Divergence, curl, and Laplacian

A vector field can behave in two very different local ways: it can spread outward from a point like water gushing from a tap, or it can swirl around a point like water circling a drain. Divergence is the scalar measurement of the first behaviour, and curl is the vector measurement of the second. Both are built from the same operator, ∇=(∂∂x,∂∂y,∂∂z)\nabla=\left(\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z}\right), applied in two different ways to a vector field F⃗\vec F. A third quantity, the Laplacian, closes the loop back to scalar fields: it is the divergence of a gradient. Together these three operators — and the two identities that link them, ∇×(∇f)=0\nabla\times(\nabla f)=\mathbf 0 and ∇⋅(∇×F⃗)=0\nabla\cdot(\nabla\times\vec F)=0 — form the working vocabulary of vector calculus that every later topic in MA10002, from conservative fields to the integral theorems, is written in.

The core insight

Divergence is ∇⋅F⃗\nabla\cdot\vec F measuring outward spreading as a scalar, curl is ∇×F⃗\nabla\times\vec F measuring local rotation as a vector, and the Laplacian ∇2f=∇⋅(∇f)\nabla^2 f=\nabla\cdot(\nabla f) ties both back to scalar fields, with gradients always curl-free and curls always divergence-free.

The core idea

For F⃗(x,y,z)=F1ı^+F2ȷ^+F3k^\vec F(x,y,z)=F_1\hat\imath+F_2\hat\jmath+F_3\hat k, the divergence is the scalar

∇⋅F⃗=∂F1∂x+∂F2∂y+∂F3∂z,\nabla\cdot\vec F=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z},

where: F1,F2,F3F_1,F_2,F_3 are the components of F⃗\vec F along ı^,ȷ^,k^\hat\imath,\hat\jmath,\hat k. It measures net outward flux per unit volume: positive at a source, negative at a sink, and zero everywhere for a solenoidal field. Think of a fluid velocity field: at a point where the fluid is being pumped in from outside (a tiny source), the divergence there is positive; at a drain, it is negative; where the fluid is neither created nor destroyed, it is zero. This is the physical picture behind the word "divergence" — it literally reports how much the field diverges away from, or converges into, the point. The curl is the vector obtained from the same three partial derivatives combined the other way,

∇×F⃗=(∂F3∂y−∂F2∂z)ı^+(∂F1∂z−∂F3∂x)ȷ^+(∂F2∂x−∂F1∂y)k^,\nabla\times\vec F=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)\hat\imath+\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)\hat\jmath+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\hat k,

where: each bracket is a difference of two partial derivatives, arranged in a cyclic pattern that is easiest to remember through the determinant ∣ı^ȷ^k^∂/∂x∂/∂y∂/∂zF1F2F3∣\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\ \partial/\partial x&\partial/\partial y&\partial/\partial z\\ F_1&F_2&F_3\end{vmatrix}. A field with ∇×F⃗=0\nabla\times\vec F=\mathbf 0 everywhere is irrotational: no tiny paddle wheel placed in the flow would ever spin. The Laplacian of a scalar field ff is the divergence of its own gradient,

∇2f=∇⋅(∇f)=∂2f∂x2+∂2f∂y2+∂2f∂z2,\nabla^2 f=\nabla\cdot(\nabla f)=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2},

where: the three terms are unmixed second partial derivatives, not the squared magnitude of the gradient. Two identities hold whenever the mixed partials of the underlying function commute (Clairaut's theorem): ∇×(∇f)=0\nabla\times(\nabla f)=\mathbf 0, so gradients are always irrotational, and ∇⋅(∇×F⃗)=0\nabla\cdot(\nabla\times\vec F)=0, so curls are always solenoidal.

Figure
Two flow patterns side by side, one spreading outward from a source point to show positive divergence and one circling around a centre to show nonzero curl
Divergence: spreading out source point, div F greater than 0 Curl: swirling around paddle wheel spins, curl F not 0
Derivation AWhy the curl of a curl-source vector field always has zero divergence

Show that ∇⋅(∇×F⃗)=0\nabla\cdot(\nabla\times\vec F)=0 for every vector field F⃗\vec F whose components have continuous second partial derivatives, by writing out both operations in full and cancelling terms two at a time using Clairaut's theorem that mixed partial derivatives taken in either order agree, which is exactly the identity used later to test whether a divergence-free field can be written as a curl.

Step 1. Write F⃗=F1ı^+F2ȷ^+F3k^\vec F=F_1\hat\imath+F_2\hat\jmath+F_3\hat k and take its curl first, using the standard component formulas: (∇×F⃗)1=∂F3∂y−∂F2∂z(\nabla\times\vec F)_1=\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z}, (∇×F⃗)2=∂F1∂z−∂F3∂x(\nabla\times\vec F)_2=\dfrac{\partial F_1}{\partial z}-\dfrac{\partial F_3}{\partial x}, (∇×F⃗)3=∂F2∂x−∂F1∂y(\nabla\times\vec F)_3=\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y}.

Step 2. Now take the divergence of this new vector field, which means differentiating the first component with respect to xx, the second with respect to yy, the third with respect to zz, and adding: ∇⋅(∇×F⃗)=∂∂x(∂F3∂y−∂F2∂z)+∂∂y(∂F1∂z−∂F3∂x)+∂∂z(∂F2∂x−∂F1∂y)\nabla\cdot(\nabla\times\vec F)=\dfrac{\partial}{\partial x}\left(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z}\right)+\dfrac{\partial}{\partial y}\left(\dfrac{\partial F_1}{\partial z}-\dfrac{\partial F_3}{\partial x}\right)+\dfrac{\partial}{\partial z}\left(\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y}\right).

Step 3. Expand into six second-partial terms: ∂2F3∂x∂y−∂2F2∂x∂z+∂2F1∂y∂z−∂2F3∂y∂x+∂2F2∂z∂x−∂2F1∂z∂y\dfrac{\partial^2 F_3}{\partial x\partial y}-\dfrac{\partial^2 F_2}{\partial x\partial z}+\dfrac{\partial^2 F_1}{\partial y\partial z}-\dfrac{\partial^2 F_3}{\partial y\partial x}+\dfrac{\partial^2 F_2}{\partial z\partial x}-\dfrac{\partial^2 F_1}{\partial z\partial y}. Each of the three functions F1,F2,F3F_1,F_2,F_3 appears exactly twice, once with each order of differentiation.

Step 4. Since the second partials are continuous, Clairaut's theorem lets us swap the order in each pair: ∂2F3∂x∂y=∂2F3∂y∂x\dfrac{\partial^2 F_3}{\partial x\partial y}=\dfrac{\partial^2 F_3}{\partial y\partial x}, ∂2F2∂x∂z=∂2F2∂z∂x\dfrac{\partial^2 F_2}{\partial x\partial z}=\dfrac{\partial^2 F_2}{\partial z\partial x}, and ∂2F1∂y∂z=∂2F1∂z∂y\dfrac{\partial^2 F_1}{\partial y\partial z}=\dfrac{\partial^2 F_1}{\partial z\partial y}. Substituting, the six terms collapse into three matched plus-minus pairs that cancel exactly, leaving ∇⋅(∇×F⃗)=0\nabla\cdot(\nabla\times\vec F)=0 for any such F⃗\vec F, regardless of what the components actually are.

Worked example

Problem. For F⃗(x,y,z)=x2z ı^−2y3z2 ȷ^+xy2z k^\vec F(x,y,z)=x^2z\,\hat\imath-2y^3z^2\,\hat\jmath+xy^2z\,\hat k — the field from the Pass-1 example, where ∇⋅F⃗=2xz−6y2z2+xy2\nabla\cdot\vec F=2xz-6y^2z^2+xy^2 and ∇×F⃗=(2xyz+4y3z) ı^+(x2−y2z) ȷ^+0 k^\nabla\times\vec F=(2xyz+4y^3z)\,\hat\imath+(x^2-y^2z)\,\hat\jmath+0\,\hat k — evaluate both at the new point (2,0,−1)(2,0,-1), and confirm ∇⋅(∇×F⃗)=0\nabla\cdot(\nabla\times\vec F)=0 using this field's own curl expression.

Step 1. Substitute x=2, y=0, z=−1x=2,\ y=0,\ z=-1 into the divergence formula: ∇⋅F⃗=2(2)(−1)−6(0)2(−1)2+(2)(0)2=−4−0+0=−4\nabla\cdot\vec F=2(2)(-1)-6(0)^2(-1)^2+(2)(0)^2=-4-0+0=-4.

Step 2. Substitute the same point into the curl formula: ı^\hat\imath-component =2(2)(0)(−1)+4(0)3(−1)=0+0=0=2(2)(0)(-1)+4(0)^3(-1)=0+0=0; ȷ^\hat\jmath-component =(2)2−(0)2(−1)=4−0=4=(2)^2-(0)^2(-1)=4-0=4; k^\hat k-component =0=0. So ∇×F⃗∣(2,0,−1)=(0, 4, 0)\nabla\times\vec F\big|_{(2,0,-1)}=(0,\,4,\,0).

Step 3. Verify the identity directly on this field's curl, G1=2xyz+4y3z, G2=x2−y2z, G3=0G_1=2xyz+4y^3z,\ G_2=x^2-y^2z,\ G_3=0: ∂G1∂x=2yz\dfrac{\partial G_1}{\partial x}=2yz, ∂G2∂y=−2yz\dfrac{\partial G_2}{\partial y}=-2yz, ∂G3∂z=0\dfrac{\partial G_3}{\partial z}=0. Adding, ∇⋅(∇×F⃗)=2yz−2yz+0=0\nabla\cdot(\nabla\times\vec F)=2yz-2yz+0=0 for every x,y,zx,y,z, matching the general result proved above.

Final answer: ∇⋅F⃗∣(2,0,−1)=−4\nabla\cdot\vec F\big|_{(2,0,-1)}=\mathbf{-4}, ∇×F⃗∣(2,0,−1)=(0,4,0)\nabla\times\vec F\big|_{(2,0,-1)}=\mathbf{(0,4,0)}, and ∇⋅(∇×F⃗)=0\nabla\cdot(\nabla\times\vec F)=\mathbf 0 identically, as the identity guarantees.

Competitive stretch

A quick way examiners extend this style of question is to ask for the Laplacian of a related scalar, built from one of the components of F⃗\vec F. Take g(x,y,z)=x2zg(x,y,z)=x^2z, which is exactly the first component F1F_1 of the same field. Its second partials are ∂2g∂x2=2z\dfrac{\partial^2 g}{\partial x^2}=2z, ∂2g∂y2=0\dfrac{\partial^2 g}{\partial y^2}=0, and ∂2g∂z2=0\dfrac{\partial^2 g}{\partial z^2}=0, so ∇2g=2z\nabla^2g=2z. At the point (2,0,−1)(2,0,-1) this gives ∇2g=2(−1)=−2\nabla^2g=2(-1)=\mathbf{-2}. Notice this is a genuinely separate computation from ∇⋅F⃗\nabla\cdot\vec F: the Laplacian acts on a scalar function, while the divergence acted on the vector field F⃗\vec F as a whole — mixing the two up is exactly the input-type trap flagged below.

Common traps

  • Applying divergence or curl to the wrong input type — divergence and curl need a vector field; a scalar field only ever takes a gradient (which is a vector), never a divergence or curl directly.
  • Dropping the minus sign on the ȷ^\hat\jmath component of curl: it is ∂F1∂z−∂F3∂x\dfrac{\partial F_1}{\partial z}-\dfrac{\partial F_3}{\partial x}, not the reverse order.
  • Computing the Laplacian as ∣∇f∣2|\nabla f|^2 (squaring or dotting the gradient with itself) instead of ∇⋅(∇f)\nabla\cdot(\nabla f), the sum of unmixed second partials.
  • Assuming ∇⋅F⃗=0\nabla\cdot\vec F=0 instantly proves F⃗\vec F is some curl ∇×A⃗\nabla\times\vec A — the identity only runs one way; constructing A⃗\vec A needs extra conditions not covered here.

Ask the tutor: Want a fast drill on the curl determinant signs, or a walkthrough of why the divergence-of-curl identity holds for any field?

Why does divergence give a scalar but curl give a vector, when both start from the same nabla and the same F?

It comes down to how ∇\nabla combines with F⃗\vec F. Divergence is a dot product, ∇⋅F⃗\nabla\cdot\vec F, and a dot product of two vectors always collapses to a single number — a scalar. Curl is a cross product, ∇×F⃗\nabla\times\vec F, and a cross product of two vectors always produces a third vector, perpendicular to the plane of rotation implied by the field. Same symbol ∇\nabla, different algebraic operation, so different output type. Checking which operation you actually need is the first thing to do in any problem.

Do I need to memorize the curl determinant, or is there an easier way to remember the signs?

Most students find it easier to remember the cyclic pattern x→y→z→xx\to y\to z\to x than to memorize six separate signs. Write the components as (F1,F2,F3)(F_1,F_2,F_3) paired with (x,y,z)(x,y,z); the ı^\hat\imath component uses y,zy,z, the ȷ^\hat\jmath component uses z,xz,x (in that order, giving the extra minus sign relative to the pattern), and the k^\hat k component uses x,yx,y. Practising the determinant a few times until the cyclic shift feels automatic beats memorizing six symbols independently.

Is the identity div of curl F equals zero only true for the specific field in the derivation, or always true?

It is always true, for any vector field whose components have continuous second partial derivatives — that is exactly what the derivation proves in general, without picking specific formulas for F1,F2,F3F_1,F_2,F_3. The worked example above only re-checks it on one concrete field as a numerical sanity test; the real content is the general cancellation argument using Clairaut's theorem, which holds for every well-behaved F⃗\vec F you will meet in this course.

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