jackiefrost said:
An April 18 newscientist.com article,
http://www.newscientist.com/article/dn18775-mysterious-radio-waves-emitted-from-nearby-galaxy.html" reports recent observations of a radio source in M82 with an apparent horizontal motion of 4x speed of light.
Could someone comment on what particular relativistic effects would cause an "optical illusion" of superluminal velocities. What specific empirical measurements are they referring to that would indicate apparent superluminal velocities?
It is a common effect that occurs when an object is moving at high velocity almost towards you. It is not a relativistic effect so much as a consequence of the finite speed of light.
Suppose, for example, that something is moving at an angle of θ away from directly towards you, at a velocity v, and you know that it is a distance away from you d.
Suppose that two signals are sent from this object, separated in time Δt. (Forget any relativistic effects. Everything is done in your own reference from.) Now over that time, it moves a distance v.Δt. We break this into two components; it is closer to you by v.Δt.cos(θ), and it has moved orthogonal to the line of sight by v.Δt.sin(θ). The latter corresponds to an apparent horizontal movement across the sky.
Note, however, the second signal has less distance to go to reach you. Therefore it arrives after a shorter period of time Δt(1 - (v/c).cos(θ)). This is simply subtracting the time it would have taken light to go that extra distance towards you, which is (v/c).Δt.cos(θ).
If you forget to consider this advantage in time, you might think that the time the object took was the time between the two signals to be observed, rather than the time between the two signals being emitted.
The apparent horizontal velocity of the object is thus a = v.sin(θ)/(1 - (v/c).cos(θ)).
Using units with c = 1, if you differentiate this with respect to θ, assuming v fixed, you get
\begin{align*}<br />
a & = \frac{v \sin \theta}{1 - v \cos \theta} \\<br />
\frac{\partial a}{\partial \theta} & = v\frac{\cos \theta ( 1 - v \cos \theta ) - \sin \theta ( v \sin \theta )}{(1 - v \cos \theta)^2} \\<br />
& = c \frac{\cos \theta - v}{(1 - v \cos \theta)^2}<br />
\end{align*}
Thus, when v = cos(θ) you get a maximum apparent velocity, which works out to be
\frac{v}{\sqrt{1-v^2}}
If I have the maths right, then you can get an apparent horizontal velocity of 4c if the v was about 97% lightspeed, and the particle was moving at about 14 degrees off directly towards you.
Cheers -- sylas
