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I Rotational Vectors not merely a bookkeeping device?

  1. Oct 7, 2016 #1
    I was told that the direction of the cross product is an arbitary convention to give rotation a "direction" + for one direction and - for the other. That it is simply a book keeping device to make sure different rotation directions are given different signs.

    It seemed to be the case as I am able to do most of the rotational problems without using these mysterious "vectors out of the page". I merely used different signs for counterclockwise vs clockwise, which doesn't need the idea of vectors at all. counterclockwise vs clockwise as a model suffices completely. So I did the "bookkeeping" without any vectors. Basically, using rotational vectors in those 2d problems really are against the principle of Ocam's Razor.


    But with gyroscopic motion in 3d, you really do need the vectors to describe the motion. So the vectors are not an imagined construct. That they are somehow real?
    Is it even possible to understand gyroscopic motion without vectors?
     
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  3. Oct 7, 2016 #2

    kuruman

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    Here is the rationale for the "mysterious vectors". Have you ever thought how you can unambiguously describe how a body is rotating? Say you want to represent angular velocity. You need a vector because there is a obviously a magnitude involved that tells you how fast in rpm perhaps the body is spinning. Now when you have a body going around in a circle, the circle is in a plane and the rotation can be clockwise or counterclockwise in that plane. So to specify the angular velocity you not only need to specify its magnitude, but also the plane of the circle that I just described and whether the rotation in that plane is "this way" or the "opposite way." So by drawing an arrow to represent the angular velocity, we essentially specify the plane in which the rotation takes place as the one perpendicular to the arrow and then use the right hand rule to establish the sense circulation of the object in that perpendicular plane. Commonly, people use curved arrows to indicate rotation as in the case of emergency exits in airplanes, but in physics arrows that represent vectors are always straight.
     
  4. Oct 7, 2016 #3

    Dale

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    The direction of the vector is still arbitrary in the sense that you could use a left hand rule as easily as a right hand rule and everything would work out the same.
     
  5. Oct 9, 2016 #4
    Well, it is actually a deep and subtle topic. The rotation vector is technically a pseudovector, or more generally a bivector, but by a lucky coincidence there is almost no difference in ordinary Euclidean 3D world ("almost" refers to the fact they don't reflect in the mirror the same way normal vectors do). However a proper geometric interpretation of a bivector is that of an "oriented plane element" (in the same sense that vector is "oriented line element"), that is a fragment of a plane (shape is not important, area is) which has a "+" side and a and "-" side, and a proper mathematical form is anti-symmetric rank 2 tensor. In 3D there is one-to-one correspondence between such plane elements and their perpendicular vectors. In 2D bivectors correspond to (signed) scalars. And vectors (and scalars) are so much easier to deal with.

    TL;DR:
    It makes perfect sense to represent rotational quantities as scalars in 2D and vectors in 3D. One can visualize them as little parallelograms or circles with direction of rotation indicated, or strait arrows along the axis of rotation or a combination of the above, whatever works. Understanding vectors is essential, for example, in order to find out total angular momentum of the system (which is conserved), one has to add together angular momenta of parts using rules of vector addition.

    400px-Exterior_calc_cross_product.svg.png
     
    Last edited: Oct 9, 2016
  6. Oct 9, 2016 #5
    The choice is set by the initial choice of coordinate system (left- or right-handed). From here X x Y = Z or it is better be otherwise a lot of formulas will need a sign change.
     
  7. Oct 9, 2016 #6
    As I understand it, the choice of a right hand rule for defining the cross product is independent of the choice of coordinate system. By convention, the right hand rule is always used in defining the vector cross product, regardless of the handedness of any particular coordinate system.
     
  8. Oct 9, 2016 #7

    Dale

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    That is my understanding also, particularly for non Cartesian coordinates.
     
  9. Oct 10, 2016 #8
    Well, that is true, but I would call it "engineering" definition. The problem with this statement is that the right hand rule is, pardon the pun, handwavy. It tells you the result points "that a-way", but how do you express it mathematically? It's kind of impossible to encode the shape of human appendage into a formula. As a result, the so-called rule does not provide a recipe of computing the cross-product unless some other assumptions are made. For example, given two vectors with coordinates (1,2,3) and (4,5,6) in some unspecified coordinate system, what is their "right-handed" cross-product?

    In fact the handedness of coordinate system and that of cross product are complimentary: a right-handed coordinate system with a right-handed cross product is indistinguishable from a left-handed coordinate system with a left-handed cross product (*). To avoid the issue, one talks instead about positively or negatively oriented basis sets, as determined by the sign of a scalar triple product [itex]\vec x \cdot (\vec y \times \vec z)[/itex]. Turns out it can be arbitrary chosen for any one basis and it automatically sets the orientation for all other bases. The "handedness" is introduced when we arbitrary designate the basis formed by one's right hand as positive and it automatically makes all other "right-handed" ones positive and it makes cross-product right-handed also.

    (*) unless we are talking about chiral symmetry breaking in QCD, but at that point we are way outside our cozy Euclidean 3D space.
     
  10. Oct 10, 2016 #9

    vanhees71

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    Indeed you can just define relative orientation of bases in vector spaces, i.e., if you have an arbitrary basis ##\vec{b}_j## (##j \in \{1,2,3 \}##) and another basis
    $$\vec{b}_k'=\vec{b}_j {T^{j}}_{k},$$
    you define ##\vec{b}_k'## (##k \in \{1,2,3, \}##) oriented in the same (opposite) sense, if ##\det T>0## (##\det T<0##).

    If you deal with geometrical vectors in the 3D affine (Euclidean) space, it's customary to call coordinate systems oriented in the same sense as thumb, index, and middle finger of the right hand (pointing in the "natural" way) as positive oriented by convention. Fortunately there seems to be no single textbook using a left-hand rule. Otherwise, you'd have the same mess with sign conventions already in Newtonian mechanics as you have in general relativity ;-)).
     
  11. Oct 10, 2016 #10

    Dale

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    That is one of the nice things about doing physics instead of pure math. We don't need to limit ourselves to strictly mathematical definitions. We must also include experimental definitions. So we can define handedness in terms of an experiment performed literally by handwaving.
     
  12. Oct 10, 2016 #11

    vanhees71

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    It's hard to conceive how to do physics otherwise. Also, if you ask me, I think that math, although a pure structural science of thought, needs some intuition from more practical problems. E.g., it's not a pure accident that the ancient egypts have found a lot of Euclidean geometry, e.g., Pythagoras's theorem, because they had to evaluate their land after each flood of the river Nile again. Great math often develops from practical problems. One example is the development of functional analysis starting from making Dirac's ##\delta## distribution rigorous.
     
  13. Oct 10, 2016 #12
    Interesting. But would that mean that it's impossible to derive gyroscopic motion using just momentum constrained to two dimensions? For example, summing up all the linear momentum of a gyroscopic wheel without considering pseudo vectors.
     
  14. Oct 10, 2016 #13

    Oh so you are saying that we need to consider direction, counterclockwise vs clockwise. Does that mean that a different formulation would get the same results as long as long as counterclockwise vs clockwise are treated as mathematical opposites?
     
  15. Oct 10, 2016 #14
    The thing is that you can analyze all sorts of rotating motions( e.g rolling without slipping, spinning disk falling on a disk at rest, speeding it up etc) using newtons laws as the basic principle.


    But then with gyroscopic motion, you can't derive from just newtons laws. You have to invent pseudo vectors just to make it work.

    This is very different. For example KE, PE, conservation, momentum, moments of inertia can all be derived from just Newton's second law.
     
  16. Oct 10, 2016 #15

    A.T.

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    That's not true. You can analyze it using linear dynamics. Rotational dynamics is derived from linear dynamics.
     
  17. Oct 10, 2016 #16

    Dale

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    No, it doesn't mean that at all. It means that at some point in your derivation you will need to pick a convention for which sense of rotation is positive. You can pick either convention, but you have to be consistent.
     
  18. Oct 10, 2016 #17
    untitled.png




    So in the image, we know that the wheel will want to go down because of gravity. Hypothetically if gravity brought the wheel down, would it not be still spinning clockwise at the same rotational speeds? I don't see why not, looking at just newtons laws.

    But looking at the directions of angular momentum and torque vectors would contradict my example.
     
  19. Oct 10, 2016 #18

    A.T.

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    You have to apply Newton's Laws to all the point masses that form the wheel, including the transmission of external forces between them.
     
  20. Oct 10, 2016 #19
    Well, gravity will try to bring the wheel down. The axle will probably create some friction on the wheel, affecting the rotational speeds, but not the direction. But lets assume frictionless axle.



    the parts the the wheel going up will be slowed down and the parts of the wheel going down will be sped up. So its like a bead constrained in a vertical loop that will go on forever.

    So the individial motion of the points are like a bunch of beads moving counterclockwise on a loop. And if I drop the loop of beads, the loop will fall. And the beads will revert back to constant cetrepetal accerlation, if at the horizontal position. It's probably somewhere in between since the loop will twist in freefall.


    Now the wheel is attached to a rope. So you might say its different. But even if the rope went taut, canceling out with the weight of the axle, if the axle is frictionless, there is no reason why the axle will spin around in the horizontal plane.

    The only way is if the wheel exerts a force on the axle in the horizontal direction. Which doesn't make sense because the rope and gravity only pulls in the vertical, so any response contact force on the axle would be up or down. Not sideways. Even if we only require constant tangental speed for horizontal movement of the axile, there at least needs to be an initial sideways force tangentaly to change bring the axle out the its rest state.
     
    Last edited: Oct 10, 2016
  21. Oct 10, 2016 #20

    A.T.

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    Handwavy reasoning won't do. All the point masses of the spinning object must obey Newton's Laws.
     
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