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Why is not angle considered as vector?

  1. Oct 22, 2011 #1
    why we can't considered angle as vector. but small angle ( such as velocity angle) is vector?
    i know, we can't write angle as a+b=b+a; but i can't understand why we can write it for small angle.
    Thx a lot...:)
  2. jcsd
  3. Oct 22, 2011 #2


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    I have absolutely no idea what you are talking about. An angle is not a vector because it does not fit the definition of a vector- a vector has both "magnitude" and "direction". An angle defines a direction but not a magnitude. I have no idea what you mean by "velocity angle" nor why it would only be defined for small angles. Could you give an example?
  4. Oct 22, 2011 #3
    as you know, we can write angular velocity (or rate of angle changes) of something in universal frame by bellow equation:
    a= angular velocity of body A on x axis
    b= angular velocity of body A on y axis
    c=angular velocity of body A on z axis
    so it is clear if body A starts to rotate by a then b then c final angular velocity is the equal of velocity when body A starts to rotate by b then c then a, or by the other word:
    but if a,b,c are angle then mentioned equation can not be true, why?
  5. Oct 22, 2011 #4


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    They're not angles, they're the rate of change of angular displacements.
  6. Oct 22, 2011 #5


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    Perhaps English isn't your first language, but you seem to be confusing "rotation" with of "angle".

    They are two different ideas, and as you said in your OP, finite rotations are NOT vectors, because they don't commute under addition.

    Because they don't commute, you have to be very careful not to make false assumptions about what is true in the limit for small angles. Some of the explanations and "proofs" about angular momentum "vectors" in beginning mechanics textbooks are horribly wrong, even though they get to the right "answer".
  7. Oct 22, 2011 #6
    absolutely yes ..English is my third (or maybe fourth) language..:( and i horrible in it....i try to improve my self..:)...Thx a lot for your answer.
    but i can't understand why rotations don't commute under addition...i know many numerical example, but analytically i can't understand it.
    furthermore why we can write angular velocity as vector?
  8. Oct 22, 2011 #7

    D H

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    There's a lot of hidden magic that lets you even start to reason this way.

    And now you are violating the terms of that hidden magic.

    Rotations in ℝ3, Euclidean three dimensional space, are not commutative. Why? Rotations in a Euclidean n-space are described by the special orthogonal group SO(n). They are not described by ℝn. Everything beyond SO(2), rotations in a plane, is not commutative. Rotation A followed by rotation B is not the same as rotation B followed by rotation A.
  9. Oct 23, 2011 #8


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    And, simply put, physical rotations aren't commutative.

    Take a book (a math book, of course!) and hold it in front of you with its front cover toward you. Rotate around the vertical axis to your left (its right side moves toward you, its left side away). Now rotate it around the axis going directly away from you, through the center of the book, clockwise (its top goes down, bottom up). You should now have the book lying on it back with front cover up.

    Now, start again with the front cover of the book facing you and do exactly those two rotations in the opposite order. After the first rotation, counter clockwise, you will still have the front cover facing you but with its top to your right. After the second rotation, the front cover will be facing to your left, the top of the book toward you. A completely different position that the first example.
  10. Oct 24, 2011 #9

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    I (and most physicists and engineers, and Euler) look at a sequence of rotations from a slightly different perspective, that of rotations about axes defined in terms of the body being rotated. So, defining some axes, let the x and y axes be on the plane defined by the cover of the book. The +x axis points from the top to the bottom while the +y points from left to right. The +z axis completes a right hand system, so it is coming out of the front cover; in the orientation described by Halls, toward your eyes. The first rotation sequence starts with a +90 degree right hand rule rotation about the book's +x axis followed by a -90 degree rotation about the book's (rotated) +y axis. Reversing that order, a -90 degree rotation about the +y axis followed by a +90 degree rotation about the rotated +x axis leaves the book with the binding down such that if you flipped the book open you could read it (the text would be oriented "the right way").

    Regardless of how you look at rotations, rotations in three space are not commutative.

    Short answer: Two reasons.
    1. In the limit of infinitesimally small rotations, rotations in three space are commutative (rotation A followed by rotation B is the same as rotation B followed by rotation A).
    2. Any sequence of rotations in three space can described in terms of a single rotation about a single axis. In other words, one can always find a single axis rotation that results in that exact same final orientation as does the sequence of rotations. (Note: This concept does not apply in other dimensions; it is special to three space.)

    Given two rotations, rotation 1 a rotation of angle θ1 about axis u1 and rotation 2 a rotation of angle θ2 about axis u2, the single axis rotation that describes rotation 1 followed by rotation 2 will be about the axis

    \vec u_{12} = \cos\frac{\theta_2}2 \sin\frac{\theta_1}2 \hat u_1
    + \cos\frac{\theta_1}2 \sin\frac{\theta_2}2 \hat u_2
    + \sin\frac{\theta_1}2 \sin\frac{\theta_2}2 \hat u_1\times\hat u_2[/tex]

    The single axis rotation corresponding to rotation 2 followed by rotation 1 will be about the axis

    \vec u_{21} = \cos\frac{\theta_2}2 \sin\frac{\theta_1}2 \hat u_1
    + \cos\frac{\theta_1}2 \sin\frac{\theta_2}2 \hat u_2
    - \sin\frac{\theta_1}2 \sin\frac{\theta_2}2 \hat u_1\times\hat u_2[/tex]

    Note that the cross product terms are of opposite sign in the two rotation sequences. The single axis rotations will be about different axes. Note also that the other two terms are identical and that this cross product term involves the product of the sines of the two half angles. As both rotation angles θ1 and θ2 tend to zero, this cross product term will tend to zero much faster than will the other terms. The two sequences look more and more like one another as the rotation angles get small. The two sequences become identical in the limit of infinitesimally small rotations.
  11. Oct 24, 2011 #10


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    If we wanted to be very rigorous, we should write angular momentum as an "anti-symmetric tensor". A tensor in three dimensions has 9 components and can be written as a 3 by 3 matrix in a given coordinate system. For an anti-symmetric tensor, [itex]a_{ij}= -a_{ji}[/itex] so [itex]a_{ii}= 0[/itex] and we have only 3 independent values. That's why angular momentum can be written like a vector.
  12. Oct 26, 2011 #11
    Thx everyone..:)
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