Refractive index, critical frequency and MUF

Refractive index of an ionospheric layer

An ionised layer contains N free electrons per m³ (charge e, mass m). Under a wave field E sin ωt, each electron moves with velocity v:

m dv/dt = e E sin ωt
v = −(eE / mω) cos ωt

The electron current J = Nev lags E by 90° (an inductive current), so it is opposite to the displacement current ε₀∂E/∂t. Total current density:

J_total = ε₀ ∂E/∂t + Nev
        = ωε₀E cos ωt − (Ne²/mω) E cos ωt
        = ωε₀ [1 − Ne² / (ε₀ m ω²)] E cos ωt

So the effective relative permittivity is εr = 1 − Ne²/(ε₀mω²), and the refractive index n = √εr. Putting e = 1.602 × 10⁻¹⁹ C, m = 9.109 × 10⁻³¹ kg, ε₀ = 8.854 × 10⁻¹² F/m and ω = 2πf:

e² / (4π² ε₀ m) = 80.6 ≈ 81
n = √(1 − 81N / f²)      (N per m³, f in Hz)

Since n < 1 and falls as N rises with height, an upward ray bends away from the normal and returns to earth.

Critical frequency

At vertical incidence the wave returns where n = 0 at the level of maximum density Nmax:

1 − 81Nmax / fc² = 0
fc = 9 √Nmax

Critical frequency is the highest frequency reflected at vertical incidence.

Maximum usable frequency (flat earth)

          ionosphere (virtual height h)
  ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~
                   / | \
                  /θi|  \
                 /   |h  \
                /    |    \
   flat earth  T-----M-----R
               <--- D ---->
  TM = MR = D/2,  θi = angle of incidence

Refractive index of an ionised layer (electron density N per m³, frequency f in Hz):

n = √(1 − 81N / f²)

By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°):

sin θi = n = √(1 − 81Nmax / f²)
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi

For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so

f_MUF = fc sec θi          (secant law)

From the flat-earth geometry (triangle TMP):

tan θi = (D/2) / h
sec θi = √(1 + tan² θi) = √(1 + D² / 4h²)

f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h)

Solving for the skip distance:

(f_MUF / fc)² = 1 + D² / 4h²
D = 2h √[(f_MUF / fc)² − 1]

Explanation: fc depends only on the layer's peak electron density; the MUF depends also on the angle of incidence, so a longer path (larger θi) can use a higher frequency. With flat earth, MUF = fc sec θi = fc√(1 + D²/4h²), always ≥ fc. Frequencies above the MUF escape the layer; in practice OWF ≈ 0.85 MUF is used.

Example: Nmax = 10¹² /m³ gives fc = 9 × 10⁶ Hz = 9 MHz; for D = 600 km and h = 300 km, f_MUF = 9 × √(1 + 1) = 12.73 MHz.