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Does anyone have an intuitive explanation of why this is the case?

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Define "rotation". To me a rotation is an action/transformation. Do you mean an actual physical quantity? –  Noldorin Nov 5 '10 at 23:33
    
Maybe he wants to know why usually angular velocities are characterized by vectors and rotations by matrices. Is that your question? –  Robert Smith Nov 5 '10 at 23:48
    
@Robert: Yes, my question is why angular velocities can be represented by vectors when rotations can't –  Casebash Nov 6 '10 at 0:09
    
@Casebash: As Noldorin said, rotations are transformations which are applied over something else. On the other hand, angular velocity is a quantity that describes a property. So it is natural that, in principle, are represented by different mathematical objects. However, there is a matrix representation for the angular velocity and a generalization to tensors (en.wikipedia.org/wiki/Angular_velocity#Angular_velocity_tensor) –  Robert Smith Nov 6 '10 at 0:33
    
By the way, I think the answer to your last question (why rotations can not be represented by vectors) is stated in en.wikipedia.org/wiki/Euler's_rotation_theorem –  Robert Smith Nov 6 '10 at 0:41

5 Answers 5

up vote 9 down vote accepted

This is a note on why angular velocities are vectors, to complement Matt and David's excellent explanations of why rotations are not.

When we say something has a certain angular velocity $\vec{\omega_1}$, we mean that each part of the thing has a position-dependent velocity

$\vec{v_1}(\vec{r}) = \vec{\omega_1} \times \vec{r}$.

We might consider another one of these motions

$\vec{v_2}(\vec{r}) = \vec{\omega_2} \times \vec{r}$

and wonder what happens when we add them. We get

$\vec{v_1}(\vec{r}) + \vec{v_2}(\vec{r}) = \vec{\omega_1} \times \vec{r} + \vec{\omega_2} \times \vec{r}$.

The cross product is linear, so this is equivalent to

$(\vec{v_1} + \vec{v_2})(\vec{r}) = (\vec{\omega_1} + \vec{\omega_2}) \times \vec{r}$,

so it makes fine sense to add angular velocities by vector addition.

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Thanks, this was exactly what I was looking for - although I don't think I expressed it clearly. I already knew that rotations were non-commutative –  Casebash Nov 6 '10 at 8:16
    
@Casebash: ah, sorry if I misjudged the level of technical detail you were looking for. I figured since your question was written in a simple manner it would probably (eventually) be read by people who didn't know about rotations not commuting. –  David Z Nov 7 '10 at 3:59
    
@David: Your answer was excellent as well –  Casebash Nov 7 '10 at 5:01
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I certainly can't argue with the OP's feeling of satisfaction, but don't you think you assumed true the very thing you were trying to explain in this answer...? "When we say something has a certain angular velocity $\vec{\omega}_1$", here you start with angular velocity as a vector, then proceed to show that it exhibits vector characteristics... This answer was bound to demonstrate its vector nature, since you assumed it... –  kηives Jan 12 '13 at 6:52
    
@kηives No, it does not assume the conclusion. That is just the definition of angular velocity. As a counterexample to your claim, one could say that a rotation vector $\vec{R}$ is defined such that displacement of a point is $\Delta \vec{r} = \vec{R} \times \vec{r}$. Nonetheless, these "rotation vectors" would not add like vectors, even though I wrote down the definition in terms of vectors. –  Mark Eichenlaub Jan 12 '13 at 14:30

There are actually several different ways to interpret that question, depending on what you mean by "vector" and "rotation". But here's a sense that I've often wondered about myself: in introductory physics, the velocity vector is defined as the time derivative of the position vector (relative to some fixed point). Why is the same not true of angular velocity - that is, why isn't there an "angular position vector" which angular velocity can be the derivative of?

As a matter of fact, sometimes there is. Think about this simple case: pick a single, fixed rotation axis and only consider rotations about that axis. (2D rotations, if you prefer to think of it that way) You'd pick a certain orientation to be the "origin," and you could actually define an angular position vector, pointing along the rotational axis, with length equal to the amount of rotation relative to that "origin" orientation.

Now, suppose your object's angular position is changing over time. You can take the derivative of the angular position vector, and hopefully you can see that what you'd get is just the good old angular velocity. No problem there.

But we live in a 3D world (relativity notwithstanding), so what happens when you try to generalize that model to 3 dimensions? That's where you run into problems. As an example, take the object from the last paragraph, which was rotating around one particular axis - say, the $\hat{z}$ axis. Now suppose it changes its motion so that it starts rotating around a different axis, maybe the $\hat{x}$ axis. How will you represent its orientation now?

You might be tempted to use an "angular position vector" pointing in the $\hat{z}$ direction, whose length represents the amount of rotation around the $\hat{z}$ axis, and another "angular position vector" pointing in the $\hat{x}$ direction, whose length represents the amount of rotation around the $\hat{x}$ axis. After all, that works for position. But it doesn't work for angular position. The reason why is that rotations do not commute, to use the technical lingo. What that means is that if you apply rotation A to an object and follow it up by rotation B which is around a different axis, you get a different result than if you apply rotation B followed by rotation A.

This little issue gets in the way if you try to combine your two $x$- and $z$- angular position vectors into one overall angular position vector. Presumably, you would write this overall vector as $(\theta_x, 0, \theta_z)$ (the zero might represent the amount of rotation around the $\hat{y}$ axis). That vector would represent the sum of $\theta_x$ times the unit rotation around the $\hat{x}$ direction and $\theta_z$ times the unit rotation around the $\hat{z}$ direction. But it's missing a critical piece of information: which of those rotations was performed first? If you gave that vector to your physicist friend, he would be unable to reproduce the orientation of the object because he doesn't know whether to perform the $x$-rotation or the $z$-rotation first. Sure, you may know that the object was rotated in the $z$-direction first, but that information needs to be contained in the vector for it to be of any use.

The point of the last paragraph is that there's no way to sensibly create linear combinations of these "angular position vectors." And that pretty much ruins their usefulness, because the ability to be linearly combined is absolutely fundamental to the definition of a vector, and it underlies a lot of the analytical techniques we use in physics.

By the way, in this view, the reason matrices work to represent rotations is that matrices offer you an additional operation, multiplication, which you can use to combine them. It happens that matrix multiplication, for certain matrices (3x3 antisymmetric with determinant 1), has the same properties as composing rotations; most notably, it's also noncommutative. Multiplying matrix A by matrix B can give you a different result from multiplying matrix B by matrix A.

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Defining properties of vectors are that you can add them and multiply them by constants. These both make sense for angular velocities. On the other hand, adding rotations doesn't make sense. What you can do with two rotations is compose them: first rotate one way, then rotate another. This operation doesn't look like addition of any sort. For one thing, it doesn't commute. Rotating something $30$ degrees around the $x$-axis, then $60$ degrees around the $y$-axis, is not the same thing as doing those two operations in the opposite order. (If you've never done so, pick up an object and try!) So the mathematical operation that corresponds to rotations has to be something that is able to express noncommutativity. Matrices work very naturally for this; for two matrices A and B, it isn't true in general that $AB = BA$.

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Particularly, the rotational group is not abelian (which express noncommutativity) and instead of regular addition (or whatever your favorite operation is), you have function composition. More info: en.wikipedia.org/wiki/Rotation_group. –  Robert Smith Nov 6 '10 at 2:50
    
+1 for making the exact same point I did in about $\frac{1}{4}$ of the space ;-) –  David Z Nov 6 '10 at 3:00

You are mixing up different things. The first is a rotation transformation. Such a transformation is linear and can therefore be written as a matrix.

$$\bf\vec x'=A\vec x $$

Now, angular velocity, is the velocity of a physical rotation. $$\vec\omega=\frac{\mbox d\vec\theta}{\mbox dt}$$ This theta is the angular displacement, or the angle of rotation. This is ALSO a vector like the angular velocity. This theta is something like the amount of rotation if you think of the displacement as the amount of motion

The $\theta$ is not the same as the A, in any case. The transformation due to a matrix as in the first case is a mathematical transformation rather than the rotation of any object. There is no time parameter associated with it unless you let $\bf A$ be a function of time (i.e. dependent on time) rather than constant. Of course, such a rotation of say, a stick CAN be modelled as such a time-dependent matrix as time evolves, i.e.

$$\vec x(t)=\mathbf{A} \vec x(0)$$

Where xis the vector pointing in the direction of the stick with the same magnitude (or length) as the length of the stick.

But still A is not the same as theta. It is just a transformation acting on vectors.

The keyword is acting on vectors. The vector describing the stick $\vec x$ is a vector, and so is the angular displacement, angular velocity, angular acceleration, angular jolt, angular jounce, etc.

Similarly, one may also describe the rotation as an affine transformation on angular displacements:

$$\vec \theta'=\vec\theta+\vec\omega t+\frac{1}{2}\vec\alpha t^2+ \frac{1}{6}\vec\varphi t^3 +...$$

This is precisely the Maclaurin series.

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Note: Here, ' does not mean derivative, it means final value. –  Dimensio1n0 Jun 2 '13 at 10:03

This may not be intuitive at first but I think it is valuable in understanding the relationship between rotation matrices and angular velocities. Also, I know it does not direction answer the question, but I sense there is confusion in the OP and this might help.

So given the rotation matrices $E_1$ and $E_2$ for two connected rigid bodies how do be establish their angular velocity kinematics? How is $\vec{\omega}_1$ related to $\vec{\omega}_2$?

Supposed the two rotation matrices are related by a single rotation about an axis $\hat{z}$ local to the first body and angle $\theta$ such that

$$ E_2 = E_1 {\rm Rot}(\hat{z}, \theta) $$

Differentiate the above equation to get to the angular velocities using the rotating frame derivative.

$$ \frac{{\rm d}}{{\rm d} t} E_1 = \vec{\omega}_1 \times E_1 $$ $$ \frac{{\rm d}}{{\rm d} t} E_2 = \vec{\omega}_2 \times E_2 $$ $$ \frac{{\rm d}}{{\rm d} t} {\rm Rot}(\hat{z},\theta)= \dot{\theta} \hat{z} \times {\rm Rot}(\hat{z},\theta) $$

Using the chain rule then

$$ \frac{{\rm d}}{{\rm d} t} E_2 =\left(\frac{{\rm d}}{{\rm d} t} E_1 \right) {\rm Rot}(\hat{z},\theta) + E_1 \left( \frac{{\rm d}}{{\rm d} t} {\rm Rot}(\hat{z},\theta)\right) $$

$$ \vec{\omega}_2 \times E_2 = \left( \vec{\omega}_1 \times E_1 \right) {\rm Rot}(\hat{z},\theta) + E_1 \left(\dot{\theta} \hat{z} \times {\rm Rot}(\hat{z},\theta)\right) $$ $$ \vec{\omega}_2 \times E_2 = \vec{\omega}_1 \times E_2 + \dot{\theta} (E_1 \hat{z}) \times (E_1 {\rm Rot}(\hat{z},\theta)) $$

$$ \vec{\omega}_2 \times E_2 = ( \vec{\omega}_1 + E_1 \hat{z} \dot{\theta} ) \times E_2 $$

which is only true when

$$ \vec{\omega}_2 = \vec{\omega}_1 + E_1 \hat{z} \dot{\theta} $$

The equation above described the rotational kinematics of the connecting joint, and it is derived from the sequence of rotations. Similarly for more complicated joints.

The time derivative of the rotation sequence yields the angular rotation kinematics.

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