Geometric intuition behind gradient, divergence and curl

Imagine a volume $V$ (with boundary $\partial V$) centered at a point $p$.

The divergence of $\nabla \cdot F$ of a vector field $F$ can be seen as the limit

$$\nabla \cdot F = \lim_{V \to 0}\frac{1}{V} \oint_{\partial V} F \cdot \hat n \, dS$$

It's not too difficult to geometrically interpret this integral. This is a flux integral--it tells us how much $F$ is normal to the surface elements $\hat n \, dS$. A function with positive divergence must be pointing mostly radially outward from a point--it diverges from that point.

The curl can be constructed in a similar way:

$$\nabla \times F = \lim_{V \to 0} \frac{1}{V} \oint_{\partial V} \hat n \times F \, dS$$

It's probably easiest to picture this in 2d: there, $\partial V$ is a circle and $\hat n$ points radially outward. The curl, then, measures how much $F$ is perpendicular to $\hat n$, or how much it curls around our central point (and if it does curl around, in what direction is it curling?).

Nothing really comes to mind for the Laplacian, but hopefully this helps.

Here are the ones I remember:

The gradient is as you described it. Also, the gradient points in the direction of "fastest increase" through the field. That gels nicely with the intuition you gave, since it seems intuitive tha the normal to the level curve (which is a region of constant value) would point either upward or downward through other values.

If you imagine a little propeller in the vector field, the curl at that point is the tendency of the propeller to be turned by the flowlines.

The divergence at a point is the tendency of the field to flow outward or inward to that point.

The Laplacian is the one I'm least familiar with, and seems to be the hardest to come up with a picture for. If you interpret it as a combination of the divergence and gradient above, it is something to do with flux of the gradient. As I read about it, it appears to be useful for studying dissolving substances.

Gradient ($\mathbf{\nabla}$)

Imagine you are hiking on a hill. If you can find the steepest way to climb up, you have found the gradient. The hill is a surface and gives you the elevation($h$) as a function of $x$ and $y$.

$s=\mathbf{\nabla}h(x,y)$

$s$ gives you the slope of the tangent on the surface.

Divergence ($\mathbf{\nabla} \cdot$) see also here ,

Imagine you have an empty bucket and you open the tap. The bucket starts filling up with water. Can you imagine where the water is going? Seeing from above the water comes outwards, and you have found the divergence.

$ \nabla \cdot \mathbf{V}(x,y) > 0 $

Positive divergence means that you have a source.

Now let's close the sink and let the bucket rest. No movement of water gives you zero divergence (no source nor sink).

$ \nabla \cdot \mathbf{V}(x,y) = 0$

Finally, drill a hole in the bottom of the bucket, such that the water starts leaving the bucket

$ \nabla \cdot \mathbf{V}(x,y) < 0$

Negative divergence means you have a sink. Let's call $\mathbf{V}$ our velocity field. Btw, I'm not considering compressibility for this case.

Laplacian ($\nabla^2 = \Delta$)

Let's make some tea. Initially you have your teacup with hot water. Now let's put a teabag inside. You will notice that the water starts getting darker, in other words the tea is diffusing.

$h=\nabla^2 C(x,y,z)$

Now your laplacian tells you how homogeneous your mixture is. The more homogeneous the smaller your laplacian becomes. Eventually, without stirring it all your tea will diffuse in your teacup.

Curl ($\mathbf{\nabla} \times$)

Let's stay with our teacup and start stirring it.

$\mathbf{S}=\mathbf{\nabla} \times \mathbf{V} $

The faster you stir your cup, the larger your curl (or vorticity) of your flow becomes.

Ref: "Theoretical Physics" by Joos.

Let ${ \vec{V}(\vec{r}) }$ be a smooth vector field.

Consider a nice closed surface ${ S . }$ Consider a nice directed loop ${ L . }$

We are often interested in the quantities

$${ {\begin{aligned} &\, \text{Flux:} \quad \quad \oint _{S} \vec{V} \cdot d\vec{A}, \\ &\, \text{Work:} \quad \quad \oint _{L} \vec{V} \cdot d\vec{\ell} . \end{aligned}} }$$

Q) Can we convert the surface and line integrals as "volume integrals"

$${ {\begin{aligned} &\, \oint _S \vec{V} \cdot d \vec{A} = \int _{\text{inside } S} (?) \, dV \\ &\, \oint _L \vec{V} \cdot d \vec{\ell} = \int _{\text{inside } L} (?) \, dA \end{aligned}} }$$

It turns out yes.

[Flux]

What is ${ \oint \vec{V} \cdot d\vec{A} }$ over a small cube

Note that

$${ {\begin{aligned} &\, \oint _{\text{small cube}} \vec{V} \cdot d\vec{A} \\ = &\, \left[ \left( V _y + \partial _y V _y \frac{\Delta y}{2} (\Delta x \Delta z) \right) - \left( V _y - \partial _y V _y \frac{\Delta y}{2} (\Delta x \Delta z) \right) \right] \\ &\, + \text{other terms} \\ = &\, \partial _x V _x \Delta x \Delta y \Delta z + \partial _y V _y \Delta x \Delta y \Delta z + \partial _z V _z \Delta x \Delta y \Delta z . \end{aligned}} }$$

Approximate the inside of ${ S }$ by a collection of axis aligned cubes.

Note that

$${ {\begin{aligned} &\, \oint _{\text{cubes}} \vec{V} \cdot d\vec{A} \\ = &\, \sum (\partial _x V _x + \partial _y V _y + \partial _z V _z ) \Delta x \Delta y \Delta z . \end{aligned}} }$$

Hence in the limit

$${ \boxed{\oint _S \vec{V} \cdot d\vec{A} = \int _{\text{inside } S} \text{div}(\vec{V}) \, dV } }$$

where

$${ \boxed{ \text{div}(\vec{V}) = \partial _x V _x + \partial _y V _y + \partial _z V _z } . }$$

[Line integral]

What is ${ \oint \vec{V} \cdot d\vec{\ell} }$ over a small rectangular loop

Note that

$${ {\begin{aligned} &\, \oint _{\text{small loop}} \vec{V} \cdot d \vec{\ell} \\ = &\, \left[ \vec{V} - {\begin{pmatrix} \sum \partial _i V _x \frac{(\Delta \vec{b}) _i}{2} \\ \vdots \\ \sum \partial _i V _z \frac{(\Delta \vec{b}) _i}{2} \end{pmatrix}} \right] \cdot \Delta \vec{a} + \left[ \vec{V} + {\begin{pmatrix} \sum \partial _i V _x \frac{(\Delta \vec{b}) _i}{2} \\ \vdots \\ \sum \partial _i V _z \frac{(\Delta \vec{b}) _i}{2} \end{pmatrix}} \right] \cdot (- \Delta \vec{a}) \\ &\, + \text{other terms} \\ = &\, - {{\begin{pmatrix} \sum \partial _i V _x (\Delta \vec{b}) _i \\ \vdots \\ \sum \partial _i V _z (\Delta \vec{b}) _i \end{pmatrix}}} \cdot \Delta \vec{a} + {{\begin{pmatrix} \sum \partial _i V _x (\Delta \vec{a}) _i \\ \vdots \\ \sum \partial _i V _z (\Delta \vec{a}) _i \end{pmatrix}}} \cdot \Delta \vec{b} \\ = &\, - \sum _{j} \sum _{i} \partial _i V _j (\Delta \vec{b}) _i (\Delta \vec{a} ) _j + \sum _{j} \sum _{i} \partial _{i} V _j (\Delta \vec{a}) _i (\Delta \vec{b}) _j \\ = &\, \sum _j \sum _i \partial _i V _j \left( (\Delta \vec{a}) _i (\Delta \vec{b}) _j - (\Delta \vec{b}) _i (\Delta \vec{a} ) _j \right) . \end{aligned}} }$$

Hence

$${ {\begin{aligned} &\, \oint _{\text{small loop}} \vec{V} \cdot d \vec{\ell} \\ = &\, \sum _j \sum _i \partial _i V _j \left( (\Delta \vec{a}) _i (\Delta \vec{b}) _j - (\Delta \vec{b}) _i (\Delta \vec{a} ) _j \right) . \end{aligned}} }$$

Note that for all distinct ${ \lbrace i, j \rbrace }$ we have the terms

$${ (\partial _i V _j - \partial _j V _i) \left( (\Delta \vec{a}) _i (\Delta \vec{b}) _j - (\Delta \vec{b}) _i (\Delta \vec{a} ) _j \right) . }$$

Hence

$${ {\begin{aligned} &\, \oint _{\text{small loop}} \vec{V} \cdot d \vec{\ell} \\ = &\, \sum _{\text{distinct } \lbrace i, j \rbrace} (\partial _i V _j - \partial _j V _i) \left( (\Delta \vec{a}) _i (\Delta \vec{b}) _j - (\Delta \vec{b}) _i (\Delta \vec{a} ) _j \right) \\ = &\, {\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \partial _1 &\partial _2 &\partial _3 \\ V _1 &V _2 &V _3 \end{vmatrix} } \cdot {\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ (\Delta \vec{a}) _1 &(\Delta \vec{a}) _2 &(\Delta \vec{a}) _3 \\ (\Delta \vec{b}) _1 &(\Delta \vec{b}) _2 &(\Delta \vec{b}) _3 \end{vmatrix} } . \end{aligned}} }$$

Hence

$${ \oint _{\text{small loop}} \vec{V} \cdot d \vec{\ell} = \text{curl}(\vec{V}) \cdot \Delta \vec{A} }$$

where

$${ \text{curl}(\vec{V}) = {\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \partial _1 &\partial _2 &\partial _3 \\ V _1 &V _2 &V _3 \end{vmatrix} } . }$$

Approximate the inside of ${ L }$ be a collection of edge aligned small loops.

Note that

$${ \oint _{\text{loops}} \vec{V} \cdot d\vec{\ell} = \sum \text{curl}(\vec{V}) \cdot \Delta \vec{A} . }$$

Hence in the limit

$${ \boxed{\oint _{L} \vec{V} \cdot d\vec{\ell} = \int _{\text{inside } L} \text{curl}(\vec{V}) \cdot d\vec{A} } }$$

where

$${ \boxed{ \text{curl}(\vec{V}) = {\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \partial _1 &\partial _2 &\partial _3 \\ V _1 &V _2 &V _3 \end{vmatrix} } }. }$$

[Summary]

Let ${ \vec{V}(\vec{r}) }$ be a smooth vector field.

Consider a nice closed surface ${ S . }$ Consider a nice directed loop ${ L . }$

Note that the flux

$${ \boxed{\oint _S \vec{V} \cdot d\vec{A} = \int _{\text{inside } S} \text{div}(\vec{V}) \, dV } }$$

where

$${ \boxed{ \text{div}(\vec{V}) = \partial _x V _x + \partial _y V _y + \partial _z V _z } . }$$

Note that the work

$${ \boxed{\oint _{L} \vec{V} \cdot d\vec{\ell} = \int _{\text{inside } L} \text{curl}(\vec{V}) \cdot d\vec{A} } }$$

where

$${ \boxed{ \text{curl}(\vec{V}) = {\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \partial _1 &\partial _2 &\partial _3 \\ V _1 &V _2 &V _3 \end{vmatrix} } }. }$$

[Nabla operator]

Let ${ \vec{V}(\vec{r}) }$ be a smooth vector field.

Note that

$${ {\begin{aligned} &\, \text{div}(\vec{V}) = \partial _1 V _1 + \partial _2 V _2 + \partial _3 V _3 \, , \\ &\, \text{curl}(\vec{V}) = {\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \partial _1 &\partial _2 &\partial _3 \\ V _1 &V _2 &V _3 \end{vmatrix} }. \end{aligned}} }$$

Hence formally defining the "operator vector"

$${ \boxed{\vec{\nabla} = {\begin{pmatrix} \partial _1 &\partial _2 &\partial _3 \end{pmatrix}} } }$$

we have

$${ \boxed{{\begin{aligned} &\, \text{div}(\vec{V}) = \vec{\nabla} \cdot \vec{V}, \\ &\, \text{curl}(\vec{V}) = \vec{\nabla} \times \vec{V} \end{aligned}}} . }$$