Do I see the moon Lorentz contracted if I rotate around myself? If an object moves at velocity $v$ relative to my frame its length $l_0$ will be Lorentz contracted to the length in my frame $l$:
$$
l = l_0 \sqrt{1-\frac{v^2}{c^2}}
$$ 
If I rotate around myself with frequency $\omega$, objects at distance $r$ will have the relative velocity $v = r \omega$. So the moon at a distance of $r=300000$ km should be Lorentz contracted to length $l= 0$, if I rotate around myself with $\omega =1$ Hz (according to the formula $l$ would even get complex, when I would rotate faster).
I haven't done the experiment, but I believe if the moon would be Lorentz contracted somebody would have already noticed that effect. So something must be wrong here. 
The funny thing is, I believe that if the moon or a spaceship would rotate around me with a speed of $v < c$, when I must see it Lorentz contracted according to the formula above. So it seems to me the two cases, I rotate around myself and something rotates around me, are not equivalent. In the first case I see no Lorentz contraction, but in the second case I do. Is my thinking correct? If so, can someone explain why these two cases cannot be equivalent and why using the formula for Lorentz contraction must be wrong when I rotate around myself?
 A: When you rotate, you are no longer in an inertial frame, so it's tricky to apply special relativity. In particular, the coordinates have a nontrivial connection, which means you cannot directly compare vectors at two different points. In particular, you cannot compute the relative velocity between you and the moon directly.
As an analogy, consider two people on the Earth, each holding compasses. It's impossible to compare if their compasses are both working unless they both meet up in the same place to compare. This process is called parallel transport.
In this case, (supposing for simplicity that the moon's velocity is zero in the nonrotating frame), the physically correct way to compute the relative velocity of the moon is to parallel transport its velocity to you, at the origin. When you do this, you'll find a velocity of zero, as expected.
In the latter case, when the moon is moving and you're not rotating, you are in an inertial frame, so the connection is trivial. So you can directly compare your velocity to the moon's, getting a nonzero result and hence a Lorentz contraction.

This problem also is common in general relativity. For example, naively calculating the coordinate velocity of distant galaxies suggests that they're moving away faster than the speed of light. And naively calculating the coordinate velocity of light falling into a black hole in the Schwarzschild metric says that the light's velocity slows down to zero as it approaches the event horizon. (The problems are worse in GR, since the nontrivial curvature of the spaces means that the result of parallel transport is nonunique.)
A: One important proviso: According to A. P. French, Special Relativity, MIT Introductory Physics Series (1968) pp 149-152 (still a gem as far as I'm concerned) who noted that there is an important difference between "observe" and "see or look at", the words see and look involve the finite time for the transit of the light from different parts of the body.
French gives references to the following papers J Terrell, Phys Rev  116, pp 1041-45 (1959), V F Weisskopf, Physics Today 13 pp 24-7 (1960) and G. D. Scott and M.R Viner, Am J Phys 33 p534 (1965) and finishes with the comment "This misconception, which must have made every physicist blush a little ....".
