Speed of light affecting appearance of galaxies viewed not face on? So this has been really bugging me over the past few days (and forgive me if the answer is so simple). Let's say we're observing the Sombrero galaxy. 
It is about 29 million light years away and 50 thousand light yyears in diameter. 
So we should be observing the "front" of it at what it looked like 29 million years ago, and the "back" of it 29.05 million years ago. 
Why doesn't this extra distance change the galaxy's shape? If, for example, the galaxy was moving directly away from us in a straight line (not that it is), wouldn't the galaxy be compressed? Hope this makes sense.
 A: Yes, the galaxy shape is changed. To explain why let me start with a somewhat odd analogy, but bear with me and you'll see how it applies to the galaxy.
Suppose at time $t = 0$ you throw a ball with some velocity $v$. The equation of motion is obviously just $d = vt$, but suppose we ask what the equation of motion appears to be if we include the time taken for light from the object to reach your eye.

Suppose the object is at some distance $d$, then the time it took the object to reach that distance is just $t_1 = d/v$. Likewise, the time it took the light from the object to get back to your eye is $t_2 = d/c$. So the total time is:
$$ t = \frac{d}{v} + \frac{d}{c} = d\frac{c + v}{cv} $$
The apparent velocity, $v'$, is just $d/t$ so:
$$ v' = \frac{cv}{c + v} $$
So the ball appears to be moving at a constant velocity $v'$ that is slower than its real velocity. The apparent distance of the ball is then just:
$$ d' = v't = \frac{cv}{c + v} t $$
Now suppose that instead of a ball we have the front edge of a galaxy, and let's graph the apparent distance of that front edge:

This is a straight line of gradient $v'$. The rear edge of the galaxy is moving at the same speed as the front edge (because the galaxy isn't changing size) so its apparent position is also a straight line of gradient $v'$, but it is displaced by the time taken for the galaxy to pass you. This time is just the true size of the galaxy, $ell$, divided by the true velocity, $v$:
$$ t = \frac{\ell}{v} $$
The apparent length of the rod, $\ell'$, is the vertical arrow marked on the graph, and this is just $\ell' = v't$, so the apparent length of the galaxy is:
$$ \ell' = \ell \frac{c}{c+v} $$
So the galaxy appears to be foreshortened by a factor of $c/(c + v)$.
But consider what this is actually going to look like. Galaxies are much too far away for us to judge distance by parallax, as we do when looking at things close to us, so we wouldn't be able to tell that the galaxy was foreshortened. Instead it would appear to be slightly rotated:

If you take some galaxy aligned at an angle $\theta$ to us then it will be forshortened along the direction of motion, $\ell' \lt \ell$, but not at right angles to it. The end result is that the apparent angle of tilt will be slightly greater than the true angle of tilt. This effect is biggest for galaxies tilted at around 45°. Galaxies that are face on or edge on to us won't appear to have changed.
