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Electromagnetic Waves, Antennas, and RF: Polarization of Electromagnetic Waves

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Electromagnetic Waves, Antennas, and RF: Polarization of Electromagnetic Waves

EM16 established that a plane electromagnetic wave in vacuum is transverse: the electric field E, magnetic field B, and propagation direction k are mutually perpendicular. EM17 and EM18 then showed that the same wave transports energy and momentum. The next question is more geometric:

|----------------------------------------------------------|
|How  does the transverse electric-field vector move in time?|
-----------------------------------------------------------
(1)

That motion defines the polarization of the wave.

For a monochromatic wave traveling in the +z direction, the most general fully polarized transverse electric field can be written as

|--------------------------------------|
|E(z,t) = ^xE0x cosψ  + ^yE0y cos(ψ + δ),|
----------------------------------------
(2)

where

ψ =  kz − ωt.
(3)

The amplitudes E0x and E0y and the relative phase δ completely determine the polarization state. Depending on those three quantities, the tip of E at a fixed point in space traces a line, a circle, or an ellipse [1, 2, 3, 4].

The central classification is

|------------------------------------------|
|linear ⊂ elliptical,    circular ⊂ elliptical.|
-------------------------------------------
(4)

Linear and circular polarization are therefore special limiting cases of the general polarization ellipse.

1 Polarization is a transverse-vector property

For propagation in the +z direction, Maxwell’s Equations require

E ⋅^z = 0,     B ⋅^z =  0.
(5)

Thus the electric field is confined to the transverse x-y plane. At a fixed spatial location z = z0, time changes the phase

ψ(t) = kz0 − ωt,
(6)

and therefore changes the two transverse components of E.

Polarization describes the curve traced by the tip of the electric-field vector in that transverse plane. It is not the path followed by a material particle, and it is not the trajectory of the wave through space.

The magnetic field is determined from

|-------------|
B  = 1-^k × E, |
-----c---------
(7)

so once the electric-field polarization is known, the magnetic-field polarization follows automatically.

2 Two perpendicular oscillations make one polarization state

Write the two transverse components as

Ex =  E0x cosψ,
(8)

and

Ey = E0y cos(ψ + δ).
(9)

The first component establishes one harmonic oscillation along x. The second establishes another along y. The polarization state is the geometric result of adding these two perpendicular oscillations.

Three pieces of information matter:

  1. the x-component amplitude E0x;
  2. the y-component amplitude E0y;
  3. the relative phase δ.

The absolute phase ψ determines where the field is in its cycle at a particular instant, but the polarization shape depends only on relative amplitudes and relative phase.

PIC

Figure. Two orthogonal amplitudes plus their relative phase determine the polarization ellipse. Linear and circular polarization are special cases.

3 Linear polarization

Suppose the two components are in phase,

δ = 0.
(10)

Then

E  = (E0x^x + E0y ^y)cos ψ.
(11)

The vector in parentheses is fixed. Only the scalar factor cos ψ changes with time. Therefore the electric field oscillates back and forth along one fixed line.

Define

     ∘ ----------
E0 =   E2  +  E2 ,
         0x    0y
(12)

and a polarization angle α such that

        E0x-             E0y-
cosα =  E   ,    sinα =  E   .
          0                0
(13)

Then

|--------------------------------|
|E =  E  (^xcos α + ^y sin α) cosψ. |
-------0-------------------------
(14)

The unit polarization vector is

----------------------
|                    |
^e-=--^xcos-α-+-^y-sin-α.-
(15)

The same linear state also occurs when the components are 180∘ out of phase,

δ = π,
(16)

because the two components still remain locked to a fixed line.

PIC

Figure. For linear polarization the electric-field tip remains on a fixed line in the transverse plane. The field reverses direction every half-cycle.

4 Example 1: a linearly polarized field at 30∘

Let

                         ∘
E0  = 10 V/m,      α = 30 .
(17)

Then

E0x = E0 cos 30∘ = 8.66 V/m, (18)
E0y = E0 sin 30∘ = 5.00 V/m. (19)

Thus

|--------------------------------------|
-E-=-(8.66^x-+-5.00^y-)cos(kz-−-ωt)V/m.---
(20)

The ratio

Ey-=  5.00 = tan 30∘
Ex    8.66
(21)

is constant whenever the common cosine factor is nonzero. That constant ratio is the mathematical signature of a fixed polarization line.

5 Relative phase bends the line into an ellipse

For general phase difference δ, begin with

Ex--= cos ψ.
E0x
(22)

Expand the y component:

Ey--
E0y = cos(ψ + δ) (23)
= cos ψ cos δ − sin ψ sin δ. (24)

Rearrange:

Ey--− -Ex-cos δ = − sinψ sinδ.
E0y   E0x
(25)

Square both sides:

(  E     E        )2
  --y- − --x-cos δ   = sin2ψ sin2δ.
  E0y    E0x
(26)

Since

                          (    )2
   2            2           Ex--
sin ψ  = 1 − cos ψ =  1 −   E0x   ,
(27)

we obtain

(                )2    [    (    )2 ]
  Ey--− -Ex- cosδ   =   1 −   Ex--    sin2 δ.
  E0y   E0x                   E0x
(28)

Expanding and collecting terms gives the polarization-ellipse equation

|--------------------------------------------|
|(    )2    (    )2                          |
|  Ex--  +    Ey--  −  2-ExEy--cos δ = sin2 δ.|
|  E0x        E0y       E0xE0y               |
----------------------------------------------
(29)

This single equation contains linear, circular, and general elliptical polarization as special cases.

6 Circular polarization

Circular polarization requires equal orthogonal component amplitudes,

E0x = E0y =  Ec,
(30)

and a quarter-cycle phase offset,

δ =  ± π.
       2
(31)

For

       π
δ =  + 2,
(32)

the field components are

Ex = Ec cosψ,
(33)

and

            (      )
E  = E  cos  ψ +  π- = − E  sinψ.
 y     c          2        c
(34)

Therefore

E2 + E2 =  E2(cos2 ψ + sin2 ψ) = E2 ,
 x     y    c                     c
(35)

so

|--------------------|
|E-| =-Ec-=-constant.-
(36)

The vector magnitude stays fixed while its direction rotates. The tip of E therefore traces a circle.

PIC

Figure. Equal component amplitudes in quadrature produce a circle. Unequal amplitudes or a nonquadrature phase offset generally produce an ellipse.

7 Handedness and convention

A rotating electric-field vector has a sense of rotation. Unfortunately, different communities have historically used different viewing conventions when assigning the labels “right-hand” and “left-hand.” This article follows the IEEE antenna convention: look in the direction of propagation. A field that appears to rotate clockwise is called right-hand circularly polarized, while counterclockwise rotation is called left-hand circularly polarized [5, 3].

Because convention errors are common, a safe technical description should state at least one of the following explicitly:

  • the propagation direction k;
  • the actual time-domain component equations;
  • the Jones vector and time convention;
  • the stated handedness convention.

The words “RHCP” or “LHCP” without a convention can be ambiguous when comparing antenna and optics literature.

8 Elliptical polarization is the general case

When the components are neither locked to the same line nor arranged as equal-amplitude quadrature components, the field tip generally traces an ellipse.

The ellipse can be described by

  • its major-axis amplitude Emaj;
  • its minor-axis amplitude Emin;
  • its orientation angle τ;
  • its sense of rotation.

The squared semi-axis amplitudes are

|-----------1-[------------∘------------------------------]--|
|E2maj,min = -- E20x + E20y ±   (E20x − E20y)2 + 4E20xE20y cos2δ .
------------2------------------------------------------------|
(37)

The plus sign gives the major axis and the minus sign gives the minor axis.

The orientation of the ellipse satisfies

|-----------------------|
|        2E0xE0y  cosδ  |
tan 2τ = ----2-----2--. |
-----------E-0x-−-E-0y----
(38)

The usual axial ratio is

|----------------|
|AR  = Emaj-≥  1.|
-------Emin-------
(39)

Thus

AR  = 1
(40)

is circular polarization, while

AR  →  ∞
(41)

approaches linear polarization.

A frequently used logarithmic measure is

|----------------------|
|AR    = 20 log  (AR  ).|
----dB---------10------
(42)

9 Example 2: determine an elliptical polarization state

Let

                                              ∘
E0x = 4.0 V/m,      E0y = 2.0V/m,       δ = 60 .
(43)

The orientation angle follows from

         2(4)(2)cos60 ∘    8    2
tan 2τ =  ----2----2----=  ---=  -.
            4  − 2        12    3
(44)

Hence

2τ ≈ 33.69∘,
(45)

so

|-----------|
τ ≈ 16.85 ∘. |
-------------
(46)

For the semi-axis amplitudes,

Emaj2 = 1-
2[     ∘ ----------------------]
 20 +   (12)2 + 4(16)(4)(0.25) (47)
= 17.211, (48)
Emin2 = 1-
2[     ∘ ----------------------]
 20 −   (12 )2 + 4(16)(4)(0.25) (49)
= 2.789. (50)

Therefore

Emaj  ≈ 4.149V/m,
(51)

and

Emin  ≈ 1.670V/m.
(52)

The axial ratio is

|--------------------|
|      4.149         |
|AR  ≈ ------≈  2.48.|
-------1.670---------
(53)

This is clearly elliptical: the axial ratio is finite but greater than unity.

10 Complex notation and the Jones vector

For monochromatic waves it is convenient to encode amplitude and relative phase in complex numbers. Using the real-field convention

            {         }
E (z,t) = ℜ  ^Eei(kz−ωt) ,
(54)

write the complex field amplitude as

^               iδ
E = E0x ^x + E0ye  ^y.
(55)

The corresponding Jones vector is

|----[------]--|
|       E0x    |
|J =  E   eiδ  .|
--------0y-----
(56)

If only polarization and not absolute field strength matters, use the normalized Jones vector

|-----------------[------]-|
|         1         E0x    |
^J =  ∘-----------  E  eiδ .|
|      E20x + E20y  0y     |
----------------------------
(57)

A common overall phase factor does not change the polarization state. Thus

J
(58)

and

eiϕ0J
(59)

represent the same polarization ellipse.

Jones vectors describe deterministic, fully polarized monochromatic fields. Partially polarized or incoherent fields require a statistical description such as Stokes parameters or a coherency matrix.

11 Useful Jones-vector special cases

Horizontal linear polarization can be represented as

|----------|
|     [1]  |
|^Jx =     .|
-------0----
(60)

Vertical linear polarization is

|----------|
|     [0]  |
|^Jy =     .|
-------1----
(61)

Linear polarization at angle α is

|------[-----]-|
|       cosα   |
^Jlin =  sin α  .|
----------------
(62)

Equal-amplitude quadrature states are

|---------[---]--|
|^     -1-- 1    |
|J± =  √ -- ±i  .|
---------2-------
(63)

The explicit sign and the adopted ei(kz−ωt) convention determine the rotation sense. This explicit complex representation is often safer than relying on a handedness label alone.

12 Stokes parameters as measurable polarization coordinates

For the Jones vector

    [      ]
J =    E0x   ,
     E0yeiδ
(64)

define the Stokes parameters in the convention used here by

S0 = E0x2 + E 0y2, (65)
S1 = E0x2 − E 0y2, (66)
S2 = 2E0xE0y cos δ, (67)
S3 = 2E0xE0y sin δ. (68)

For a fully polarized wave,

|-------------------|
S20 = S21 + S22 + S23. |
---------------------
(69)

The ellipse orientation and ellipticity angle χ can be written as

          S2-
tan 2τ =  S1,
(70)

and

        S3
sin 2χ = ---.
        S0
(71)

The limiting values

χ = 0
(72)

and

        ∘
|χ| = 45
(73)

correspond to linear and circular polarization, respectively. The sign of S3 and the associated handedness label must always be interpreted together with the stated convention.

13 Polarization does not change the basic plane-wave energy relation

For a vacuum plane wave,

     1
B  = --^k × E.
     c
(74)

Therefore at every instant

|B | = |E|.
       c
(75)

The instantaneous Poynting vector is

S = -1-E × B.
    μ0
(76)

For propagation in +z,

     E2
S =  --^z,
     Z0
(77)

where

      ∘ ---
Z0 =    μ0-.
        𝜖0
(78)

For the general two-component harmonic field,

  2     2    2
E   = E x + Ey.
(79)

The cycle averages satisfy

          2              E2
⟨E2x⟩ = E-0x-,    ⟨E2y⟩ = -0y.
         2                2
(80)

Hence

|----------------------|
|           E20x + E20y  |
|I = ⟨S ⟩ = ----------.|
---------------2Z0-----
(81)

The relative phase changes the polarization geometry, but for fixed component amplitudes it does not change the time-averaged intensity.

14 Example 3: intensity of a circularly polarized wave

Suppose

E0x =  E0y = 3.0V/m
(82)

with a 90∘ phase difference. Then

     32 +-32   ---18-----
I =   2Z0   =  2(376.73).
(83)

Therefore

|-------------−2-----2-|
I-≈--2.39-×-10---W/m---.-
(84)

Notice that the magnitude |E| = 3.0 V/m is constant in time for this circular state. Consequently the instantaneous power flow is also constant, unlike a single-component linearly polarized sinusoid whose instantaneous E2 oscillates between zero and its maximum.

15 Polarization and antennas

An antenna is sensitive not only to frequency and direction of arrival but also to polarization. A receiving antenna couples most strongly when its polarization state matches that of the incident field.

Let

^et
(85)

be the normalized polarization vector of the incoming wave and

^er
(86)

be the normalized polarization state to which the receiving antenna is matched. The polarization loss factor is

|----------------|
|         ∗    2 |
PLF--=--|^e-r ⋅^et|-.-
(87)

The complex conjugate is required because polarization states can contain relative phase.

The accepted received Power due to polarization alone is reduced by this factor:

|--------------------|
|Pr = Pr,matchedPLF.  |
---------------------
(88)

PIC

Figure. For two linear polarization directions separated by Δα, the polarization loss factor becomes cos 2Δα.

16 Linear-polarization mismatch

For linearly polarized transmit and receive states,

^et = ^x cosαt + ^y sin αt,
(89)

and

^er = ^x cosαr + ^y sinαr.
(90)

Their dot product is

er ⋅et = cos αr cos αt + sin αr sin αt (91)
= cos(αt − αr). (92)

Thus

|----------------|
|PLF  = cos2Δ α, |
-----------------
(93)

where

Δα =  αt − αr.
(94)

Important special cases are

Δα = 0∘ : PLF = 1, (95)
Δα = 45∘ : PLF = 1
--
2, (96)
Δα = 90∘ : PLF = 0. (97)

The corresponding mismatch loss in decibels is

|--------------------------|
|Lpol,dB = − 10 log10(PLF  ).|
---------------------------
(98)

17 Example 4: a 30∘ polarization mismatch

Suppose a linearly polarized wave reaches a linear receiving antenna with

Δα  = 30∘.
(99)

Then

           2  ∘
PLF  =  cos 30  = 0.75.
(100)

Thus the receiver captures 75% of the power that it would capture under perfect polarization alignment, all else being equal.

The mismatch loss is

Lpol,dB = − 10 log10(0.75) ≈ 1.25 dB.
(101)

Therefore

|------------------|
|Lpol,dB ≈ 1.25 dB. |
-------------------
(102)

18 Circular polarization and linear antennas

Take a normalized circular state

^ec = √1--(^x + i^y)
       2
(103)

and a normalized linear state at arbitrary angle α,

^eℓ = ^x cosα + ^y sinα.
(104)

Then

eℓ∗⋅e c = -1--
√2--(cos α + isinα ) . (105)

Its magnitude squared is

PLF  =  1-(cos2α + sin2α ) = 1.
        2                    2
(106)

Therefore

|----------------------|
|                   1  |
|PLFcircular- to- linear = --,|
--------------------2--
(107)

which corresponds to

|---------------|
Lpol ≈ 3.01 dB. |
----------------
(108)

The result is independent of the linear antenna’s orientation angle. A circularly polarized wave always contains equal power in any pair of orthogonal linear basis directions.

19 Matched and opposite circular polarization

For two identical normalized circular Jones vectors,

^er = ^et,
(109)

so

PLF  = 1.
(110)

For the opposite circular state, the two normalized Jones vectors are orthogonal in the complex inner-product sense. Therefore

|--------|
PLF--=--0-
(111)

for ideal opposite-handed circular polarizations.

Real antennas are never perfect. Their finite axial ratio, multipath, reflections, radome effects, platform blockage, and propagation medium can all convert one polarization state into another, so practical rejection is finite rather than infinite.

20 How circular polarization can be produced

The component picture immediately suggests a physical method. Use two orthogonal radiating elements that produce equal field amplitudes and drive them with a 90∘ phase difference.

Symbolically,

                        ∘
E0x =  E0y,    δ =  ±90 .
(112)

The same principle appears in crossed dipoles, turnstile antennas, quadrature-fed patches, and many other antenna structures. The antenna geometry creates two orthogonal transverse field components; the feed network or structural mode relationship establishes the relative phase.

This is an important bridge from plane-wave physics to antenna engineering: polarization is not an additional property pasted onto Maxwell’s equations. It emerges directly from the relative amplitude and phase of transverse field components generated by the antenna currents.

21 Propagation direction and polarization must be stated together

Handedness reversals can appear when the propagation direction is reversed or when the field is viewed from the opposite side. Therefore a polarization statement is incomplete unless the propagation direction is known.

For a plane wave,

^k ∥ E ×  B.
(113)

If k changes sign while the coordinate axes are held fixed, the relationship between phase progression and observed field rotation must be reconsidered. This is one reason polarization bookkeeping becomes important in reflected waves, radar, multipath channels, and satellite links.

22 Common mistakes

  1. Treating polarization as the path of the wave. Polarization is the motion of the transverse field vector at a point.
  2. Ignoring relative phase. Two perpendicular components with the same amplitudes can produce linear, circular, or elliptical polarization depending on δ.
  3. Calling every equal-amplitude state circular. Equal amplitudes are not enough; the phase difference must also be ±90∘.
  4. Forgetting that linear polarization is a limiting ellipse. The minor axis tends to zero and the axial ratio tends to infinity.
  5. Forgetting the viewing convention for handedness. Always specify propagation direction and convention.
  6. Using an ordinary dot product for complex polarization vectors. The receiving polarization overlap uses the complex-conjugate inner product.
  7. Confusing amplitude loss with power loss. Polarization loss factor is a power ratio, so decibel loss uses 10 log 10.
  8. Assuming polarization changes vacuum wave speed. In isotropic vacuum all polarization states propagate at the same speed c.

23 A compact polarization hierarchy

The main ideas can be summarized as

two transverse components →relative amplitude and phase (114)
→polarization ellipse (115)
→linear, circular, or elliptical state (116)
→Jones/Stokes representation (117)
→antenna polarization matching. (118)

The electromagnetic-wave sequence has now progressed from field generation and Maxwell’s equations to wave propagation, energy, momentum, and finally the internal transverse geometry of the wave itself.

24 What EM19 adds to the series

EM19 adds the following pieces to the electromagnetic-wave framework:

  • polarization defined as transverse electric-field-vector motion;
  • linear, circular, and elliptical polarization derived from two orthogonal harmonic components;
  • the polarization ellipse derived algebraically;
  • ellipse orientation and axial ratio;
  • an explicit handedness convention and warning about convention differences;
  • Jones vectors for fully polarized monochromatic fields;
  • a first Stokes-parameter representation;
  • the fact that average intensity depends on component amplitudes but not relative phase;
  • polarization loss factor and its antenna interpretation;
  • the cos 2Δα law for linear mismatch;
  • the 3 dB circular-to-linear mismatch result;
  • the connection between quadrature-fed orthogonal antenna modes and circular polarization.

A natural next step is to study how polarization changes at interfaces and in matter: reflection, transmission, birefringence, wave plates, and polarization-dependent propagation. In the RF branch, the same machinery leads directly to antenna polarization purity, axial ratio, multipath polarization change, and link-budget mismatch losses.

References

References

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

[2]   J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1999.

[3]   C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed., Wiley, 2016.

[4]   W. L. Stutzman and G. A. Thiele, Antenna Theory and Design, 3rd ed., Wiley, 2012.

[5]   IEEE, IEEE Standard for Definitions of Terms for Antennas, IEEE Std 145-2013, 2014.


"Electromagnetic Waves, Antennas, and RF: Polarization of Electromagnetic Waves" is owned by bloftin.
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Other names:  EM19
Also defines:  electromagnetic polarization, linear polarization, circular polarization, elliptical polarization, polarization ellipse, polarization handedness, axial ratio, Jones vector, polarization loss factor
Keywords:  electromagnetic polarization, linear polarization, circular polarization, elliptical polarization, polarization ellipse, Jones vector, relative phase, axial ratio, handedness, polarization loss factor, antenna polarization, RF, plane wave

Cross-references: speed, inner product, vectors, dot product, Power, instantaneous power, representation, matrix, parameters, phase factor, magnitude, square, unit, scalar, oscillation, particle, vector, Maxwell's Equations, traces, momentum, energy, EM18, EM17, magnetic field, electric field, electromagnetic wave, EM16

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Classification:
Physics Classification: 42.25.Ja (Polarization)
 41.20.Jb (Electromagnetic wave propagation; radiowave propagation )
 42.25.Bs (Wave propagation, transmission and absorption radiation interactions with plasma and 52.38-r Laser-plasma interactions-in pla)
 84.40.Ba (Antennas: theory, components and accessories )
 07.57.-c (Infrared, submillimeter wave, microwave and radiowave instruments and equipment )

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