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Nabla Operator and its 3 Most Important Applications

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:

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:

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.

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.

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:

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

#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:

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

1. Form the derivative of the first component $$F_{\text x}(x,y,z)$$ to $$x$$.

2. Form the derivative of the second component $$F_{\text y}(x,y,z)$$ to $$y$$.

3. Form the derivative of the third component $$F_{\text z}(x,y,z)$$ to $$z$$.

4. 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!

#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:

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

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:

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:

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:

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 $$f$$:

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:

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:

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):

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:

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.

1. Distributivity in multiplication.

2. Distributivity with scalar product

3. Distributivity with cross product

4. Product rule for scalar functions

5. Product rule for scalar and vector function

6. Triple product with Nabla

7. Double cross product with Nabla

8. Gradient of a scalar product.

9. Rotation of a scaled vector field

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.

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise: Curl of the Curl of a Vector Field

Show that the following relationship between the Nabla operator and a vector field $$\boldsymbol{F}$$: $$\nabla ~\times~ \left(\nabla \times \boldsymbol{F}\right) ~=~ \nabla \, \left(\nabla ~\cdot~ \boldsymbol{F}\right) ~-~ \left(\nabla \cdot \nabla \right) \, \boldsymbol{F}$$

Hint: First, rewrite the two rotation operators in index notation using the Levi-Civita symbol. Then apply the identity for the product of two Levi-Civita symbols.

Solution to the Problem

Since this is a double cross product, this problem can be solved more easily in index notation! Here, we will use the Einstein summation convention.

First, rewrite the outer cross product using the Levi-Civita symbol $$\varepsilon_{ijk}$$: 1 $\boldsymbol{\hat{e}}_i \, \varepsilon_{ijk} \, \partial_j \, \left( \nabla \times \boldsymbol{F} \right)_k$

Then rewrite the second cross product inside the parenthesis: 2 $\boldsymbol{\hat{e}}_i \, \varepsilon_{ijk} \, \partial_j \, \varepsilon_{kmn} \, \partial_m \, F_n$

Now, rewrite the product of two Levi-Civita symbols using Kronecker delta: 3 $\varepsilon_{ijk} \, \varepsilon_{kmn} ~=~ \delta_{jn} \, \delta_{im} ~-~ \delta_{in} \, \delta_{jm}$

Substituting equation 3 into equation 2 yields: 4 $\boldsymbol{\hat{e}}_i \, \left( \delta_{jn} \, \delta_{im} ~-~ \delta_{in} \, \delta_{jm} \right) \, \partial_j \, \partial_m \, F_n$

Multiply out the bracket in 4 and use the rules of Kronecker delta: 5 $\boldsymbol{e}_m \, \partial_n \, \partial_m \, F_n ~-~ \boldsymbol{e}_n \, \partial_m \, \partial_m \, F_n$

Then rewrite equation 5 in vector notation (swap the factors to see better what can be combined): 6 $\boldsymbol{e}_m \, \partial_m \, \partial_n \, F_n ~-~ \partial_m \, \partial_m \, \boldsymbol{e}_n \, F_n$

Apply the definition of the dot product to $$\partial_n \, F_n$$ and $$\partial_m \, \partial_m$$, and rewrite the sums $$\boldsymbol{e}_m \, \partial_m$$ and $$\boldsymbol{e}_n \, F_n$$ in vectors. Then you will get exactly the relationship to be derived: $\nabla \, \left(\nabla \cdot \boldsymbol{F}\right) ~-~ \left(\nabla \cdot \nabla \right) \, \boldsymbol{F}$

Exercise #3: Calculate Magnetic Field from Vector Potential

Calculate the magnetic flux density $$\boldsymbol{B}(\boldsymbol{r}) ~=~ \nabla ~\times~ \boldsymbol{A}(\boldsymbol{r})$$ from the following vector potential $$\boldsymbol{A}(\boldsymbol{r})$$: $$\boldsymbol{A}(\boldsymbol{r}) ~=~ \frac{\boldsymbol{m} ~\times~ \boldsymbol{r}}{r^3}$$

Solution to Exercise #4

To calculate the magnetic flux density from the given vector potential, proceed as follows: 1 $\boldsymbol{B}(\boldsymbol{r}) ~=~ \nabla ~\times~ \boldsymbol{A}(\boldsymbol{r})$

First, substitute the given vector potential into Eq. 1: 2 $\boldsymbol{B}(\boldsymbol{r}) ~=~ \nabla ~\times~ \left( \frac{ \boldsymbol{m} ~\times~ \boldsymbol{r} }{ r^3 } \right)$

As seen in Eq. 2, it is a double cross product that you need to evaluate to obtain the magnetic flux density. You can handle this problem most elegantly using index notation. Therefore, rewrite the outer cross product in 2 using the Levi-Civita symbol $$\varepsilon_{ijk}$$, and represent the nabla operator $$\nabla$$ as a differential operator with an index $$\partial_j$$: 3 $\boldsymbol{B}(\boldsymbol{r}) ~=~ \boldsymbol{\hat{e}}_i \, \varepsilon_{ijk} \, \partial_j \, r^{-3} \, \left( \boldsymbol{m} ~\times~ \boldsymbol{r} \right)_k$

You can represent the magnitude $$r ~=~ (x_1 x_1 + x_2 x_2 + x_3 x_3)^{1/2}$$ of the position vector $$\boldsymbol{r}$$ as follows using the Einstein summation convention, where the index $$s$$ is summed over when it appears twice: 4 $r ~=~ (x_s \, x_s)^{1/2}$ thus, the term $$r^{-3}$$ in the vector potential becomes: 5 $r^{-3} ~=~ (x_s \, x_s)^{-3/2}$

If you express the other cross product in Eq. 3 using the Levi-Civita symbol $$\varepsilon_{klm}$$ and substitute Eq. 5, you get: 6 $\boldsymbol{B}(\boldsymbol{r}) ~=~ \boldsymbol{\hat{e}}_i \, \varepsilon_{ijk} \, \partial_j \, (x_s \, x_s)^{-3/2} \, \varepsilon_{klm} \, m_l \, x_m$

Now you have to apply the differential operator $$\partial_j$$ (derivative with respect to the spatial coordinate $$x_j$$), considering the product rule for derivatives. You can factor out the two Levi-Civita symbols and the unit vector because they do not depend on spatial coordinates and thus are constants from the perspective of the differential operator: 7 $\boldsymbol{B}(\boldsymbol{r}) ~=~ \boldsymbol{\hat{e}}_i \, \varepsilon_{ijk} \, \varepsilon_{klm} \left( m_l \, x_m \, \partial_j \, (x_s \, x_s)^{-3/2} ~+~ (x_s \, x_s)^{-3/2} \, \partial_j \, m_l \, x_m \right)$

The $$m_l$$ is also only a constant from the perspective of partial integration with respect to $$x_s$$. The derivative $$\partial_j \, x_m$$ equals 1 only if $$j = m$$, and the derivative is 0 if $$j \neq m$$. Therefore, you can represent the derivative using the Kronecker delta: $$\partial_j \, x_m ~=~ \delta_{jm}$$, which is either 1 or 0 depending on whether the indices are equal or not. With this consideration, Eq. 7 becomes: 8 $\boldsymbol{B}(\boldsymbol{r}) ~=~ \boldsymbol{\hat{e}}_i \, \varepsilon_{ijk} \, \varepsilon_{klm} \left( m_l \, x_m \, \partial_j \, (x_s \, x_s)^{-3/2} ~+~ (x_s \, x_s)^{-3/2} \, m_l \, \delta_{jm} \right)$

Multiply out the brackets and combine using Kronecker delta rule: $$\varepsilon_{ijk}\delta_{jm} = \varepsilon_{imk}$$. From the Exercises for Levi-Civita symbol, we know the following relation: $$\varepsilon_{imk}\varepsilon_{klm} = 2\delta_{il}$$. Use it: 9 $\boldsymbol{B}(\boldsymbol{r}) ~=~ \boldsymbol{\hat{e}}_i \, \varepsilon_{ijk} \, \varepsilon_{klm} m_l \, x_m \, \partial_j \, (x_s \, x_s)^{-3/2} ~+~ \boldsymbol{\hat{e}}_i \, 2\delta_{il} \, m_l \, (x_s \, x_s)^{-3/2}$

Then combine $$\delta_{il}\,m_l = m_i$$ and compute the derivative $$\partial_j \, (x_s \, x_s)^{-3/2} = -3x_j \, (x_s \, x_s)^{-5/2}$$. Next, apply the identity $$\varepsilon_{ijk}\,\varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{jl}\delta_{im}$$: 10 $\boldsymbol{B}(\boldsymbol{r}) ~=~ -3\boldsymbol{\hat{e}}_i \, (\delta_{il}\delta_{jm} - \delta_{jl}\delta_{im}) \, m_l \, x_j \, x_m \, (x_s \, x_s)^{-5/2} ~+~ 2\boldsymbol{\hat{e}}_i \, \, m_i \, (x_s \, x_s)^{-3/2}$

Multiply out the bracket and combine Kronecker deltas: 11 $\boldsymbol{B}(\boldsymbol{r}) ~=~ -3\boldsymbol{e}_l \, m_l \, x_m \, x_m \, (x_s \, x_s)^{-5/2} ~+~ 3\boldsymbol{e}_m \, m_j \, x_j \, x_m \, (x_s \, x_s)^{-5/2} ~+~ 2 \boldsymbol{\hat{e}}_i \, \, m_i \, (x_s \, x_s)^{-3/2}$

At this point, you can convert the indices back into vectors. With $$\boldsymbol{\hat{e}}_i \, m_i ~=~ \boldsymbol{m}$$ and $$x_m \, x_m = r^2$$, you have: 12 $\boldsymbol{B}(\boldsymbol{r}) ~=~ -3 \boldsymbol{m} r^2 \, r^{-5} ~+~ 3\boldsymbol{r}(\boldsymbol{m} \cdot \boldsymbol{r}) \, r^{-5} ~+~ 2 \boldsymbol{m} r^{-3}$

Because $$r^2 \cdot r^{-5} = r^{-3}$$, the last term cancels out with two of the three preceding terms. What remains is the expression for the magnetic flux density: 13 $\boldsymbol{B}(\boldsymbol{r}) ~=~ -3 \boldsymbol{r} \, \frac{\boldsymbol{m} \cdot \boldsymbol{r}}{r^5} ~-~ \frac{\boldsymbol{m}}{r^3}$