Can an angle be considered to be a vector?

[TonyWallace]

Hello

I have been thinking about dimensional analysis with respect to computer systems. It has become obvious to me that to be meaningful such an analysis has to distinguish dot product from vector products. An area can be considered the cross product of vectors. The question that arose is "Can an angle be considered a vector". Let us consider rotating a vector from one line of action to another line of action. Such a rotation will occur around an axis which is orthogonal to both vectors just like a cross product. Furthermore an, just as an area can only be viewed accurately from a normal to the plane, so to can an angle only be viewed accurately from that axis of rotation.

If an angle was a vector defined in this way:
WorkDone = Torque . RotationalAngle

which is completely analoguous with:
WorkDone = Force . Displacement

Your comments are invited.

 

[tiny-tim]

Welcome to PF!

Hello Tony! Welcome to PF!

Yes, angular velocity (angle per time), ω, and angular acceleration, α, are well-known vectors,

equal to the time derivatives of the angle vector that you've described.

(but we don't usually call angle a vector)

 

[AlchemistK]

When we were taught about angular velocity , we were told that angle is not a vector since it does not follow the rules of vector operations.
However, a very small angle CAN be considered as an angle.

 

[Simon Bridge]

An angular displacement would be a vector - it is related to "angle" like displacement is related to "length".

iirc: the "angular" vectors are a special case: "pseudo-vectors".
http://en.wikipedia.org/wiki/Pseudovector

 

[D H]

When we were taught about angular velocity , we were told that angle is not a vector since it does not follow the rules of vector operations.

Correct.

However, a very small angle CAN be considered as an angle.

Only infinitesimally small angles can be treated as a vector (better said, as a pseudovector), and even then, only in ℝ³.The reason angles cannot be treated as vectors is because rotation is not a commutative operation. Pick up a book. Imagine a set of coordinate axis based on the front cover. The x and y axes lie on the plane of the front cover, with +x pointing from the top to the bottom and +y pointing left to right. The +z axis completes a right hand system. Now rotate the book by 90 degrees about the book's +x axis, then another 90 degrees about the rotated +y axis. Reversing the order of the operations yields a very different orientation.

There's nothing special about 90 degrees here. Define sequence 1 as a rotation about +x by some angle θ followed by a rotation about +y by some angle ϕ. Define sequence 2 as a rotation about +y by some angle ϕ followed by a rotation about +x by some angle θ. The orientations that result from these two sequences will only be the same if at least one of θ and ϕ is zero.

The difference between the orientations becomes progressively smaller as the angles θ and ϕ get smaller. This difference in orientations goes to zero much faster than does either θ or ϕ. In the limit θ→0 and ϕ→0 (infinitesimal rotations), the sequences do commute. Infinitesimally small rotations in ℝ³ look somewhat like vectors, which is why we can describe angular velocity in terms of a vector.

Angular velocities scale according to the rules of vector multiplication by a scalar, add according to the rules of vector addition, and for a proper transformation, they transform according to the rules of vector transformation. They behave oppositely from vectors under reflection (an improper transformation). Things that exhibit these characteristics are called pseudovectors.

Angular velocities in general are co-vectors. Only one parameter is needed to describe an angular velocity in ℝ². Six are needed for ℝ⁴, and in general N*(N-1)/2 are needed for ℝ^N. In ℝ³, this N*(N-1)/2 is three, which is why we can get away with treating angular velocity in ℝ³ as a vector (better said, pseudovector).

 

[TonyWallace]

I do not expect the application of a series of rotations of an object in 3d space to be commutative because they are cross product operations and cross products are not commutative. Therefore I do not accept your argument as it seems to be based on commutativity.

 

[TonyWallace]

If we agree that angular velocity and angular acceleration are vectors, the it can be seen that the differentiation by time results in a colinear vector with the one differentiated by. It would follow that integration would also result in a colinear vector as is true from angular acceleration to angular velocity. Why should the integration with respect to time of angular velocity result in anything but a vector, colinear with the angular velocity?

 

[TonyWallace]

Okay I have done some experiments and rotation does not follow cross product rules either. For a cross product (a X b)=-(b X a) and it does not work.

I will read up pseudovectors on wikipedia as you suggest.

 

[TonyWallace]

I have gone to Hamiltons "Elements of Quarternions" and angles the way I have defined them are vectors according to his definition, having direction and magnitude.

 

[D H]

Therefore I do not accept your argument as it seems to be based on commutativity.

That argument is essential. Rotations cannot be viewed as vectors precisely because rotations do not commutative.

An even better reason is that vectors in three space and rotations in three space are creatures from very different mathematical spaces. Vectors are elements of ℝ³ while rotations are elements of SO(3), the group of special orthogonal 3x3 matrices. You wrote later about quaternions. Quaternions (unit quaternions) are one of many charts on SO(3). Other ways to represent rotations in three space include Euler angles and variations on that theme, Cayley-Klein parameters, Rodrigues parameters, modified Rodrigues parameters, ..., and of course the 3x3 special orthogonal matrices themselves.

 

[TonyWallace]

The earlier post about commutivity and turning the book was defined by operations in eucledian space. I suggest that the reason that the operations were not commutative was the choice of a rectangular coordinate system not appropriate for the discussion. Instead we should consider a spherical geometry when discussing operations on angles in three dimensions.

Consider the earth. Let us take the north pole as our origin. If we take an equatorial axis such that any positive angle describes a line down the prime meridian, and we rotate through 130 degrees we get to 40 degrees south, in the region of South Africa. If we then rotate along the polar axis by 174 degrees we get to New Zealand. Now this second rotation has not changed the arc length as the resulting arc is from the North Pole to New Zealand. This is because we are working off a sphere.

What now happens if we reverse the order of operations? Standing on the North Pole, we are rotated along the polar axis, next we rotate along a different meridian and end up at New Zealand. We have our commutivity back by changing our coordinate system. If you try this with the book it works, the book faces the same way after both rotations.

 

[TonyWallace]

Is that the essential argument, rotation is in the wrong type of space to be a vector, that vectors exist only in eucledian spaces? Rotation is obviously not a scalar because of its axial direction, if it is not a vector, scalar or field, then what is it?

 

[D H]

Rotation in ℝ^N is well known to be non Abelian (non commutative) for all N≥3. I really don't want to argue about a known fact. I suggest you read about group theory in general and the special orthogonal groups in general.

 

[TonyWallace]

I accept that in general rotation is non-commutative. My question involves a number of different aspects. (1) The calculus aspect: if the integration of an angular acceleration to an angular velocity results in a colinear vector, why would the same operation, that is integration with respect to time when done an angular velocity result in anything but a vector? (2) Angles have obvious similarities with vectors, that is a magnitude and a direction. (3) If not a vector or a scalar or a field, what is an angle in terms of such aggregate classes? (4) The operation of a dimensionless number upon a physical quantity results in a new number of the same dimension. For example the area of a circle is πr2. r is in the dimension of length (vector) r2 is result giving an area (vector). The application of π leaves the answer still as an area (vector). This is to be contrasted with applying an torque to an angle and getting work. This change of physical expression is a characteristic of dimensional numbers, not dimensionless ones.

I saw a comment time back about angular vectors being pseudo-vectors. Is a pseudo-vector a vector? Examples of cross-product pseudo-vectors abound in physics. Everyone would agree that torque for example is a vector, but it is defined by F X R, and is by the wikipedia definition a pseudo vector. If we agree that angle is a pseudo vector in that way, then we have agreement.

I will do some more reading as you suggest.

 

[D H]

My question involves a number of different aspects. (1) The calculus aspect: if the integration of an angular acceleration to an angular velocity results in a colinear vector, why would the same operation, that is integration with respect to time when done an angular velocity result in anything but a vector? (2) Angles have obvious similarities with vectors, that is a magnitude and a direction. (3) If not a vector or a scalar or a field, what is an angle in terms of such aggregate classes?

Integrating angular velocity with respect to time obviously does result in a vector. The problem is that this integrated vector in general does not have any physical meaning.

The problem is that rotations are not vectors. Having direction and magnitude is not enough to qualify as a "vector". There are rules. Amongst other things, vectors must add commutatively to form another vector. Rotations in 2 space are commutative, but beyond that they are not.

Rotations in N-space are instead members of a different kind of mathematical structure, the group of the NxN special orthogonal matrices. An orthogonal NxN matrix is one in which each row, or each column, is a unit vector, and any two different rows, or any two different columns, are orthogonal to one another. The qualifier "special" means that the determinant is 1 rather than -1. (Note: Another name for an orthogonal matrix with a determinant of 1 is a proper rotation matrix. Those with a negative determinant are "improper".)

While NxN matrices don't commute in general, a matrix does commute with itself. These means that one can unambiguously define Aⁿ, where A is some NxN matrix, and n is any non-negative integer. This in turn means that one can unambiguously define the matrix exponential exp(A) by using the standard power series representation of the exponential function. This concept introduces another way to represent an NxN proper orthogonal matrix: As the matrix exponential of an NxN skew symmetric matrix. The matrix exponential of any NxN skew symmetric matrix is always a proper rotation matrix, and any proper rotation matrix can be represented as the matrix exponential of some NxN skew symmetric matrix. (Proving these statements is non-trivial.)

As there are N*(N-1)/2 independent members of a skew symmetric matrix, the number of degrees of freedom in a N-dimensional rotation is N*(N-1)/2. The number of degrees of freedom is equal to the dimensionality of the space for N=3 only. It is this fact that make some people think that rotations in 3-space can be represented by vectors. They can't. Just because they can be represented by three numbers does not mean they are vectors. These three numbers don't obey the rules of what constitutes a vector.

 

[TonyWallace]

In general a rotational vector may not have physical meaning, but in rotational kinematics it is the rotational analogue of distance, and can be used in equations accordingly. WorkDone = Torque.Rotation. I accept that this is a special case, and that the more normal understanding of rotation is a mapping function. A mapping function has no physical significance.

In rotational kinematics we are not working with an eucledian geometry, but rather a cylindrical or spherical geometry. In this case a vector from the centre which I will call a radial vector, is at right angles to a vector running in a circular loci around that centre which I will call a tangential vector. An angle is the quotient of arc length to radial length (which is an angle in radians). Because these vectors are at right angles to each other this quotient is not dimensionless as they are vectors and not scalars and do not cancel as their directions differ. For a rotating system this angle is integral of angular velocity with respect to time, and is colinear with the angular velocity vector. As a torque is applied over an angle, the radial vector in the torque cancels with the radial vector in the angle leaving the tangential vector of distance and the tangential vector of force resulting in work done. Thus if follows that WorkDone = Torque.Rotation. It also follows that this rotational vector has physical significance within this area of physics.

 

[D H]

In general a rotational vector may not have physical meaning, but in rotational kinematics it is the rotational analogue of distance, and can be used in equations accordingly. WorkDone = Torque.Rotation. I accept that this is a special case, and that the more normal understanding of rotation is a mapping function. A mapping function has no physical significance.

In the special case of rotation about an inertially fixed axis, yes rotation is commutative. This is because what you are talking about is just a plane (2D) rotation, and rotation in ℝ² is commutative.

Your final sentence makes no sense. Are you here to learn, or just argue nonsense?

 

[TonyWallace]

You said

Integrating angular velocity with respect to time obviously does result in a vector. The problem is that this integrated vector in general does not have any physical meaning.

The comment about physical meaning was yours, my comment was agreeing with you. I have shown quite clearly that this integrated vector does have physical meaning in this particular case. As the the general case, it may well have no meaning. I perceived that you were taking a general case of rotation in general rather than the specific case that I am referring to.

I have been following up on your posts, primarily with wikipedia with regard to rotational matrices. I think in general the discussion has made good ground. Originally we were discussing rotation with regard to Cartesian coordinates, and we agreed that this was not the best. We have agreed that the integration of angular velocity with respect to time is a vector. I believe that I have shown in a single case that the integrated angular velocity vector has physical significance.

 

[TonyWallace]

I think I understand. A vector must follow the laws of vector addition. If we had 3 axes, representing the three possible orthogonal axes of rotation. each axis marked out with angle of rotation along its length, each point in that three dimensional space would be a specific rotation from the origin. For rotation to be a vector space then any combination of rotational vectors within that space would have to follow the laws of vectors. That means to rotate 1/4 rotations along the x-axis followed by 1/4 rotations along the y-axis must be the same as 1/4 rotations along y-axis followed by 1/4 rotations along x axis. If this does not follow then this is not a vector space. If it is not a vector space, then these rotations are not vectors.

 

[lugita15]

D H, let me put it this way. If you didn't already know that rotations don't commute, what reason would you have to suspect that the integral of the angular velocity is not the rotation vector?

The reason I ask this is that the non-commutativity rotation has been a deeply counterintuitive fact to me for many years (and yes, I've done the book thing countless times, and I'm still startled every time), and I want a better understanding of it. I know about rotation matrices not commuting, I know about the anti-commuting Lie algebra which gives rise to the non-commuting rotation group SO(3), and I know about quaternions not commuting, etc., but this all seems to be designed to capture the fact we already know that rotations don't commute. But what is the fundamental reason for this fact? Why don't rotation vectors add according to the parallelogram rule, or since rotations can be represented as arcs on the surface of a sphere, why don't such arcs obey a "spherical parallelogram rule"? I suspect the answer to my last question has somehing to do with the double cover property of the rotation group, and I have some inklings that the answer to the general question of how to geometrically understand compositions of rotations purely in terms of rotation vectors (i.e. axis-angle representations) can be found in the works of Olinde Rodrigues on rotations, but this requires more investigation.

 

[lugita15]

Does anyone have any thoughts on my post #20?

 

[TonyWallace]

I think the small angle commutativity of rotation allows that change between the vector world of angular velocity, and the non vector world of rotation. By definition differentiation is about deltas approaching zero. Which makes the differential linear and commutative where the parent function is not.

 

[lugita15]

I think the small angle commutativity of rotation allows that change between the vector world of angular velocity, and the non vector world of rotation. By definition differentiation is about deltas approaching zero. Which makes the differential linear and commutative where the parent function is not.

I'm aware that infinitesimal rotations commute to first order and not to second order, but I'm looking for a more rigorous explanation. The proof of commutativity of angular velocity can be done without any reference to infinitesimal rotations or matrices, so in particular we don't even invoke any arguments about second-order noncommutativity of infinitesimal rotations. So where in principle, without knowing in advance about the noncommutativity of rotation, would defining a rotation vector as the integral of the angular velocity vector break down?

 

[TonyWallace]

The answer to this question is precisely the same reason we can't predict the weather. Differentiation essentially makes things into linear functions. Just because we understand all there is to know about the laws of motion in a tiny piece of fluid, does not imply we can understand the entire system. This was the change of thinking that came in with chaos theory. A segment of a circle may appear to be a function, but a circle is not a function (because for any x there is more than one y). So integrating, or adding together, circle segment functions will not leave us with a function for the circle itself. The circle is not a functions, but differentiation at any point acts like functions.

 

[D H]

So where in principle, without knowing in advance about the noncommutativity of rotation, would defining a rotation vector as the integral of the angular velocity vector break down?

Rotations in ℝ² are commutative. In ℝ³ and large dimensional spaces that they aren't. The concept would break down as soon as one left the regime of plane rotations. Use the wrong description of reality and at some point you'll get the wrong answer.

The answer to this question is precisely the same reason we can't predict the weather.

Nonsense.

We can't predict the weather beyond seven days or so because weather is a chaotic system. Rotations are not chaotic in and of themselves. They aren't sensitive to initial conditions. A small change in initial conditions does not lead to a marked divergence.


A better analogy comes from special relativity. Velocities appear to add as vectors in Euclidean 3 space in our everyday, slow-moving world. Using Euclidean vector addition for fast moving objects yields incorrect results because velocities don't add as Euclidean 3-vectors. Once again, using the wrong description of reality and at some point you'll get the wrong answer.

 

[TonyWallace]

Dear DH. We in general cannot know a system by knowing its parts. Systems take on properties of their own. This is why we can build and program computers. Yes at one level they are arrays of transitors, but understanding the transitor arrays does not help us understand the accounting system that runs on it. Yes sensitivity to initial conditions is an important part of chaotic systems, but you have taken it out of context and deliberately misinterpreted what I was saying.

 

[lugita15]

Where would the following argument for commutativity of rotations break down? From the proof shown in the attached excerpt from Taylor's Classical Mechanics, which doesn't use any arguments about infinitesimal rotations approximately commuting to first order, we conclude that angular velocity adds vectorially. Therefore, since any infinitesimal rotation can be represented as ω(t)dt, we conclude that infinitesimal rotations add vectorially and thus commute not just to first order but to all orders.

 

[D H]

Just because angular velocities adds vectorially does not mean you can integrate angular velocity to yield a meaningful quantity. Integrating angular velocity is in general meaningless. The one exception is a unidirectional angular velocity (i.e., plane rotation rather than 3D rotation), which can be integrated. This special case works because the rotation is restricted to a plane, and rotations in ℝ² do commute.

 

[lugita15]

Just because angular velocities adds vectorially does not mean you can integrate angular velocity to yield a meaningful quantity. Integrating angular velocity is in general meaningless. The one exception is a unidirectional angular velocity (i.e., plane rotation rather than 3D rotation), which can be integrated. This special case works because the rotation is restricted to a plane, and rotations in ℝ² do commute.

But the whole point of my question is to try to get at the fundamental reason rotations don't commute. See my post #20.

 

[D H]

But the whole point of my question is to try to get at the fundamental reason rotations don't commute. See my post #20.

First off, your analysis in post #27 is incorrect. Differential angle does not commute to all orders.

The fundamental reason rotations don't commute is because they aren't vectors. They are instead rotation matrices. You've done the experiment with the book. Believe your eyes.

One way to look at rotation is that a proper orthogonal matrix is the matrix exponential of a skew symmetric matrix, $R=\exp(S)$. This works in ℝ², ℝ³, ℝ⁴, in fact, in any finite dimension N>1. (Note: The requirement that the dimensionality be greater than one arises from the fact that there is no such thing as rotation in ℝ¹, and there is no such thing as a skew symmetric 1x1 matrix.)

Note that this says that the number of degrees of freedom in an N-dimensional rotation is N*(N-1)/2. This happens to be equal to N in ℝ³ only. This is ultimately the reason one can represent rotation in ℝ³ as a rotation about an axis (the axis-angle representation of rotation in ℝ³) and why one can treat angular velocity in ℝ³ as a vector. Angular velocity is better treated as a skew symmetric matrix, a bivector.

This matrix exponential form suggests using the matrix logarithm to find the underlying skew symmetric matrix: $S = \log(R)$. This does work in a sense. One problem that arises is that the matrix logarithm, like the complex logarithm, is not unique. Another problem that arises is that adding these skew symmetric matrices does not do what you would naively think it would do. While logarithms are additive for the non-negative reals, $\log(\exp(a)\exp(b)) = a+b$, this doesn't work with matrix exponential / matrix logarithm for N>2. Proper orthogonal matrices don't commute for N>2.

 

[lugita15]

First off, your analysis in post #27 is incorrect. Differential angle does not commute to all orders.

Yes, it is certainly true that infinitesimal rotations commute to first order but not to second order. But I would like to find the error in my proof that they DO commute to all orders.

The fundamental reason rotations don't commute is because they aren't vectors. They are instead rotation matrices.

But the reason why we represent rotations as noncommuting matrices is because we know in advance that they do not commute. Surely we would represent them in an abelian group if we happened to believe that they do commute. So this does not strike me as a fundamental reason.

To me, it seems like the reason rotations form a nonabelian Lie group is because they are exponentials of an anti-commuting Lie algebra. So perhaps the real question is why do we make the Lie bracket anti-symmetric? Or to put it another way, why are the cross product of vectors and the wedge product of differential forms anti-commuting? (And no, I wouldn't consider the fact that the cross product can be represented in terms of a skew-symmetric matrix to be a good enough explanation, because why don't we define a product of vectors that is not represented by a skew-symmetric matrix?)

You've done the experiment with the book. Believe your eyes.

The book demonstration is sufficient to convince me THAT rotations don't commute, but it does not suffice as an explanation for WHY I'm seeing what I'm seeing.

 

[D H]

Yes, it is certainly true that infinitesimal rotations commute to first order but not to second order. But I would like to find the error in my proof that they DO commute to all orders.

You didn't state your proof mathematically. Try doing so. It won't work.

But the reason why we represent rotations as noncommuting matrices is because we know in advance that they do not commute.

Exactly. That's physics for ya. A physicist's job is to describe reality.

To me, it seems like the reason rotations form a nonabelian Lie group is because they are exponentials of an anti-commuting Lie algebra. So perhaps the real question is why do we make the Lie bracket anti-symmetric?

To me it seems that the best place to start is linear algebra. Euler did quite a bit studying rotation without the benefit of linear algebra. Linear algebra just makes Euler's work easier. The connection between the transformation matrices in linear algebra and Lie groups is an after-the-fact development and in a sense makes things harder. You won't see much on Lie algebras and Lie groups in physics until you get to grad school. The mathematics is not easy. The mathematics of linear algebra can be taught to some extent at the high school level.

The time derivative of a time-varying transformation matrix can be represented as the matrix product of the transformation matrix and a skew symmetric matrix or as the matrix product of a different skew symmetric matrix and the transformation matrix. A 3x3 skew symmetric matrix can be represented by an axial vector, and this is the reason why we can represent angular velocity in three space as a vector. Note well: This only works in ℝ³ only. For example, a 2x2 skew symmetric matrix has one free parameter, so angular velocity in ℝ² can be represented by a scalar (a pseudoscalar to be precise) rather than a two vector. A 4x4 skew symmetric matrix has six free parameters, so angular velocity in ℝ⁴ cannot be represented by a four vector.

Or to put it another way, why are the cross product of vectors and the wedge product of differential forms anti-commuting?

We use the cross product of vectors in ℝ³ (the vector cross product as-is is unique to ℝ³) because it is a useful concept. If it wasn't useful we wouldn't use it. Physicists use mathematical tools that help in the endeavor of describing reality.