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In a curvilinear system with coordinates: $u_1, u_1, u_3$.

I want to understand why: $$\nabla\cdot \vec{V} \neq \frac{1}{h_1} \frac{\partial V_1}{\partial u_1} +\frac{1}{h_2} \frac{\partial V_2}{\partial u_2} + \frac{1}{h_3}\frac{\partial V_3}{\partial u_3}. \tag1$$

My reasoning to explain why eq.(1) is not true is the following:

Taking the first term of the dot product:

$$\frac{\hat{e_1} }{h_1}\frac{\partial}{\partial u_1} (V_1 \hat{e_1}) = \frac{\hat{e_1}}{h_1} \big( \hat{e_1} \frac{\partial V_1}{\partial u_1} + V_1 \frac{\partial \hat{e_1}}{\partial u_1} \big)$$

This seems rare to me since the definition of the dot product is:

$$A \cdot B = A_1B_1 + A_2B_2 + A_3B_3 $$

And using that definition the divergence should be eq.(1) (with an equality).

I would appreciate any references that explain this, since all I can find is the formula for the divergence in curvilinear systems but without a deduction.

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  • $\begingroup$ What is $h_i$? What does This seems rare to me mean? $\endgroup$
    – Kyle Kanos
    Commented Mar 4, 2017 at 19:40
  • $\begingroup$ Maybe rare means weird, like in dutch $\endgroup$
    – Libertas
    Commented Apr 22, 2020 at 17:15

4 Answers 4

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Defining a "del" vector $\nabla=\mathbf{e}_i\frac{\partial}{\partial x^i}$ (summation on repeated indices assumed) is sort-of a mistake. This isn't a vector (field). It is just a nice notational device, because in this case, in cartesian coordinates, we have $$ \text{grad}(f)=\mathbf{e}_i\frac{\partial f}{\partial x^ i},$$ which looks like $\nabla f$, where $\nabla$ is the "vector" defined above and we have $$ \text{div}(\mathbf{A})=\frac{\partial A^ i}{\partial x^i}, $$ which sort of looks like $\nabla\cdot\mathbf{A}$, etc. But this is just a useful way to remember these formulas.


Now let's note several things. We can define a matrix whose $i,j$-th element is $\partial A^ i/\partial x^ j$. Let's call this matrix $\nabla\otimes\mathbf{A}$. We clearly have $$ \text{div}(\mathbf{A})=\text{Tr}(\nabla\otimes\mathbf{A})=\sum_i\frac{\partial A^ i}{\partial x^ i}. $$ (I'm gonna suspend the automatic summation from now on.)

If we have an orthogonal curvilinear coordinate system $(u^1,u^2,u^3)$, we can define the following quantities:

  • Coordinate basis vectors $$ \mathbf{g}_i=\frac{\partial \mathbf{r}}{\partial u^ i}, $$

  • "Reciprocal" coordinate basis vectors $$ \mathbf{g}^i=\nabla u^ i, $$

  • An orthonormal frame $$ \hat{\mathbf{e}}_i=\frac{1}{h_i}\mathbf{g}_i, $$ where $h_i=\sqrt{\mathbf{g}_i\cdot\mathbf{g}_i}$.

These quantities have the property that $$ \mathbf{g}^ i\cdot\mathbf{g}_j=\delta^ i_j, $$ since ($x^ i$ are cartesian coordinates and $\mathbf{e}_i$ are cartesian basis vectors) we have $$ \mathbf{g}^i\cdot\mathbf{g}_j=\nabla u^i\cdot\frac{\partial\mathbf{r}}{\partial u^ j}=\sum_k\frac{\partial u^ i}{\partial x^k}\mathbf{e}_k\cdot\sum_l\frac{\partial x^l}{\partial u^ j}\mathbf{e}_l=\sum_{kl}\frac{\partial u^ i}{\partial x^k}\frac{\partial x^l}{\partial u_j}\delta_{kl}=\sum_k \frac{\partial u^ i}{\partial x^k}\frac{\partial x^k}{\partial u_j}=\delta^i_j.$$

We further introduce a matrix $g_{ij}$ given as $g_{ij}=\mathbf{g}_i\cdot\mathbf{g}_j$. Because of orthogonality, we have $g_{ij}=h_i^2\delta_{ij}$. We also introduce $g^{ij}=\mathbf{g}^i\cdot\mathbf{g}^j$, feel free to check that $g^{ij}=\frac{1}{h_i^2}\delta_{ij}$, thus $g_{ij}$ and $g^{ij}$ are inverse matrices.

If $\mathbf{A}$ is a vector field, we associate three kinds of components with $\mathbf{A}$. We can have

  • $\mathbf{A}=\sum_i A^ i\mathbf{g}_i$, and the $A^ i$ are called contravariant components,

  • $\mathbf{A}=\sum_i A_i\mathbf{g}^i$, and the $A_i$ are called covariant components and

  • $\mathbf{A}=\sum_i \hat{A}_i\hat{\mathbf{e}}_i$ and the $\hat{A}_i$ are called physical components (here whether you put the index upwards or downwards doesn't matter!).

It is clear that we have $A^ i=\mathbf{g}^i\cdot\mathbf{A}$ and $A_i=\mathbf{g}_i\cdot\mathbf{A}$, and what is less clear is that differentiation with respect to the coordinates $\partial/\partial u^ i$ produce covariant components, eg. we have for any scalar function $f$, $\nabla f=\sum_i\mathbf{g}^i\frac{\partial f}{\partial u^ i}$ (check the definition of $\mathbf{g}^i$!).

We define Christoffel symbols of the second kind as $$ \Gamma^ k_{ij}=\frac{1}{2}\sum_l g^{kl}\left(\frac{\partial g_{jl}}{\partial u^ i}+\frac{\partial g_{il}}{\partial u^ j}-\frac{\partial g_{ij}}{\partial u^ l}\right). $$ Check that $\Gamma^k_{ij}=\mathbf{g}^k\cdot\frac{\partial \mathbf{g}_i}{\partial u^ j}$ (I really don't want to derive this identity right now).

In curvilinear coordinates, the basis vectors also depend on positions, so every time you differentiate a vector field, you need to make sure to take the variation of the basis vectors also into account, so we calculate the divergence as $$ \text{div}(\mathbf{A})=\sum_i\mathbf{g}^i\cdot\frac{\partial\mathbf{A}}{\partial u^i}=\sum_{ij}\mathbf{g}^i\left(\frac{\partial A^j}{\partial u^ i}\mathbf{g}_j+A^j\frac{\partial \mathbf{g}_j}{\partial u^ i}\right)=\sum_{ij}\left(\frac{\partial A^ i}{\partial u^i}+\Gamma^i_{ji}A^j\right). $$

We now calculate $$\Gamma^i_{ji}=\sum\frac{1}{2}g^{il}\left(\frac{\partial g_{jl}}{\partial u^i}+\frac{\partial g_{il}}{\partial u^j}-\frac{\partial g_{ij}}{\partial u^l}\right)=\sum\frac{1}{2}g^{il}\frac{\partial g_{il}}{\partial u^j},$$ which, in matrix notation ($(g_{ij})=g$) can be written as $$ \Gamma^i_{ji}=\frac{1}{2}\text{Tr}\left(g^{-1}\frac{\partial g}{\partial u^j}\right), $$ which, using Jacobi's formula, can be written as $$ \sum_i\Gamma^ i_{ji}=\frac{1}{2}\frac{1}{\det g}\frac{\partial}{\partial u^ j}\det g=\frac{1}{2}\frac{\partial}{\partial u^ j}\ln\det g=\frac{\partial}{\partial u^j}\ln\sqrt{\det g}=\frac{1}{\sqrt{\det g}}\frac{\partial}{\partial u^j}\sqrt{\det g}. $$

Plugging this back into the divergence formula gives $$\text{div}(\mathbf{A})=\sum_{ij}\left(\frac{\partial A^ i}{\partial u^i}+\Gamma^i_{ji}A^j\right)=\sum_{ij}\left(\frac{\partial A^i}{\partial u^i}+\frac{1}{\sqrt{\det g}}\frac{\partial}{\partial u^j}\sqrt{\det g}A^j\right)= \\ =\sum_{i}\frac{1}{\sqrt{\det g}}\frac{\partial}{\partial u^i}(\sqrt{\det g}A^i). $$

But since $g=\text{diag}(h_1^2,h_2^2,h_3^2)$

, we have $\det g=(h_1h_2h_3)^2$, so $$ \text{div}(\mathbf{A})=\frac{1}{h_1h_2h_3}\left\{\frac{\partial}{\partial u^1}(h_1h_2h_3A^1)+\frac{\partial}{\partial u^2}(h_1h_2h_3A^2)+\frac{\partial}{\partial u^3}(h_1h_2h_3A^3)\right\}= \\ =\frac{1}{h_1h_2h_3}\left\{\frac{\partial}{\partial u^1}(h_2h_3\hat{A}_1)+\frac{\partial}{\partial u^2}(h_1h_3\hat{A}_2)+\frac{\partial}{\partial u^3}(h_1h_2\hat{A}_3)\right\}, $$

which is the formula for divergence in terms of the physical components.

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  • $\begingroup$ My fingers fell of typing all this, so I don't want to edit the answer right now, but if it is unclear to you why $\text{div}(\mathbf{A})=\sum_i\mathbf{g}^i\cdot(\partial\mathbf{A}/\partial u^i)$, ask me here and I'll edit later. $\endgroup$ Commented Mar 4, 2017 at 20:11
  • $\begingroup$ In your statement "we have an orthogonal curvilinear (u1,u2,u3)", can't you remove "orthogonal". I do not see where you sue the assumption later on? $\endgroup$
    – pluton
    Commented Oct 20, 2017 at 21:02
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    $\begingroup$ @pluton The formula $\text{div}(\mathbf{A})=\frac{1}{\sqrt{\det g}}\sum_i\partial_i(\sqrt{\det g}A^i)$ is valid without orthogonality as well. However, if your coordinates are not orthogonal, the metric $g_{ij}$ does not take the simple form $g_{ij}=h_i^2\delta_{ij}$, and the definition of "physical components" is not unique (you can Gram-Schmidt orthogonalize the oblique basis, but the result depends on which basis vector you take to be the "first one"). So the answer as a whole depends on orthogonality. $\endgroup$ Commented Oct 21, 2017 at 7:46
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By definition $\nabla \cdot \vec{E}=\lim_{\Delta V\to 0 }\frac{\int \vec{E}\cdot \hat{n}dA}{\Delta V} =\lim_{\Delta V\to 0} \frac{\int E_in_ih_jh_kdx_jdx_k +\int E_jn_jh_ih_kdx_idx_k+ \int E_kn_kh_jh_idx_jdx_i}{\int h_ih_jh_kdx_idx_jdx_k}$

$$= \frac{1}{h_ih_jh_k}(\frac{\partial(E_ih_jh_k) }{\partial x_i }+\frac{\partial(E_jh_ih_k) }{\partial x_j }+ \frac{\partial(E_kh_jh_i) }{\partial x_k } )$$

So for example in cylindrical coordinates:

Cancel all the $dx_i$'s in the numerator with the denominator,term by term and differentiate with the respect to the remaing differential. Treat where applicable the $h_i$'s as constants.

$$\nabla \cdot \vec{E}=\frac{1}{r}\frac{\partial (rE_r)}{\partial r}+\frac{1}{r}\frac{\partial E_\theta }{\partial \theta}+\frac{\partial E_z}{\partial z}$$

The scaling factors can make things complicated. Sometimes they don't effect the differentiation and so cancel upon division.

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Any good book on vector or tensor analysis should provide you with a first principles derivation for some specific curvilinear coordinate systems (spherical and cylindrical systems for example). For the more general case, you need to understand that your definition for the dot product of two vectors is valid only for Cartesian coordinates. Why is that? Because in general the dot product is defined as the multiplicative operation that converts two vector quantities into a scalar quantity. Likewise the divergence is defined as the first order differential operator that yields a scalar quantity when operating on a vector. In tensor analysis this operator is called the covariant derivative and for curvilinear coordinate systems the usual partial derivatives are augmented by terms called Christoffel symbols. These extra terms render the divergence a scalar quantity. This answer should provide you with keywords that will take you to a complete understanding.

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From my college text Field and Wave Electromagnetics by Cheng, he defines:

$$ \vec{\nabla} \overset{\nabla}= \left(\vec a_{u1}\frac1{h_1}\frac{\partial}{u_1} +\vec a_{u2}\frac1{h_2}\frac{\partial}{u_2} +\vec a_{u3}\frac1{h_3}\frac{\partial}{u_3}\right) $$ whereas: $$ \vec{\nabla}\bullet\vec{A}=\frac1{h_1 h_2 h_3} \left( \frac{\partial}{\partial u_1}h_2 h_3 A_1 +\frac{\partial}{\partial u_2}h_3 h_1 A_2 +\frac{\partial}{\partial u_3}h_1 h_2 A_3 \right) $$ $$ \vec{\nabla}\bullet\vec{A}= \frac1{h_2 h_3} \left(\frac1{h_1}\frac{\partial}{\partial u_1}h_2 h_3 A_1\right) +\frac1{h_3 h_1} \left(\frac1{h_2}\frac{\partial}{\partial u_2}h_3 h_1 A_2\right) +\frac1{h_1 h_2} \left(\frac1{h_3}\frac{\partial}{\partial u_3}h_1 h_2 A_3\right) $$ $$ \vec{\nabla_1}\bullet h_2 h_3 \vec{ A}= \left(\frac1{h_1}\frac{\partial}{\partial u_1}h_2 h_3 A_1\right) , \vec{\nabla_2}\bullet h_3 h_1 \vec{A}= \left(\frac1{h_2}\frac{\partial}{\partial u_2}h_3 h_1 A_2\right) , \\ \vec{\nabla_3}\bullet h_1 h_2\vec{A}= \left(\frac1{h_3}\frac{\partial}{\partial u_3}h_1 h_2 A_3\right) $$ Now it is becoming more apparent why $\vec{\nabla}\bullet \vec{A}$ seems to be different than $\vec{\nabla}\bullet\vec{A}=\nabla_1*A_1+\nabla_2*A_2+\nabla_3*A_3$ because the coordinate system imposes surface area coordinates only, with for instance $h_2\hat{u_2}$ and $h_3\hat{u_3}$ being perpendicular to $u_1$ to calculate the unit perpendicular flux surface of $h_2*h_3$ that corresponds to a unit length $\hat{u_1}$ which is $\left|\hat{u_1}\right|=1$.

As referenced by Richard Feynman (when he taught at Cal Tech in his Lecture 18 The Maxwell Equations):

Feynman Maxwell Equations Quote

The first equation—that the divergence of E is the charge density over $ϵ_0$ —is true in general. In dynamic as well as in static fields, Gauss’ law is always valid. The flux of $\vec{E}$ through any closed surface is proportional to the charge inside.

Thus to restate Richard Feynman, when considering the coordinate $A_1$, which in the direction of $\hat{u_1}$ has a magnitude of $A_1$, one really needs to consider the entire coordinate system, where the flux of $A_1 \hat{a_1}$ goes through a perpendicular closed surface areas $h_2 h_3$ which are the length metrics of the coordinate system under consideration.

These surface areas are coming into play because of the definition of the Gradient as a surface quantity. From "13.1 Differential Form of Gauss' Law":

Definition of Del Dot in Differential Form

From the previously referenced Equation:

$$ \vec{\nabla}\bullet\vec{A}=\frac1{h_1 h_2 h_3} \left( \frac{\partial}{\partial u_1}h_2 h_3 A_1 +\frac{\partial}{\partial u_2}h_3 h_1 A_2 +\frac{\partial}{\partial u_3}h_1 h_2 A_3 \right) $$ Then $$ \int_{\partial Volume} \vec{\nabla}\bullet\vec{A}= \int_{\partial Surfaces} \hat{n}\bullet\left(\vec{A} \partial S\right) = \int_{\partial Volume} \frac1{h_1 h_2 h_3} \left( \frac{\partial}{\partial u_1}h_2 h_3 A_1 +\frac{\partial}{\partial u_2}h_3 h_1 A_2 +\frac{\partial}{\partial u_3}h_1 h_2 A_3 \right) \left(h_1 \partial u_1\right) \left(h_2 \partial u_2\right) \left(h_3 \partial u_3\right)$$ $$= \int_{\partial Volume} \left( \frac{\partial}{\partial u_1}h_2 h_3 A_1 +\frac{\partial}{\partial u_2}h_3 h_1 A_2 +\frac{\partial}{\partial u_3}h_1 h_2 A_3 \right) \left(\partial u_1\right)\left(\partial u_2\right) \left(\partial u_3\right)$$ $$= \int_{\partial Volume} \left(\frac1{h_1} \frac{\partial}{\partial u_1} \left(h_2 \partial u_2\right) \left(h_3 \partial u_3\right) A_1\right) \left(h_1 \partial u_1\right) $$ $$ +\int_{\partial Volume} \left(\frac1{h_2} \frac{\partial}{\partial u_2} \left(h_3 \partial u_3\right) \left(h_1 \partial u_1\right) A_2\right) \left(h_2 \partial u_2\right) $$ $$ +\int_{\partial Volume} \left(\frac1{h_3} \frac{\partial}{\partial u_3} \left(h_3 \partial u_3\right) \left(h_1 \partial u_1\right) A_3\right) \left(h_3 \partial u_3\right) $$ In particular consider $\vec{\nabla}\bullet\vec{A_1}$ as follows: $$ \int_{\partial Volume} \vec{\nabla}\bullet\vec{A_1}\left(\partial Vol\right)= \int_{\partial Volume}\frac1{h_1 h_2 h_3} \left( \frac{\partial}{\partial u_1}h_2 h_3 A_1 \right)\left(\partial Vol\right) $$ $$ =\int_{\partial Volume} \vec{\nabla}\bullet\vec{A_1}\left(\partial Vol\right)= \int_{\partial Volume}\frac1{h_1 h_2 h_3} \left( \frac{\partial}{\partial u_1}h_2 h_3 A_1 \right) \left(h_1 \partial u_1\right)\left(h_2 \partial u_2\right) \left(h_3 \partial u_3\right) $$ $$= \int_{\partial Volume} \left(\frac1{h_1} \frac{\partial}{\partial u_1} \left(h_2 \partial u_2\right) \left(h_3 \partial u_3\right) A_1\right) \left(h_1 \partial u_1\right) $$ Here, $\vec{\nabla_1}=\frac1{h_1}\frac{\partial}{\partial u_1}\hat{u_1}$ as is to be expected. Further the differential surface area is $\partial S=\left(h_2 \partial u_2\right)\left(h_3 \partial u_3\right)$ as to be expected. Also, $\left(\partial l_1\right)=\left(h_1 \partial u_1\right)$ as is to be expected, because the length traveled has to be the coordinate value $u_1$ times its length metric $h_1$. And the volume element is also as to be expected as $\left(h_2 \partial u_2\right)\left(h_3 \partial u_3\right)\left(h_1 u_1\right)$ being the differential surface area times traveled length as perpendicular width times perpendicular breadth times length.

In this form the surface integral equivalent becomes visible. The point being that what is being calculated in Gauss's Law that defines the gradient is a flux $\vec{A}$ going through a perpendicular surface area that has a surface area metric of $h_1*h_2$. So because of this it seems that $\vec{\nabla}\bullet \vec{A}$ is not a simple dot product like in Cartesian Coordinates. However, when broken down in the volume integration of a differential element, everything makes sense. The gradient $\vec{\nabla_1}=\frac1{h_1}\frac{\partial}{\partial u_1}\hat{u_1}$ as is to be expected. Further the differential surface area is $\partial S=\left(h_2 \partial u_2\right)\left(h_3 \partial u_3\right)$ as to be expected. Also, $\left(\partial l_1\right)=\left(h_1 \partial u_1\right)$ as is to be expected, resulting in the corresponding differential volume element by inspection.

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