Electricity-And-Magnetism · Unit 2 · Video 2 · Interactive Practice

Scalar Fields: Contour, Color and Relief Maps

IKey Formulas

Formula	Name	What it says
$T(r,\theta,\phi) = \big[T_0 + T_1\sin^2\theta + T_2(1+\sin\phi)\big]\,e^{-\alpha(r-R)}$	Planet temperature model	$r$ appears only in the exponential, $\theta$ only in $\sin^2\theta$ , $\phi$ only in $1+\sin\phi$
$\phi(x,y,z) = \dfrac{1}{\sqrt{x^2+(y+d)^2+z^2}} - \dfrac{1/3}{\sqrt{x^2+(y-d)^2+z^2}}$	A field given by an equation	Infinite at $(0,-d,0)$ and $(0,d,0)$
$\phi(x,y,0) = c$	Contour of the slice $z = 0$	One curve per constant value $c$
$f(x,y) = \dfrac{1}{\sqrt{x^2+(y-1)^2}} - \dfrac{1}{\sqrt{x^2+(y+1)^2}}$	Dipole relief	Peak at $(0,1)$ , well at $(0,-1)$ , $f = 0$ on the $x$ -axis

Key Insight: A contour map, a color map and a relief map of one slice carry the same information: a contour is a line of constant color, and a horizontal cut through the relief surface.

IIThe Planet Model, One Factor at a Time

On the equator, $T$ depends on $r$ only through $e^{-\alpha(r-R)}$ and on $\phi$ only through $1+\sin\phi$ .

💡 The panel uses $T_0 = 150\ \text{K}$ , $T_1 = 50\ \text{K}$ , $T_2 = 40\ \text{K}$ and $\alpha = 1/R$ ; on the equator $\theta = \pi/2$ , so $\sin^2\theta = 1$ throughout the slice.

IIIContours and Colors Carry the Same Information

Along each contour of $\phi(x,y,0)$ for $d = 1$ , the value of $\phi$ , and so its color, never changes.

Picture

IVRelief Map: Height Is the Value

A horizontal plane at height $c$ cuts the relief surface exactly along the contour $\phi(x,y,0) = c$ .

Field Height c 0.5 View angle 30°

💡 Heights beyond $\pm 2.5$ are cut flat, since each field is infinite at its singular points. The vertical direction is the value of the field, not the coordinate $z$ , so a three-dimensional field needs a stack of such slices.

VQuiz Questions

Problem 1 · The Point Under the Star

Given: $T(r,\theta,\phi) = \big[T_0 + T_1\sin^2\theta + T_2(1+\sin\phi)\big]e^{-\alpha(r-R)}$ with $T_0 = 150\ \text{K}$ , $T_1 = 50\ \text{K}$ , $T_2 = 40\ \text{K}$ . Find the ground-level temperature at the point directly under the star, $(r,\theta,\phi) = (R, \pi/2, \pi/2)$ .

✅ Correct!

T_0 + T_1 + 2T_2 = 150 + 50 + 80 = 280\ \text{K}

❌ Close, but… at

\phi = \pi/2

the factor

1 + \sin\phi

equals

2

, not

1

, so

T_2

counts twice.

❌ Not quite.

T_0 + T_1 = 200\ \text{K}

is the opposite point,

\phi = 3\pi/2

, where

1 + \sin\phi = 0

❌ Not quite.

230\ \text{K}

drops the

T_1

term, as at a pole (

\theta = 0

); on the equator

\sin^2\theta = 1

❌ Not quite. Evaluate each factor:

\sin^2(\pi/2) = 1

1 + \sin(\pi/2) = 2

e^{0} = 1

Show solution

At ground level $r = R$ , so the exponential is $e^{-\alpha \cdot 0} = 1$ . On the equator $\sin^2(\pi/2) = 1$ , and under the star $1 + \sin(\pi/2) = 2$ :

T(R,\tfrac{\pi}{2},\tfrac{\pi}{2}) = T_0 + T_1 + 2T_2 = 150 + 50 + 2(40) = 280\ \text{K}.

The opposite point has $1 + \sin(3\pi/2) = 0$ , so it sits at $T_0 + T_1 = 200\ \text{K}$ , colder by $2T_2 = 80\ \text{K}$ .

Problem 2 · Polar Angle, Not Latitude

Given: the same model and constants. Find the ground-level temperature at polar angle $\theta = \pi/6$ (measured down from the $z$ axis) and azimuth $\phi = \pi$ .

✅ Correct!

150 + 50\cdot\tfrac14 + 40\cdot 1 = 202.5\ \text{K}

❌ Close, but… that treats

\pi/6

as a latitude, giving

\sin^2(\pi/3) = \tfrac34

. The polar angle is measured from the

z

axis, so

\sin^2(\pi/6) = \tfrac14

❌ Not quite. The term is

\sin^2\theta = (\tfrac12)^2 = \tfrac14

, not

\sin\theta = \tfrac12

❌ Not quite. The model uses

1 + \sin\phi

, and

\sin\pi = 0

, so this factor is

1

, not

1 + \cos\pi = 0

❌ Not quite. Evaluate

\sin^2(\pi/6)

and

1 + \sin\pi

separately, with

r = R

Show solution

Ground level: $r = R$ , so the exponential factor is $1$ . Polar angle: $\sin(\pi/6) = \tfrac12$ , so $\sin^2\theta = \tfrac14$ . Azimuth: $1 + \sin\pi = 1 + 0 = 1$ .

T = 150 + 50\cdot\tfrac14 + 40\cdot 1 = 150 + 12.5 + 40 = 202.5\ \text{K}.

The point is $30\degree$ from the pole, close to it, so the $T_1$ term contributes only a quarter of its equatorial value.

Problem 3 · The Zero Contour

Given: $\phi(x,y,0) = \dfrac{1}{\sqrt{x^2+(y+1)^2}} - \dfrac{1/3}{\sqrt{x^2+(y-1)^2}}$ (so $d = 1$ ). Find where the contour $\phi = 0$ crosses the $y$ -axis.

Crossing between the two points ( $-1 < y < 1$ )

Crossing above the upper point ( $y > 1$ )

✅ Correct! The zero contour is the circle through

y = \tfrac12

and

y = 2

: centre

(0, \tfrac54)

, radius

\tfrac34

❌ Close, but… at the origin

\phi = 1 - \tfrac13 = \tfrac23 \ne 0

. The weights are unequal, so the zero is not at the midpoint.

❌ Not quite. That solves

r_1 = \tfrac13 r_2

. Setting

\phi = 0

gives

\tfrac{1}{r_1} = \tfrac{1}{3r_2}

, i.e.

r_1 = 3r_2

: the far point is three times as distant.

❌ Close, but…

(0, \tfrac54)

is the centre of the zero circle, not a point on it: there

r_1 = \tfrac94

and

r_2 = \tfrac14

, so

\phi = \tfrac49 - \tfrac43 \ne 0

❌ Check the first crossing. On the axis between the points,

r_1 = y + 1

and

r_2 = 1 - y

; solve

r_1 = 3r_2

❌ Check the second crossing. Above

y = 1

r_1 = y + 1

and

r_2 = y - 1

; solve

r_1 = 3r_2

Show solution

Write $r_1 = \sqrt{x^2+(y+1)^2}$ and $r_2 = \sqrt{x^2+(y-1)^2}$ . Then $\phi = 0$ means

\frac{1}{r_1} = \frac{1}{3r_2} \quad\Longleftrightarrow\quad r_1 = 3r_2.

Between the points ( $-1 < y < 1$ , $x = 0$ ): $r_1 = y + 1$ , $r_2 = 1 - y$ .

y + 1 = 3(1 - y) \;\Rightarrow\; 4y = 2 \;\Rightarrow\; y = \tfrac12.

Above the upper point ( $y > 1$ ): $r_1 = y + 1$ , $r_2 = y - 1$ .

y + 1 = 3(y - 1) \;\Rightarrow\; 2y = 4 \;\Rightarrow\; y = 2.

Check: at $y = 2$ , $\phi = \tfrac13 - \tfrac{1/3}{1} = 0$ . The whole zero contour is the circle with these two crossings as a diameter, centre $(0, \tfrac54)$ and radius $\tfrac34$ , enclosing the point $(0, 1)$ where $\phi$ is negative.

Problem 4 · Reading the Dipole Relief

Given: the dipole relief $f(x,y) = \dfrac{1}{\sqrt{x^2+(y-1)^2}} - \dfrac{1}{\sqrt{x^2+(y+1)^2}}$ . Find the height of the surface above $(0, 2)$ , and the curve that the horizontal plane at height $0$ cuts from it.

Height above $(0, 2)$

The cut at height $0$

✅ Correct!

f(0,2) = 1 - \tfrac13 = \tfrac23

, and

f = 0

wherever the two distances are equal: the whole

x

-axis.

❌ Not quite.

(0,2)

is nearer the positive term's point

(0,1)

, so the first term wins and

f > 0

❌ Not quite. The two terms are subtracted:

1 - \tfrac13

, not

1 + \tfrac13

❌ Not quite. That holds for

1/r

, which is positive everywhere. The dipole has a peak and a well, so between them it passes through

0

❌ Check the height. From

(0,2)

the distances are

1

(0,1)

and

3

(0,-1)

❌ Check the cut.

f = 0

requires

\sqrt{x^2+(y-1)^2} = \sqrt{x^2+(y+1)^2}

; square both sides.

Show solution

Height. From $(0,2)$ : distance to $(0,1)$ is $1$ , distance to $(0,-1)$ is $3$ , so

f(0,2) = \frac11 - \frac13 = \frac23.

Cut at height 0. $f = 0$ means the two distances are equal:

x^2 + (y-1)^2 = x^2 + (y+1)^2 \;\Rightarrow\; -2y = 2y \;\Rightarrow\; y = 0.

The plane at height $0$ meets the surface along the entire $x$ -axis, the contour $f = 0$ of the base map. Compare the two-point field of Problem 3, whose unequal weights bend the zero contour into a circle.

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