Rotational Vectors not merely a bookkeeping device?

In summary, the direction of the cross product is an arbitrary convention used to give rotation a "direction" (+ for one direction and - for the other). It serves as a bookkeeping device to ensure that different rotation directions are given different signs. However, in 2D rotational problems, it is possible to use counterclockwise vs. clockwise as a model without the use of vectors. But in 3D gyroscopic motion, vectors are necessary to describe the motion. The direction of the vector is still arbitrary, depending on the initial choice of coordinate system. The right hand rule is commonly used to define the cross product, but it is more of an engineering definition and does not provide a mathematical formula for computing it. The handedness
  • #1
FallenApple
566
61
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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
FallenApple said:
But with gyroscopic motion in 3d, you really do need the vectors to describe the motion.
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.
 
  • #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:
  • #5
Dale said:
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.
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.
 
  • #6
Delta Kilo said:
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.
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.
 
  • #7
TonyS said:
As I understand it, the choice of a right hand rule for defining the cross product is independent of the choice of coordinate system.
That is my understanding also, particularly for non Cartesian coordinates.
 
  • #8
TonyS said:
By convention, the right hand rule is always used in defining the vector cross product, regardless of the handedness of any particular coordinate system.

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 anyone 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.
 
  • #9
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 ;-)).
 
  • #10
Delta Kilo said:
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?
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.
 
  • Like
Likes Nugatory and vanhees71
  • #11
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.
 
  • #12
Dale said:
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.
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.
 
  • #13
kuruman said:
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.
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?
 
  • #14
vanhees71 said:
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.
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.
 
  • #15
FallenApple said:
But then with gyroscopic motion, you can't derive from just Newtons laws. You have to invent pseudo vectors just to make it work.
That's not true. You can analyze it using linear dynamics. Rotational dynamics is derived from linear dynamics.
 
  • #16
FallenApple said:
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.
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.
 
  • #17
A.T. said:
That's not true. You can analyze it using linear dynamics. Rotational dynamics is derived from linear dynamics.
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.
 
  • #18
FallenApple said:
I don't see why not, looking at just Newtons laws.
You have to apply Newton's Laws to all the point masses that form the wheel, including the transmission of external forces between them.
 
  • #19
A.T. said:
You have to apply Newton's Laws to all the point masses that comprise the wheel, including the transmission of external forces between them.
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 let's 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:
  • #20
FallenApple said:
...there is no reason...
Handwavy reasoning won't do. All the point masses of the spinning object must obey Newton's Laws.
 
  • #21
A.T. said:
Handwavy reasoning won't do. All the point masses of the spinning object must obey Newton's Laws.

I'm not trying to be handwavy. I'm trying to figure out from a freebody diagram.
 
  • #22
FallenApple said:
I'm trying to figure out from a freebody diagram.
For a qualitative linear dynamic explanation see this.



 
  • Like
Likes Dale
  • #23
A.T. said:
For a qualitative linear dynamic explanation see this.
The satellite example of the first one really hit the key misconception that makes everything else fall into place. The place where the force is applied deflects the velocity, not the position. Precession follows directly.
 
  • #24
Dale said:
The place where the force is applied deflects the velocity, not the position.
That's also why the satellite doesn't fall to Earth despite Earth's gravity, which is also an often asked question. The gyroscope confusion is based on the same misconception, that "force direction = subsequent movement direction".
 
  • Like
Likes Dale
  • #25
A.T. said:
For a qualitative linear dynamic explanation see this.




Oh I see. That makes sense. So in the first video, the orbit got tilted.

So the wheel(i.e orbit) when looked from the top down will constantly tilt in the counter clockwise direction. But in order for this to happen, the twisting force from gravity must always be perpendicular to the path of orbit. Which make sense because once the orbit tilts, the axle tilts with it, maintaining the twisting force from gravity as always perpendicular.
 
  • #26
FallenApple said:
the twisting force from gravity
The "twisting force" (corresponding to the green arrows in the first video) comes from the support force, not from gravity. It is transmitted to the outer masses via the spokes (which have different tension depending on the side).
 
  • #27
A.T. said:


Similar but longer explanation of gyros here:

 
  • Like
Likes FactChecker and FallenApple
  • #28
In 3D, there is more than clockwise/counter-clockwise; there is orientation. So rotations have an orientation and a magnitude. That is a vector. There is another way of looking at these problems that associates a rotation with an oriented (distinguishing clockwise/counter-clockwise) element of area. Geometric algebra takes that approach with some advantage.
 
  • #29
A.T. said:
Similar but longer explanation of gyros here:


Wow, that is a really good video. I especially like the part of expanation of the force moving around constantly on the maximum tilt, making the tilt glide around.
 
  • #30
Yeah! :smile: Science-toy shops are almost as dangerous for my budget as book shopd :nb).
 
  • Like
Likes Dale
  • #31
FallenApple said:
Wow, that is a really good video. I especially like the part of expanation of the force moving around constantly on the maximum tilt, making the tilt glide around.
Yes, I like it too. Although I personally find it more intuitive to consider the torques around the center of mass of an object, rather than the pivot. Gravity creates no such torque, it also creates no differential internal stresses, because it's uniform on such small scales. The support force is the one applied non-uniformly and off center, which is transmitted with differential stresses to the outer rim. In the blowing-on-spining-disk-model from the video, the blow force is also such a local off center force.
 

Related to Rotational Vectors not merely a bookkeeping device?

1. What are rotational vectors?

Rotational vectors are mathematical quantities that represent the magnitude and direction of rotation in three-dimensional space. They are often used to describe the orientation of an object or the movement of a rigid body.

2. How are rotational vectors different from regular vectors?

While regular vectors represent translation or movement in a straight line, rotational vectors represent rotation around a fixed point. They have both magnitude and direction, but their direction is perpendicular to the plane of rotation.

3. What is the significance of rotational vectors?

Rotational vectors are important in physics, engineering, and other scientific fields because they allow us to describe and analyze rotational motion and its effects. They are also used in computer graphics and animation to simulate realistic movement.

4. Can rotational vectors be used in any coordinate system?

Yes, rotational vectors can be used in any coordinate system as long as the axes of rotation are defined. They can be represented in Cartesian coordinates, polar coordinates, or any other coordinate system.

5. How do rotational vectors relate to angular velocity and acceleration?

Rotational vectors are closely related to angular velocity and acceleration, which describe how quickly an object is rotating and how quickly its rotation is changing, respectively. The direction of the rotational vector is the same as the axis of rotation, while its magnitude is equal to the angular velocity or acceleration multiplied by the distance from the axis of rotation.

Similar threads

Replies
1
Views
538
Replies
6
Views
9K
Replies
10
Views
2K
Replies
10
Views
2K
Replies
9
Views
1K
Replies
8
Views
2K
Back
Top