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Worked longitudinal Doppler shift at beta = 3/5

Physical scope

Worked longitudinal Doppler computation of the ideal plane-wave model at relative speed (3/5)c in the two propagation directions along the boost axis. Conditional on the model; no measurement, statistical inference or empirical claim.

Example

Worked example in the model of sr-light-and-doppler, on the page sr-light-and-doppler-examples. Let $x$ and $x^{'}$ be inertial charts with $x^{'} = B (β) x + b$ , $β = (3 / 5) e_{1}$ , so that $γ = 5 / 4$ and the frame $x^{'}$ moves with velocity $(3 / 5) c e_{1}$ relative to $x$ . Let an ideal plane wave have frame- $x$ data $(ω, s)$ with $ω > 0$ and (a) $s = e_{1}$ , (b) $s = - e_{1}$ . Then Doppler shift and aberration for an ideal plane wave between inertial descriptions gives:

(a) $ω^{'} = γ (1 - 3 / 5) ω = (5 / 4) (2 / 5) ω = ω / 2$ , $ν^{'} = ν / 2$ , $λ^{'} = 2 λ$ and $s^{'} = e_{1}$ : an observer at rest in $x^{'}$ moving in the direction of propagation reads half the frequency and twice the wavelength.

(b) $ω^{'} = γ (1 + 3 / 5) ω = (5 / 4) (8 / 5) ω = 2 ω$ , $ν^{'} = 2 ν$ , $λ^{'} = λ / 2$ and $s^{'} = - e_{1}$ : an observer at rest in $x^{'}$ moving opposite to the propagation direction reads twice the frequency and half the wavelength.

In both cases $∣ s^{'} ∣ = 1$ and $ω^{'} > 0$ , in agreement with the general theorem; the verification uses only the exact rational values $γ = 5 / 4$ , $1 - 3 / 5 = 2 / 5$ and $1 + 3 / 5 = 8 / 5$ . Scope: a conditional computation in the ideal plane-wave model; no measurement is performed and no empirical claim is made.

Facts & Assumptions

Given: Inertial charts $x, x^{'}$ with $x^{'} = B (β) x + b$ , $β = (3 / 5) e_{1}$ ; an ideal plane wave with frame- $x$ data $(ω, s)$ , $ω > 0$ , $∣ s ∣ = 1$ , and $s = \pm e_{1}$ .

[A1]

Adopted ideal plane-wave model: one ideal monochromatic plane wave with future null wave covector; the rest observers of $x$ and $x^{'}$ read the frequencies $ω$ and $ω^{'}$ (Adopted ideal plane-wave and frequency-reading model, Null wave covectors, frame frequency data and the observer frequency of an ideal plane wave).

[L1]

For the chart pair of the Given the theorem gives $ω^{'} = γ (1 - β \cdot s) ω > 0$ , $ν^{'} = γ (1 - β \cdot s) ν$ , $λ^{'} = λ / [γ (1 - β \cdot s)]$ , $∣ s^{'} ∣ = 1$ , and in the axis case $β = b e_{1}$ the direction formulas $s_{1}^{'} = (s_{1} - b) / (1 - b s_{1})$ , $s_{2}^{'} = s_{2} / [γ (1 - b s_{1})]$ , $s_{3}^{'} = s_{3} / [γ (1 - b s_{1})]$ (Doppler shift and aberration for an ideal plane wave between inertial descriptions, clauses (i) and (ii)).

[L2]

Longitudinal clause: for $s = β_{h a t}$ the factor is $γ (1 - ∣ β ∣) = \sqrt{(1 - ∣ β ∣) / (1 + ∣ β ∣)}$ and for $s = - β_{h a t}$ it is $γ (1 + ∣ β ∣) = \sqrt{(1 + ∣ β ∣) / (1 - ∣ β ∣)}$ ; here $β_{h a t} = e_{1}$ (Doppler shift and aberration for an ideal plane wave between inertial descriptions, clause (iii)).

[L3]

$γ = (1 - ∣ β ∣^{2})^{- 1 / 2}$ , so for $β = (3 / 5) e_{1}$ : $γ = (1 - 9 / 25)^{- 1 / 2} = 5 / 4$ ; and the boost transformation of the frame data is the one of Boost transformation of a future null wave covector and of its frame data (The passive Lorentz boost with parameter beta).

Verification

technique · direct

1.1A1L3given

Setup. Here $β_{h a t} = e_{1}$ , $∣ β ∣ = 3 / 5$ and $γ = (1 - 9 / 25)^{- 1 / 2} = (16 / 25)^{- 1 / 2} = 5 / 4$ by [L3]; both witness directions $s = \pm e_{1}$ have $∣ s ∣ = 1$ , so the wave is in the model [A1].

2.1step 1.1L1L2

Case (a), $s = e_{1}$ . Then $β \cdot s = 3 / 5$ , so [L1] gives $ω^{'} = γ (1 - 3 / 5) ω = (5 / 4) (2 / 5) ω = ω / 2$ and hence $ν^{'} = ω^{'} / (2 π) = ν / 2$ and $λ^{'} = c / ν^{'} = 2 c / ν = 2 λ$ ; the axis formulas of [L1] with $b = 3 / 5$ , $s_{1} = 1$ , $s_{2} = s_{3} = 0$ give $s_{1}^{'} = (1 - 3 / 5) / (1 - 3 / 5) = 1$ and $s_{2}^{'} = s_{3}^{'} = 0$ , so $s^{'} = e_{1}$ ; the frequency factor agrees with the longitudinal clause [L2], $ω^{'} / ω = \sqrt{(1 - 3 / 5) / (1 + 3 / 5)} = \sqrt{(2 / 5) / (8 / 5)} = 1 / 2$ , since $s = β_{h a t}$ .

3.1step 1.1L1L2∎

Case (b), $s = - e_{1}$ . Then $β \cdot s = - 3 / 5$ , so [L1] gives $ω^{'} = γ (1 + 3 / 5) ω = (5 / 4) (8 / 5) ω = 2 ω$ , $ν^{'} = 2 ν$ and $λ^{'} = c / ν^{'} = λ / 2$ ; the axis formulas of [L1] with $b = 3 / 5$ , $s_{1} = - 1$ give $s_{1}^{'} = (- 1 - 3 / 5) / (1 + 3 / 5) = - 1$ and $s_{2}^{'} = s_{3}^{'} = 0$ , so $s^{'} = - e_{1}$ ; the frequency factor agrees with the longitudinal clause [L2], $ω^{'} / ω = \sqrt{(1 + 3 / 5) / (1 - 3 / 5)} = \sqrt{(8 / 5) / (2 / 5)} = 2$ , since $s = - β_{h a t}$ . In both cases $ω^{'} > 0$ and $∣ s^{'} ∣ = 1$ as the theorem asserts. This proves (a) and (b) by exact rational arithmetic; the computation is conditional on the ideal plane-wave model, no measurement is performed and no empirical claim is made; no choice principle is used.

Worked longitudinal Doppler shift at beta = 3/5

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources