In Lecture 9, Leonard Susskind reviews gradient, curl and divergence. He says that the curl of a gradient is zero; and, furthermore, if the curl of a vector field is 0, then that vector field is a gradient.

Why is that?

The wiki page for curl explains this indirectly. Let’s spell it out.

We start with the operator \(\begin{align} \vec{\nabla} = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}) \end{align}\)

This is an operator on functions \(f \colon \mathbb{R}^3 \rightarrow \mathbb{R}\). The operator \(\vec{\nabla}\) is called nabla, or, abbreviated, del, and is written as an upside-down triangle.

In what follows, we will assume all functions are smooth.

Gradient

Gradient is defined for functions \(f \colon \mathbb{R}^3 \rightarrow \mathbb{R}\) as \(\begin{align} grad(f) = (f_x, f_y, f_z) \end{align}\) which is a vector field on \(\mathbb{R}^3\). Here, \(f_x, f_y, f_z\) are the \(x, y, z\) partial derivatives of the function \(f\).

The gradient is intepreted as the direction and rate of fastest increase.

Another way to write the gradient is \(grad(f) = \vec{\nabla} f\), as the operator \(\vec{\nabla}\) applied to \(f\): \(\begin{align} \vec{\nabla} f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) \end{align}\)

Curl

Curl is defined for vector fields \(F \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3\) to be the \(\mathbb{R}^3\) vector field \(\begin{align} curl(F) = (\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}) \end{align}\) Here \(F_x, F_y, F_z\) are the \(x, y, x\) coordinates of the vector field \(F\).

Expressed as a determinant of operators \(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\), and denoting \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) the unit vectors along the \(x, y, z\) axis, the curl can be expressed in the more easily memorized form

\[\begin{align} curl(F) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix} \end{align}\]

Here we abuse somewhat the notation, in that the middle row of the determinant represents operators, not functions (and the operators act on the third row, rather than being multiplied with the third row). Using the \(\vec{\nabla}\) operator, we can express this, also by abuse of notation, as the cross product \(\begin{align} curl(F) = \vec{\nabla} \times F \end{align}\)

At a point \(x, y, z\), if we draw parallels to the axes at that point, the curl represents rotation of an infinitesimal area around these parallel lines to the x-axis, y-axis, z-axis.

Divergence

The divergence of a vector field \(F \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3\) is the function \(div(F) \colon \mathbb{R}^3 \rightarrow \mathbb{R}\) given by

\[\begin{align} div(F) = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \end{align}\]

The divergence at a point represents the extent to which the vector field behaves like a source at that point. If the vector field corresponds to the flow of an incompressible fluid, for example, its divergence will be zero.

Lecture 7 in leonard Susskind’s classical mechanics course proves that the gradient of motion given by a Hamiltonian in phase space is zero (this is the Liouville Theorem).

In terms of \(\vec{\nabla}\), again by abuse of notation, we can express the divergence as a dot product

\[\begin{align} div(F) = \vec{\nabla} \cdot F \end{align}\]

where \(\vec{\nabla}\) acts on, rather than multiplies with the components of \(F\).

Relations between gradient, curl, divergence

Bref:

  • div of curl is zero
  • curl of grad is zero
  • and, conversely, if div of a vector field is zero, the field is a curl
  • if curl of a vector field is zero, the field is a grad

This can be seen by expressing div, curl, grad in terms of differentiation of forms, noting that \(\mathbb{R}^3\) is contractible, so any closed 1- and 2-form is exact.

Recollection: differential forms

More precisely, denote \(\Omega^k(\mathbb{R}^n)\) the set of smooth differential \(k\)-forms in \(n\) variables.

In coordinates, a differential form of order \(k\) is given by \(\begin{align} \omega^{(k)}=\sum_{1 \le i_1 \lt... \lt i_k \le n} a_{i_1,...,i_k} dx_{i_1} ... dx_{i_k} \end{align}\)

The product of differential forms is anticommutative, i.e. \(dx_i dx_j = - dx_j dx_i\). And the differential operator \(d \colon \Omega^k(\mathbb{R}^n) \rightarrow \Omega^{k+1} (\mathbb{R}^n)\) is given by \(\begin{align} d\omega^{(k)}=\sum_{1 \le j \le n, \, i_1 \lt ... \lt i_k}^n\frac{\partial a_{i_1 ... i_k}}{\partial x_j}\,dx_j dx_{i_1} ... dx_{i_k} \end{align}\)

The sqare of the differential is zero: \(d^2 \omega = d(d \omega) = 0\). Differentiation of forms \(d\) gives us a sequence of vector spaces

\[\begin{align} \bf{0} \overset{d}{\longrightarrow} \Omega^0(\mathbb{R}^n) \overset{d}{\longrightarrow} \Omega^1(\mathbb{R}^n) \overset{d}{\longrightarrow} \Omega^2(\mathbb{R}^n) \overset{d}{\longrightarrow} \Omega^3(\mathbb{R}^n) ... \overset{d}{\longrightarrow} \Omega^n(\mathbb{R}^n) \overset{d}{\longrightarrow} \bf{0} \end{align}\]

The composite of consecutive maps in this sequence of vector spaces is zero. Contractibility of \(\mathbb{R}^n\) ensures, by DeRham theory, that, in \(\Omega^k(\mathbb{R}^n)\), the image of \(d\) is the kernel of \(d\), for \(k \ge 1\).

In \(\Omega^0(\mathbb{R}^n)\), the kernel of \(d\) is the set of functions \(f \colon \mathbb{R}^n \rightarrow \mathbb{R}\) with zero differential \(df = 0\), thus is the set of constant functions.

Closed forms, exact forms

Recall a bit more language: a form \(\omega^{(k)}\) is called closed if its differential is zero: \(d \omega^{(k)} = 0\). And the form is called exact if it is in the image of \(d\), meaning that there exists a form \(\omega^{(k-1)}\) with \(d \omega^{(k-1)} = \omega^{(k)}\).

Forms can, more generally, be defined on smooth manifolds. Exact forms are always closed, because \(d^2 = 0\). Closed forms are not necessarily exact.

By DeRham theory, the quotient vector space of \(k\)-closed over \(k\)-exact forms is the \(k\)-th cohomology group with real coefficients of the manifold, and depends only on the homotopy type of the manifold.

If the manifold is contractible, which is the case of \(\mathbb{R}^n\), then all \(k > 0\) cohomology groups are \(0\), and the \(0\)-th cohomology group has dimension 1. The closed forms of dimension 0 are, as explained above, the constant functions.

Gradient, curl, divergence as differential forms

Let us use the standard notation \(\mathcal{C}^{\infty}(\mathbb{R}^m, \mathbb{R}^n)\) for the set of smooth functions \(\mathbb{R}^m \rightarrow \mathbb{R}^n\). And, furthermore, let’s limit the discussion to dimension \(3\).

Gradient

The functions \(\mathbb{R}^3 \rightarrow \mathbb{R}\) are 0-forms. We can naturally identify \(\mathcal{C}^{\infty}(\mathbb{R}^3, \mathbb{R}) \cong \Omega^0(\mathbb{R}^3)\).

The vector fields \(\mathbb{R}^3 \rightarrow \mathbb{R}^3\) can be identified as 1-forms by \((F_x, F_y, F_z) \rightarrow F_x dx + F_y dy + F_z dz\). This gives us a natural identification \(\mathcal{C}^{\infty}(\mathbb{R}^3, \mathbb{R}^3) \cong \Omega^1(\mathbb{R}^3)\)

By this identification, the gradient \((f_x, f_y, f_z)\) of a function \(f : \mathbb{R}^3 \rightarrow \mathbb{R}\) corresponds to differentiation \(df = f_x dx + f_y dy + f_x dz\).

Curl

The fields \(\mathbb{R}^3 \rightarrow \mathbb{R}^3\) can also be identified as 2-forms by \((F_x, F_y, F_z) \rightarrow F_x dydz + dx F_y dy + dxdy F_z\). This gives us a natural identification \(\mathcal{C}^{\infty}(\mathbb{R}^3, \mathbb{R}^3) \cong \Omega^2(\mathbb{R}^3)\)

With this last two indentifications, \(curl(F)\) corresponds also to differentiation:

\[\begin{align} d(F_x dx + F_y dy + F_z dz) = (\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}) dydz + (\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x})dzdx + ( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y})dxdy \end{align}\]

Divergence

Finally, functions \(f \colon \mathbb{R}^3 \rightarrow \mathbb{R}\) can be identified with 3-forms \(fdxdydz\). This gives us a natural identification \(\mathcal{C}^{\infty}(\mathbb{R}^3, \mathbb{R}) \cong \Omega^3(\mathbb{R}^3)\)

With these last identifications, \(div(F)\) corresponds also to differentiation:

\[\begin{align} d(F_x dydz + F_y dzdx + F_z dxdy) = (\frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z})dxdydz \end{align}\]

Putting it together

The sequence of vector space maps

\[\begin{align} \bf{0} \longrightarrow \mathcal{C}^{\infty} (\mathbb{R}, \mathbb{R}) \overset{grad}{\longrightarrow} \mathcal{C}^{\infty} (\mathbb{R}^3, \mathbb{R}) \overset{curl}{\longrightarrow} \mathcal{C}^{\infty} (\mathbb{R}^3, \mathbb{R}) \overset{div}{\longrightarrow} \mathcal{C}^{\infty} (\mathbb{R}, \mathbb{R}) \longrightarrow \bf{0} \end{align}\]

is naturally identified with the sequence

\[\begin{align} \bf{0} \overset{d}{\longrightarrow} \Omega^0(\mathbb{R}^3) \overset{d}{\longrightarrow} \Omega^1(\mathbb{R}^3) \overset{d}{\longrightarrow} \Omega^2(\mathbb{R}^3) \overset{d}{\longrightarrow} \Omega^3(\mathbb{R}^3) \overset{d}{\longrightarrow} \bf{0} \end{align}\]

Since any exact form is closed, div of curl and curl of grad are zero. And since any form of degree \(1\) or higher that is closed is also exact, any vector field with zero divergence is a curl, and any vector field with zero curl is a gradient, completing our proof.

Higher dimensions

The definition of \(grad\), \(div\) extends naturally to \(n\) dimensions. For \(curl\), only the definition in terms of \(d\) of a \(1\)-differential form extends to \(n\) dimensions.