Divergence formula derivation ?

[Outrageous]

Homework Statement

How to get equation 1 from the thumbnail?
h1 h2 h3 doesn't have to be constant.
The most I can try is equation 2 .
Please guide thanks.

Homework Equations

The Attempt at a Solution

 

[vanhees71]

Use the coordinate-independent definition of the divergence as the limit of an appropriate integral over the surface of an infinitesimal cuboid spanned by the tangent vectors of the coordinate lines at the point in question.

 

[verty]

Is it not just factoring out $h_1 h_2 h_3$? As a check, multiplying out (1) gives (2).

Hmm, clearly it's not that simple.

 

[vanhees71]

It's not that simple, because the curvilinear basis vectors are position dependent! Rather use the definition of the divergence,
$$\vec{\nabla} \cdot \vec{a}(\vec{x})=\lim_{\mathrm{d} V \rightarrow \{ \vec{x} \}} \frac{1}{\mathrm{d} V} \int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{a}.$$
Hint: Prove first that $\mathrm{d}V=h_1 h_2 h_3 \mathrm{d}^3 u$.

 

[Outrageous]

This is what I found from a book, not really understand how the methods mean,
1) why the negative of the first equation be ignored?
2) keeping u1 constant, then why second equation come out?

Thanks

 

[Outrageous]

Sorry. The first and second equations are in this page.

 

[vanhees71]

The idea is to use Gauß's Integral Theorem to an infinitesimal box spanned by the coordinate lines of your orthogonal curvilinear coordinates. It's a little cuboid with 6 surfaces giving the boundary of the volume, and you have to approximately evaluate the surface integral.

To that end we must remember the definition of the basis vectors. Let $(q_1,q_2,q_3)$ be the coordinates and $\vec{r}(q_1,q_2,q_3)$ the position vector as functions of them. Then the unit-basis vectors are defined by
$$\hat{e}_j=\frac{1}{h_j} \frac{\partial \vec{r}}{\partial q_j}.$$
Since the coordinates are assumed to be orthogonal this means that $\hat{e}_j \cdot \hat{e}_k = \delta_{jk}$. Also we assume that the order of the coordinates are chosen such that the $\hat{e}_j$ build a postively oriented basis, i.e., $\hat{e}_1 \times \hat{e}_2=\hat{e}_3$.

Now the surface-normal vectors of your cuboid are easily determined. Take the surface parallel to the $q_2 q_3$-plane at $q_1+\mathrm{d} q_1$. The surface-normal vector is
$$\mathrm{d} \vec{F}=\mathrm{d} q_2 \mathrm{d} q_3 \left . \left (\frac{\partial \vec{r}}{\partial q_2} \times \frac{\partial \vec{r}}{\partial q_3} \right ) \right |_{q_1+\mathrm{d} q_1,q_2,q_3} = \mathrm{d} q_2 \mathrm{d} q_3 (h_2 h_3 \hat{e}_2 \times \hat{e}_3) = \mathrm{d} q_2 \mathrm{d} q_3 (h_2 h_3 \hat{e}_1)_{q_1+\mathrm{d} q_1,q_2,q_3}.$$
The corresponding contribution to the surface integral thus reads
$$\mathrm{d} q_2 \mathrm{d} q_3 (h_2 h_3 a_1)|_{q_1+\mathrm{d} q_1,q_2,q_3} = \mathrm{d} q_2 \mathrm{d} q_3 \left [(h_2 h_3 a_1)|_{q_1,q_2,q_3} + \mathrm{d} q_1 \left . \left ( \frac{\partial(h_2 h_3 a_1)}{\partial q_1} \right ) \right |_{q_1,q_2,q_3} + \mathcal{O}(\mathrm{d} q_1^2) \right ].$$
The contribution of the cuboid's edge parallel to the $q_2 q_3$ plane at $(q_1,q_2,q_3)$ gives in the same way
$$-\mathrm{d} q_2 \mathrm{d} q_3 (h_2 h_3 a_1)|_{q_1,q_2,q_3}$$
Thus taking these two contributions together you get
$$\mathrm{d}q_1 \mathrm{d} q_2 \mathrm{d} q_3 \left (\frac{\partial(h_2 h_3 a_1)}{\partial q_1} \right )+\mathcal{O}(\mathrm{d} q_1^2 \mathrm{d} q_2 \mathrm{d} q_3).$$
Now the volume of the cuboid is $\mathrm{d} q_1 \mathrm{d} q_2 \mathrm{d} q_3 h_1 h_2 h_3$.
Dividing the above calculated contribution to the surface integral by this and letting the $\mathrm{d} q_j \rightarrow 0$ gives the following contribution to the divergence:
$$\vec{\nabla} \cdot \vec{a}=\frac{1}{h_1 h_2 h_3} \frac{\partial(h_2 h_3 a_1)}{\partial q_1} + \text{contributions from other surface elements}.$$
Now you can think about the other surface elements. You'll see that this gives a proof for the correct form of the divergence in curvilinear orthonormal coordinates and also a very intuitive picture of why this must be the correct formula :-).

 

[Outrageous]

The idea is to use Gauß's Integral Theorem to an infinitesimal box spanned by the coordinate lines of your orthogonal curvilinear coordinates. It's a little cuboid with 6 surfaces giving the boundary of the volume, and you have to approximately evaluate the surface integral.

To that end we must remember the definition of the basis vectors. Let $(q_1,q_2,q_3)$ be the coordinates and $\vec{r}(q_1,q_2,q_3)$ the position vector as functions of them. Then the unit-basis vectors are defined by
$$\hat{e}_j=\frac{1}{h_j} \frac{\partial \vec{r}}{\partial q_j}.$$
Since the coordinates are assumed to be orthogonal this means that $\hat{e}_j \cdot \hat{e}_k = \delta_{jk}$. Also we assume that the order of the coordinates are chosen such that the $\hat{e}_j$ build a postively oriented basis, i.e., $\hat{e}_1 \times \hat{e}_2=\hat{e}_3$.

This is to Define the coordinate on curvilinear basis, and the direction of $\hat{e}_j$ .
For orthogonal, $\hat{e}_j \cdot \hat{e}_k = \delta_{jk}$? Is not zero?

Now the surface-normal vectors of your cuboid are easily determined. Take the surface parallel to the $q_2 q_3$-plane at $q_1+\mathrm{d} q_1$. The surface-normal vector is
$$\mathrm{d} \vec{F}=\mathrm{d} q_2 \mathrm{d} q_3 \left . \left (\frac{\partial \vec{r}}{\partial q_2} \times \frac{\partial \vec{r}}{\partial q_3} \right ) \right |_{q_1+\mathrm{d} q_1,q_2,q_3} = \mathrm{d} q_2 \mathrm{d} q_3 (h_2 h_3 \hat{e}_2 \times \hat{e}_3) = \mathrm{d} q_2 \mathrm{d} q_3 (h_2 h_3 \hat{e}_1)_{q_1+\mathrm{d} q_1,q_2,q_3}.$$
The corresponding contribution to the surface integral thus reads
$$\mathrm{d} q_2 \mathrm{d} q_3 (h_2 h_3 a_1)|_{q_1+\mathrm{d} q_1,q_2,q_3} = \mathrm{d} q_2 \mathrm{d} q_3 \left [(h_2 h_3 a_1)|_{q_1,q_2,q_3} + \mathrm{d} q_1 \left . \left ( \frac{\partial(h_2 h_3 a_1)}{\partial q_1} \right ) \right |_{q_1,q_2,q_3} + \mathcal{O}(\mathrm{d} q_1^2) \right ].$$

The surface-normal vector dF is the value of surface area with direction pointing to the direction ei. ?
Why there are dq1 and dq2?
Thanks for answering. My mathematics is not very good, do you have any link for answer or any book?

 

[vanhees71]

The Kronecker-$\delta$ symbol is defined as
$$\delta_{jk}=\begin{cases} 1 & \text{for} \quad j=k,\\<br /> 0 & \text{for} \quad j \neq k.<br /> \end{cases}$$
The area element in question is spanned by the coordinate lines, and thus the surface-normal vector is given by
$$\mathrm{d} \vec{F}=\frac{\partial \vec{r}}{\partial q_2} \mathrm{d} q_2 \times \frac{\partial \vec{r}}{\partial q_3} \mathrm{d} q_3.$$
It's magnitude is the area of the surface, and it's direction is perpendicular on the surface. In Gauß's Theorem the orientation has to be chosen such that the normal vectors point out of the enclosed volume, over which you integrate:
$$\int_{V} \mathrm{d}^3 \vec{r} \vec{\nabla} \cdot \vec{a}=\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{a}.$$

 

[Outrageous]

$q_2 q_3$-plane at $q_1+\mathrm{d} q_1$. The surface-normal vector is
$$\mathrm{d} \vec{F}=\mathrm{d} q_2 \mathrm{d} q_3 \left . \left (\frac{\partial \vec{r}}{\partial q_2} \times \frac{\partial \vec{r}}{\partial q_3} \right ) \right |_{q_1+\mathrm{d} q_1,q_2,q_3} = \mathrm{d} q_2 \mathrm{d} q_3 (h_2 h_3 \hat{e}_2 \times \hat{e}_3)$$

Should it be like this? dF is the change of the surface area.
$$\mathrm{d} \vec{F}=\mathrm{d} q_2 \mathrm{d} q_3 \left . \left (\frac{\partial \vec{r}}{\partial q_2} \times \frac{\partial \vec{r}}{\partial q_3} \right ) \right |_{\mathrm{d} q_1,q_2,q_3}$$