Math 344: Calculus III

16.1 Vector Fields

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16.1.1 What is a Vector Field?

A vector field is a function of several variables that assigns a each point in its domain to a vector. Formally,

Definition

Let $D$ be a set in $\mathbb{R}^n$ (usually $n=2$ or $n=3$). A vector field on $\ \mathbb{R}^n$ is a function $\mathbf{F}$ that assigns each point $\mathbf{x}\in D$ to an $n$-dimensional vector $\mathbf{F}(\mathbf{x})$.

Typically, a vector field is written in terms of its component functions

$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pzi}[2][]{\dfrac{\partial #1}{\partial #2}} \begin{align*} \mathbf{F}(x,y) &= f(x,y)\,\ihat + g(x,y)\,\jhat = \langle\,f(x,y),g(x,y)\,\rangle \\ \\ \mathbf{F}(x,y,z) &= f(x,y,z)\,\ihat + g(x,y,z)\,\jhat + h(x,y,z)\,\khat = \langle\,f(x,y,z),g(x,y,z),h(x,y,z)\,\rangle \end{align*} $$
where the scalar component functions $f$, $g$, and $h$ are sometimes called scalar fields to separate them from vector fields.

Vector fields are best understood visually, but drawing the required pictures can be cumbersome. The procedure is to select several points and draw the vectors given by $\mathbf{F}(\mathbf{x})$ starting at those points.

Example 16.1.1

This is the vector field corresponding to the equation

$$ \mathbf{F}(x,y) = x\,\ihat + \frac{1}{2}y\,\jhat $$

Here, several values of $(x,y)$ are plotted and associated vectors are plotted, starting at each of those ordered pairs. The vectors are determined by $\mathbf{F}(x,y)$ at each point, with differing lengths due determined by $\left\langle\, x,\frac{1}{2}y\,\right\rangle$ at each point.

Example 16.1.2

In this example, the vector field is given by

$$ \mathbf{F}(x,y) = -y\,\ihat + x\,\jhat $$

The circles are not part of the vector field, but meant to give an visual indicator that makes the field easier to interpret. Specifically, we can see that each of the vectors appears to be tangent to the a circle centered at the origin. The position vector for each point $(x,y)$ is given by $\mathbf{x} = x\,\ihat + y\,\jhat$, and to check for perpendicularity we can compute the dot product

$$ \mathbf{x}\cdot\mathbf{F}(\mathbf{x}) = \left(x\,\ihat + y\,\jhat\right)\cdot \left(-y\,\ihat + x\,\jhat\right) = -xy + yx = 0 $$
Therefore, $\mathbf{F}(\mathbf{x})$ is perpendicular to $\mathbf{x}$ and hence tangent to the circle centered at the origin with radius $|\mathbf{x}| = \sqrt{x^2 + y^2}$. Additionally,

$$ \left|\mathbf{F}(\mathbf{x})\right| = \sqrt{(-y)^2 + x^2} = \sqrt{x^2+y^2} = |\mathbf{x}| $$
so the magnitude of the vector at each point is the same as the magnitude of the position vector.

Try out typing in these vector fields for yourself into this vector field generator made by Juan Carlos Ponce Campuzano on GeoGebra

16.1.2 Three-Dimensional Examples

It is often completely impractical to plot vector fields by hand, especially when the equations are complicated. You may be able to produce a reasonable sketch if the vector field equation is something simple like

$$ \mathbf{F}(x,y,z) = z\,\ihat $$
where the plot looks like

We see that the image here might have been able to be drawn by hand, since the vectors point along the positive or negative $x$-axis according to the value of $z$, but anything more complicated than this could easily be extremely difficult to sketch. Therefore, the use of technology is prevalent in creating these images.

Example 16.1.3

The vector field

$$ \mathbf{F}(x,y,z) = y\,\ihat + z\,\jhat + x\,\khat $$
cannot easily be visualized. Even with the help of technology, the picture may be confusing to interpret without the ability to change perspective or rotate the field.

Therefore, it may be necessary to use more sophisticated tools to get an impression of the vector field. The same GeoGebra creator as above has made a tool that will allow for the creation of 3D vector fields . Try changing the input to the function from Example 3 and rotate to get a better impression of the vector field's properties.

16.1.3 Force Fields

Force fields are vector fields in which the vectors of note are physical forces. There are two important physics examples that you will see often. These are gravity fields and electric fields .

Gravity Fields

The magnitude of the force of gravity on a body is given by Newton's Law of Universal Gravitation .

$$ \left|\mathbf{F}\right| = \dfrac{mMG}{r^2} $$
The values $m$ and $M$ are the masses of two objects, $G$ is the gravitational constant

$$ G = 6.67430\times 10^{-11} \dfrac{\text{N}\cdot \text{m}^2}{\text{kg}^2} $$
and $r$ is the distance between the centers of mass of the $m$ and $M$ bodies. If we want to write this as a vector equation in 3D, the standard approach is to imagine the mass $M$ to be at the origin and for $m$ to be located at a point with position vector $ \mathbf{x} = \langle\,x,y,z\,\rangle $. This means that $ r = |\mathbf{x}| $ and thus $r^2 = |\mathbf{x}|^2$. The direction of the gravitational force is towards the origin, and the unit vector at point $(x,y,z)$ pointing towards the origin is given by

$$ -\dfrac{\mathbf{x}}{|\mathbf{x}|} $$
so the gravitational force acting on the mass $m$ object at $\mathbf{x}$ is

$$ \mathbf{F}(\mathbf{x}) = -\dfrac{mMG}{|\mathbf{x}|^3}\mathbf{x} $$
The magnitude of this force is an example of the inverse square law , where the force is proportional to the reciprocal of the distance from the origin squared. A gravitational field looks like

where the magnitudes of the forces are greater the nearer you are to the origin.

The function $\mathbf{F}$ above is written in a compact form, but the component form looks like

$$ \mathbf{F}(\mathbf{x}) = -\dfrac{mMGx}{\left(x^2+y^2+z^2\right)^{3/2}}\,\ihat - \dfrac{mMGy}{\left(x^2+y^2+z^2\right)^{3/2}}\,\jhat - \dfrac{mMGz}{\left(x^2+y^2+z^2\right)^{3/2}}\,\khat$$
where we used $\mathbf{x} = x\,\ihat + y\,\jhat + z\,\khat$ and $|\mathbf{x}| = \sqrt{x^2 + y^2 + z^2}$ to expand the expression.

Electric Fields

Another very common force field in applications is an electric field. For an electric field, the mathematical setup is precisely the same as for gravitational fields. You have two charges $q$ and $Q$, with $q$ located at $\mathbf{x}$ and $Q$ at the origin. The electric constant $\varepsilon$, also called the permittivity of free space , is given by

$$ \varepsilon = 8.85419\times 10^{-12} \dfrac{\text{F}}{\text{m}} $$
The force of attraction (or repulsion) on a particle with charge $q$ at position $\mathbf{x}$ is given by

$$ \mathbf{F}(\mathbf{x}) = -\dfrac{\varepsilon qQ}{|\mathbf{x}|^3}\mathbf{x} $$
Commonly, physicists want to talk about the force per unit charge, and so $q$ is considered to be a "test charge" at position $\mathbf{x}$. The formula for this is

$$ \mathbf{E}(\mathbf{x}) = \dfrac{1}{q}\mathbf{F}(\mathbf{x}) = -\dfrac{\varepsilon Q}{|\mathbf{x}|^3}\mathbf{x} $$
and this vector field is called the electric field generated by $Q$.

16.1.4 Gradient Fields and Potentials

If $f$ is a scalar function of two or three variables, then its gradient $\nabla f$ is given by

$$ \begin{align*} \nabla f(x,y) &= f_x(x,y)\,\ihat + f_y(x,y)\,\jhat \\ \\ \nabla f(x,y,z) &= f_x(x,y,z)\,\ihat + f_y(x,y,z)\,\jhat + f_z(x,y,z)\,\khat \end{align*} $$
Clearly, this is a vector field, and it is referred to as a gradient vector field .

Exercise 16.1.1

Find the gradient field for

$$ f(x,y) = \sqrt{2x + 3y} $$

Check Your Work

$$ \nabla f(x,y) = \dfrac{1}{\sqrt{2x + 3y}}\,\ihat + \dfrac{3}{2\sqrt{2x + 3y}}\,\jhat $$

Follow Along

We need to compute the gradient of $f$, which has the form

$$ \nabla f(x,y) = f_x(x,y)\,\ihat + f_y(x,y)\,\jhat $$
Evaluating the gradient,

$$ \begin{align*} \nabla f(x,y) &= \pzi{x}\Big(\sqrt{2x + 3y}\Big)\,\ihat + \pzi{y}\Big(\sqrt{2x + 3y}\Big)\,\jhat \\ \\ &= \dfrac{(2x+3y)_x}{2\sqrt{2x + 3y}}\,\ihat + \dfrac{(2x+3y)_y}{2\sqrt{2x + 3y}}\,\jhat \\ \\ &= \dfrac{2}{2\sqrt{2x + 3y}}\,\ihat + \dfrac{3}{2\sqrt{2x + 3y}}\,\jhat \\ \\ &= \dfrac{1}{\sqrt{2x + 3y}}\,\ihat + \dfrac{3}{2\sqrt{2x + 3y}}\,\jhat \end{align*} $$

The most important theoretical detail concerning gradient fields is that any vector field that is also a gradient field is called conservative . This means that the vector field $\mathbf{F}$ is generated by the gradient of a scalar function $f$ and

$$ \mathbf{F} = \nabla f $$
In this case, we call $f$ a potential function for $f$. The use of the word "potential" in this case is a reference to potential energy, and will be discussed more deeply later.

Both gravitational fields and electric fields are conservative, as we can consider their potentials to be

$$ f(x,y,z) = \dfrac{mMG}{\sqrt{x^2 + y^2 + z^2}} $$
and

$$ g(x,y,z) = \dfrac{\varepsilon Q}{\sqrt{x^2 + y^2 + z^2}} $$
respectively.

Here we use the electric potential that results in the force per unit charge electric field $\,\mathbf{E}(\mathbf{x})$.

Exercise 16.1.2

Verify that the gravitational field $\mathbf{F} = \nabla f$ and the electric field $\mathbf{E} = \nabla g$ using the potential functions above.

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