# Why is angular velocity perpendicular to the plane of rotation?

1. May 22, 2014

### Maxo

Could someone explain how angular velocity points perpendicular to the plane of rotation? I mean what is a physical explanation of this? (not mathematical)

I mean for a rotating object it's easy to see that is has a tangential velocity that always points tangential to the direction of movement. That makes sense, since the object is moving in the same direction as the velocity vector points to.

But with angular velocity, how can this be understood physically? A rotating object never moves in a direction perpendicular to the plane of rotation, so why does the angular velocity point in this direction?

2. May 22, 2014

### Staff: Mentor

Note that every point on the rotating body has a different tangential velocity. Yet the body rotates with a single angular velocity.

If you wanted to describe the rotation of the body as a whole, does it not make sense to describe the axis about which it is rotating?

3. May 22, 2014

### A.T.

Because it is angular velocity, not linear velocity. The vector defines the axis around which the angle changes with time. A normal vector is the simplest way to define the orientation of the plane of rotation.

It is a matter of convention, and this one proved to be practical. But you can propose a better one.

4. May 22, 2014

### Maxo

Yeah, that makes sense.

So the direction of the angular velocity vector is simply to show a vector that is parallel to the axis of rotation? Nothing else?

Could someone please also explain why angular velocity is a "pseudovector"? As I understood pseudovectors, it means that if you "mirror" a system, the pseudo vector switches sign compared to how a normal vector would be mirrored. But why/when would you ever "mirror" a physical system? (Unless we're talking about optics?)

5. May 22, 2014

### D H

Staff Emeritus
Mathematically, it's easy: We live in three dimensional space. I'll have more to say on the importance of three dimensional space later.

Physically, the velocity due to rotation is proportional to the distance to the axis of rotation. The axis of rotation is the set of points that aren't affected by the rotation. There's a deep mathematical reason why this works: It's because we live in three dimensional space.

A much better explanation comes from looking at rotation in something other than three dimensional space. How many degrees of freedom are there to a rotation in some N-dimensional space?

If N=2, there's only one degree of freedom. Rotation can be treated as a scalar (better: a pseudoscalar) in two dimensional space. If N=3, there are three degrees of freedom. What about higher dimensions? The answer is six for four dimensional space, ten for five dimensional space, and in general, it's $\frac {N(N-1)} 2$. Three dimensional space is the only one for which the number of rotational degrees of freedom is the same as the dimensionality of the space.

There is a way to represent rotation generically, and that is with a 2-form. One way to represent a 2-form is via a skew symmetric NxN matrix. The number of independent parameters in a skew symmetric NxN matrix is exactly that quantity specified above, $\frac {N(N-1)} 2$. 2-forms transform a bit differently than do vectors. It turns out that in three dimensions, they transform just like vectors for proper rotations but differently for improper rotations (i.e., transformations involving reflections).

Note that as far as the rules of what makes a thing qualify as a vector, angular velocity is a vector. They can be scaled by a scalar, and they can be summed to form a new angular velocity vector. It's the addition of the concept of how they transform that makes some call them pseudovectors. How things transform is however a concept from tensor algebra, not vector algebra.

It's only three dimensional space where the normal to a plane is a line. In four dimensional space, the normal to a plane is another plane. That one can always describe a rotation as having an axis unperturbed by the rotation is something else that's unique to three dimensional space. The only thing that's unperturbed by the rotation in two dimensional space is the center of rotation, a point. In four dimensions, the unperturbed space due to a primitive rotations is a plane. There's a weird twist to four dimensional rotations: A combination of two simultaneous primitive rotations can result in a non-primitive rotation. The only thing that remains unperturbed is the center. Clifford rotations are truly bizarre.

We don't get the utter bizarreness of rotations in higher dimensional spaces in three dimensional space. Rotation in three dimensional space does have one weird aspect to it: It is not commutative. Rotate a book about axis A, then about axis B. You'll get a different orientation if you reverse the order, rotating about axis B first and then about axis A.

6. May 22, 2014

### Khashishi

This is a common question and I think it should go into some kind of FAQ. Actually, people often ask, "why do we use a right hand rule and not a left hand rule?" but the answer is the same. The essential point is that physics is just a model for reality, and not reality itself, so we are allowed to make certain choices in how we represent things in mathematics out of convenience. It is convenient to represent a rotation in a plane as an arrow in the direction perpendicular to the plane, because we only need to keep track of one direction instead of two. This trick only works in 3D.

7. May 22, 2014

### Maxo

D H: Thanks for the interesting information. Unfortunately I don't have to go in to N-dimensional physics at the moment (but it's nice that you shared this information anyway). For now I would just like to understand it in three dimensions.

Now without going in to N-dimensions or 2-forms etc. but just considering 3 dimensions. Why would you ever want to "mirror"/"reflect"/"rotate"/"transform" etc such a system?

Let's say we have a system involving angular velocity, just for simplicitys sake let's say we have a car driving around in a circle. In this system, what would be the "physical property", so to speak, of the pseudovector? I mean in this case, if we would consider the angular velocity as a "regular" vector instead of a pseudovector, would that change ANYTHING about how you would look at this system? I mean what's the physical property of the pseudovector (which makes it different from a regular vector) in this case?

8. May 22, 2014

### D H

Staff Emeritus
We use different coordinate systems because one size does not fit all. At least not very well. That means we need to be able to transform from one to another.

You appear to be overly hung up on the term "pseudovector", making it so that you cannot see the big picture. Which way the angular velocity vector points is a convention. Everything would be consistent if people used a left-hand rule rather than the near universal right-hand rule.

Note that I wrote "near universal". Things get dicey, for example, when you find out that someone is using a north-east-up frame. (Try to contort your fingers on your right hand so your thumb points north, your index finger east, and your middle finger up. You can't. You have to use your left hand.) Just as there's nothing conceptually wrong with a north-east-up frame, there's nothing wrong with using a left-handed screw rule for rotation or a left-handed rule for cross product.

Perhaps that's a better way of thinking about pseudovectors, that they are things that act just like vectors except that there's an arbitrariness regarding the direction in which they point.

But thinking that there's anything physical to the distinction? You're making a left-handed mountain out of a right-handed molehill.

9. May 22, 2014

### A.T.

It's not about physics, but about conventions. The cross product used to define angular velocity, torque etc. uses the right hand rule by convention. But if you want to mirror everything, your right hand becomes a left hand. So if you want to stick to the right-hand rule, you have to negate all cross products.

10. May 22, 2014

### my2cts

Angular velocity is just angular momentum divided by the moment of inertia, I. It is called "velocity" because of the similarity with the relation between linear momentum and velocity. This is imho a misnomer, as the angular "velocity" is oriented perpendicularly to the material velocity. Angular momentum is a pseudovector because it is even (keeps its sign) under space inversion and odd (changes sign) under time reversal.

Last edited: May 22, 2014
11. May 22, 2014

### sophiecentaur

I could ask you to choose a better direction. That direction would have to be the same, whichever part of a rigid object you were considering. I think that more or less fixes a choice of perpendicular - just on the grounds of symmetry.

12. May 22, 2014

### my2cts

What puzzles is why the angular "velocity" has a direction perpendicular to all linear velocities.
The origin of this confusion is that angular velocity is a misnomer, as explained above.

13. May 22, 2014

### my2cts

A vector changes sign/inverts direction under space inversion, but not under time inversion. For a pseudovector it is the other way around. Really, a pseudovector is not a vector but an antisymmetric tensor of rank 2. In 3 dimensions this object happens to be similar to a vector apart from its properties under inversion.
The physics. In this case if you invert all the coordinates with respect to the center of the circle (any point on the rotation axis will do) , the angular momentum (and thus the unfortunate quantity "angular velocity") remains the same. If you make a movie and run it backwards it changes sign.

Last edited: May 22, 2014
14. May 22, 2014

### Staff: Mentor

Because we are lazy.

If we have solved a problem once and are then confronted with a problem that is similar but mirrored, then we would like to use our previous solution and some easy transformation rules to get the new answer without having to rework the whole problem.

15. May 24, 2014

### D H

Staff Emeritus
Lower level physics classes treat moment of inertia as a scalar. It isn't. It's a second order tensor, typically represented as a 3x3 symmetric matrix. It's much better to think of the angular momentum of
a rigid body as being the matrix-vector product $\vec L = I\vec\omega$ rather than angular velocity being given by $\vec\omega = I^{-1}\vec L$. Angular momentum is hard to measure. Angular velocity? That's easy to measure. Even your cell phone might have a MEMS gyro in it if its new enough.

It is not a misnomer. It describes something rather different from yet in a way analogous to translational velocity. Angular velocity, or rotational velocity, describes how fast some object is rotating. Translational velocity, or plain old velocity, describes how fast some object is translating.

16. May 24, 2014

### rcgldr

Getting back to the original question:

I'm not sure there is a physical explanation. The mathematical reason for this is that doing math using vectors to represent angular velocity (or angular acceleration or angular force == torque) is fairly straight forward, while some mathematical invention would be needed to perform math based on rotating planes.

17. May 25, 2014

### my2cts

I am aware of all that.Still I think it is a misnomer.