Vector fields — definition, visualization: Question bank

Before we start, one reminder of the vocabulary this page leans on, so no symbol is used unseen:

True or false — justify

A vector field is the same object as a single vector.

False. A single vector is one arrow; a field is a rule giving a possibly different arrow at every point. The single vector is only the special case of a constant field.

F (x, y) = (3, 3)

is a vector field.

True. Constant fields are still fields — the rule just happens to output the same arrow

(3, 3)

everywhere. Same-dimension in/out still holds.

Every function

R^{2} \to R^{3}

can be drawn as arrows attached to their points.

False. It's a legitimate function, but the "arrow rooted at its point" picture only works cleanly when input and output share a dimension, since a 3D arrow doesn't live in the 2D plane of the point.

If two fields have the same magnitude

∣ F ∣

at every point, they are the same field.

False. Magnitude ignores direction; e.g.

(x, y)

and

(- y, x)

both have magnitude

r

everywhere yet one is radial and the other rotational.

The field

F (x, y) = (- y, x)

points in the same direction as the position vector

(x, y)

False. Their dot product is

- y x + x y = 0

, so

F

is perpendicular to the radius — tangent to the circle, not along it (see Divergence and Curl).

Any radial-looking field is automatically a gradient (conservative) field.

False.

(x, y) = \nabla (\frac{1}{2} (x^{2} + y^{2}))

is conservative, but "points outward" alone doesn't guarantee it; conservativeness is a condition on the components, tested via Conservative Fields and Potential Functions.

F = \nabla f

, then reversing all arrows still gives a gradient field.

True.

- F = \nabla (- f)

, so the reversed field is the gradient of

- f

— still conservative.

A constant nonzero field can be a gradient field.

True.

(a, b) = \nabla (a x + b y)

, so uniform fields are gradients of linear scalar functions.

Longer arrows in a drawing always mean the field is "more important" there.

False. Length encodes magnitude (strength), not importance; and in same-length plots we deliberately drop length and use color, so length can be meaningless.

Spot the error

F (x, y) = (x, y)

has magnitude

x + y

, since you add the components."

Wrong: magnitude is

x^{2} + y^{2} = r

, not

x + y

. You take the square root of the sum of squares, never the raw sum (which can even be negative).

"At the origin the rotation field

(- y, x)

still points somewhere, say up."

(0, 0)

the field is

(- 0, 0) = (0, 0)

— the zero vector, which has no direction. The origin is a degenerate point where the arrow vanishes.

"The inverse-square field

c r /∣ r ∣^{3}

is defined everywhere in space."

It blows up at

r = 0

where

∣ r ∣ = 0

makes division undefined. The origin must be excluded from the domain

D

F /∣ F ∣

is a safe way to always extract direction."

It fails wherever

∣ F ∣ = 0

(division by zero), e.g. the center of a radial or rotational field. Zero-magnitude points have no unit direction.

"Since

\nabla (x^{2} + y^{2}) = (2 x, 2 y)

and

(x, y)

point the same way, they are literally equal fields."

Same direction at each point, but different magnitude:

(2 x, 2 y)

is twice as long. Equal direction is not equal vector.

"A field with all arrows the same length must be a constant field."

No — the rotation field drawn in same-length style has equal drawn arrows but they point different directions at each point. Equal magnitude

\neq =

equal vector.

"Because gravity 'pulls in', the inverse-square field must have

c > 0

Inward pull needs the arrow to oppose

r

, so

c < 0

; a positive

c

would push outward (like a repulsive charge in Flux and the Divergence Theorem).

Why questions

Why must the output dimension of

F

equal the input dimension for the arrow picture?

So the output arrow can be attached at the input point in the same space — a plane-point needs a plane-arrow, otherwise "arrow rooted here" is meaningless.

Why do we often draw all arrows the same length and use color instead?

Because near strong regions long arrows overlap and hide the pattern; fixing length keeps direction legible and offloads magnitude to color.

Why does a dot product of zero prove the rotation field is "pure swirl"?

F \cdot (x, y) = 0

means

F

is perpendicular to the radius, hence tangent to the circle through that point, so the flow circulates without moving toward or away from the center.

Why is velocity along a moving curve naturally a vector field?

At each point of the path there is both a location and a direction+speed of motion, exactly the "arrow at every point" data — see Parametric Curves and Velocity.

Why does the inverse-square form use

r /∣ r ∣^{3}

rather than

r /∣ r ∣^{2}

r /∣ r ∣

is the unit outward direction; dividing that by

∣ r ∣^{2}

gives magnitude

\propto 1/ r^{2}

, and combining the two divisions yields

∣ r ∣^{3}

in the denominator.

Why can two different scalar functions never give genuinely different gradient fields if they differ by a constant?

\nabla (f + C) = \nabla f

because the derivative of a constant is zero; adding a constant shifts the "height" but not the slope, so the arrow field is identical.

Why is

∣ F ∣ = P^{2} + Q^{2}

rather than

∣ P ∣ + ∣ Q ∣

It's the Pythagorean length of the arrow with legs

P

and

Q

; the straight-line arrow length uses squares and a square root, not a sum of absolute values (which measures a different, "taxicab" distance).

Edge cases

At a point where

F = 0

, what does the drawn arrow look like?

It shrinks to a point — zero length, undefined direction. These are the critical/equilibrium points of the field (centers of radial and rotational fields).

What happens to the radial field

(x, y)

as you move very far from the origin?

Its magnitude

r = x^{2} + y^{2}

grows without bound, so arrows get longer and longer — the field is unbounded at infinity.

What happens to the inverse-square field

∣ F ∣ = ∣ c ∣/ r^{2}

r \to 0^{+}

and as

r \to \infty

r \to 0^{+}

it diverges to infinity (singular at the source); as

r \to \infty

it decays to zero — strong up close, negligible far away.

Is a field defined on a domain with a hole (like

R^{2}

minus the origin) still a valid vector field?

Yes —

D

may be any region; excluding bad points (where components are undefined) is exactly how we handle singular fields like inverse-square.

Can a gradient field have a zero vector somewhere?

Yes — wherever

\nabla f = 0

, i.e. at a maximum, minimum, or saddle of

f

; e.g.

\nabla (x^{2} + y^{2}) = (2 x, 2 y)

vanishes at the origin, the minimum.

Along the positive

x

-axis where

y = 0

, what does the rotation field

(- y, x)

output?

(- 0, x) = (0, x)

, a purely vertical arrow (up for

x > 0

, down for

x < 0

) — perpendicular to the axis, confirming the tangent-to-circle behaviour even on the axes.

Recall One-line summary

A field is a rule of arrows, not one arrow; magnitude is Pythagorean; direction dies at zero-vector points; and singular fields simply drop the bad points from their domain.

Connections

Gradient and Directional Derivatives — which fields are gradients.
Divergence and Curl — swirl vs. spread diagnostics behind the traps.
Conservative Fields and Potential Functions — the real test for "is it $\nabla f$ ?".
Flux and the Divergence Theorem — sign of $c$ and outward/inward flow.
Parametric Curves and Velocity — velocity as a field along a path.
Line Integrals — reversing arrows flips the sign of work.