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Linear, circular and elliptical amplitude choices

Physical scope

Explicit ideal plane-wave amplitude choices on $R^{3} \times R$ in a named inertial frame with SI units: linear, circular (both handedness signs), elliptical with orthogonal axes, the zero wave and one dependent-amplitude linear edge case. Excluded: partially polarized light, superposition with other waves, and any empirical or measurement claim.

Example

Use the frame basis $(e_{1}, e_{2}, e_{3})$ and the SI conventions of Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields and Framework-local SI units, the vacuum constants and the exact/measured status of their values. Fix $K > 0$ , put $ω : = c K$ and $k : = K e_{3}$ , so that $∣ k ∣ = K = ω / c$ and $\hat{k} = e_{3}$ , and let $E_{0} > 0$ carry the unit $V m^{- 1}$ . For each amplitude choice below put $b : = k \times e / ω$ ; then the amplitude conditions of Dispersion, transversality and the two polarization conditions for plane-wave solutions hold and the corresponding fields $(E, B)$ of Real plane-wave fields encoded by complex amplitudes are smooth source-free vacuum Maxwell solutions on $R^{3} \times R$ , with $θ (x, t) = K x_{3} - ω t$ .

(i) [Linear] $e : = E_{0} e_{1}$ , $b = (E_{0} / c) e_{2}$ . Here $a = E_{0} e_{1}$ and $d = 0$ , the trace is the segment $[- E_{0}, E_{0}] e_{1} = {s e_{1} : - E_{0} \leq s \leq E_{0}}$ traversed twice per period, the polarization is linear and $H = ⟨ a \times d, \hat{k} ⟩ = 0$ ; moreover $E = E_{0} \cos θ e_{1}$ , $B = (E_{0} / c) \cos θ e_{2}$ and $∣ B ∣ = ∣ E ∣ / c$ .

(ii) [Circular] $e : = E_{0} (e_{1} + i e_{2})$ , $b = (E_{0} / c) (e_{2} - i e_{1})$ . Here $a = E_{0} e_{1}$ and $d = E_{0} e_{2}$ , so $∣ a ∣ = ∣ d ∣ = E_{0}$ and $⟨ a, d ⟩ = 0$ : the polarization is circular of radius $E_{0}$ , with $E = E_{0} (\cos θ e_{1} - \sin θ e_{2})$ , $B = (E_{0} / c) (\sin θ e_{1} + \cos θ e_{2})$ , $∣ E ∣ = E_{0}$ and $∣ B ∣ = E_{0} / c$ everywhere, and $H = ⟨ a \times d, \hat{k} ⟩ = E_{0}^{2} > 0$ , so this polarization is left-handed in the convention of Polarization of a transverse plane wave: linear, circular and elliptical traces, and handedness. The reversed choice $e = E_{0} (e_{1} - i e_{2})$ gives $H = - E_{0}^{2} < 0$ and is right-handed.

(iii) [Elliptical] $e : = E_{0} (e_{1} + 2 i e_{2})$ . Here $a = E_{0} e_{1}$ and $d = 2 E_{0} e_{2}$ are orthogonal with unequal lengths, so by clause (b)(iii) of Polarization of a transverse plane wave: linear, circular and elliptical traces, and handedness the trace is the axis-aligned ellipse $E_{0} \cos θ e_{1} - 2 E_{0} \sin θ e_{2}$ with semiaxes $E_{0}$ and $2 E_{0}$ along $e_{1}$ and $e_{2}$ ; $∣ E ∣^{2} = E_{0}^{2} (\cos^{2} θ + 4 \sin^{2} θ)$ varies between $E_{0}^{2}$ and $4 E_{0}^{2}$ ; and $b = (E_{0} / c) (e_{2} - 2 i e_{1})$ , $H = 2 E_{0}^{2} > 0$ .

(iv) [Degenerate and edge cases] $e = 0$ gives the zero wave, for which $a = d = 0$ and the trace is the single point ${0}$ . For $e : = E_{0} (1 + i) e_{1}$ one has $a = d = E_{0} e_{1}$ , so the polarization is linear with trace the segment $[- \sqrt{2} E_{0}, \sqrt{2} E_{0}] e_{1}$ and $H = 0$ ; the classification (b) of Polarization of a transverse plane wave: linear, circular and elliptical traces, and handedness covers this case without requiring $a$ and $d$ to be orthogonal.

(v) [Verification and scope] In every case $b = k \times e / ω$ with $∣ k ∣ = ω / c$ , so the amplitude equations hold and the fields are smooth source-free solutions by Dispersion, transversality and the two polarization conditions for plane-wave solutions; the amplitudes and the semiaxis lengths carry the unit $V m^{- 1}$ , and $∣ B ∣ = ∣ E ∣ / c$ holds pointwise. The examples assert no empirical realization and no measured polarization.

Facts & Assumptions

Given: The frame basis $(e_{1}, e_{2}, e_{3})$ , the constants $c > 0$ and $K > 0$ , $ω = c K$ , $k = K e_{3}$ , the amplitude $E_{0} > 0$ in $V m^{- 1}$ , the six amplitude choices $e \in {E_{0} e_{1}, E_{0} (e_{1} + i e_{2}), E_{0} (e_{1} - i e_{2}), E_{0} (e_{1} + 2 i e_{2}), E_{0} (1 + i) e_{1}, 0}$ with the corresponding $b = k \times e / ω$ , and the phase $θ (x, t) = K x_{3} - ω t$ .

[F1]

The polarization definition fixes the trace $E = a \cos θ - d \sin θ$ with $a = Re e$ , $d = Im e$ , the linear, circular and elliptical clauses with their criteria, and the helicity sign $H = ⟨ a \times d, \hat{k} ⟩$ , with $H = 0$ exactly in the linear case, right-handed meaning $H < 0$ and left-handed meaning $H > 0$ (Polarization of a transverse plane wave: linear, circular and elliptical traces, and handedness).

[F2]

For a plane wave with $k \neq 0$ , $ω > 0$ , if $∣ k ∣ = ω / c$ and $e$ satisfies $k \cdot e = 0$ then $b = k \times e / ω$ makes the amplitude equations hold, so the pair lies in the source-free smooth Maxwell branch; all solution fields are $C^{\infty}$ and satisfy $∣ B (x, t) ∣ = ∣ E (x, t) ∣ / c$ pointwise (Dispersion, transversality and the two polarization conditions for plane-wave solutions).

[F3]

The plane-wave pair is $E = Re (e \exp (i θ))$ , $B = Re (b \exp (i θ))$ with $θ = k \cdot x - ω t$ , $b$ the complex magnetic amplitude, and $\hat{k} = k / ∣ k ∣$ (Real plane-wave fields encoded by complex amplitudes).

[F4]

The cross product has the coordinate formula $u \times v = (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1})$ on the standard basis (The cross product in $R^{3}$ ).

[F5]

The Euclidean inner product is $⟨ u, v ⟩ = \sum_{k} u_{k} v_{k}$ , additive and real-homogeneous in each argument, so that $∣ u_{1} a - u_{2} d ∣^{2} = u_{1}^{2} ∣ a ∣^{2} + u_{2}^{2} ∣ d ∣^{2} - 2 u_{1} u_{2} ⟨ a, d ⟩$ for real $u_{1}, u_{2}$ and $a, d \in R^{3}$ ; in particular the frame basis is orthonormal (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F6]

The SI conventions provide the propagation constant $c > 0$ with its unit and the units of $k$ , $ω$ , $E$ and $B$ (Framework-local SI units, the vacuum constants and the exact/measured status of their values).

[F7]

The laboratory region carries the frame basis $(e_{1}, e_{2}, e_{3})$ in which components are read (Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields).

Verification

technique · direct

1.1givenF3F4F5F7algebra

Common data. With $k = K e_{3}$ one has $∣ k ∣ = K$ , $∣ k ∣ = ω / c$ because $ω = c K$ , and $\hat{k} = e_{3}$ ; by [F4] and the coordinate formula, $e_{3} \times e_{1} = e_{2}$ and $e_{3} \times e_{2} = - e_{1}$ . Hence for each choice $b = k \times e / ω = (K / ω) (e_{3} \times e)$ : for $e = E_{0} e_{1}$ , $b = (E_{0} / c) e_{2}$ ; for $e = E_{0} (e_{1} + i e_{2})$ , $b = (E_{0} / c) (e_{2} - i e_{1})$ ; for $e = E_{0} (e_{1} - i e_{2})$ , $b = (E_{0} / c) (e_{2} + i e_{1})$ ; for $e = E_{0} (e_{1} + 2 i e_{2})$ , $b = (E_{0} / c) (e_{2} - 2 i e_{1})$ ; and for $e = E_{0} (1 + i) e_{1}$ , $b = (E_{0} / c) (1 + i) e_{2}$ . Also $k \cdot e = K e_{3} \cdot e = 0$ for each of these amplitudes, since only the first two coordinates are nonzero.

2.1step 1.1F1F2F3F4algebra

Case (i), linear. For $e = E_{0} e_{1}$ one has $a = E_{0} e_{1}$ , $d = 0$ , so $\dim span {a, d} = 1$ and clause (b)(i) of [F1] gives a linear trace, the segment $[- E_{0}, E_{0}] e_{1}$ traversed twice per period; $a \times d = E_{0} e_{1} \times 0 = 0$ by [F4], so $H = 0$ . The fields are $E = Re (E_{0} e_{1} e^{i θ}) = E_{0} \cos θ e_{1}$ and $B = (E_{0} / c) \cos θ e_{2}$ , whence $∣ B ∣ = ∣ E ∣ / c$ pointwise; this also matches $b = (E_{0} / c) e_{2}$ from step 1.1 and clause (c) of [F2].

2.2step 1.1F1F2F3F4F5algebra

Case (ii), circular. For $e = E_{0} (e_{1} + i e_{2})$ one has $a = E_{0} e_{1}$ , $d = E_{0} e_{2}$ , so $∣ a ∣ = ∣ d ∣ = E_{0} > 0$ and $⟨ a, d ⟩ = E_{0}^{2} ⟨ e_{1}, e_{2} ⟩ = 0$ by the orthonormality of [F5]; clause (b)(ii) of [F1] gives a circular polarization of radius $E_{0}$ . The trace is $E = E_{0} (\cos θ e_{1} - \sin θ e_{2})$ with $∣ E ∣^{2} = E_{0}^{2} (\cos^{2} θ + \sin^{2} θ) = E_{0}^{2}$ by [F5], and $B = Re ((E_{0} / c) (e_{2} - i e_{1}) e^{i θ}) = (E_{0} / c) (\sin θ e_{1} + \cos θ e_{2})$ with $∣ B ∣ = E_{0} / c$ ; finally $H = ⟨ a \times d, e_{3} ⟩ = E_{0}^{2} ⟨ e_{1} \times e_{2}, e_{3} ⟩ = E_{0}^{2} > 0$ by [F4] and [F5], so this polarization is left-handed by [F1]. For the reversed choice $e = E_{0} (e_{1} - i e_{2})$ one has $d = - E_{0} e_{2}$ and $H = E_{0}^{2} ⟨ e_{1} \times (- e_{2}), e_{3} ⟩ = - E_{0}^{2} < 0$ , which is right-handed.

2.3step 1.1F1F2F3F4F5algebra

Case (iii), elliptical. For $e = E_{0} (e_{1} + 2 i e_{2})$ one has $a = E_{0} e_{1}$ , $d = 2 E_{0} e_{2}$ , which are orthogonal by [F5] and have unequal lengths, and $\dim span {a, d} = 2$ ; since $∣ a ∣ = E_{0} \neq 2 E_{0} = ∣ d ∣$ , clause (b)(ii) fails, so clause (b)(iii) of [F1] applies and, because $⟨ a, d ⟩ = 0$ , the trace is the axis-aligned ellipse with semiaxes $E_{0}$ and $2 E_{0}$ along $e_{1}$ and $e_{2}$ . Explicitly $E = E_{0} \cos θ e_{1} - 2 E_{0} \sin θ e_{2}$ and $∣ E ∣^{2} = E_{0}^{2} \cos^{2} θ + 4 E_{0}^{2} \sin^{2} θ$ by [F5], which lies between $E_{0}^{2}$ and $4 E_{0}^{2}$ ; $b = (E_{0} / c) (e_{2} - 2 i e_{1})$ by step 1.1, and $H = ⟨ E_{0} e_{1} \times 2 E_{0} e_{2}, e_{3} ⟩ = 2 E_{0}^{2} > 0$ .

2.4step 1.1givenF1F3F4F5algebra

Case (iv), degenerate and edge cases. For $e = 0$ the fields vanish identically by [F3] (the zero wave), $a = d = 0$ , and clause (b)(i) of [F1] covers the trace ${0}$ ; for $e = E_{0} (1 + i) e_{1}$ one has $a = d = E_{0} e_{1}$ , so $\dim span {a, d} = 1$ , the polarization is linear, and the trace is $E = E_{0} (\cos θ - \sin θ) e_{1}$ , whose coefficient $\cos θ - \sin θ = \sqrt{2} \cos (θ + π / 4)$ ranges over $[- \sqrt{2}, \sqrt{2}]$ ; the trace is therefore the segment $[- \sqrt{2} E_{0}, \sqrt{2} E_{0}] e_{1}$ , and $H = ⟨ a \times a, e_{3} ⟩ = 0$ by alternation of the coordinate formula [F4]. No orthogonality of $a$ and $d$ is used, matching the covering clause of [F1]; the corresponding $b$ is the one computed in step 1.1.

3.1step 1.1F2F6given∎

Scope check (v). In every case step 1.1 gives $b = k \times e / ω$ with $k \cdot e = 0$ and $∣ k ∣ = ω / c$ , so by clause (b) of [F2] the amplitude equations hold and the corresponding fields are smooth source-free Maxwell solutions; the amplitudes $e$ and the semiaxis lengths $E_{0}$ , $2 E_{0}$ carry $V m^{- 1}$ by [F6], and the pointwise relation $∣ B ∣ = ∣ E ∣ / c$ is clause (c) of [F2]. The item computes only ideal field quantities and asserts no measurement or empirical realization.

Remarks

Classification, not evidence. The six choices illustrate clauses of Polarization of a transverse plane wave: linear, circular and elliptical traces, and handedness; they neither predict nor report any laboratory measurement, and the handedness names are the conventional labels fixed there, not physical claims.

Linear, circular and elliptical amplitude choices

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

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