\partial u}{\partial x}.\vec i+\frac{\partial v}{\partial x}.\vec j+\frac{\partial w}{\partial x}.\vec k)+dy.(\frac{\partial u}{\partial y}.\vec i+\frac{\partial v}{\partial y}.\vec j+\frac{\partial w}{\partial y}.\vec k)+dz.(\frac{\partial u}{\partial z}.\vec i+\frac{\partial v}{\partial z}.\vec j+\frac{\partial w}{\partial z}.\vec k)$$

$$= dx.\frac{\partial \vec X}{\partial x}+dy.\frac{\partial \vec X}{\partial y}+dz.\frac{\partial \vec X}{\partial z}.$$

$\frac{\partial \vec X}{\partial x}$ is the directional derivative of vector $\vec X$ in the $\vec {Ox}$ direction. $\frac{\partial \vec X}{\partial x}$ is vector with 3 components, each component is the directional derivative of the corresponding scalar component of vector $\vec X$ in the $\vec {Ox}$ direction. $\frac{\partial \vec X}{\partial y}$ and $\frac{\partial \vec X}{\partial z}$ are similar to $\frac{\partial \vec X}{\partial x}$

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