Divergence and curl
Before you start: you'll need the gradient and the \(\nabla\) operator.
A vector field \(\vec F = F_1\mathbf{i} + F_2\mathbf{j} + F_3\mathbf{k}\) attaches a vector to every point: the velocity of a fluid, or an electric or magnetic field. Divergence and curl are the two ways of differentiating it.
\[\operatorname{div}\vec F = \nabla\cdot\vec F = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\]
Curl (a vector)
\[\operatorname{curl}\vec F = \nabla\times\vec F = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ F_1 & F_2 & F_3 \end{vmatrix}\]
Expanded: \(\left(\dfrac{\partial F_3}{\partial y} - \dfrac{\partial F_2}{\partial z}\right)\mathbf{i} + \left(\dfrac{\partial F_1}{\partial z} - \dfrac{\partial F_3}{\partial x}\right)\mathbf{j} + \left(\dfrac{\partial F_2}{\partial x} - \dfrac{\partial F_1}{\partial y}\right)\mathbf{k}\)
What they mean
| Name | Condition | Meaning |
|---|---|---|
| Solenoidal | \(\nabla\cdot\vec F = 0\) | no sources or sinks; what flows in flows out |
| Irrotational | \(\nabla\times\vec F = \vec 0\) | no swirling; the field has a scalar potential |
Scalar potential
If \(\nabla\times\vec F = \vec 0\), there is a scalar function \(\phi\), the scalar potential, with \[\vec F = \nabla\phi\]
How to find \(\phi\)
- Check \(\nabla\times\vec F = \vec 0\) first. If not, no potential exists.
- Integrate \(F_1\) with respect to \(x\), treating \(y\) and \(z\) as constants.
- Add the terms of \(\displaystyle\int F_2\,dy\) that don't contain \(x\), and the terms of \(\displaystyle\int F_3\,dz\) that contain neither \(x\) nor \(y\).
- Check by differentiating: \(\nabla\phi\) must give back \(\vec F\). Add a constant \(c\).
Identities worth knowing
- \(\nabla\times(\nabla\phi) = \vec 0\): every gradient field is irrotational.
- \(\nabla\cdot(\nabla\times\vec F) = 0\): every curl is solenoidal.
- \(\nabla\cdot\vec r = 3\), \(\;\nabla\times\vec r = \vec 0\), \(\;\nabla\cdot(r^n\vec r) = (n + 3)\,r^n\).
Solved examples
Example 1: Divergence and curl
EasyFind \(\nabla\cdot\vec F\) and \(\nabla\times\vec F\) for \(\vec F = x^2y\,\mathbf{i} + xz\,\mathbf{j} + 2yz\,\mathbf{k}\).
- \(\nabla\cdot\vec F = 2xy + 0 + 2y\).
- \(\mathbf{i}\): \(\dfrac{\partial(2yz)}{\partial y} - \dfrac{\partial(xz)}{\partial z} = 2z - x\).
- \(\mathbf{j}\): \(\dfrac{\partial(x^2y)}{\partial z} - \dfrac{\partial(2yz)}{\partial x} = 0\). \(\;\mathbf{k}\): \(\dfrac{\partial(xz)}{\partial x} - \dfrac{\partial(x^2y)}{\partial y} = z - x^2\).
Answer\(\nabla\cdot\vec F = 2xy + 2y\), \(\;\nabla\times\vec F = (2z - x)\,\mathbf{i} + (z - x^2)\,\mathbf{k}\)
Example 2: Making a field solenoidal
EasyFind \(a\) so that \(\vec F = (x + 3y)\,\mathbf{i} + (y - 2z)\,\mathbf{j} + (x + az)\,\mathbf{k}\) is solenoidal.
- \(\nabla\cdot\vec F = 1 + 1 + a = 0\).
Answer\(a = -2\)
Example 3: Irrotational, with its potential
MediumShow that \(\vec F = (y + z)\,\mathbf{i} + (z + x)\,\mathbf{j} + (x + y)\,\mathbf{k}\) is irrotational, and find its scalar potential.
- \(\nabla\times\vec F = (1 - 1)\,\mathbf{i} + (1 - 1)\,\mathbf{j} + (1 - 1)\,\mathbf{k} = \vec 0\).
- \(\displaystyle\int (y + z)\,dx = xy + xz\). From \(\displaystyle\int (z + x)\,dy\), the new term without \(x\) is \(yz\). Nothing new comes from \(F_3\).
Answer\(\phi = xy + yz + zx + c\)
Example 4: A classic potential
Exam levelShow that \(\vec F = (6xy + z^3)\,\mathbf{i} + (3x^2 - z)\,\mathbf{j} + (3xz^2 - y)\,\mathbf{k}\) is irrotational and find \(\phi\) with \(\vec F = \nabla\phi\).
- \(\mathbf{i}\): \(-1 - (-1) = 0\). \(\;\mathbf{j}\): \(3z^2 - 3z^2 = 0\). \(\;\mathbf{k}\): \(6x - 6x = 0\). So \(\nabla\times\vec F = \vec 0\).
- \(\displaystyle\int F_1\,dx = 3x^2y + xz^3\). Its \(y\)-derivative is \(3x^2\), but \(F_2 = 3x^2 - z\), so add \(-yz\).
- Check: \(\dfrac{\partial}{\partial z}(3x^2y + xz^3 - yz) = 3xz^2 - y = F_3\). ✓
Answer\(\phi = 3x^2y + xz^3 - yz + c\)
Example 5: Finding constants for an irrotational field
Exam levelFind \(a, b, c\) so that \(\vec F = (x + 2y + az)\,\mathbf{i} + (bx - 3y - z)\,\mathbf{j} + (4x + cy + 2z)\,\mathbf{k}\) is irrotational, and find its potential.
- \(\mathbf{i}\): \(c - (-1) = 0\), so \(c = -1\). \(\;\mathbf{j}\): \(a - 4 = 0\), so \(a = 4\). \(\;\mathbf{k}\): \(b - 2 = 0\), so \(b = 2\).
- Potential: \(\displaystyle\int (x + 2y + 4z)\,dx = \frac{x^2}{2} + 2xy + 4xz\); add \(-\dfrac{3y^2}{2} - yz\) from \(F_2\), and \(z^2\) from \(F_3\).
Answer\(a = 4,\; b = 2,\; c = -1\); \(\;\phi = \dfrac{x^2}{2} - \dfrac{3y^2}{2} + z^2 + 2xy + 4xz - yz + c'\)
Example 6: A result about r
Exam levelShow that \(\nabla\cdot(r^n\vec r) = (n + 3)\,r^n\). Hence show that \(\dfrac{\vec r}{r^3}\) is solenoidal.
- Product rule: \(\nabla\cdot(r^n\vec r) = \nabla(r^n)\cdot\vec r + r^n\,\nabla\cdot\vec r\).
- \(\nabla r^n = n\,r^{n-2}\vec r\) (see gradient) and \(\nabla\cdot\vec r = 3\). So \(= n\,r^{n-2}\,r^2 + 3r^n = (n + 3)r^n\).
- With \(n = -3\): \(\nabla\cdot\dfrac{\vec r}{r^3} = 0\).
Answer\(\dfrac{\vec r}{r^3}\) is solenoidal: this is the inverse-square field of gravity and electrostatics.
For curl, use the cyclic pattern \(x \to y \to z \to x\). The \(\mathbf{i}\) part is \(\dfrac{\partial F_3}{\partial y} - \dfrac{\partial F_2}{\partial z}\); shift every letter one step to get the \(\mathbf{j}\) part, and again for \(\mathbf{k}\). No determinant expansion needed.
Common mistakes
1. Mixing up which is a scalar
Divergence is a scalar (a dot product); curl is a vector (a cross product).
2. Getting the middle component of curl the wrong way round
The \(\mathbf{j}\) component is \(\dfrac{\partial F_1}{\partial z} - \dfrac{\partial F_3}{\partial x}\). Expanding the determinant, students often forget its minus sign.
3. Looking for a potential before checking the curl
If \(\nabla\times\vec F \ne \vec 0\), no scalar potential exists, however you integrate.
4. Counting a term twice in the potential
Only add the terms from \(F_2\) and \(F_3\) that haven't already appeared. Always check by differentiating.
How it's asked in exams
- Find \(\nabla\cdot\vec F\) and \(\nabla\times\vec F\) at a point.
- Show a field is solenoidal or irrotational, or find constants that make it so.
- Find the scalar potential of an irrotational field.
- Prove identities such as \(\nabla\cdot(r^n\vec r) = (n + 3)r^n\) or \(\nabla\times(\nabla\phi) = \vec 0\).
Practice questions
Try each one on paper before you open the answer.
Q1Find \(\nabla\cdot\vec r\) and \(\nabla\times\vec r\), where \(\vec r = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\).
\(\nabla\cdot\vec r = 3\), \(\;\nabla\times\vec r = \vec 0\).
Q2Find \(\nabla\cdot\vec F\) at \((1, -1, 1)\) for \(\vec F = x^2z\,\mathbf{i} - 2y^3z^2\,\mathbf{j} + xy^2z\,\mathbf{k}\).
\(\nabla\cdot\vec F = 2xz - 6y^2z^2 + xy^2 = 2 - 6 + 1 = -3\).
Q3Find \(\nabla\times\vec F\) for \(\vec F = xy\,\mathbf{i} + yz\,\mathbf{j} + zx\,\mathbf{k}\).
\(-y\,\mathbf{i} - z\,\mathbf{j} - x\,\mathbf{k}\).
Q4Show that \(\vec F = (2xy + z^3)\,\mathbf{i} + x^2\,\mathbf{j} + 3xz^2\,\mathbf{k}\) is irrotational and find its potential.
Curl components: \(0 - 0\), \(3z^2 - 3z^2\), \(2x - 2x\), all zero. \(\phi = x^2y + xz^3 + c\).
Q5Find \(a\) so that \(\vec F = (axy - z^3)\,\mathbf{i} + (a - 2)x^2\,\mathbf{j} + (1 - a)xz^2\,\mathbf{k}\) is irrotational.
\(\mathbf{j}\): \(-3z^2 - (1 - a)z^2 = (a - 4)z^2\). \(\mathbf{k}\): \(2(a - 2)x - ax = (a - 4)x\). Both vanish when \(a = 4\).
Q6Show that \(\nabla\times(\nabla\phi) = \vec 0\) for every scalar field \(\phi\).
The \(\mathbf{i}\) component is \(\phi_{zy} - \phi_{yz} = 0\), since mixed partials are equal; likewise for \(\mathbf{j}\) and \(\mathbf{k}\).
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Frequently asked questions
What is the physical meaning of divergence?
The net rate at which the field flows out of a tiny volume around a point. Positive means a source, negative means a sink, and zero means as much flows in as out.
What is the physical meaning of curl?
How much the field rotates around a point. For a rigidly rotating body, the curl of the velocity is twice the angular velocity.
What is a solenoidal field?
A field with zero divergence everywhere, such as a magnetic field.
What is an irrotational field?
A field with zero curl. Such a field is the gradient of a scalar potential, such as a gravitational or electrostatic field.
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.