Electromagnetic Waves: Deriving the Electromagnetic Wave Equation from Maxwell's Equations

Electromagnetic Waves, Antennas, and RF: Deriving the Electromagnetic Wave Equation from Maxwell’s Equations

EM15 assembled Maxwell’s Equations into one coupled field theory. The next question is the central one for electromagnetic-wave physics:

|------------------------------------------------------------------------------------| |Can Maxwell ’s equations support disturbances that propagate through empty space? | -------------------------------------------------------------------------------------

(1)

The answer is yes. In a source-free vacuum, the two curl equations combine into second-order partial differential equations for both the electric field E and magnetic field B. Those equations have exactly the mathematical form of wave equations, and their propagation speed is

|------------| | ---1-- | |c = √ μ-𝜖-. | ---------00--

(2)

This article derives that result carefully and then specializes it to a one-dimensional sinusoidal plane wave. Along the way we will see why electromagnetic waves are transverse, why E and B are perpendicular to each other, why they oscillate in phase in vacuum, and why their amplitudes satisfy E₀ = cB₀ [1, 2, 3, 4, 5].

1 Start from the source-free Maxwell system

In a region containing no free charge and no conduction current,

ρ = 0, J = 0.

(3)

The four differential Maxwell equations reduce to

∇ ⋅ E = 0,

(4)

∇ ⋅ B = 0,

(5)

∂B-- ∇ × E = − ∂t ,

(6)

and

∂E ∇ × B = μ0 𝜖0 ---. ∂t

(7)

The first two equations are divergence constraints. The last two are dynamical curl equations. The wave equations emerge when the two curl equations are combined.

2 The strategy: take one more curl

Faraday’s law already couples spatial variation of E to time variation of B:

∇ × E = − ∂B--. ∂t

(8)

To obtain an equation involving E alone, take the curl of both sides:

( ) ∂B ∇ × (∇ × E) = − ∇ × ---- . ∂t

(9)

Assuming the fields are sufficiently smooth, spatial and temporal differentiation commute, so

( ∂B ) ∂ ∇ × ---- = ---(∇ × B ). ∂t ∂t

(10)

Therefore,

|------------------------------| | ∂-- | |∇ × (∇ × E ) = − ∂t(∇ × B ).| -------------------------------

(11)

Now Ampere–Maxwell can eliminate B from the right-hand side.

Figure. The electric-field wave equation is obtained by taking the curl of Faraday’s law, using the curl–curl identity, imposing the vacuum divergence constraint, and substituting the vacuum Ampere–Maxwell equation.

3 The curl–curl vector identity

The key vector-calculus identity is

|------------------------------2---| -∇-×--(∇--×-A-)-=-∇-(∇-⋅ A-)-−-∇-A.-|

(12)

Here ∇²A means the vector Laplacian applied component by component in Cartesian coordinates:

∇2A = (∇2Ax )ˆx + (∇2Ay )ˆy + (∇2Az )ˆz.

(13)

Apply the identity to E:

∇ × (∇ × E) = ∇ (∇ ⋅ E ) − ∇2E.

(14)

In source-free vacuum,

∇ ⋅ E = 0,

(15)

∇ (∇ ⋅ E ) = 0.

(16)

Hence,

|-------------------2---| ∇--×-(∇-×--E)-=-−-∇--E.--

(17)

This is the step where Gauss’s Law for electricity enters the wave-equation derivation.

4 Derivation of the electric-field wave equation

Return to

∇ × (∇ × E ) = − ∂-(∇ × B ). ∂t

(18)

Use the vacuum Ampere–Maxwell law,

∂E- ∇ × B = μ0 𝜖0 ∂t .

(19)

Then

−(∇× B)	= −	(20)
	= −μ₀𝜖₀.	(21)

Using the result from the previous section,

∂2E − ∇2E = − μ0𝜖0--2-. ∂t

(22)

Cancel the minus signs:

∂2E ∇2E = μ0𝜖0---2 . ∂t

(23)

Equivalently,

|--------------------| | 2 ∂2E- | ∇ E − μ0𝜖0∂t2 = 0.| ----------------------

(24)

This is the electromagnetic wave equation for the electric field in source-free vacuum.

5 Derivation of the magnetic-field wave equation

The same reasoning can be repeated starting from the vacuum Ampere–Maxwell law:

∇ × B = μ0 𝜖0 ∂E-. ∂t

(25)

Take the curl:

-∂- ∇ × (∇ × B ) = μ0𝜖0∂t (∇ × E ).

(26)

Use Faraday’s law,

∂B ∇ × E = − ----, ∂t

(27)

2 ∇ × (∇ × B ) = − μ0𝜖0∂-B-. ∂t2

(28)

The curl–curl identity gives

∇ × (∇ × B ) = ∇ (∇ ⋅ B ) − ∇2B.

(29)

Since

∇ ⋅ B = 0,

(30)

we obtain

2 ∂2B-- − ∇ B = − μ0𝜖0∂t2 .

(31)

Therefore,

|------------2--------| ∇2B − μ0𝜖0∂-B--= 0. | ------------∂t2--------

(32)

Thus both fields obey the same propagation equation.

6 Comparison with the generic wave equation

A scalar wave field ψ(r,t) propagating at speed v satisfies

|------------------| | 2 1 ∂2ψ | ∇ ψ − -2---2 = 0.| --------v--∂t-------

(33)

Compare this with the electric-field equation

2 ∇2E − μ0𝜖0∂-E- = 0. ∂t2

(34)

The coefficients must correspond:

1--= μ 𝜖 . v2 0 0

(35)

Therefore,

|------------| | 1 | |v = √------.| -------μ0-𝜖0--

(36)

For electromagnetic fields in vacuum this speed is denoted by c:

|-------1----------------------| |c = √------≈ 2.998 × 108 m/s. | -------μ0𝜖0--------------------|

(37)

Maxwell’s field equations therefore predict propagating waves whose speed is the speed of light. Historically, the agreement between the electromagnetic speed scale and measured optical propagation speed was one of the decisive clues that light is an electromagnetic phenomenon [6, 4].

7 Dimensional check of the speed

It is worth checking the units. Using

[μ0] = H/m

(38)

and

[𝜖0] = F/m,

(39)

we have

[μ₀𝜖₀]

(40)

Since

H = V--s, F = A-s, A V

(41)

then

2 H F = s.

(42)

Therefore,

-s2- [μ0𝜖0] = m2 ,

(43)

and

[ ] √-1---- = m-, μ0𝜖0 s

(44)

as required for a speed.

8 What source-free vacuum really means

The condition

ρ = 0, J = 0

(45)

inside the region of interest does not mean that no source exists anywhere in the universe. An antenna can create a disturbance near its Conductors, and after that disturbance leaves the source region it can propagate through a region where

ρ = 0, J = 0.

(46)

The vacuum wave equation describes that propagation region.

This distinction is essential in radio physics: sources launch the wave, but the wave does not need local charges and currents at every point along its later path.

9 One-dimensional reduction

Suppose the fields vary only with position z and time t. For one Cartesian component, the three-dimensional wave equation reduces to

|-2--------2-------| |∂-u-− 1-∂-u-= 0.| -∂z2----c2∂t2-------

(47)

Here u can represent a Cartesian component such as E_x or B_y.

The standard traveling-wave solutions are

|-------------------------------| u-(z,-t)-=-F-(z −-ct) +-G-(z +-ct).-

(48)

The first term moves in the +z direction at speed c. The second moves in the −z direction at speed c.

10 Why F(z − ct) moves without changing shape

Consider

u(z,t) = F (z − ct).

(49)

Define

ξ = z − ct.

(50)

Then

∂u- ′ ∂z = F (ξ),

(51)

and

2 ∂-u- = F ′′(ξ). ∂z2

(52)

For the time derivatives,

∂u ---= − cF ′(ξ), ∂t

(53)

2 ∂--u = c2F ′′(ξ). ∂t2

(54)

Substituting into the one-dimensional wave equation gives

−	= F′′(ξ) −c²F′′(ξ)	(55)
	= 0.	(56)

Therefore any sufficiently smooth waveform F can propagate rigidly at speed c.

Figure. A function of z − ct translates in the +z direction at speed c without changing its mathematical shape.

11 Sinusoidal solution and the dispersion relation

Now consider a monochromatic electric field of the form

Ex(z,t) = E0 cos(kz − ωt).

(57)

The second spatial derivative is

∂2Ex- 2 ∂z2 = − k E0 cos(kz − ωt).

(58)

The second time derivative is

∂2Ex- 2 ∂t2 = − ω E0 cos(kz − ωt).

(59)

Insert these into

∂2Ex 1 ∂2Ex ---2-− -2---2- = 0. ∂z c ∂t

(60)

Then

ω2 − k2Ex + --2 Ex = 0. c

(61)

For a nonzero field,

|----------| -ω2-=-c2k2.-

(62)

For a wave traveling in the +z direction we take positive ω and k, giving

|--------| -ω-=-ck.-|

(63)

Thus the phase velocity is

|ω-----| |--= c.| -k------

(64)

Using

ω = 2πf, k = 2π-, λ

(65)

we recover

|--------| -c-=-fλ.-|

(66)

12 Worked example: GPS L1 wavelength in vacuum

The GPS L1 carrier frequency is

f = 1.57542 × 109 Hz. L1

(67)

Using

c λ = f,

(68)

we obtain

λ_L1	= m	(69)
	≈ 0.1903 m.	(70)

Therefore,

|--------------| -λL1 ≈-19.0cm.--

(71)

The wave equation therefore connects the field theory of Maxwell directly to the wavelength scales encountered in RF and GNSS engineering.

13 Recovering the magnetic field from Faraday’s law

Consider an electromagnetic wave traveling in the +z direction with

E (z,t) = E cos(kz − ωt)ˆx. 0

(72)

Assume the magnetic field has the form

B (z,t) = B0 cos(kz − ωt)ˆy.

(73)

For this electric field,

∂Ex ∇ × E = ---- ˆy. ∂z

(74)

Since

∂Ex ---- = − kE0 sin(kz − ωt), ∂z

(75)

Faraday’s law gives

− kE0 sin(kz − ωt ) = − ∂By-. ∂t

(76)

But

∂By- ∂t = ωB0 sin(kz − ωt).

(77)

Therefore,

kE0 = ωB0.

(78)

B = k-E . 0 ω 0

(79)

Using ω∕k = c,

|--------| | E | B0 = --0| -------c--

(80)

or equivalently

|----------| -E0-=-cB0.--

(81)

This relation is specific to plane electromagnetic waves in vacuum.

14 The electric and magnetic fields are in phase

The electric and magnetic fields just derived are

ˆ E = E0 cos(kz − ωt )x,

(82)

and

B = B0 cos(kz − ωt )ˆy.

(83)

Both contain the same phase

ϕ = kz − ωt.

(84)

Therefore their maxima, minima, and zero crossings occur at the same positions and times.

Figure. For a vacuum plane wave, normalized E_x and B_y have the same phase dependence. The plotted curves coincide; the dashed curve is shown only to identify the magnetic field separately.

15 Why a plane electromagnetic wave is transverse

Let a monochromatic plane electric field be written in the general form

E (r,t) = E0 cos(k ⋅ r − ωt ).

(85)

Take its divergence:

∇ ⋅ E = − (k ⋅ E )sin(k ⋅ r − ωt). 0

(86)

In source-free vacuum,

∇ ⋅ E = 0

(87)

for all positions and times. Therefore,

|----------| k-⋅ E0-=-0.-

(88)

Hence

|--------| -E0-⊥-k.-|

(89)

Applying the same argument to

∇ ⋅ B = 0

(90)

gives

|--------| |B0 ⊥ k.| ----------

(91)

Both fields are therefore transverse to the propagation direction.

16 Why E, B, and k form an orthogonal triad

For a plane wave, Faraday’s law gives the vector amplitude relation

|--------------| k × E0 = ωB0. | ----------------

(92)

Therefore,

1 B0 = --k × E0. ω

(93)

Since

ω = ck,

(94)

we can write

|---------------| | 1-^ | B0 = c k × E0. | ----------------

(95)

This immediately shows that B₀ is perpendicular to both k and E₀.

For a wave traveling in the +z direction,

E ∥ ˆx, B ∥ ˆy, k ∥ ˆz,

(96)

and

xˆ× ˆy = ˆz.

(97)

Thus the propagation direction agrees with

|--------| -E-×--B.-|

(98)

Figure. A vacuum plane wave is transverse: E, B, and the propagation vector k are mutually perpendicular, and E × B points in the propagation direction.

17 Checking the same plane wave with Ampere–Maxwell

For

B = B0 cos(kz − ωt )ˆy,

(99)

the x component of the curl is

(∇ × B) = − ∂By- . x ∂z

(100)

Since

∂By -∂z- = − kB0 sin(kz − ωt),

(101)

we obtain

∇ × B = kB0 sin(kz − ωt )xˆ.

(102)

Meanwhile,

∂E --- = ωE0 sin(kz − ωt)ˆx. ∂t

(103)

Ampere–Maxwell requires

kB0 = μ0𝜖0ωE0.

(104)

Using

k- B0 = ω E0,

(105)

we obtain

k2 ---E0 = μ0 𝜖0ωE0. ω

(106)

For E₀≠0,

k2 = μ0𝜖0ω2.

(107)

Therefore,

|------------------| | k2 | |ω2 = -----= c2k2.| ------μ0-𝜖0--------

(108)

The two curl equations are therefore mutually consistent with the same propagation speed and amplitude relation.

18 A useful hierarchy of equations

The derivation can now be organized into four levels.

At the field-law level:

∂B-- ∂E- ∇ × E = − ∂t , ∇ × B = μ0 𝜖0∂t .

(109)

At the wave-equation level:

1 ∂2E ∇2E − -2---2 = 0, c ∂t

(110)

2 1-∂2B-- ∇ B − c2 ∂t2 = 0.

(111)

At the monochromatic plane-wave level:

E = E0 cos(k ⋅ r − ωt),

(112)

B = B0 cos(k ⋅ r − ωt).

(113)

And the geometry and dispersion relations are

ω = ck,

(114)

k ⋅ E0 = 0, k ⋅ B0 = 0,

(115)

1-^ B0 = c k × E0.

(116)

These are not separate assumptions. They are progressively specialized consequences of Maxwell’s equations.

19 Common mistakes

Forgetting that the wave-equation derivation shown here assumes a source-free region, so ρ = 0 and J = 0.
Taking the curl of Faraday’s law but forgetting that the right-hand side becomes a time derivative of ∇× B.
Using ∇× (∇× E) = −∇²E without first invoking ∇⋅ E = 0.
Dropping a minus sign in the curl–curl identity.
Confusing the scalar Laplacian ∇² with the divergence or curl operators.
Assuming that source-free means the wave has never been created by a source; it means only that the local propagation region contains no charge or conduction-current source terms.
Writing E₀ = B₀. In SI units their numerical amplitudes differ by the factor c: E₀ = cB₀.
Assuming every possible electromagnetic field configuration is transverse. The transverse relations derived here are for source-free plane waves.
Treating k as a frequency. The angular frequency is ω; the Wavenumber k = 2π∕λ measures spatial phase variation.
Forgetting that F(z − ct) moves in the +z direction, while G(z + ct) moves in the −z direction.

20 What EM16 adds to the series

EM15 showed that Maxwell’s equations form a self-consistent coupled system. EM16 now shows that the same system contains wave propagation intrinsically.

The central chain is

Maxwell curl equations −→ curl–curl operation −→ wave equations − → c = √-1---. μ0𝜖0

(117)

For a monochromatic plane wave, Maxwell’s equations further imply

|--------| -ω-=-ck,-|

(118)

|----------| |E0 = cB0, | ------------

(119)

and

|------------------------------| E ⊥ B, E ⊥ k, B ⊥ k.| --------------------------------

(120)

The next natural step is to study electromagnetic-wave energy flow: energy density, the Poynting vector, intensity, and how power spreads through space.

References

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

[2] Edward M. Purcell and David J. Morin, Electricity and Magnetism, 3rd ed., Cambridge University Press, 2013.

[3] Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 2, OpenStax, 2016, sections on Maxwell’s equations and electromagnetic waves.

[4] Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Volume II, Addison-Wesley, 1964, chapters on electromagnetic waves and Maxwell’s equations.

[5] Massachusetts Institute of Technology, 8.02 Physics II: Electricity and Magnetism, MIT OpenCourseWare, materials on Maxwell’s equations and electromagnetic waves.

[6] James Clerk Maxwell, “A Dynamical Theory of the Electromagnetic Field,” Philosophical Transactions of the Royal Society of London, vol. 155, pp. 459–512, 1865.

"Electromagnetic Waves: Deriving the Electromagnetic Wave Equation from Maxwell's Equations" is owned by bloftin.
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