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Addition of angles are non-commutative in three dimensions. Hence some other angular vector quantities like angular velocity, momentum become non-commutative. What is the physical significance of this non-commutative property?

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Comment to the question(v1): Rotations form a group, and are hence associative. Did you mean non-commutative instead of non-associative? –  Qmechanic Jan 4 '13 at 17:07
    
angular displacement does`t obey the rule a+b=b+a......@ Qmechanic –  Arafat Jan 4 '13 at 17:39
    
Yes, but the rule $a+b=b+a$ is called the commutative law. –  Qmechanic Jan 4 '13 at 18:38
    
oh.... you are right... –  Arafat Jan 4 '13 at 18:47
    
For physical significance of non-commutative rotations, see e.g. this Phys.SE answer. –  Qmechanic Jan 4 '13 at 22:23

1 Answer 1

I'll compare the equations of motion of a point particle and a rigid body.

To understand a free point particle, once I solve its motion in one dimension ($x(t) = v_0 t + x_0$) I can immediately solve the three dimensional problem. This is because the equations are separable and commutative: since translations along $x$ (label this symmetry $T_x$) implies a conserved quantity $p_x$, and translations along different axes are commutative ($T_x T_y = T_y T_x$), then (1) I can write any translation as $T_x T_y T_z$, and (2) I have the conserved quantities $p_x$, $p_y$, and $p_z$, which don't have anything to do with each other.

Even if I lose translational symmetry, the equations I need to solve generally look like $$ \ddot x = f(x, y, z) $$ $$ \ddot y = g(x, y, z) $$ $$ ... $$

Now, I know that rotation about one axis (say $R_x$ is a rotation about the $x$ axis) has a conserved quantity $L_x$. That's true for all three axes. But it doesn't mean I can immediately write down the equations of motion; since $R_x R_y \neq R_y R_x$, I can't just write down three angular momenta and be done with it. The actual equations of motion are confusing because the coordinates are all mixed up. Even for a rigid body in free space, the equations we need to solve use Euler angles (this might be a better reference), which are the correct representation to find constants of motion. As you can see (when you look up Euler angles), they're coordinates called spin, nutation, precession that are defined in some funny way that seems to mix up the usual angles. However, that's the way to write down coordinates that will simplify the equations of motion and gives you conserved quantities.

In quantum mechanics, things get even more interesting. For translations, I have a quantum mechanical operator $\hat x$ that describes position. I know that $\hat p_x$ is the conjugate operator and that they don't commute ($[\hat x, \hat p_x] = i \hbar$). However, commutativity of the translational group means that most everything else you write down does commute, e.g. $[x, y] = [p_x, p_y] = [x, p_y] = 0$. This greatly simplifies the problem of solving the evolution of a point particle.

Now, the lack of commutativity of rotations gives rise to the Lie group in quantum mechanics, where the momentum operators $\hat L_x$, $\hat L_y$, and $\hat L_z$ don't commute. This leads to interesting symmetry groups in quantum mechanics, e.g. SU(N). SU(2), the symmetry of a spin-1/2 particle, turns out to be very related to SO(3), the rotational symmetries of a rigid body.

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