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Worked transverse Doppler protocols at beta = 3/5

Physical scope

Worked comparison of the two transverse Doppler protocols of the ideal plane-wave model at relative speed (3/5)c: gamma in the frame where the wave is transverse to the motion and 1/gamma in the frame where it is transverse in the observer frame. 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, showing that the transverse Doppler shift depends on the frame in which transversality is required. Let $x$ and $x^{'}$ be inertial charts with $x^{'} = B (β) x + b$ , $β = (3 / 5) e_{1}$ and $γ = 5 / 4$ , and let an ideal plane wave have frame- $x$ data $(ω, s)$ with $ω > 0$ .

(a) If $s = e_{2}$ , so $β \cdot s = 0$ and the propagation direction is perpendicular to the relative velocity in $x$ , then Doppler shift and aberration for an ideal plane wave between inertial descriptions gives $ω^{'} = γ ω = (5 / 4) ω > ω$ and $s^{'} = (- 3 / 5, 4 / 5, 0)$ , which has $∣ s^{'} ∣ = 1$ ; the frequency read by the $x^{'}$ -rest observer is blue-shifted by the factor $5 / 4$ .

(b) If $s = (3 / 5, 4 / 5, 0)$ , so $β \cdot s = 9 / 25 = ∣ β ∣^{2}$ and equivalently $s^{'} \cdot e_{1} = 0$ , the propagation direction is perpendicular to the relative velocity in $x^{'}$ , and the theorem gives $ω^{'} = γ (1 - 9 / 25) ω = (5 / 4) (16 / 25) ω = (4 / 5) ω = ω / γ < ω$ and $s^{'} = e_{2}$ ; the frequency read by the $x^{'}$ -rest observer is red-shifted by the factor $4 / 5$ .

In neither case is $ω^{'} = ω$ for the nonzero $β$ , and the two exact factors differ; a statement of a transverse Doppler shift is therefore incomplete until the frame in which the direction is transverse is specified. Scope: a conditional computation in the ideal plane-wave model; no measurement and no empirical claim.

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 $(ω, s)$ , $ω > 0$ , $∣ s ∣ = 1$ .

[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 $ω^{'} = γ (1 - β \cdot s) ω > 0$ with $γ = (1 - ∣ β ∣^{2})^{- 1 / 2}$ , 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, clause (i)).

[L2]

Transverse protocols of the theorem: if $β \cdot s = 0$ then $ω^{'} = γ ω$ and this is greater than $ω$ when $β \neq 0$ ; if $β \cdot s = ∣ β ∣^{2}$ , equivalently $s^{'} \cdot β_{h a t} = 0$ for $β \neq 0$ , then $ω^{'} = ω / γ$ , which is less than $ω$ when $β \neq 0$ (Doppler shift and aberration for an ideal plane wave between inertial descriptions, clause (iv)).

[L3]

$γ = (1 - ∣ β ∣^{2})^{- 1 / 2}$ , so for $β = (3 / 5) e_{1}$ : $γ = (16 / 25)^{- 1 / 2} = 5 / 4$ , and $∣ β ∣^{2} = 9 / 25$ (The passive Lorentz boost with parameter beta, Boost transformation of a future null wave covector and of its frame data).

Verification

technique · direct

1.1A1L1L2L3

Case (a). Here $s = e_{2}$ has $∣ s ∣ = 1$ and $β \cdot s = (3 / 5) (0) = 0$ , so the propagation direction is perpendicular to the relative velocity in $x$ , and [L2] gives $ω^{'} = γ ω = (5 / 4) ω$ , which is greater than $ω$ because $γ = 5 / 4 > 1$ by [L3]; the axis formulas of [L1] with $b = 3 / 5$ , $s_{1} = 0$ , $s_{2} = 1$ , $s_{3} = 0$ give $s_{1}^{'} = (0 - 3 / 5) / (1 - 0) = - 3 / 5$ , $s_{2}^{'} = 1 / [(5 / 4) (1 - 0)] = 4 / 5$ and $s_{3}^{'} = 0$ , and $∣ s^{'} ∣^{2} = 9 / 25 + 16 / 25 = 1$ .

1.2A1L1L2L3

Case (b). Here $s = (3 / 5, 4 / 5, 0)$ has $∣ s ∣^{2} = 9 / 25 + 16 / 25 = 1$ and $β \cdot s = (3 / 5) (3 / 5) = 9 / 25 = ∣ β ∣^{2}$ ; by [L2] this is equivalent to $s^{'} \cdot β_{h a t} = 0$ with $β_{h a t} = e_{1}$ , the propagation direction is perpendicular to the relative velocity in $x^{'}$ , and $ω^{'} = γ (1 - 9 / 25) ω = (5 / 4) (16 / 25) ω = (4 / 5) ω = ω / γ$ , which is less than $ω$ ; the axis formulas of [L1] give $s_{1}^{'} = (3 / 5 - 3 / 5) / (1 - 9 / 25) = 0$ , $s_{2}^{'} = (4 / 5) / [(5 / 4) (16 / 25)] = (4 / 5) / (4 / 5) = 1$ , $s_{3}^{'} = 0$ , so $s^{'} = e_{2}$ .

2.1step 1.1step 1.2∎

Conclusion. In case (a) the shift is by $5 / 4$ , in case (b) by $4 / 5$ ; both configurations are called transverse because the propagation direction is perpendicular to the relative velocity in one of the two frames, and the two factors are different and neither is $1$ for the stated nonzero $β$ . Hence a transverse Doppler statement is incomplete until the frame in which transversality holds is specified. The computation is conditional on the ideal plane-wave model and makes no measurement or empirical claim; no choice principle is used.

Worked transverse Doppler protocols at beta = 3/5

Example

Facts & Assumptions

Verification

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Dependency tree · 2 levels

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