Nabla Operator and its 3 Most Important Applications

by Alexander FufaeV

Table of contents

3 basic ways to apply Nabla to functions Here you will learn how to form the gradient, divergence and rotation of a function using the Nabla operator.
Apply Nabla to Nabla Here you apply the Nabla operator to Nabla operator using the scalar and cross product.
5 useful ways to apply Nabla twice Here you will learn how to get divergence of the gradient, divergence of the rotation and so on when the Nabla operator is applied twice to a function.
The 9 most useful rules with Nabla Here you will learn some important mathematical rules that you can use to simplify or rewrite expressions with Nabla.

Nabla operator $\nabla$ is notationally similar to a vector and looks like this in the three-dimensional case when we express it with Cartesian coordinates:

Nabla operator in 3d

$$ \begin{align} \nabla ~=~ \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} \end{align} $$

The three components of the Nabla operator are partial derivatives with respect to $x$, $y$ or $z$. Derivatives that occur alone are called differential operators. You can apply a differential operator to a function. The result is the derivative of the function.

The Nabla operator unfolds its effect only if it is applied to a scalar or vector function. The result is a multidimensional derivative of the function.

3 basic ways to apply Nabla to functions

The Nabla operator can be applied to scalar functions $ f(x,y,z) $ as well as to vector functions. A 3d vector function has three components:

$$ \begin{align} \boldsymbol{F}(x,y,z) ~=~ \begin{bmatrix}F_{1}(x,y,z)\\ F_{2}(x,y,z)\\F_{3}(x,y,z)\end{bmatrix} \end{align} $$

The components of a vector function are scalar functions like $ f(x,y,z) $. So you can think of a scalar function as a 1d vector function that has exactly one component. If a vector function, as in Eq. 2, depends on the space coordinates $ x $, $y$, and $z$, then we call it a vector field. A component of the vector field, such as the first component $ F_{1}(x,y,z) $, is also sometimes written down with index '$\text{x}$' instead of index '1' to indicate that it is the component of the vector field that points along the $x$ direction: $ F_{\text{x}}(x,y,z) $.

So a vector field is nothing else than a vector $ \boldsymbol{F} $ which can change its length and direction depending on its position in space.

You can manipulate a vector in different ways:

Scalar multiplication - you can multiply a vector by a real number, let's call it $ a \in \mathbb{R} $: $ \boldsymbol{F} \, a $. For example, the number $ a $ could be the scalar function $ f $ and $ \boldsymbol{F} $ could be the Nabla operator.
You can form a scalar product of the vector with another vector $ \boldsymbol{R} $: $ \boldsymbol{R} \cdot \boldsymbol{F} $. For example, the nabla operator could be this vector $ \boldsymbol{R} $.
You can form a cross product of the vector with another vector $ \boldsymbol{R} $: $ \boldsymbol{R} \times \boldsymbol{F} $. For example, the nabla operator could be this vector $ \boldsymbol{R} $.

#1: Scalar multiplication with Nabla

Let's calculate what we would get if we apply the Nabla operator $ \nabla $ to a scalar function $ f(x,y,z) $ that depends on three variables.

Scalar multiplication with Nabla = Gradient

$$ \begin{align} \nabla \, f(x,y,z) ~=~ \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} \, f(x,y,z) ~=~ \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z} \end{bmatrix} \end{align} $$

In the second step, we just used the definition of the Nabla operator, and in the last step, we pulled the scalar function (a number, so to speak) into the vector and omitted the dependence of $x$, $y$, and $z$ to make the result more compact.

The result of Nabla applied to $ f $ is called a gradient and obviously represents a three-dimensional vector field with three components:

The 1st component contains the slope $ \frac{\partial f}{\partial x} $ in $ x $ direction.
The 2nd component contains the slope $ \frac{\partial f}{\partial y} $ in $ y $ direction.
The 3rd component contains the slope $ \frac{\partial f}{\partial z} $ in $ z $ direction.

Note: Such scalar multiplication as in 3 is not commutative, so mathematically speaking, the nabla operator is not a proper vector.

What is the gradient of a function?

If you apply the Nabla operator $ \nabla $ to a scalar function $ f $, then the resulting vector $ \nabla \, f $ is called the gradient of $ f $.

Of course, you may also use a two-dimensional Nabla operator, which has only two (and not three) components. You could then calculate a two-dimensional gradient as follows:

$$ \begin{align} \nabla \, f(x,y) ~=~ \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix} \end{align} $$

And the Nabla operator in one dimension is simply a partial derivative $ \frac{\partial f}{\partial x} $.

Example: Calculate gradient of a function

Given a scalar function $ f(x,y,z) = x^2 + 5xy + z $. Apply the Nabla operator - as explained above - to $ f $:

$$ \begin{align} \nabla \, f(x,y,z) &~=~ \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z} \end{bmatrix} \\\\
&~=~ \begin{bmatrix} 2x+5y \\ 5x \\ 1 \end{bmatrix} \end{align} $$

#2: Scalar product with Nabla

Another way to combine the Nabla operator this time with a vector field $\boldsymbol{F}(x,y,z)$ as in 2 is to form a scalar product:

Scalar product with Nabla = divergence

$$ \begin{align} \nabla \cdot \boldsymbol{F} &~=~ \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} ~\cdot~ \begin{bmatrix}F_{\text x}(x,y,z)\\ F_{\text y}(x,y,z)\\F_{\text z}(x,y,z)\end{bmatrix} \\\\
&~=~ \frac{\partial F_{\text x}}{\partial x} + \frac{\partial F_{\text y}}{\partial y} + \frac{\partial F_{\text z}}{\partial z} \end{align} $$

In the scalar product, you apply the derivatives to the scalar functions (vector field components) componentwise:

Form the derivative of the first component $ F_{\text x}(x,y,z) $ to $x$.
Form the derivative of the second component $ F_{\text y}(x,y,z) $ to $y$.
Form the derivative of the third component $ F_{\text z}(x,y,z) $ to $z$.
Add the three derivatives together.

The result of the scalar product of Nabla with $ \boldsymbol{F} $ is called divergence and represents a three-dimensional scalar function $ f(x,y,z) $. In the case of gradient, a vector function $ \boldsymbol{F} $ was obtained from a scalar function $f$. In the case case of divergence, we make a scalar function out of a vector function. So exactly the other way around!

What is the divergence of a vector field?

If you apply the nabla operator $ \nabla $ to a vector function $\boldsymbol{F}$ using the scalar product, then the result $ \nabla \cdot \boldsymbol{F} $ is called divergence of $\boldsymbol{F}$.

Example: Calculate divergence of a vector field

The following three-dimensional vector field is given:

$$ \begin{align} \boldsymbol{F}(x,y,z) ~=~ \begin{bmatrix} 2x^3 \\ zy \\ 5xy \end{bmatrix} \end{align} $$

Form the scalar product of the Nabla operator with the vector field $ \boldsymbol{F} $:

$$ \begin{align} \nabla \cdot \boldsymbol{F} &~=~ \frac{\partial F_{\text x}}{\partial x} + \frac{\partial F_{\text y}}{\partial y} + \frac{\partial F_{\text z}}{\partial z} \\\\
&~=~ 6x^2 + z \end{align} $$

#3: Cross product with Nabla

As with the scalar product 6, you apply the nabla operator to a vector function $ \boldsymbol{F}(x,y,z) $ to form the cross product:

Cross product with Nabla = rotation

$$ \begin{align} \nabla \times \boldsymbol{F} &~=~ \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} ~\times~ \begin{bmatrix}F_{x}(x,y,z)\\ F_{y}(x,y,z)\\F_{z}(x,y,z)\end{bmatrix} \\\\
&~=~ \begin{bmatrix} \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \\ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \\ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \end{bmatrix} \end{align} $$

The result of the cross product 9 is a vector again! Nabla leaves here the vector function $\boldsymbol{F}$ as vector function.

What is the rotation of a vector field?

If you apply the nabla operator $\nabla$ to a vector function $\boldsymbol{F}$ using the cross product, the resulting vector $\nabla \times \boldsymbol{F} $ is called the rotation of $\boldsymbol{F}$.

Example: Calculate rotation of a vector field

Consider again the vector field as in Eq. 7:

$$ \begin{align} \boldsymbol{F}(x,y,z) ~=~ \begin{bmatrix} 2x^3 \\ zy \\ 5xy \end{bmatrix} \end{align} $$

Apply the nabla operator to $ \boldsymbol{F} $ by forming the cross product:

$$ \begin{align} \nabla \times \boldsymbol{F} ~=~ \begin{bmatrix} \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \\ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \\ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \end{bmatrix} ~=~ \begin{bmatrix} 5x - y \\ - 5y \\ 0 \end{bmatrix} \end{align} $$

Apply Nabla to Nabla

Of course you can also form the scalar and cross product of Nabla with Nabla. The scalar product gives a new operator $ \nabla \cdot \nabla $, which we call the Laplace operator:

Laplace operator: scalar product of two Nabla operators

$$ \begin{align} \nabla \cdot \nabla &~=~ \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} ~\cdot~ \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} \\\\
&~=~ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \\\\
&~=:~ \nabla^2 \end{align} $$

The Laplace operator is the sum of the second derivatives with respect to $x$, $y$ and $z$. The scalar product $ \nabla \cdot \nabla $ is written as $ \nabla^2 $ for short (sometimes also as $\Delta$).

The cross product of two Nabla operators is not interesting because it always gives the zero vector. Why? Because the partial derivatives are commutative and thus cancel each other out inside the Nabla-Nabla cross product:

$$ \begin{align} \nabla \times \nabla ~=~ \begin{bmatrix} \frac{\partial}{\partial z}\frac{\partial}{\partial y} - \frac{\partial}{\partial y}\frac{\partial}{\partial z} \\ \frac{\partial}{\partial x}\frac{\partial}{\partial z} - \frac{\partial}{\partial z}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y}\frac{\partial}{\partial x} - \frac{\partial}{\partial x}\frac{\partial}{\partial y} \end{bmatrix} ~=~ \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \end{align} $$

So if you apply the Nabla cross product 12 to any vector function: $ (\nabla \times \nabla) ~\cdot~ \boldsymbol{F} = \boldsymbol{0} $ or $ (\nabla \times \nabla) ~\times~ \boldsymbol{F} = \boldsymbol{0}$ then you always get the zero vector.

Note! Associativity does not apply:

$$ \begin{align} (\nabla \times \nabla) ~\cdot~ \boldsymbol{F} ~\neq~ \nabla \times (\nabla ~\cdot~ \boldsymbol{F}) \\\\
0 ~=~ (\nabla \times \nabla) ~\times~ f ~\neq~ \nabla \times (\nabla \times f) \end{align} $$

5 useful ways to apply Nabla twice

In electrodynamics, fluid dynamics, and other areas of physics, there are equations in which Nabla operator is applied twice to a vector field or scalar field. There are exactly five useful different possibilities to apply Nabla twice:

1st possibility: divergence of the gradient

If you apply the Laplace operator 11 to a scalar function $ f(x,y,z) $, then you get the divergence of the gradient of $ f$:

$$ \begin{align} \nabla^2 \, f &~=~ \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \, f \\\\
&~=~ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \end{align} $$

By the way, associativity applies here, so parentheses are redundant: $ (\nabla \cdot \nabla) \, f ~=~ \nabla \cdot (\nabla \, f) ~=~ \nabla \cdot \nabla \, f$.

2nd possibility: divergence of rotation

For this you need of course a vector function $ \boldsymbol{F} $, because the rotation is defined only for a vector function. You have already calculated the rotation of $ \boldsymbol{F} $ in 9. FJust form the scalar product with the result 9 of the rotation:

$$ \begin{align} \nabla \cdot (\nabla \times \boldsymbol{F}) &~=~ \frac{\partial}{\partial x}\left( \frac{\partial F_z}{\partial y} ~-~ \frac{\partial F_y}{\partial z} \right) + \frac{\partial}{\partial y}\left( \frac{\partial F_x}{\partial z} ~-~ \frac{\partial F_z}{\partial x} \right) \\\\
&~+~ \frac{\partial}{\partial z}\left( \frac{\partial F_y}{\partial x} ~-~ \frac{\partial F_x}{\partial y} \right) ~=~ 0 \end{align} $$

As you can see from the result: Divergence of rotation is always zero. A physical example is the magnetic field $ \boldsymbol{B} = \nabla \times \boldsymbol{A} $ (with $ \boldsymbol{A} $ as vector potential). Divergence of the magnetic field vanishes: $ \nabla \cdot \boldsymbol{B} = 0 $.

3rd possibility: rotation of the gradient

Here you first form the gradient of a scalar function $ f $ as in 3 and then you form the cross product of the Nabla operator with the resulting vector $ \nabla \, f $ as in 9:

$$ \begin{align} \nabla \times (\nabla \, f) &~=~ \nabla \times \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z} \end{bmatrix} \\\\
&~=~ \begin{bmatrix} \frac{\partial}{\partial y}\frac{\partial f}{\partial z} - \frac{\partial}{\partial z}\frac{\partial f}{\partial y} \\ \frac{\partial}{\partial z}\frac{\partial f}{\partial x} - \frac{\partial}{\partial x}\frac{\partial f}{\partial z} \\ \frac{\partial}{\partial x}\frac{\partial f}{\partial y} - \frac{\partial}{\partial y}\frac{\partial f}{\partial x} \end{bmatrix} ~=~ 0 \end{align} $$

The rotation of the gradient is always zero. A physical example: The electrostatic field can be written as a gradient of a scalar potential $V$: $ \boldsymbol{E} = \nabla \, V $. Thanks to our result we can conclude that the rotation of the E-field vanishes: $ \nabla \times \boldsymbol{E} = 0 $. We say: Electrostatic E-fields are therefore vortex-free!

4th possibility: rotation of the rotation

You can apply the Nabla operator twice as a cross product to a vector function (see Eq. 8 for how to do this):

$$ \begin{align} \nabla \times (\nabla \times \boldsymbol{F}) &~=~ \nabla ~\times~ \begin{bmatrix} \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \\ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \\ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \end{bmatrix} \\\\
&~=~ \begin{bmatrix} \frac{\partial}{\partial y}\frac{\partial F_y}{\partial x} - \frac{\partial^2 F_x}{\partial y^2} - \frac{\partial^2 F_x}{\partial z^2} + \frac{\partial}{\partial z}\frac{\partial F_z}{\partial x} \\ \frac{\partial}{\partial z}\frac{\partial F_z}{\partial y} - \frac{\partial^2 F_y}{\partial z^2} - \frac{\partial^2 F_y}{\partial x^2} + \frac{\partial}{\partial x}\frac{\partial F_x}{\partial y} \\ \frac{\partial}{\partial x}\frac{\partial F_x}{\partial z} - \frac{\partial^2 F_z}{\partial x^2} - \frac{\partial^2 F_z}{\partial y^2} + \frac{\partial}{\partial y}\frac{\partial F_y}{\partial z} \end{bmatrix} \end{align} $$

5th possibility: gradient of divergence

The last useful way is to form the divergence $ \nabla \cdot \boldsymbol{F} $ first. The result is a scalar function. And then apply the nabla operator to this scalar function:

$$ \begin{align} \nabla \left( \nabla \cdot \boldsymbol{F} \right) &~=~ \nabla \, \left( \frac{\partial F_{\text x}}{\partial x} + \frac{\partial F_{\text y}}{\partial y} + \frac{\partial F_{\text z}}{\partial z} \right) \\\\
&~=~ \begin{bmatrix} \frac{\partial^2 F_x}{\partial x^2} + \frac{\partial }{\partial x}\frac{\partial F_y}{\partial y} + \frac{\partial }{\partial x}\frac{\partial F_z}{\partial z} \\ \frac{\partial }{\partial y}\frac{\partial F_x}{\partial x} + \frac{\partial^2 F_y}{\partial y^2} + \frac{\partial }{\partial y}\frac{\partial F_z}{\partial z} \\ \frac{\partial }{\partial z}\frac{\partial F_x}{\partial x} + \frac{\partial }{\partial z}\frac{\partial F_y}{\partial y} + \frac{\partial^2 F_z}{\partial z^2} \end{bmatrix} \end{align} $$

All the other conceivable cases: "Rotation of divergence", "Gradient of rotation", "Divergence of divergence" and "Gradient of gradient" are not defined and do not occur in physics at all.

The 9 most useful rules with Nabla

With the knowledge you have just acquired, you can derive the following mathematical rules with the Nabla operator. You will encounter these rules in electrodynamics, for example, because they can be used to simplify and to transform certain expressions.

Distributivity in multiplication.

Rule: Distributivity for multiplication
Formula anchor $$ \begin{align} \nabla \, (f + g ) ~=~ \nabla f ~+~ \nabla g \end{align} $$
Distributivity with scalar product

Rule: Distributivity with scalar product
Formula anchor $$ \begin{align} \nabla \cdot (\boldsymbol{F} + \boldsymbol{R} ) ~=~ \nabla \cdot \boldsymbol{F} ~+~ \nabla \cdot \boldsymbol{R} \end{align} $$
Distributivity with cross product

Rule: Distributivity with cross product
Formula anchor $$ \begin{align} \nabla \times (\boldsymbol{F} + \boldsymbol{R} ) ~=~ \nabla \times \boldsymbol{F} ~+~ \nabla \times \boldsymbol{R} \end{align} $$
Product rule for scalar functions

Rule: Product rule for scalar functions
Formula anchor $$ \begin{align} \nabla (f \, g) ~=~ f \, \nabla g ~+~ g \nabla f \end{align} $$
Product rule for scalar and vector function

Rule: Product rule for scalar and vector function
Formula anchor $$ \begin{align} \nabla \, \cdot \, \left( f \, \boldsymbol{R} \right) ~=~ f \, \left( \nabla \cdot \boldsymbol{R} \right) ~+~ \boldsymbol{R} \cdot \left( \nabla f \right) \end{align} $$
Triple product with Nabla

Rule: Triple product with Nabla
Formula anchor $$ \begin{align} \nabla \cdot \left( \boldsymbol{F} \times \boldsymbol{R} \right) ~=~ \boldsymbol{R} \cdot \left( \nabla \times \boldsymbol{F} \right) ~-~ \boldsymbol{F} \cdot \left( \nabla \times \boldsymbol{R} \right) \end{align} $$
Double cross product with Nabla

Rule: Double cross product with Nabla
Formula anchor $$ \begin{align} \nabla \times \left( \boldsymbol{F} \times \boldsymbol{R} \right) &~=~ \left( \boldsymbol{R} \cdot \nabla \right) \, \boldsymbol{F} ~-~ \left( \boldsymbol{F} \cdot \nabla \right) \, \boldsymbol{R} \\\\
&~+~ \boldsymbol{F} \, \left( \nabla \cdot \boldsymbol{R} \right) ~-~ \boldsymbol{R} \, \left( \nabla \cdot \boldsymbol{F} \right) \end{align} $$
Gradient of a scalar product.

Rule: Divergence of a scalar product
Formula anchor $$ \begin{align} \nabla \, \left( \boldsymbol{F} \cdot \boldsymbol{R} \right) &~=~ \boldsymbol{F} \times \left( \nabla \times \boldsymbol{R} \right) ~+~ \boldsymbol{R} \times \left( \nabla \times \boldsymbol{F} \right) \\\\
&~+~ \left( \boldsymbol{R} \cdot \nabla \right) \, \boldsymbol{F} ~+~ \left( \boldsymbol{F} \cdot \nabla \right) \, \boldsymbol{R} \end{align} $$
Rotation of a scaled vector field

Rule: Rotation of a scaled vector field
Formula anchor $$ \begin{align} \nabla \times \left( f \, \boldsymbol{R} \right) ~=~ f \, \left( \nabla \times \boldsymbol{R} \right) ~-~ \boldsymbol{R} \times \left( \nabla f \right) \end{align} $$

Swap nabla with function?

For example, if you swap the function $ \boldsymbol{F} $ with Nabla in the scalar product 6, you get a new operator:

$$ \begin{align} \boldsymbol{F} \cdot \nabla ~=~ F_x\frac{\partial}{\partial x} + F_y\frac{\partial }{\partial y} + F_z\frac{\partial }{\partial z} \end{align} $$

Just like the nabla operator, you can apply the operator 27 to any function $ f $: $ (\boldsymbol{F} \cdot \nabla)\,f$.

Now you have learned what the nabla operator is and how to apply it to scalar and vector functions. In the next lesson, we'll take a closer look at the gradient $ \nabla \, f $ and learn about the directional derivative.