Electromagnetic Waves, Antennas, and RF: Vector Fields for Wave Mechanics

EM01 introduced the idea that Electromagnetism is naturally written in the language of fields. A scalar field assigns one number to every point in space and time, while a vector field assigns a vector. The next step is to make that vector language precise enough that later electromagnetic equations can be read without ambiguity.

This article therefore develops only the vector mathematics needed immediately for wave mechanics:

|--------------------------------------------------------------| | components − → magnitude and direction − → dot product | | dot product −→ cross product −→ transverse wave geometry | ---------------------------------------------------------------|

(1)

Gradient, divergence, curl, and Maxwell’s equations are intentionally deferred. The purpose of EM02 is to establish the algebra and geometry that those later ideas will use [1, 2, 3, 4].

1 One vector versus a vector field

A single vector might be written

A.

(2)

It has a magnitude and a direction.

A vector field assigns such a vector to every point in space and time:

|------------| |A = A(r,t).| --------------

(3)

Using Cartesian coordinates,

r = xˆx + yyˆ+ zˆz,

(4)

so the field may also be written

A = A (x,y,z, t).

(5)

The distinction is important:

A can mean one particular vector;
A(r,t) emphasizes that the vector can change from place to place and from time to time.

For an electric field, for example,

E (r,t)

(6)

returns one electric-field vector when a position and time are specified.

2 Cartesian basis vectors

In three-dimensional Cartesian coordinates, we use the unit basis vectors

ˆx, ˆy, ˆz.

(7)

Each has magnitude one:

|ˆx| = |ˆy| = |ˆz| = 1.

(8)

They point in mutually perpendicular directions.

A vector may therefore be decomposed as

|-----------------------| A = Ax ˆx + Ay ˆy + Azˆz. | -------------------------

(9)

The quantities

Ax, Ay, Az

(10)

are scalar components. They may be positive, negative, or zero.

The basis vectors carry the direction information.

Figure. A vector can be reconstructed by adding its Cartesian component vectors. The scalar numbers A_x and A_y specify signed amounts along the basis directions.

In two dimensions,

A = Ax ˆx + Ay ˆy.

(11)

In three dimensions, the z component is added in exactly the same way.

3 Components can themselves be fields

For a vector field,

A (r,t) = Ax(r,t)ˆx + Ay (r,t)ˆy + Az (r,t)zˆ.

(12)

Each component is therefore a scalar field.

This point is extremely useful. A vector field can be viewed as three coupled scalar fields:

|----------------------------------------| |A (r,t) ← → {Ax (r,t),Ay(r,t),Az(r,t)}. | -----------------------------------------

(13)

For example,

E(z,t) = ˆxE0 cos(kz − ωt )

(14)

has components

E_x(z,t)	= E₀ cos(kz − ωt),	(15)
E_y(z,t)	= 0,	(16)
E_z(z,t)	= 0.	(17)

The field varies with z and t, but every field vector points along either +x or −x.

4 Magnitude of a vector

For

A = Ax ˆx + Ay ˆy + Azˆz,

(18)

the magnitude is

|----------------------| | ∘ --2----2----2-| |A | = A x + A y + Az. ------------------------

(19)

This is the three-dimensional Pythagorean theorem.

The magnitude is never negative:

|A | ≥ 0.

(20)

A negative component does not mean a negative magnitude. It means the vector points partly in the negative direction of that coordinate axis.

5 Unit vector in an arbitrary direction

Suppose A≠0. Dividing the vector by its magnitude produces a unit vector pointing in the same direction:

|----------| |ˆA = -A--.| | |A | | -----------

(21)

By construction,

|Aˆ| = 1.

(22)

Therefore every nonzero vector can be written as

|-----------| A = |A |Aˆ. | -------------

(23)

This separation into magnitude and direction will recur throughout electromagnetism.

6 The Cartesian basis is orthonormal

The word orthonormal combines two ideas:

each basis vector has unit magnitude;
different basis vectors are perpendicular.

Thus

ˆx ⋅ ˆx = ˆy ⋅ ˆy = ˆz ⋅ ˆz = 1,

(24)

while

ˆx ⋅ ˆy = ˆy ⋅ ˆz = ˆz ⋅ ˆx = 0.

(25)

The dot product makes these statements precise.

7 The dot product

For two vectors

A = Ax ˆx + Ay ˆy + Az ˆz

(26)

and

B = Bx ˆx + By ˆy + Bzˆz,

(27)

the Cartesian dot product is

|------------------------------| A ⋅ B = AxBx + AyBy + AzBz. | --------------------------------

(28)

The result is a scalar.

The same dot product also has the geometric form

|--------------------| A--⋅ B-=-|A-||B-|cos𝜃,-

(29)

where 𝜃 is the angle between the vectors.

Figure. The dot product measures alignment. The factor |A| cos 𝜃 is the scalar projection of A onto the direction of B when B is used only to define a direction.

8 Interpreting the sign of the dot product

Because

A ⋅ B = |A ||B |cos𝜃,

(30)

the sign tells us about relative orientation.

0 ≤ 𝜃 < π, 2

(31)

then

cos 𝜃 > 0

(32)

and the dot product is positive.

π 𝜃 = 2,

(33)

then

|----------| -A-⋅ B-=-0.-

(34)

π- < 𝜃 ≤ π, 2

(35)

then the dot product is negative.

Thus the dot product provides a direct test for perpendicularity of nonzero vectors.

9 Projection onto a unit direction

Let n be a unit vector. The scalar component of A along n is

|------------| |A ∥ = A ⋅ ˆn.| -------------

(36)

The corresponding vector projection is

|---------------| A ∥ = (A ⋅ ˆn)ˆn. | ----------------

(37)

The part perpendicular to n is

|--------------| A = A − A .| --⊥----------∥--

(38)

This decomposition will later be useful for separating field components normal and tangential to surfaces and for distinguishing polarization components.

10 The wave vector

The Wave Mechanics series used the scalar Wavenumber

2π k = -λ-.

(39)

In three dimensions we also need a propagation direction. We therefore define the wave vector

|--------| -k-=-knˆ,-|

(40)

where n points in the propagation direction.

The magnitude is

2π |k| = k = ---. λ

(41)

The vector k therefore contains two pieces of information:

its magnitude gives spatial phase rate;
its direction gives the direction of propagation.

For propagation along +z,

k = k ˆz.

(42)

11 Why the plane-wave phase contains a dot product

A three-dimensional plane-wave phase is written

|----------| -k-⋅ r −-ωt.

(43)

To see why, write

k = kxˆx + kyˆy + kz ˆz

(44)

and

r = xˆx + yyˆ+ zˆz.

(45)

Then

|------------------------| -k-⋅ r-=-kxx-+-kyy-+-kzz.|

(46)

For a wave traveling only along +z,

kx = 0, ky = 0, kz = k,

(47)

k ⋅ r = kz.

(48)

The familiar one-dimensional phase

kz − ωt

(49)

is therefore a special case of the three-dimensional phase.

12 Surfaces of constant phase

At a fixed time, points satisfying

k ⋅ r = C

(50)

for some constant C lie on a plane perpendicular to k.

Thus k is normal to a plane wavefront.

This will later make Huygens’ principle, apertures, phased arrays, and antenna far-field phase much easier to express mathematically.

13 The cross product

The cross product of two vectors is another vector:

|------------| |C = A × B. | --------------

(51)

Its magnitude is

|----------------------| |A × B | = |A ||B |sin 𝜃. ------------------------

(52)

Its direction is perpendicular to both A and B.

The right-hand rule determines which of the two possible perpendicular directions is chosen.

Figure. The cross product is perpendicular to the plane containing A and B. The dot-in-circle symbol indicates a vector pointing out of the page.

14 Cross product in Cartesian components

For

A = Ax ˆx + Ay ˆy + Az ˆz

(53)

and

B = Bx ˆx + By ˆy + Bzˆz,

(54)

the cross product is

|------------------------------------------------------------------| -A--×-B-=--(AyBz-−--AzBy-)ˆx-+-(AzBx--−-AxBz-)ˆy-+-(AxBy--−-AyBx-)ˆz.-|

(55)

A useful memory pattern comes from the right-handed Cartesian basis:

xˆ× ˆy = ˆz,

(56)

yˆ× ˆz = ˆx,

(57)

ˆz × xˆ= ˆy.

(58)

Reversing the order changes the sign:

|------------------| A × B = − B × A. | --------------------

(59)

This is called anticommutativity.

15 Parallel and perpendicular cases

If A and B are parallel, then

𝜃 = 0

(60)

and

|------------| |A × B = 0. | -------------

(61)

If they are perpendicular, then

𝜃 = π- 2

(62)

and

|A × B | = |A ||B |.

(63)

The dot and cross products therefore respond differently to angle:

|----------------------------------| | | | | |---geometry----|-A-⋅ B-||A--×-B-|-| | parallel ||A ||B || 0 | | perpendicular | 0 | |A ||B | | -----------------------------------

(64)

16 A first electromagnetic transverse triad

Later, Maxwell’s equations will show that an ideal uniform plane electromagnetic wave in a simple isotropic medium has mutually perpendicular electric field, magnetic field, and propagation direction.

For the moment, we use that only as a geometric preview:

|-------------------------------| E--⊥-H,-----E--⊥-k,-----H--⊥-k.--

(65)

The dot-product tests are therefore

|----------------------------------------| -E-⋅ H-=-0,-----E-⋅ k-=-0,----H--⋅ k-=-0.|

(66)

Their orientation also satisfies

|------------------------------------------| |E × H points in the propagation direction.| --------------------------------------------

(67)

Figure. A right-handed transverse field triad. Here E points along +x, H points out of the page along +y, and k points along +z. Detailed electromagnetic field relations are deferred; EM02 uses this only to establish the geometry.

This geometry is central to radio waves. It will later connect the field vectors to energy transport through the Poynting vector.

17 Field direction is not propagation direction

Consider

E (z,t) = ˆxE0 cos(kz − ωt).

(68)

The vector x tells us the electric-field direction.

The phase

kz − ωt

(69)

tells us the pattern propagates toward increasing z.

Therefore

|--------------------------------------| |field direction ⁄= propagation direction.| ----------------------------------------

(70)

For this example,

E ⊥ k.

(71)

This distinction is one of the most important geometric habits to establish before studying electromagnetic waves.

18 Worked Example 1: resolve a vector into components

Let

A = 3ˆx − 4ˆy.

(72)

The components are

A = 3, A = − 4. x y

(73)

The negative y component means the vector points partly toward −y.

Its magnitude is

\|A\|	=	(74)
	=	(75)
	= 5.	(76)

Thus

|--------| ||A | = 5.| ----------

(77)

The corresponding unit vector is

A	=	(78)
	= x −y.	(79)

|-----------------| Aˆ = 0.6ˆx − 0.8ˆy. | -------------------

(80)

19 Worked Example 2: dot product and perpendicularity

Let

A = 2ˆx + 3 ˆy

(81)

and

B = 3ˆx − 2ˆy.

(82)

Then

A ⋅ B	= (2)(3) + (3)(−2)	(83)
	= 6 − 6	(84)
	= 0.	(85)

Therefore

|--------| -A-⊥--B.-|

(86)

We did not need to draw the vectors or explicitly compute the angle. The dot product provided the test directly.

20 Worked Example 3: projection onto a measurement axis

Suppose

E = 4ˆx + 3ˆy V/m

(87)

and a sensor measures only the component along

1 ˆn = √---(ˆx + ˆy). 2

(88)

The scalar measured component is

E_∥	= E ⋅n	(89)
	= ⋅	(90)
	= V/m.	(91)

Therefore

|---------------------------| | -7-- | E ∥ = √2--V/m ≈ 4.95 V/m. | -----------------------------

(92)

The dot product extracts the part of the field aligned with the sensor axis.

21 Worked Example 4: evaluate a three-dimensional plane-wave phase

Let

k = 2ˆx + 3ˆz rad/m

(93)

and

r = 4ˆx + 1ˆy + 5ˆz m.

(94)

Then

k ⋅ r	= (2)(4) + (0)(1) + (3)(5)	(95)
	= 8 + 15	(96)
	= 23 rad.	(97)

Thus the spatial phase at that point is

|--------------| |k ⋅ r = 23 rad.| ---------------

(98)

The absent y component of k means displacement along y does not change the phase in this example.

22 Worked Example 5: compute a cross product

Let

A = 2ˆx

(99)

and

B = 3 ˆy.

(100)

Using

xˆ× ˆy = ˆz,

(101)

we obtain

A × B	= (2)(3)(x ×y)	(102)
	= 6z.	(103)

Therefore

|------------| A × B = 6ˆz.| --------------

(104)

Reversing the order gives

|--------------| B × A = − 6ˆz.| ----------------

(105)

23 Worked Example 6: verify a transverse triad

Suppose

E = E0ˆx,

(106)

H = H0yˆ,

(107)

and

k = k ˆz.

(108)

The dot products are

E ⋅ H = E0H0 ˆx ⋅ ˆy = 0,

(109)

E ⋅ k = 0,

(110)

and

H ⋅ k = 0.

(111)

Thus the three directions are mutually perpendicular.

The cross product is

E × H	= E₀H₀x ×y	(112)
	= E₀H₀z.	(113)

Therefore

|-----------| E × H ∥ k. | -------------

(114)

24 Worked Example 7: identify field and propagation directions

Consider

E (r,t) = ˆyE0 cos(4x − 3z − ωt).

(115)

The electric field points along

|----| -±yˆ.--

(116)

The spatial phase is

4x − 3z.

(117)

Therefore

|------------| k = 4ˆx − 3ˆz.| --------------

(118)

Its magnitude is

∘ ----------- |k | = 42 + (− 3)2 = 5 rad/m.

(119)

The propagation unit vector is

|--------------| | 4 3 | |ˆk = --ˆx − --ˆz.| -----5-----5---

(120)

Finally,

ˆ ˆy ⋅k = 0,

(121)

so the electric field is transverse to the propagation direction.

25 Common mistakes

Mistake: treating A_xx as the same thing as the scalar A_x. One is a vector component contribution; the other is a scalar coefficient.
Mistake: assuming a negative component means a negative magnitude. Magnitude is nonnegative; component signs encode direction.
Mistake: confusing k with k. The scalar k is the magnitude of the wave vector, while k also contains propagation direction.
Mistake: forgetting that a dot product produces a scalar while a cross product produces a vector.
Mistake: assuming the cross product is commutative. Reversing the order reverses the sign.
Mistake: confusing field direction with propagation direction. A transverse wave can propagate in one direction while its field vectors point in perpendicular directions.
Mistake: using the plane-wave relation E ⊥ H ⊥ k as though it described every electromagnetic field everywhere. EM02 uses it only as a preview of the ideal uniform plane-wave geometry that will be derived later.

26 What EM02 adds to the field picture

EM01 introduced

A (r,t)

(122)

as a vector-valued field.

EM02 now gives that notation operational meaning:

|----------------------| A--=-Ax-ˆx-+-Ay-ˆy-+-Az-ˆz-

(123)

with magnitude

|------∘---------------| |A | = A2x + A2y + A2z. ------------------------

(124)

The dot product measures alignment and projection:

|--------------------| A ⋅ B = |A ||B |cos𝜃.| ----------------------

(125)

The cross product measures oriented perpendicular geometry:

|----------------------| |A--×-B-| =-|A-||B-|sin-𝜃.

(126)

The three-dimensional plane-wave phase becomes

|----------| -k-⋅ r −-ωt.

(127)

And the electromagnetic-wave geometry we will later derive is summarized by

|----------------------------| |E ⊥ H ⊥ k, E × H ∥ k.| -----------------------------

(128)

EM03 next uses this vector foundation to introduce spatial derivatives of fields: gradient, divergence, and curl.

References

[1] David J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017.

[2] Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 2, OpenStax, 2016, chapters on electric and magnetic fields and electromagnetic waves.

[3] Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Volume II, Addison-Wesley, 1964, chapters introducing vector electromagnetic fields.

[4] Massachusetts Institute of Technology, 8.02 Physics II: Electricity and Magnetism, MIT OpenCourseWare, materials on vector fields and electromagnetic phenomena.

[5] H. M. Schey, Div, Grad, Curl, and All That, 4th ed., W. W. Norton & Company, 2005.

[6] Frank S. Crawford, Jr., Waves, Berkeley Physics Course, Volume 3, McGraw-Hill, 1968.

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