4. Variations of Fields#

4.1. Divergence#

If we start with a vector field $A (r) = A_{x} (r) \hat{x} + A_{y} (r) \hat{y} + A_{z} (r) \hat{z}$ , then we can define the quantity $\nabla \cdot A (r)$ , i.e. taking the scalar product of the gradient operator with vector field $A (r)$ which since the gradient vector is a differential operator, means the expression has the form:

\nabla \cdot A (r) = \partial_{x} A_{x} + \partial_{y} A_{y} + \partial_{z} A_{z}

This is often known as the Divergence of $A$ or $div A$ . Note that in the case of operators we need to be a little careful of the ordering of the terms, since $A (r) \cdot \nabla$ would give:

A (r) \cdot \nabla = A_{x} \partial_{x} + A_{y} \partial_{y} + A_{z} \partial_{z}

which is also a differential operator, waiting to act on another term, which we apply on the right.

Lets see the effect of this on two vector fields $A_{1} (r) = (\begin{matrix} x \\ y \\ 0 \end{matrix}), A_{2} (r) = (\begin{matrix} y \\ - x \\ 0 \end{matrix})$ :

\begin{aligned} \nabla \cdot A_{1} & = \partial_{x} x + \partial_{y} y + \partial_{z} 0 = 2 \\ \nabla \cdot A_{2} & = \partial_{x} y - \partial_{y} x + \partial_{z} 0 = 0 \end{aligned}

which we can visualise in Fig. 4.1.

../_images/div.png — Fig. 4.1 Plotting the vector field $A_{1}$ which has non-zero divergence.#

We can see that the origin here really plays the role of a centre of the divergence, the field lines all appear to flow outwards, becuase here $\nabla \cdot A > 0$ . In the case of $\nabla \cdot A < 0$ , field lines would flow into a point and there would be a convergence of the field.

4.1.1. Cylindrical coordinate system#

In cylindrical polar coordinates, we have the additional complication that the unit vectors:

\begin{array}{r} \hat{r} = (\begin{array}{c} \cos (θ) \\ \sin (θ) \\ 0 \end{array}), \hat{θ} = (\begin{array}{c} - \sin (θ) \\ \cos (θ) \\ 0 \end{array}), \hat{z} = (\begin{array}{c} 0 \\ 0 \\ 1 \end{array}) \end{array}

are now not all constant, so in fact when we apply the derivative operator, we need to do any derivatives ahead of the dot product:

\begin{aligned} \nabla \cdot A & = (\hat{r} \frac{\partial}{\partial r} + \frac{\hat{θ}}{r} \frac{\partial}{\partial θ} + \hat{z} \frac{\partial}{\partial z}) \cdot (A_{r} \hat{r} + A_{θ} \hat{θ} + A_{z} \hat{z}) \\ = \hat{r} \cdot (\frac{\partial (A_{r} \hat{r})}{\partial r} + \frac{\partial (A_{θ} \hat{θ})}{\partial r} + \frac{\partial (A_{z} \hat{z})}{\partial r}) \\ + \frac{\hat{θ}}{r} \cdot (\frac{\partial (A_{r} \hat{r})}{\partial θ} + \frac{\partial (A_{θ} \hat{θ})}{\partial θ} + \frac{\partial (A_{z} \hat{z})}{\partial θ}) \\ + \hat{z} \cdot (\frac{\partial (A_{r} \hat{r})}{\partial z} + \frac{\partial (A_{θ} \hat{θ})}{\partial z} + \frac{\partial (A_{z} \hat{z})}{\partial z}) \end{aligned}

So whereas the first and third sets of terms here will not have additional terms, the middle set of terms will be more complicated, because $\hat{r} = \hat{r} (θ), \hat{θ} = \hat{θ} (θ)$ .

We find that:

\begin{array}{r} \frac{\partial \hat{r}}{\partial θ} = (\begin{array}{c} - \sin (θ) \\ \cos (θ) \end{array}) = \hat{θ}, \frac{\partial \hat{θ}}{\partial θ} = (\begin{array}{c} - \cos (θ) \\ - \sin (θ) \end{array}) = - \hat{r} \end{array}

which means that:

\begin{aligned} \nabla \cdot A & = \hat{r} \cdot (\frac{\partial (A_{r})}{\partial r} \hat{r} + \frac{\partial (A_{θ})}{\partial r} \hat{θ} + \frac{\partial (A_{z})}{\partial r} \hat{z}) \\ + \frac{\hat{θ}}{r} \cdot (\frac{\partial (A_{r})}{\partial θ} \hat{r} + A_{r} \hat{θ} + \frac{\partial (A_{θ})}{\partial θ} \hat{θ} - A_{θ} \hat{r} + \frac{\partial (A_{z})}{\partial θ} \hat{z}) \\ + \hat{z} \cdot (\frac{\partial (A_{r})}{\partial z} \hat{r} + \frac{\partial (A_{θ})}{\partial z} \hat{θ} + \frac{\partial (A_{z})}{\partial z} \hat{z}) \\ = \frac{\partial (A_{r})}{\partial r} + \frac{1}{r} \frac{\partial (A_{θ})}{\partial θ} + \frac{A_{r}}{r} + \frac{\partial (A_{z})}{\partial z} \\ \Rightarrow \nabla \cdot A & = \frac{1}{r} \frac{\partial (r A_{r})}{\partial r} + \frac{1}{r} \frac{\partial (A_{θ})}{\partial θ} + \frac{\partial (A_{z})}{\partial z} \end{aligned}

If we consider a vector in cylindrical coordinates, a vector with constant coefficients, e.g. $A = (\begin{matrix} a \\ b \\ c \end{matrix})$ will have non-zero divergence if and only if the $A_{r} \neq 0$ , since:

\nabla \cdot A = \frac{1}{r} \frac{\partial (r a)}{\partial r} + \frac{1}{r} \frac{\partial b}{\partial θ} + \frac{\partial c}{\partial z} = \frac{a}{r}

4.1.2. Spherical coordinate system#

In spherical polar coordinates, we also have the additional complication that the unit vectors:

\begin{array}{r} \hat{r} = (\begin{array}{c} \cos (θ) \sin (ϕ) \\ \sin (θ) \sin (ϕ) \\ \cos (ϕ) \end{array}), \hat{ϕ} = (\begin{array}{c} \cos (θ) \cos (ϕ) \\ \sin (θ) \cos (ϕ) \\ - \sin (ϕ) \end{array}), \hat{θ} = (\begin{array}{c} - \sin (θ) \\ \cos (θ) \\ 0 \end{array}) \end{array}

are now not all constant, so we need to do any derivatives ahead of the dot product:

\begin{aligned} \nabla \cdot A & = (\hat{r} \frac{\partial}{\partial r} + \frac{\hat{ϕ}}{r} \frac{\partial}{\partial ϕ} + \frac{\hat{θ}}{r \sin (ϕ)} \frac{\partial}{\partial θ}) \cdot (A_{r} \hat{r} + A_{ϕ} \hat{ϕ} + A_{θ} \hat{θ}) \\ = \hat{r} \cdot (\frac{\partial (A_{r} \hat{r})}{\partial r} + \frac{\partial (A_{ϕ} \hat{ϕ})}{\partial r} + \frac{\partial (A_{θ} \hat{θ})}{\partial r}) \\ + \frac{\hat{ϕ}}{r} \cdot (\frac{\partial (A_{r} \hat{r})}{\partial ϕ} + \frac{\partial (A_{ϕ} \hat{ϕ})}{\partial ϕ} + \frac{\partial (A_{θ} \hat{θ})}{\partial ϕ}) \\ + \frac{\hat{θ}}{r \sin (ϕ)} \cdot (\frac{\partial (A_{r} \hat{r})}{\partial θ} + \frac{\partial (A_{ϕ} \hat{ϕ})}{\partial θ} + \frac{\partial (A_{θ} \hat{θ})}{\partial θ}) \end{aligned}

So whereas the first set of terms here will not have additional terms, the second and third set of terms will be more complicated, because $\hat{r} = \hat{r} (ϕ, θ), \hat{ϕ} = \hat{ϕ} (ϕ, θ), \hat{θ} = \hat{θ} (ϕ, θ)$ .

We find that:

\begin{aligned} \frac{\partial \hat{r}}{\partial ϕ} & = (\begin{array}{c} \cos (θ) \cos (ϕ) \\ \sin (θ) \cos (ϕ) \\ - \sin (ϕ) \end{array}) = \hat{ϕ}, \frac{\partial \hat{ϕ}}{\partial ϕ} = (\begin{array}{c} - \cos (θ) \sin (ϕ) \\ - \sin (θ) \sin (ϕ) \\ - \cos (ϕ) \end{array}) = - \hat{r}, \frac{\partial \hat{θ}}{\partial ϕ} = (\begin{array}{c} 0 \\ 0 \\ 0 \end{array}) \\ \frac{\partial \hat{r}}{\partial θ} & = (\begin{array}{c} - \sin (θ) \sin (ϕ) \\ \cos (θ) \sin (ϕ) \\ 0 \end{array}) = \sin (ϕ) \hat{θ}, \frac{\partial \hat{ϕ}}{\partial θ} = (\begin{array}{c} - \sin (θ) \cos (ϕ) \\ \cos (θ) \cos (ϕ) \\ 0 \end{array}) = \cos (ϕ) \hat{θ}, \\ \frac{\partial \hat{θ}}{\partial θ} & = (\begin{array}{c} - \cos (θ) \\ - \sin (θ) \\ 0 \end{array}) = - (\sin (ϕ) \hat{r} + \cos (ϕ) \hat{ϕ}), \end{aligned}

which means that:

\begin{aligned} \nabla \cdot A & = \hat{r} \cdot (\frac{\partial (A_{r})}{\partial r} \hat{r} + \frac{\partial (A_{ϕ})}{\partial r} \hat{ϕ} + \frac{\partial (A_{θ})}{\partial r} \hat{θ}) \\ + \frac{\hat{ϕ}}{r} \cdot (\frac{\partial (A_{r})}{\partial ϕ} \hat{r} + A_{r} \hat{ϕ} + \frac{\partial (A_{ϕ})}{\partial ϕ} \hat{ϕ} - A_{ϕ} \hat{r} + \frac{\partial (A_{θ})}{\partial ϕ} \hat{θ}) \\ + \frac{\hat{θ}}{r \sin (ϕ)} \cdot (\frac{\partial (A_{r})}{\partial θ} \hat{r} + A_{r} \sin (ϕ) \hat{θ} + \frac{\partial (A_{ϕ})}{\partial θ} \hat{ϕ} \\ + A_{ϕ} \cos (θ) \hat{θ} + \frac{\partial (A_{θ})}{\partial θ} \hat{θ} - A_{θ} (\hat{r} \sin (ϕ) + \hat{ϕ} \cos (ϕ))) \\ = \frac{\partial (A_{r})}{\partial r} + \frac{A_{r}}{r} + \frac{2}{r} \frac{\partial (A_{ϕ})}{\partial ϕ} + \frac{A_{ϕ} \cos (ϕ)}{r \sin (ϕ)} + \frac{1}{r \sin (ϕ)} \frac{\partial (A_{θ})}{\partial θ} \\ \Rightarrow \nabla \cdot A & = \frac{1}{r^{2}} \frac{\partial (r^{2} A_{r})}{\partial r} + \frac{1}{r \sin (ϕ)} \frac{\partial (\sin (ϕ) A_{ϕ})}{\partial ϕ} + \frac{1}{r \sin (ϕ)} \frac{\partial (A_{θ})}{\partial θ} \end{aligned}

4.2. Curl#

We can likewise take a vector field $A (r)$ and here take the vector product with the gradient operator $\nabla \times A (r)$ , which has components:

\begin{array}{r} \nabla \times A (r) = | \begin{array}{c} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_{x} & A_{y} & A_{z} \end{array} | = (\begin{array}{c} \partial_{y} A_{z} - \partial_{z} A_{y} \\ \partial_{z} A_{x} - \partial_{x} A_{z} \\ \partial_{x} A_{y} - \partial_{y} A_{x} \end{array}) \end{array}

which is know as the Curl or Rotation of the vector field $A (r)$ or $curl A (r)$ .

To see the effects of these, lets consider again $A_{1} (r) = (\begin{matrix} x \\ y \\ 0 \end{matrix}), A_{2} (r) = (\begin{matrix} y \\ - x \\ 0 \end{matrix})$ :

\begin{aligned} \nabla \times A_{1} & = (\begin{array}{c} \partial_{y} 0 + \partial_{z} x \\ \partial_{z} y - \partial_{x} 0 \\ \partial_{x} y - \partial_{y} x \end{array}) = (\begin{array}{c} 0 \\ 0 \\ 0 \end{array}) \\ \nabla \times A_{2} & = (\begin{array}{c} \partial_{y} 0 + \partial_{z} x \\ \partial_{z} y - \partial_{x} 0 \\ \partial_{x} x + \partial_{y} y \end{array}) = (\begin{array}{c} 0 \\ 0 \\ 2 \end{array}) \end{aligned}

which we can visualise in Fig. 4.2.

../_images/curl.png — Fig. 4.2 Plotting the vector field $A_{2}$ which has non-zero curl.#

We note that “centre” of this rotation is at the origin and that $\nabla \times A_{2}$ points in the $z$ direction. We define this rotation as following the right hand rule, depicted in Fig. 4.3, curling the fingers on our right hand around and the thumb pointing in the direction of the vector.

../_images/Rhandrule.png — Fig. 4.3 The right hand rule for a curl field#

We can see that a linear combination of vector fields $A_{1} + A_{2}$ would produce a vector field with both divergence and curl, which is shown in Fig. 4.4.

../_images/divcurl.png — Fig. 4.4 The effect of adding a divergence (curl free) field (left hand figure) to a curl (divergence free) field (middle figure) is shown in the right hand figure.#

We can use the product rule as well as the rules following scalar and vector products to find a vareity of vector calculus relations:

\begin{aligned} \nabla (ϕ ψ) & = ψ (\nabla ϕ) + ϕ (\nabla ψ) \\ \nabla \cdot (ϕ A) & = (\nabla ϕ) \cdot A + ϕ (\nabla \cdot A) \\ \nabla \times (ϕ A) & = (\nabla ϕ) \times A + ϕ (\nabla \times A) \\ \nabla \cdot (A \times B) & = (\nabla \times A) \cdot B - A \cdot (\nabla \times B) \\ \nabla \times (A \times B) & = (\nabla \cdot B + B \cdot \nabla) A - (\nabla \cdot A + A \cdot \nabla) B \end{aligned}

4.2.1. Cylindrical coordinate systems#

As with the divergence, the variation of the unit vectors causes additional complexity in calculating the curl:

\nabla \times A = (\hat{r} \frac{\partial}{\partial r} + \frac{\hat{θ}}{r} \frac{\partial}{\partial θ} + \hat{z} \frac{\partial}{\partial z}) \times (A_{r} \hat{r} + A_{θ} \hat{θ} + A_{z} \hat{z})

which given our expressions for variation of unit vectors with coordinate variables means:

\begin{aligned} \nabla \times A & = \hat{r} \times (\frac{\partial (A_{r})}{\partial r} \hat{r} + \frac{\partial (A_{θ})}{\partial r} \hat{θ} + \frac{\partial (A_{z})}{\partial r} \hat{z}) \\ + \frac{\hat{θ}}{r} \times (\frac{\partial (A_{r})}{\partial θ} \hat{r} + A_{r} \hat{θ} + \frac{\partial (A_{θ})}{\partial θ} \hat{θ} - A_{θ} \hat{r} + \frac{\partial (A_{z})}{\partial θ} \hat{z}) \\ + \hat{z} \times (\frac{\partial (A_{r})}{\partial z} \hat{r} + \frac{\partial (A_{θ})}{\partial z} \hat{θ} + \frac{\partial (A_{z})}{\partial z} \hat{z}) \end{aligned}

Given the cylic nature of unit vectors and their orthongality:

\begin{aligned} \hat{r} \times \hat{θ} & = \hat{z} \\ \hat{θ} \times \hat{z} & = \hat{r} \\ \hat{z} \times \hat{r} & = \hat{θ} \\ \hat{r} \times \hat{r} & = \hat{θ} \times \hat{θ} = \hat{z} \times \hat{z} = 0 \end{aligned}

means we find that:

\begin{aligned} \nabla \times A & = (\frac{\partial (A_{θ})}{\partial r} \hat{z} - \frac{\partial (A_{z})}{\partial r} \hat{θ}) + (- \frac{1}{r} \frac{\partial (A_{r})}{\partial θ} \hat{z} + \frac{A_{θ}}{r} \hat{z} + \frac{1}{r} \frac{\partial (A_{z})}{\partial θ} \hat{r}) \\ + (\frac{\partial (A_{r})}{\partial z} \hat{θ} - \frac{\partial (A_{θ})}{\partial z} \hat{r}) \\ \Rightarrow \nabla \times A & = (\frac{1}{r} \frac{\partial A_{z}}{\partial θ} - \frac{\partial A_{θ}}{\partial z}) \hat{r} + (\frac{\partial A_{r}}{\partial z} - \frac{\partial A_{z}}{\partial r}) \hat{θ} + \frac{1}{r} (\frac{\partial (r A_{θ})}{\partial r} - \frac{\partial A_{r}}{\partial θ}) \hat{z} \end{aligned}

4.2.2. Spherical coordinate systems#

In a similar fashion to the cylindrical polar coordinates, we need to consider how the unit vectors change under differentiation first before applying the cross product, therefore:

\begin{aligned} \nabla \times A & = (\hat{r} \frac{\partial}{\partial r} + \frac{\hat{ϕ}}{r} \frac{\partial}{\partial ϕ} + \frac{\hat{θ}}{r \sin (ϕ)} \frac{\partial}{\partial θ}) \times (A_{r} \hat{r} + A_{ϕ} \hat{ϕ} + A_{θ} \hat{θ}) \\ = \hat{r} \times (\frac{\partial (A_{r})}{\partial r} \hat{r} + \frac{\partial (A_{ϕ})}{\partial r} \hat{ϕ} + \frac{\partial (A_{θ})}{\partial r} \hat{θ}) \\ + \frac{\hat{ϕ}}{r} \times (\frac{\partial (A_{r})}{\partial ϕ} \hat{r} + A_{r} \hat{ϕ} + \frac{\partial (A_{ϕ})}{\partial ϕ} \hat{ϕ} - A_{ϕ} \hat{r} + \frac{\partial (A_{θ})}{\partial ϕ} \hat{θ}) \\ + \frac{\hat{θ}}{r \sin (ϕ)} \times (\frac{\partial (A_{r})}{\partial θ} \hat{r} + A_{r} \sin (ϕ) \hat{θ} + \frac{\partial (A_{ϕ})}{\partial θ} \hat{ϕ} \\ + A_{ϕ} \cos (θ) \hat{θ} + \frac{\partial (A_{θ})}{\partial θ} \hat{θ} - A_{θ} (\hat{r} \sin (ϕ) + \hat{ϕ} \cos (ϕ))) \end{aligned}

Given the cylic nature of unit vectors and their orthongality:

\begin{aligned} \hat{r} \times \hat{ϕ} & = \hat{θ} \\ \hat{ϕ} \times \hat{θ} & = \hat{r} \\ \hat{θ} \times \hat{r} & = \hat{ϕ} \\ \hat{r} \times \hat{r} & = \hat{ϕ} \times \hat{ϕ} = \hat{θ} \times \hat{θ} = 0 \end{aligned}

means we find that:

\begin{aligned} \nabla \times A & = (\frac{\partial (A_{ϕ})}{\partial r} \hat{θ} - \frac{\partial (A_{θ})}{\partial r} \hat{ϕ}) + (- \frac{1}{r} \frac{\partial (A_{r})}{\partial ϕ} \hat{θ} + \frac{A_{ϕ}}{r} \hat{θ} + \frac{1}{r} \frac{\partial (A_{θ})}{\partial ϕ} \hat{r}) \\ + (\frac{1}{r \sin (ϕ)} \frac{\partial (A_{r})}{\partial θ} \hat{ϕ} - \frac{1}{r \sin (ϕ)} \frac{\partial (A_{ϕ})}{\partial θ} \hat{r} - \frac{A_{θ}}{r} \hat{ϕ} + \frac{A_{θ} \cos (ϕ)}{r \sin (ϕ)} \hat{r}) \\ \Rightarrow \nabla \times A & = \frac{1}{r \sin (ϕ)} (\frac{\partial (A_{θ} \sin (ϕ))}{\partial ϕ} - \frac{\partial A_{ϕ}}{\partial θ}) \hat{r} \\ + \frac{1}{r} (\frac{1}{\sin (ϕ)} \frac{\partial A_{r}}{\partial θ} - \frac{\partial (r A_{θ})}{\partial r}) \hat{ϕ} \\ + \frac{1}{r} (\frac{\partial (r A_{ϕ})}{\partial r} - \frac{\partial A_{r}}{\partial ϕ}) \hat{θ} \end{aligned}

4.3. Second order variations of fields#

We can combine two or more gradients in a vector expression, one of the most useful is to find the divergence of the gradient of a scalar field $ϕ$ ,

div grad ϕ = \nabla \cdot (\nabla ϕ) = \nabla^{2} ϕ = (\frac{\partial}{\partial_{x}^{2}} + \frac{\partial}{\partial_{y}^{2}} + \frac{\partial}{\partial_{z}^{2}}) ϕ

This is sometimes also written as $Δ ϕ = \nabla^{2} ϕ$ and is known as the Laplacian of $ϕ$ .

We can also find the divergence of the curl of a vector field:

div curl A = \nabla \cdot (\nabla \times A) = 0

which holds for all vector fields. Thinking again about the fields shown in Fig. 4.4, we can think of these two processes as complementary, rotation around a point compared with emergence from / convergence to a point.

Likewise if we look at the curl of a gradient field:

curl grad ϕ = \nabla \times \nabla ϕ = 0

which is true for all scalar fields.

Also we sometimes find the curl of the curl a useful quantity:

\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla^{2} A

In general we can write a vector field as having two sets of components, one curl free and one divergence free, this is known as the Helmholtz Decomposition of a vector field:

B = - \nabla ϕ + \nabla \times A

4.4. Conservative vector fields#

Recall the concept of a perfect (or exact) differential where the following property holds:

P (x, y) d x + Q (x, y) d y = 0 ⟺ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

where this holds because there is some function $f (x, y)$ which is constant such that $d f = 0$ and so this expression is really just:

d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y = 0

and the expression calculated is just:

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = \frac{\partial^{2} f}{\partial x \partial y}

There is a vector calculus generalisation of this sort of scalar function, known as the conservative vector field.

Definition

A conservative vector field $v$ is one which is formed solely from the gradient of some function $f (x, y, z)$ :

v = \nabla f

Such a vector field is therefore curl free:

\nabla \times v = \nabla \times \nabla f = 0

Such a vector field also has a path independent line integral (more on this property later).

Lets examine the case of a vector field $v$ formed from the function $f (x, y, z) = x y z$ , the gradient of this function is given by:

\begin{array}{r} v = \nabla f = (\begin{array}{c} y z \\ x z \\ x y \end{array}) \end{array}

and if we take the curl of such a vector field:

\begin{aligned} \nabla \times v & = | \begin{array}{c} \hat{x} & \hat{y} & \hat{z} \\ \partial_{x} & \partial_{y} & \partial_{z} \\ y z & x z & x y \end{array} | \\ = (x - x) \hat{x} + (y - y) \hat{y} + (z - z) \hat{z} = 0 \end{aligned}

Likewise if we started with some vector field and wanted to work out if it is conservative (and if so which scalar field sourced it), we could examine each component and integrate, for instance if:

\begin{array}{r} v = (\begin{array}{c} 2 x \sin (y) \\ x^{2} \cos (y) \\ 2 \end{array}) \end{array}

then examining each component we find:

\begin{aligned} v_{x} = 2 x \sin (y) & \Rightarrow ϕ = \int 2 x \sin (y) d x = x^{2} \sin (y) + a (y, z) \\ v_{y} = x^{2} \cos (y) & \Rightarrow ϕ = \int 2 x \sin (y) d x = x^{2} \sin (y) + b (x, z) \\ v_{z} = 2 & \Rightarrow ϕ = \int 2 d z = 2 z + c (x, y) \end{aligned}

Comparing all three expressions we find that $f = x^{2} \sin (y) + 2 z$ and hence $v = \nabla f$ .

Mathematics for Natural Sciences 2

Variations of Fields

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