Faster than light galaxies/clusters? A few years ago in an astronomy course, we calculated some (transverse?) velocity of a moving object and got super luminal results. The answer was apparent and not physical velocity of the object. Hence no problem. But at the moment, I don't recall the solution to this apparent issue. Anyone? 
 A: This question already has two good answers, but I needed an excuse to learn tikz, so here is my answer.
This "transverse" speed can be faster than the speed of light $c$ if the object is coming towards you sufficiently fast. To see how this works, look at the diagram below.

Here the object is moving at a speed $\beta c$, but it is not moving perpendicular to your line of sight. It is moving at an angle $\theta$ to the line of sight. 
How would the speed of this object be estimated? A naive way would be to divide the observed transverse displacement by the observed time duration. 
Let's see what these two quantities are when the actual time duration is $ T_{\mathrm{actual}}$. In this time the object will move an actual distance $v  T_{\mathrm{actual}}$. Its apparent transverse displacement will be $d=v  T_{\mathrm{actual}} \sin \theta$.
What will the observed time duration be? It won't be $ T_{\mathrm{actual}}$ because the light rays from the final position get a head start relative to the light rays from the initial position. The length of this head start is $v  T_{\mathrm{actual}} \cos \theta$. Since light moves at $c$, this translates into a time speed-up of $v  T_{\mathrm{actual}} \cos \theta /c$. That is how much faster than it "should" the final light gets there. Thus the observed time difference is shorter than the actual time difference by this amount. We get that the observed time difference is $T_{\mathrm{observed}} = T_{\mathrm{actual}} - v  T_{\mathrm{actual}} \cos \theta /c = T_{\mathrm{actual}}(1 - v   \cos \theta /c)$.
The observed speed is then $\frac{v T_{\mathrm{actual}} \sin \theta}{T_{\mathrm{actual}}(1 - v   \cos \theta /c)} =\frac{v  \sin \theta}{1 - \frac{v}{c}   \cos \theta}$. Plugging in $v=\beta c$, we get that the observed speed is $c \frac{\beta \sin \theta}{1-\beta \cos \theta}$. To get an observed speed greater than $c$, we need $\frac{\beta \sin \theta}{1-\beta \cos \theta}>1$. Picking $\theta = 45^\circ$, this becomes $\frac{\beta } {\sqrt{2}-\beta }>1$. This inequality is achieved for $\beta > \frac{1}{\sqrt{2}}$, so superluminal speeds can be "observed".
A: The velocity calculation was distance*(change in angle). However, this does not take into account the changing time-delay of light: we see it sped-up because the time delay is decreasing, like a TV recording where you are fast-forwarding as you gradually catch up with real time. Fortunately, all we need to do to calculate the real speed is to account for the time delay, no weird relativity is necessary.
Suppose a far away object is approaching at $0.8c$ and has a transverse velocity of $0.25c$ (total velocity of $0.84c$). It emits a burst of light (in our frame) at time $t$ and $t+5$ seconds. It is $d$ light seconds from us at time $t$ and $d-4$ at $t+5$. Accounting for the time delay of light, we see flashes at time $t+d$ and $t+5+(d-4) = t+d+1$; we see them only $1$ second apart. In those $5$ seconds, it moved transversely a distance of $5\times0.25 = 1.25$ light seconds. Since it appeared to move $1.25$ light seconds in $1$ second, we see apparent superluminal motion.
A: Great answer @Kevin, I'm just going to rewrite it in different letters. 
Note: Only classical mechanics is applied below.
Suppose a far away object has a velocity satisfying (in units where velocity of light $c:=1$)
$$v^2 = v_{\perp}^2+v_{\|}^2<1^2\tag1$$ 
where $v_{\perp}$ denotes its perpendicular (to our line of sight) velocity and $v_{\|}$ its velocity parallel to our line of sight. 
It emits a burst of light (in our frame) at times $t$ and $t+\tau.$
If $d_t$ denotes the distance to the object at time $t$, then its distance at time $t+\tau$ is $$d_{t+\tau} = d_t-v_{\|}\tau.$$ 
When will the light reach us?
The light emitted at time $t$ will reach us at time 
$$t+d_t$$
(at the least, since its distance was $d_t$ and nothing travels faster than light). 
The light emitted at time $t+\tau$ will reach us at time 
$$t+\tau+d_{t+\tau}$$ but this is equal to 
$$ t+\tau +(d_t-v_{\|}\tau). $$
Thus, we will see two light flashes seperated by a time 
$$\Delta t = t+\tau +(d_t-v_{\|}\tau) - \big[t+d_t\big]\\=\tau\cdot(1-v_{\|}).$$
During the time of size $\tau$, the object has moved a distance of 
$$d_{\perp} = v_{\perp}\cdot \tau,$$ perpendicular to our line of sight. But since we see the flashes appearing with time $\Delta t$ apart, we will calculate the (apparent) transverse velocity of the object to be 
$$v_{\perp}^a(v_\perp,v_\|) = \frac{d_{\perp}}{\Delta t}\\ =\frac{v_{\perp}}{1-v_{\|}}$$
where $v_\perp,v_\|$ (should in principle) satisfy condition $(1).$ 
Now plugging in some numbers e.g. $v_\perp=\frac{1}{4}$ and $v_\|=\frac{4}{5}$ we get 
$$v_{\perp}^a(1/4,4/5) = \frac{5}{4}>1.$$ This being larger than 1 (i.e. larger than the speed of light in vacuum $c$) could then be interpreted as superluminal motion. 
