Divergence of Vector Function in Cartesian Cooridenates System
Definition
For a vector function $\mathbf{F}(x,y,z)=F_{x}\hat{\mathbf{x}}+F_{y}\hat{\mathbf{y}} + F_{z}\hat{\mathbf{z}}$, the following scalar function is defined as the divergence $\mathbf{F}$ of $\mathbf{F}(x,y,z)=F_{x}\hat{\mathbf{x}}+F_{y}\hat{\mathbf{y}} + F_{z}\hat{\mathbf{z}}$ and is denoted by $\nabla \cdot \mathbf{F}$.
$$ \begin{equation} \nabla \cdot \mathbf{F} := \frac{ \partial F_{x}}{ \partial x} + \frac{ \partial F_{y}}{ \partial y }+ \frac{ \partial F_{z}}{ \partial z} \label{divergence} \end{equation} $$
Explanation
Geometrically, if $\nabla \cdot \mathbf{F}>0$, it means that $\mathbf{F}$ is spreading out or diverging. If $\nabla \cdot \mathbf{F}<0$, it means that $\mathbf{F}$ is converging or moving inward. If $\nabla \cdot \mathbf{F}=0$, it means that $\mathbf{F}$ is neither diverging nor converging, indicating an equilibrium where the amount going out equals the amount coming in.
Divergence is translated as “divergence”. To maintain consistency with using gradient for slope and curl for rotation, it is denoted as divergence instead of using the literal translation.
It is important to note that the value defined as $\dfrac{ \partial F_{x}}{ \partial x} + \dfrac{ \partial F_{y}}{ \partial y }+ \dfrac{ \partial F_{z}}{ \partial z}$ is denoted by $\nabla \cdot \mathbf{F}$. While $\nabla$ is referred to as the del operator, considering it as inherently meaningful can lead to confusion with $\nabla \cdot \mathbf{F}$ or $\nabla \times \mathbf{F}$ being mistaken for dot and cross products. Therefore, $\nabla$ should be understood merely as a convenient notation, and it might be better to think of the del operator as equivalent to gradient. The details continue below.
Important Points
$\nabla \cdot \mathbf{F}$ is not the dot product of $\nabla$ and $\mathbf{F}$
A dot product is fundamentally an operation between two vectors. Thinking of $\nabla \cdot \mathbf{F}$ as a dot product implies considering $\nabla$ as a vector.
$$ \nabla \overset{?}{=}\dfrac{ \partial }{ \partial x}\hat{\mathbf{x}} +\dfrac{ \partial }{ \partial y}\hat{\mathbf{y}}+\dfrac{ \partial }{ \partial z}\hat{\mathbf{z}} = \left( \dfrac{ \partial }{ \partial x},\ \dfrac{ \partial }{ \partial y},\ \dfrac{ \partial }{ \partial z} \right) $$
Indeed, thinking in this way conveniently matches the definition of divergence $(1)$.
$$ \nabla \cdot \mathbf{F} = \left( \dfrac{ \partial }{ \partial x},\ \dfrac{ \partial }{ \partial y},\ \dfrac{ \partial }{ \partial z} \right) \cdot \left( F_{x}, F_{y}, F_{z} \right) = \frac{ \partial F_{x}}{ \partial x} + \frac{ \partial F_{y}}{ \partial y }+ \frac{ \partial F_{z}}{ \partial z} $$
However, if this were truly a dot product, the commutative property would imply an odd conclusion as follows:
$$ \mathbf{F} \cdot \nabla = F_{x} \dfrac{\partial }{\partial x} + F_{y} \dfrac{\partial }{\partial y} + F_{z} \dfrac{\partial }{\partial z} \overset{?}{=} \frac{ \partial F_{x}}{ \partial x} + \frac{ \partial F_{y}}{ \partial y }+ \frac{ \partial F_{z}}{ \partial z} = \nabla \cdot \mathbf{F} $$
In reality, $\nabla \cdot$ represents an operator that maps a vector function $\mathbf{F}(x,y,z)$ to a scalar function $\frac{ \partial F_{x}(x,y,z)}{ \partial x} + \frac{ \partial F_{y}(x,y,z)}{ \partial y }+ \frac{ \partial F_{z}(x,y,z)}{ \partial z}$. This means that $\operatorname{div}$, when defined as follows, simply yields $\dfrac{ \partial F_{x}}{ \partial x} + \dfrac{ \partial F_{y}}{ \partial y }+ \dfrac{ \partial F_{z}}{ \partial z}$ for every substitution of $\mathbf{F}$. This intuitive understanding explains why $\operatorname{div}(\mathbf{F})$ is denoted as $\nabla \cdot \mathbf{F}$ instead of a direct interpretation. In advanced mathematics, divergence is often denoted differently, reflecting a less intuitive handling of 3D vectors than in physics.
Then, why can’t $\nabla \cdot \mathbf{F}$ be considered a non-commutative dot product?
It’s because in $\nabla \cdot \mathbf{F}$, $\nabla \cdot$ itself is a function (operator), and $\mathbf{F}$ is a variable. In contrast, $\mathbf{F} \cdot \nabla$ is an operator by itself. Therefore, $\nabla \cdot \mathbf{F}$ represents the function value of $\nabla \cdot$, while $\mathbf{F} \cdot \nabla$ is a function awaiting variable substitution. The notation $\mathbf{F} \cdot \nabla$ intuitively represents function $f$. $f$ is an operator that takes the vector function $\mathbf{A}$ as a variable and applies $\left( F_{x}\dfrac{\partial }{\partial x} + F_{y}\dfrac{\partial }{\partial y} + F_{z}\dfrac{\partial }{\partial z} \right)$ to each component.
$$ \begin{align*} f (\mathbf{A}) \overset{\text{definition}}{=}& \left( F_{x}\dfrac{\partial A_{x}}{\partial x} + F_{y}\dfrac{\partial A_{x}}{\partial y} + F_{z}\dfrac{\partial A_{x}}{\partial z} \right)\hat{\mathbf{x}} \\ &\quad+ \left( F_{x}\dfrac{\partial A_{y}}{\partial x} + F_{y}\dfrac{\partial A_{y}}{\partial y} + F_{z}\dfrac{\partial A_{y}}{\partial z} \right)\hat{\mathbf{y}} \\ &\quad+ \left( F_{x}\dfrac{\partial A_{z}}{\partial x} + F_{y}\dfrac{\partial A_{z}}{\partial y} + F_{z}\dfrac{\partial A_{z}}{\partial z} \right)\hat{\mathbf{z}} \\ \overset{\text{notation}}{=}& (\mathbf{F}\cdot \nabla) (\mathbf{A}) \end{align*} $$
When a scalar function $\phi$ exists as a variable, it is considered as follows:
$$ \begin{align*} f (\phi) \overset{\text{definition}}{=}& F_{x}\dfrac{\partial \phi}{\partial x} + F_{y}\dfrac{\partial \phi}{\partial y} + F_{z}\dfrac{\partial \phi}{\partial z} \\ \overset{\text{notation}}{=}& (\mathbf{F}\cdot \nabla) (\phi) \end{align*} $$
Thus, $\nabla \cdot \mathbf{F}$ and $\mathbf{F}\cdot \nabla$ should not be understood as the dot product of $\nabla$ and $\mathbf{F}$, but as functions in themselves. This explanation applies not only to divergence but also to gradient $\nabla f$ and curl $\nabla \times \mathbf{F}$.
Derivation
First, consider a small volume in 3D space as shown below.
Our goal is to understand how $\mathbf{F}$ appears at each point within this small volume. Analogously, if $\mathbf{F}$ represents heat, we aim to determine the direction and speed of flow, or if $\mathbf{F}$ represents water, whether it is entering or exiting through a tap or drain. Let’s calculate for the $x$ axis direction first. The amount of $\mathbf{F}$ passing through $d\mathbf{a}_{1}$ can be calculated by the dot product.
$$ \begin{align} \mathbf{F}(x+dx) \cdot d\mathbf{a}_{1} &= \left( F_{x}(x+dx)\hat{\mathbf{x}}+F_{y}(x+dx)\hat{\mathbf{y}}+F_{z}(x+dx)\hat{\mathbf{z}} \right) \cdot dydz\hat{\mathbf{x}} \nonumber \\ &= F_{x}(x+dx)dydz \end{align} $$
If $F_{x}(x+dx)dydz >0$, it indicates the amount of $\mathbf{F}$ exiting the small volume, and if $F_{x}(x+dx)dydz<0$, the amount entering. Similarly, the amount of $\mathbf{F}$ exiting through $d\mathbf{a}_{2}$ is as follows.
$$ \begin{equation} \mathbf{F}(x) \cdot d \mathbf{a}_{2} = F_{x}(x)\hat{\mathbf{x}} \cdot(-dydz\hat{\mathbf{x}})=-F_{x}(x)dydz \end{equation} $$
Therefore, $(2) + (3)$ represents the net flux of $\mathbf{F}$ in the $x$ direction.
$$ \begin{align*} (2) + (3) &=\left[ F_{x}(x+dx) -F_{x}(x)\right]dydz \\ &= \frac{F_{x}(x+dx) -F_{x}(x) }{dx}dxdydz \end{align*} $$
Since $dx$ is a small length, it can be approximated as follows. Thus, the amount of $\mathbf{F}$ entering or exiting the small volume in the $x$ direction is expressed as:
$$ \frac{ \partial F_{x}}{ \partial x}dxdydz $$ Calculating similarly for the $y$ and $z$ directions yields:
$$ \frac{ \partial F_{y}}{ \partial y}dxdydz \quad \text{and} \quad \frac{ \partial F_{z}}{ \partial z}dxdydz $$
Summing these gives the total flux of $\mathbf{F}$ entering or exiting the small volume, and dividing by $dxdydz$ gives the flux per unit volume.
$$ \frac{ \partial F_{x}}{ \partial x}+\frac{ \partial F_{y}}{ \partial y}+\frac{ \partial F_{z}}{ \partial z} $$
From now on, this will be referred to as the divergence of $\mathbf{F}$ and denoted by $\nabla \cdot \mathbf{F}$.
$$ \nabla \cdot \mathbf{F} := \frac{ \partial F_{x}}{ \partial x}+\frac{ \partial F_{y}}{ \partial y}+\frac{ \partial F_{z}}{ \partial z} $$
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As derived, it’s important to remember that $\nabla \cdot \mathbf{F}$ is not the dot product of $\nabla$ and $\mathbf{F}$.
Related Formulas
Linearity:
$$ \nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f) $$ $$ \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}) $$
$$ \nabla \cdot (\nabla T) = \dfrac{\partial^{2} T}{\partial x^{2}} + \dfrac{\partial ^{2} T} {\partial y^{2}} + \dfrac{\partial ^{2} T}{\partial z^{2}} $$ $$ \nabla (\nabla \cdot \mathbf{A} ) $$ $$ \nabla \cdot (\nabla \times \mathbf{A})=0 $$
Gauss’s Theorem (Divergence Theorem)
$$ \int_\mathcal{V} \nabla \cdot \mathbf{ F} dV = \oint _\mathcal{S} \mathbf{F} \cdot d \mathbf{S} $$
$$ \int_{\mathcal{V}} \left[ T \nabla^{2} U + (\nabla T) \cdot (\nabla U) \right] d \tau = \oint_{\mathcal{S}} (T \nabla U) \cdot d \mathbf{a} $$ $$ \int_{\mathcal{V}} \left( T \nabla^{2} U - U \nabla^{2} T \right) d \tau = \oint_{\mathcal{S}} \left( T \nabla U - U \nabla T \right) \cdot d \mathbf{a} $$
$$ \int_{\mathcal{V}}\mathbf{A} \cdot (\nabla f)d\tau = \oint_{\mathcal{S}}f\mathbf{A} \cdot d \mathbf{a}-\int_{\mathcal{V}}f(\nabla \cdot \mathbf{A})d\tau $$