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We recently went over some rotational motion in my freshman physics class, I had a question for the professor on your frame of reference, and how that would effect the rotation of the object.

Say you have a glass clock. You are facing the clock and you note that the clock is rotating clock-wise. You then walk behind the clock, and notate that the clock is now rotating counter-clock-wise. But something seems off about that?

When I asked the professor, he didn't immediately have an answer, instead he did some scratch work and mumbled to himself for a bit before saying "I don't know." He also said something along the lines that is was weird philosophically, and that your things (including direction of rotation) shouldn't depend on your frame of reference from which you view. He then left the classroom while he quietly mumbled "I'll have to think about that..." to himself several times while scratching his chin. He has me worried.

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  • $\begingroup$ The right-hand-rule produces the same vector from either side (pointing out the back of the clock). I don't see how one side or the other actually differs. $\endgroup$
    – Brandon Enright
    Commented Nov 29, 2014 at 6:50
  • $\begingroup$ Can you clarify what is worrying you? Your description just says that rotation transforms like a vector under coordinate transformations, which is hardly surprising as it is a vector (well, pseudo-vector). If you're asking about the transformation of angular momentum under Lorentz transformations this is indeed quite subtle and I'm not surprised your professor couldn't give you the details off the top of his head (I couldn't either :-). $\endgroup$
    – John Rennie
    Commented Nov 29, 2014 at 8:33
  • $\begingroup$ @JohnRennie Just his erratic behavior at what should be a simple scenario. $\endgroup$
    – abrahimladha
    Commented Dec 1, 2014 at 22:20

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Angular speed is a vector (a pseudovector actually), and as such it changes relative to you under a rotation of coordinates. Angular speed is defined as the vector $\boldsymbol \omega =\boldsymbol r$ x $\boldsymbol v/|r|^2$, and it will change direction in an angle of $\pi$ as, expected, if your system of coordinates rotates by $\pi$ too (as it was in your example). What this means, is that, while the vector angular speed will keep pointying in the same direction relative to the clock, in your new system of coordinates it will point in the opposite direction. That is, away from you if you see the clock from the front, and towards you if you are looking at the clock from behind.

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