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Conducting-medium propagation: the exact attenuation constant, the skin depth and its good- and poor-conductor regimes

Physical scope

Exact conditional statement about a monochromatic plane-wave field in a homogeneous isotropic Ohmic medium at rest with constant scalar eps,mu>0 and sigma>0: the exact complex propagation constant and decay, the closed-form skin depth, explicit bounded remainder inequalities in the good-conductor regime epsomega/sigma<=1/4 and the poor-conductor regime sigma/(epsomega)<=1/2, and the phase relation. Excluded: sigma=0 (no attenuation, infinite skin depth), the ideal-conductor limit, frequency-dependent or tensor conductivity, anisotropy, dispersion, nonlinearity, any laboratory material and all empirical claims.

Statement

Assume the elementary inertial-rest branch of Declared material-closure branches: linear nondispersive Ohmic response and the causal memory branch in the isotropic scalar case at rest in the laboratory frame with constants $ε, μ > 0$ and $σ > 0$ , the fields of The electric and magnetic fields as fixed-time spatial vector fields on the laboratory region of Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields and the plane-wave structure of Plane-wave fields in a homogeneous Ohmic medium: amplitude conditions, dispersion, averaged flux and the evanescent branch: let $ω > 0$ , $e \in C^{3} ∖ {0}$ , $n \in R^{3}$ a unit vector and $k = λ n$ , $λ \in C$ , and let $E (x, t) = Re (e e^{i (k \cdot x - ω t)})$ be a plane-wave field satisfying the amplitude conditions of Plane-wave fields in a homogeneous Ohmic medium: amplitude conditions, dispersion, averaged flux and the evanescent branch(b) with $k \cdot e = 0$ . Put $a : = μ ε ω^{2} > 0$ , $b : = μ σ ω > 0$ , $r : = \sqrt{a^{2} + b^{2}}$ and let $p, q > 0$ be the quantities of Homogeneous isotropic medium parameters: phase speed, impedance, refractive index, complex propagation constant and skin depth. Then:

(a) [Exact propagation constant and decay] $λ^{2} = k \cdot k = ε μ ω^{2} + i μ σ ω = a + i b$ , so $λ = \pm (p + i q)$ ; if $x_{3}^{'} : = n \cdot x$ , then for $λ = p + i q$ $E (x, t) = e^{- q x_{3}^{'}} Re (e e^{i (p x_{3}^{'} - ω t)}),$ so the field oscillates with wavenumber $p$ and its amplitude decays in the direction of increasing $x_{3}^{'}$ at the exact rate $q$ ; with $λ = - (p + i q)$ the exact formula is $E (x, t) = e^{q x_{3}^{'}} Re (e e^{i (- p x_{3}^{'} - ω t)})$ , which grows in that direction, so the decaying branch is the one with $λ = p + i q$ , in agreement with the selection rule of Homogeneous isotropic medium parameters: phase speed, impedance, refractive index, complex propagation constant and skin depth(b).

(b) [Exact skin depth] $q^{2} = (\sqrt{a^{2} + b^{2}} - a) / 2$ , $p^{2} = (\sqrt{a^{2} + b^{2}} + a) / 2$ , $p q = b / 2 = μ σ ω / 2$ , and the skin depth of Homogeneous isotropic medium parameters: phase speed, impedance, refractive index, complex propagation constant and skin depth equals the exact closed expression $δ (ω) = \frac{1}{q} = \frac{\sqrt{2 (r + a)}}{b} = \frac{\sqrt{2 (\sqrt{a^{2} + b^{2}} + a)}}{μ σ ω} .$

(c) [Good-conductor regime] put $t : = a / b = ε ω / σ > 0$ . For every $t \leq 1 / 4$ the exact value satisfies the explicit bounds $∣ q - \sqrt{b / 2} ∣ \leq 2 t \sqrt{b / 2} = \frac{2 ε ω}{σ} \sqrt{\frac{μ σ ω}{2}}, ∣ δ - \sqrt{2 / b} ∣ \leq 4 t \sqrt{2 / b} = \frac{4 ε ω}{σ} \sqrt{\frac{2}{μ σ ω}};$ equivalently $δ = \sqrt{2 / (μ σ ω)} (1 + O (ε ω / σ))$ in the declared regime $ε ω / σ \leq 1 / 4$ , with the displayed explicit remainder bounds.

(d) [Poor-conductor regime] put $t^{'} : = b / a = σ / (ε ω) > 0$ . For every $t^{'} \leq 1 / 2$ the exact value satisfies $∣ q - \frac{σ}{2} \sqrt{\frac{μ}{ε}} ∣ \leq \frac{t^{' 2}}{4} \cdot \frac{σ}{2} \sqrt{\frac{μ}{ε}}, ∣ δ - \frac{2}{σ} \sqrt{\frac{ε}{μ}} ∣ \leq \frac{t^{' 2}}{2} \cdot \frac{2}{σ} \sqrt{\frac{ε}{μ}},$ so in this regime $δ$ is frequency-independent up to the displayed relative error $O ((σ / (ε ω))^{2})$ .

(e) [Phase relation for a real amplitude] if $e \in R^{3}$ and $λ = p + i q$ , then $E (x, t) = e^{- q x_{3}^{'}} e \cos (p x_{3}^{'} - ω t), H (x, t) = \frac{e^{- q x_{3}^{'}}}{μ ω} (p \cos (p x_{3}^{'} - ω t) - q \sin (p x_{3}^{'} - ω t)) (n \times e),$ and the ratio of the quadrature to the in-phase coefficient is $q / p = \sqrt{1 + t^{2}} - t$ with $1 - 2 t \leq q / p \leq 1$ for every $t \leq 1 / 2$ ; in particular $q / p \to 1$ as $t ↓ 0$ , so the magnetic scalar factor is $\sqrt{p^{2} + q^{2}} \cos (p x_{3}^{'} - ω t + η)$ with $η : = \arctan (q / p) \to π / 4$ (45 degrees), not a 90-degree quadrature phase shift. At fixed position with the $e^{- i ω t}$ convention this corresponds to the stated signed $+ η$ phase in the decreasing phase variable.

(f) [Limits and scope] for $σ = 0$ one has $b = 0$ , $q = 0$ , no attenuation and $δ = + \infty$ , and the approximations of (c) and (d) are not applicable; the ideal-conductor limit $σ \to \infty$ is not covered by this theorem, which works inside the finite-Ohmic branch of Declared material-closure branches: linear nondispersive Ohmic response and the causal memory branch (a separate boundary model is needed for an ideal conductor); the frequency dependence of $σ$ (Drude or other) is not supplied, and $σ$ is the declared constant of the branch; the conclusions are exact identities for the ideal plane-wave field and carry no empirical content, no measured conductivity and no claim about any laboratory field.

Facts & Assumptions

Given: The declared Ohmic branch with constants $ε, μ > 0$ , $σ > 0$ ; the unit vector $n$ , $k = λ n$ , $e \neq 0$ with $k \cdot e = 0$ ; the plane-wave field $E = Re (e e^{i (k \cdot x - ω t)})$ with $H$ determined by the amplitude conditions; the numbers $a = μ ε ω^{2} > 0$ , $b = μ σ ω > 0$ , $r = \sqrt{a^{2} + b^{2}}$ and the parameters $p, q > 0$ of Homogeneous isotropic medium parameters: phase speed, impedance, refractive index, complex propagation constant and skin depth.

[F1]

Concrete propagation data: $k \cdot k = ε μ ω^{2} + i μ σ ω$ and $h = k \times e / (ω μ)$ by Plane-wave fields in a homogeneous Ohmic medium: amplitude conditions, dispersion, averaged flux and the evanescent branch(b)-(c); and by Homogeneous isotropic medium parameters: phase speed, impedance, refractive index, complex propagation constant and skin depth(b), $κ = p + i q$ is the unique complex number with $Re κ \geq 0$ , $Im κ \geq 0$ and $κ^{2} = a + i b$ , where $p = \sqrt{(r + a) / 2}$ and $q = \sqrt{(r - a) / 2}$ ; also $v = (ε μ)^{- 1 / 2}$ and $Z = (μ / ε)^{1 / 2}$ .

[F2]

The complex exponential in Cartesian form: for real $x, y$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ and $∣ \exp (x + i y) ∣ = e^{x}$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[F3]

Complex arithmetic and modulus: $C$ is a field, a product vanishes only if a factor does, and from $z^{2} = w$ and $κ^{2} = w$ one gets $(z - κ) (z + κ) = 0$ , hence $z = \pm κ$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ); the modulus satisfies $z \overline{z} = ∣ z ∣^{2}$ and $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ (Real and imaginary parts, complex conjugation, and modulus, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F4]

Real roots and order: every nonnegative real has a unique nonnegative square root, and $\sqrt{a} \sqrt{b} = \sqrt{a b}$ for $a, b \geq 0$ follows from squaring and uniqueness (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ); for $0 \leq a \leq b$ one has $a^{2} \leq b^{2}$ , and for $0 \leq a, b$ with $a^{2} \leq b^{2}$ one has $a \leq b$ (Squaring is monotone on the nonnegatives); if $0 \leq a \leq b$ and $0 \leq c \leq d$ then $a c \leq b d$ , and if $0 < c < d$ then $1 / d < 1 / c$ in the sense $0 < 1 / d \leq 1 / c$ (Multiplying inequalities of positives, Sign rules for products and monotonicity of multiplication: multiplication by a positive preserves order, and $1 / c - 1 / d = (d - c) / (c d) > 0$ ).

[F5]

Every complex number has a square root (Every complex number has a square root, by an explicit Cartesian formula).

[F6]

The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ is continuous and strictly increasing with $\tan (\arctan y) = y$ (The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ ), and the addition formulas $\cos (θ + η) = \cos θ \cos η - \sin θ \sin η$ hold for real angles (The addition formulas for sine and cosine).

[F7]

Trigonometry: $\sin^{2} θ + \cos^{2} θ = 1$ and $\cos (- θ) = \cos θ$ (Pythagorean and parity identities for all six trigonometric functions on their natural domains); $\tan θ = \sin θ / \cos θ$ where defined (Tangent, cotangent, secant, and cosecant on their exact natural domains); $\sin (θ + π / 2) = \cos θ$ and $\cos (π / 2) = 0$ (Quarter-turn values and shifts by pi/2 and pi). Cosine is strictly decreasing on $[0, π]$ and hence positive on $(0, π / 2)$ (Signs, monotonicity intervals, and ranges of sine and cosine). Consequently $\sin (π / 4) = \cos (- π / 4) = \cos (π / 4) > 0$ , so $\tan (π / 4) = 1$ and $\arctan 1 = π / 4$ by [F6].

[F8]

The laboratory setting and units are those of Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields, Framework-local SI units, the vacuum constants and the exact/measured status of their values and The electric and magnetic fields as fixed-time spatial vector fields; the branch assumption is that of Declared material-closure branches: linear nondispersive Ohmic response and the causal memory branch.

Proof

technique · direct; substitute the concrete data into the medium plane-wave theorem, extract the decaying root via Euler's Cartesian form, and bound the exact radicals with elementary square-root inequalities

1.1givenF1F2F3F5algebra

[Propagation constant and decay, part (a)] By [F1], $k \cdot k = ε μ ω^{2} + i μ σ ω = a + i b$ , and since $k = λ n$ with $n \cdot n = 1$ the bilinear expansion gives $λ^{2} = k \cdot k = a + i b$ ; by [F1] and [F5] with [F3], the solutions of $z^{2} = a + i b$ are exactly $z = \pm (p + i q)$ . Writing $x_{3}^{'} = n \cdot x$ , so that $k \cdot x = λ x_{3}^{'}$ , the exponent is $i (k \cdot x - ω t) = i λ x_{3}^{'} - i ω t$ ; for $λ = p + i q$ this is $- q x_{3}^{'} + i (p x_{3}^{'} - ω t)$ , and the Cartesian form [F2] gives $e^{i (k \cdot x - ω t)} = e^{- q x_{3}^{'}} (\cos (p x_{3}^{'} - ω t) + i \sin (p x_{3}^{'} - ω t))$ , whence $E = e^{- q x_{3}^{'}} Re (e e^{i (p x_{3}^{'} - ω t)})$ componentwise; the amplitude factor $e^{- q x_{3}^{'}}$ decreases as $x_{3}^{'}$ increases because $q > 0$ by [F1]. For $λ = - (p + i q)$ the same computation gives $i λ x_{3}^{'} - i ω t = q x_{3}^{'} + i (- p x_{3}^{'} - ω t)$ and $E = e^{q x_{3}^{'}} Re (e e^{i (- p x_{3}^{'} - ω t)})$ , which grows in the direction of increasing $x_{3}^{'}$ ; the decaying branch is therefore $λ = p + i q$ , matching the root selection of [F1].

2.1step 1.1F1F4algebra

[Exact skin depth, part (b)] By [F1], $q^{2} = (r - a) / 2$ and $p^{2} = (r + a) / 2$ with $r = \sqrt{a^{2} + b^{2}}$ ; multiplying, $(p q)^{2} = (r^{2} - a^{2}) / 4 = b^{2} / 4$ , and since $p, q \geq 0$ and $b > 0$ the uniqueness of the nonnegative square root in [F4] gives $p q = b / 2 = μ σ ω / 2$ . For the skin depth $δ = 1 / q$ of Homogeneous isotropic medium parameters: phase speed, impedance, refractive index, complex propagation constant and skin depth(c): $δ^{2} = 1 / q^{2} = 2 / (r - a) = 2 (r + a) / ((r - a) (r + a)) = 2 (r + a) / b^{2}$ , where $(r - a) (r + a) = r^{2} - a^{2} = b^{2}$ by [F4]; since $δ > 0$ and $b > 0$ , taking the nonnegative square root gives $δ = \sqrt{2 (r + a)} / b$ , and substituting $a = μ ε ω^{2}$ and $b = μ σ ω$ gives the displayed closed form.

3.1step 2.1F1F4algebra

[Ratio identities] With $t = a / b > 0$ one has $r = b \sqrt{1 + t^{2}}$ because $r^{2} = a^{2} + b^{2} = b^{2} (t^{2} + 1)$ and both sides are nonnegative, by [F4]; hence $q^{2} = (r - a) / 2 = \frac{b}{2} (\sqrt{1 + t^{2}} - t)$ and $\sqrt{1 + t^{2}} - t = (\sqrt{1 + t^{2}} + t)^{- 1}$ , the second identity because $(\sqrt{1 + t^{2}} - t) (\sqrt{1 + t^{2}} + t) = 1 + t^{2} - t^{2} = 1$ with all factors positive by [F4] and [F1]. Moreover $q / p = \sqrt{1 + t^{2}} - t$ : indeed $q^{2} / p^{2} = (r - a) / (r + a)$ , and the same rationalisation gives $(r - a) / (r + a) = (\sqrt{1 + t^{2}} - t)^{2}$ , so the positive quantities $q / p$ and $\sqrt{1 + t^{2}} - t$ have equal squares and are equal by [F4]. With $s = t^{'} = b / a > 0$ one has analogously $r = a \sqrt{1 + s^{2}}$ and $q^{2} = \frac{a}{2} (\sqrt{1 + s^{2}} - 1)$ . Writing $q_{t} : = \frac{σ}{2} \sqrt{\frac{μ}{ε}}$ , the identity $q_{t} = \frac{\sqrt{a} s}{2}$ holds: expanding $\sqrt{a} = ω (ε μ)^{1 / 2}$ and $s = \frac{σ}{ε ω}$ by [F1] gives $\frac{\sqrt{a} s}{2} = \frac{ω (ε μ)^{1 / 2} σ}{2 ε ω} = \frac{σ}{2} \sqrt{\frac{μ}{ε}}$ ; hence $q^{2} / q_{t}^{2} = \frac{2 (\sqrt{1 + s^{2}} - 1)}{s^{2}} = \frac{2}{1 + \sqrt{1 + s^{2}}}$ by the same rationalisation.

4.1step 3.1F1F4algebra

[Good-conductor bounds, part (c)] For $t \geq 0$ one has $1 \leq \sqrt{1 + t^{2}} \leq 1 + t$ : the left inequality because $1 \leq 1 + t^{2}$ and both sides are nonnegative, the right because $1 + t^{2} \leq (1 + t)^{2} = 1 + 2 t + t^{2}$ , both by [F4]. Hence $\frac{1}{1 + 2 t} \leq \sqrt{1 + t^{2}} - t \leq \frac{1}{1 + t} \leq 1$ by step 3.1 and [F4], and $\frac{1}{1 + 2 t} \geq 1 - 2 t$ because $0 \leq 1 - 2 t \leq 1$ and $(1 - 2 t) (1 + 2 t) = 1 - 4 t^{2} \leq 1$ for $t \leq 1 / 2$ ; therefore $\sqrt{1 - 2 t} \leq \frac{q}{\sqrt{b / 2}} \leq 1$ , that is $\sqrt{b / 2} \sqrt{1 - 2 t} \leq q \leq \sqrt{b / 2}$ . For $t \leq 1 / 4$ one has $1 - 2 t \geq 0$ and $(1 - 2 t)^{2} \leq 1 - 2 t$ , so $\sqrt{1 - 2 t} \geq 1 - 2 t$ by [F4], giving $∣ q - \sqrt{b / 2} ∣ \leq 2 t \sqrt{b / 2}$ with the stated equivalent form since $\sqrt{b / 2} = \sqrt{μ σ ω / 2}$ . For $δ = 1 / q$ one has $\sqrt{2 / b} \leq δ \leq \frac{\sqrt{2 / b}}{1 - 2 t}$ from the bound on $q$ and [F4]; and $\frac{1}{1 - 2 t} \leq 1 + 4 t$ for $t \leq 1 / 4$ because $1 \leq (1 + 4 t) (1 - 2 t) = 1 + 2 t - 8 t^{2}$ , i.e. $0 \leq 2 t (1 - 4 t)$ , by [F4]; hence $∣ δ - \sqrt{2 / b} ∣ \leq 4 t \sqrt{2 / b}$ with the stated equivalent form.

4.2step 3.1F4algebra

[Poor-conductor bounds, part (d)] By step 3.1, $q^{2} = \frac{a}{2} (\sqrt{1 + s^{2}} - 1)$ and $q^{2} / q_{t}^{2} = \frac{2}{1 + \sqrt{1 + s^{2}}}$ with $q_{t} = \frac{σ}{2} \sqrt{\frac{μ}{ε}}$ , $s = t^{'} = b / a > 0$ . Since $\sqrt{1 + s^{2}} \geq 1$ , the ratio is at most $1$ , so $q \leq q_{t}$ ; and $\sqrt{1 + s^{2}} \leq 1 + \frac{s^{2}}{2}$ because $1 + s^{2} \leq (1 + \frac{s^{2}}{2})^{2} = 1 + s^{2} + \frac{s^{4}}{4}$ , so $q^{2} / q_{t}^{2} \geq \frac{2}{2 + \frac{s^{2}}{2}} = \frac{1}{1 + \frac{s^{2}}{4}}$ ; for $s^{2} \leq 1$ (which holds when $s \leq 1 / 2$ ) one has $\frac{1}{1 + s^{2} / 4} \geq 1 - \frac{s^{2}}{4}$ since $(1 - \frac{s^{2}}{4}) (1 + \frac{s^{2}}{4}) = 1 - \frac{s^{4}}{16} \leq 1$ with $1 - \frac{s^{2}}{4} \geq 0$ ; therefore $q / q_{t} \geq 1 - \frac{s^{2}}{4}$ , i.e. $∣ q - q_{t} ∣ \leq \frac{s^{2}}{4} q_{t}$ . For $δ = 1 / q$ and $δ_{t} = 1 / q_{t}$ we get $δ / δ_{t} = q_{t} / q \leq \frac{1}{1 - s^{2} / 4} \leq 1 + \frac{s^{2}}{2}$ whenever $s^{2} \leq 1$ (since $\frac{1}{1 - s^{2} / 4} \leq 1 + \frac{s^{2}}{2}$ is equivalent to $1 \leq 1 + \frac{s^{2}}{4} - \frac{s^{4}}{8}$ , true for $s^{2} \leq 2$ ), whence $∣ δ - δ_{t} ∣ \leq \frac{s^{2}}{2} δ_{t}$ ; substituting $δ_{t} = \frac{2}{σ} \sqrt{\frac{ε}{μ}}$ gives both displayed bounds, by [F4] throughout.

5.1givenstep 1.1step 3.1step 4.1F1F2F4F6F7algebra

[Phase relation, part (e)] Assume $e \in R^{3}$ and $λ = p + i q$ ; by [F1], $h = k \times e / (ω μ) = λ (n \times e) / (ω μ)$ because $k = λ n$ and $n \times e$ is real, and the exponent is computed as in step 1.1. Writing $θ : = p x_{3}^{'} - ω t$ and using $λ / (μ ω) = (p + i q) / (μ ω)$ with real $p, q$ , the Cartesian form [F2] gives $Re ((p + i q) (\cos θ + i \sin θ)) = p \cos θ - q \sin θ$ , so $E = e^{- q x_{3}^{'}} e \cos θ$ and $H = e^{- q x_{3}^{'}} (p \cos θ - q \sin θ) (n \times e) / (μ ω)$ , the two displayed identities. By step 3.1, $q / p = \sqrt{1 + t^{2}} - t$ , and the bounds $1 - 2 t \leq q / p \leq 1$ for $t \leq 1 / 2$ are the ones established in step 4.1 for the same quantity; as $t ↓ 0$ the inequality $0 \leq \sqrt{1 + t^{2}} - 1 = t^{2} / (\sqrt{1 + t^{2}} + 1) \leq t^{2} / 2$ gives $∣ \sqrt{1 + t^{2}} - t - 1 ∣ \leq t + t^{2} / 2 \to 0$ , so $η : = \arctan (q / p)$ tends to $\arctan 1 = π / 4$ by the continuity of $\arctan$ in [F6]. Put $r : = q / p > 0$ and $η = \arctan r \in (0, π / 2)$ ; by [F6], $\sin η = r \cos η$ , and the Pythagorean identity $\sin^{2} η + \cos^{2} η = 1$ with $\cos η > 0$ gives $\cos η = 1 / \sqrt{1 + r^{2}}$ and $\sin η = r / \sqrt{1 + r^{2}}$ ; equivalently $\cos η = p / \sqrt{p^{2} + q^{2}}$ and $\sin η = q / \sqrt{p^{2} + q^{2}}$ , since $\sqrt{1 + r^{2}} = \sqrt{p^{2} + q^{2}} / p$ with $p > 0$ by [F4]. The addition formula of [F6] then gives $\sqrt{p^{2} + q^{2}} \cos (θ + η) = p \cos θ - q \sin θ$ , which is the stated form of the magnetic scalar factor.

6.1givenstep 4.2F1F8∎

[Limits and scope, part (f)] Clause (f): for $σ = 0$ the definition of Homogeneous isotropic medium parameters: phase speed, impedance, refractive index, complex propagation constant and skin depth gives $b = 0$ , $q = 0$ , no attenuation and $δ = + \infty$ , and the divisions and bounds of steps 3.1-4.2 use $b > 0$ , $q > 0$ and are not evaluated at zero; the limit $σ \to \infty$ is outside the finite-Ohmic branch and the frequency dependence of $σ$ is not supplied; all identities are exact for the plane-wave field and involve no measurement, material data or empirical statement, and no choice principle is used since every quantity is defined by explicit formulas in the given scalars.

Conducting-medium propagation: the exact attenuation constant, the skin depth and its good- and poor-conductor regimes

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources