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Next: Scalar and Vector Products Up: Introduction Previous: Introduction

Vector Fields and Fieldlines

Many processes in the real world involve continuous media with their properties varying continuously in 3-D space and time. An example would be the density of a gas or fluid which is written as $\rho
(x,y,z,t)$. This is different from MT2003 where the dynamics section only considered point masses or particles. Examples of continuous media are (i) fluids, such as water, beer,$ \cdots$, (ii) gases, for example air, clouds,$ \cdots$, (iii) electromagnetic fields, namely electric field, magnetic field and (iv) plasmas like the solar atmosphere, magnetosphere. A plasma is the fourth state of matter consisting of an ionised gas. In a plasma the temperature is so high that the electrons are no longer bound to a particular atom and the gas consists of electrons and positive ions.

A field is a function that describes a physical quantity at all points in space. For a scalar field this quantity is specified by a single variables at each point such as

For a vector field the physical quantity is specified by a vector, giving a direction as well as a magnitude. Examples are Note that we use bold face type for vectors but when writing a vector you MUST remember to underline vectors. Thus,

\begin{displaymath}
{{\bf i}} = \underline{i}.
\end{displaymath}

If you forget to underline vectors, I will mark you wrong!

Example 1. .1If

\begin{displaymath}
{\bf v} = y \sin x {\bf i} - {y^{2}\over 2}\cos x {\bf j},
\end{displaymath}

then

\begin{displaymath}
\vert{\bf v}\vert = ({\bf v}\cdot {\bf v})^{1/2} = \sqrt{y^{2}\sin^{2}x +
{y^{4}\over 4} \cos^{2}x}.
\end{displaymath}

A vector field line, such as a magnetic field line (for B) or a streamline (fluid or plasma velocity v) is drawn such that the tangent to the vector field line at any point is in the direction of the vector. Thus, in 2-D
\begin{displaymath}
{dy\over dx} = {B_{y}\over B_{x}},
\end{displaymath} (1.1)

as illustrated in Figure 1.1.

Figure 1.1: The field line and the tangent to the field line produces a vector in the direction of the magnetic field ${\bf B} =
B_{x}{\bf i} + B_{y}{\bf j}$.
\includegraphics [scale=0.7]{fundfig1.ps}

An alternative derivation of the equation of the field lines is given by the use of similar triangles, as shown in Figure 1.2. Here the distance along the curve is $s$ and the triangle on the left shows how the small distance $ds$ is related to the small horizontal and vertical distances $dx$ and $dy$ respectively. Using similar triangles we get

\begin{displaymath}
{dy\over dx} = {B_{y}\over B_{x}},
\end{displaymath}

as in Equation (1.1). Now, we may use the other relationships to get the parametric form of the field line as
\begin{displaymath}
{dy\over ds} = {B_{y}\over B}, \qquad {dx\over ds} = {B_{x}\over B}.
\end{displaymath} (1.2)

Figure 1.2: Using similar triangles we can obtain the equation of field lines in parametric form.
\includegraphics [scale=0.7]{fundfig2.ps}

In using the form (1.2) we make use of

\begin{displaymath}
B = \vert{\bf B}\vert = \sqrt{B_{x}^{2} + B_{y}^{2}}.
\end{displaymath}

In 3-D the equation of a field line is obtained by solving the equations

\begin{displaymath}
{dx\over ds} = {B_{x}\over B}, \qquad {dy\over ds} = {B_{y}\over
B}, \qquad {dz\over ds}= {B_{z}\over B},
\end{displaymath} (1.3)

and

\begin{displaymath}
B = \vert{\bf B}\vert = \sqrt{B_{x}^{2} + B_{y}^{2} + B_{z}^{2}}.
\end{displaymath}

The field line separation gives an indication of the strength of the field. When the field lines are closer together then the field is stronger and sketches of the field lines should reflect this.

Example 1. .2For ${\bf B} = y {\bf i}$, so that $B_{x}=y$, and $B_{y}=0$, the field lines are given by

\begin{displaymath}
{dy\over dx} = {B_{y}\over B_{x}} = {0\over y} = 0.
\end{displaymath}

Hence, the field lines are given by simply $y = \hbox{const.}$ and

\begin{displaymath}
\vert{\bf B}\vert = B = y.
\end{displaymath}

So the field is stronger for larger values of y. Note that the direction of the field lines is easily worked out by looking at the field components. We see that $B_{x}=y$ so that $B_{x}$ is positive when y is positive and negative when y is negative. This is shown in Figure 1.3.

Figure 1.3: Sketch of the field lines in Example 1.2.
\includegraphics [scale=0.7]{fundfig3.ps}

Example 1. .3For a fluid velocity ${\bf v} = -y{\bf i} + x {\bf j}$, the streamlines are given by

\begin{displaymath}
{dy\over dx} = {x\over -y} \qquad \Rightarrow \qquad \int x dx +
\int y dy = 0.
\end{displaymath}

Integrating we obtain

\begin{displaymath}
{1\over 2} x^{2} + {1\over 2} y^{2} = \hbox{const.} \qquad
\Rightarrow \qquad x^{2} + y^{2} = r^{2} = \hbox{const.}
\end{displaymath}

So the streamlines are given by concentric circles as shown in Figure 1.4.

Figure 1.4: The streamlines of the fluid velocity given in Example 1.3.
\includegraphics [scale=0.7]{fundfig4.ps}

Note the direction of the flow is given by the components of ${\bf v}$. For $x$ and $y$ positive, we have $v_{x} <0$ and $v_{y} > 0$.

We can calculate how individual fluid elements move in response to this flow from the fact that $v_{x} = dx/dt$ and so on. Hence,

\begin{displaymath}
v_{x} = {dx\over dt} = -y,\qquad v_{y} = {dy\over dt} = x,
\end{displaymath}

and so

\begin{displaymath}
{d^{2}x\over dt^{2}} = - {dy\over dt} = - x, \qquad \Rightarrow
\qquad {d^{2}x\over dt^{2}} + x = 0.
\end{displaymath}

The solution is

\begin{displaymath}
x = A \cos t + B \sin t \hbox{ and } y = A \sin t - B \cos t .
\end{displaymath}

There are some problems where a change of coordinate system simplifies the problem. For example, in 2-D polar coordinates $(r, \theta )$ the magnetic field lines are given by

\begin{displaymath}
{dr\over r d\theta} = {B_{r}\over B_{\theta}},
\end{displaymath} (1.4)

or in terms of the parameter $s$ we have

\begin{displaymath}
{dr\over ds} = {B_{r}\over B}, \hbox{ } r{d\theta\over ds} =
{B_{\theta}\over B},
\end{displaymath}

where $s$ is the distance along a field line.
next up previous
Next: Scalar and Vector Products Up: Introduction Previous: Introduction
Prof. Alan Hood
2000-11-06