Editing Superluminal motion (section)

==Derivation of the apparent velocity==

{{Unreferenced section|date=August 2023}}
A [[relativistic jet]] coming out of the center of an [[active galactic nucleus]] is moving along AB with a velocity ''v'', and is observed from the point O. At time <math>t_1</math> a light ray leaves the jet from point A and another ray leaves at time <math>t_2=t_1+\delta t </math> from point B. An observer at O receives the rays at time <math>t_1^\prime</math> and <math>t_2^\prime</math> respectively. The angle <math>\phi</math> is small enough that the two distances marked <math>D_L</math> can be considered equal.
[[File:Superluminal motion in AGN jets.png|right|400px]]
: <math>AB = v \, \delta t</math>
: <math>AC = v \, \delta t \cos\theta</math>
: <math>BC = v \, \delta t \sin\theta</math>
: <math>t_2-t_1 = \delta t</math>

: <math>t_1' = t_1 + \frac{D_L + v \, \delta t \cos\theta}{c}</math>
: <math>t_2' = t_2 + \frac{D_L}{c}</math>

: <math>\delta t' = t_2' - t_1' = t_2 - t_1 - \frac{v \, \delta t \cos\theta}{c} = \delta t - \frac{v \, \delta t \cos\theta}{c} = \delta t (1-\beta \cos\theta)</math>, where <math>\beta=v/c</math>

: <math>\delta t = \frac{\delta t'}{1-\beta\cos\theta}</math>

: <math>BC = D_L\sin\phi \approx \phi D_L = v \, \delta t \sin\theta \Rightarrow \phi D_L = v\sin\theta\frac{\delta t'}{1-\beta\cos\theta}</math>

Apparent transverse velocity along <math>CB</math>, <math>v_\text{T} = \frac{\phi D_L}{\delta t'}=\frac{v\sin\theta}{1-\beta\cos\theta}</math>

: <math>\beta_\text{T} = \frac{v_\text{T}}{c} = \frac{\beta\sin\theta}{1-\beta\cos\theta}.</math>

The apparent transverse velocity is maximal for angle (<math>0 <\beta < 1</math> is used)

: <math>\frac{\partial\beta_\text{T}}{\partial\theta} = \frac{\partial}{\partial\theta} \left[\frac{\beta\sin\theta}{1-\beta\cos\theta}\right] = \frac{\beta\cos\theta}{1-\beta\cos\theta} - \frac{(\beta\sin\theta)^2}{(1-\beta\cos\theta)^2} = 0</math>

: <math>\Rightarrow \beta\cos\theta(1-\beta\cos\theta)^2 = (1 - \beta\cos\theta) (\beta\sin\theta)^2</math>

: <math>\Rightarrow \beta\cos\theta (1-\beta\cos\theta) = (\beta\sin\theta)^2 \Rightarrow \beta\cos\theta - \beta^2\cos^2\theta = \beta^2\sin^2\theta \Rightarrow \cos\theta_\text{max} = \beta</math>

: <math>\Rightarrow \sin\theta_\text{max} = \sqrt{1-\cos^2\theta_\text{max}} = \sqrt{1-\beta^2} = \frac{1}{\gamma}\,</math>, where <math>\gamma=\frac{1}{\sqrt{1-\beta^2}}</math>

: <math>\therefore \beta_\text{T}^\text{max} = \frac{\beta\sin\theta_\text{max}}{1-\beta\cos\theta_\text{max}} = \frac{\beta/\gamma}{1-\beta^2} = \beta\gamma</math>

If <math>\gamma \gg 1</math> (i.e. when velocity of jet is close to the velocity of light) then <math>\beta_\text{T}^\text{max} > 1</math> despite the fact that <math>\beta < 1</math>. And of course <math>\beta_\text{T} > 1</math> means that the apparent transverse velocity along <math>CB</math>, the only velocity on the sky that can be measured, is larger than the velocity of light in vacuum, i.e. the motion is apparently superluminal.