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I am a high schooler so I don't know a lot of fancy maths but I do know some of calculus and multiplication of vectors as dot or cross product. I am learning about Angular velocity. And I am confused that is the direction of angular velocity just a definition or has a physical significance. I looked and searched for this on the internet and several other places and of course, I found the answer but its too diverse as someone says that it is defined and others say that it has some significance. I was amazed and confused much more when I saw gyroscopes in action.

Here are some of the search work I did:

And there are several other pages on the internet that I tried but this remains the same all over. So what I want is not only the answer but also its validity. Thanks and appreciation to anyone who answers or puts his/her effort into this question.

Edit

Many people were getting confused by what I mean by Physical Significance. Here's what I mean If a thing has physical significance then its effects will be real and you will be able to see them. As a force, although the force itself is not visible its effects are and that too in the same direction in which a force is said to be in. So a direction is real but a quantity assigned in that direction can be just to help us solve some problems or fix some glitches and it could very well be a mathematical trick like a pseudo force in an accelerated frame. Hence for this question, has the direction which is told to be the direction of angular velocity something physical that is happening in that direction? Like a motion, you can not say that a car is moving in $-X$ direction if it is moving in $+X$ direction if the coordinate system is already defined of course.

Edit 2

Everyone confused due to a lot of ambiguity in the question. Here's the final Edit and this is the actual question whose answer would be indirectly the answer to this entire title- Could we have defined the direction of Angular Velocity to any other direction if we had more options or let's say we had 4-dimensions reality?

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  • $\begingroup$ The stackexchange answer doesn't say that the direction has no physical significance. The direction is perpendicular to the plane. But there are two directions perpendicular to the plane. The convention is which direction we equate to clockwise and anticlockwise rotation. The fact that it's perpendicular to the plane is not a convention. $\endgroup$ Oct 3, 2020 at 6:35
  • $\begingroup$ @BrainStrokePatient Please refer to this link, it also states thats its just a mere defenition $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 6:39
  • $\begingroup$ @BrainStrokePatient Where can I get a perfect answer which is true, any ideas $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 6:40
  • $\begingroup$ @BrainStrokePatient It indirectly says the same that it doesn't have physical significance, we just did it to remove confusion, we could have also found any other method to do so it means $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 6:41
  • $\begingroup$ I don't really understand what you mean by physical significance now. The answer you linked does connect the direction of angular velocity to something physical: which direction the rotation is in. $\endgroup$ Oct 3, 2020 at 6:45

7 Answers 7

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Since the question mentioned higher-dimensional space, I wanted to give an answer that works in any-dimensional space, not just 3. I'll start with formal, mathematical definitions and then connect them to physical intuition.

Rotations in $n$-dimensional space form a group. Specifically, they form a group called the special orthogonal group, which is denoted by $\mathrm{SO}(n)$. $\mathrm{SO}(n)$ is also a smooth manifold, so we call it a Lie group.

Every point on a manifold has a tangent space. The elements of this tangent space are called tangent vectors. Intuitively, a tangent vector tells us which direction to move in and how fast to move in that direction. That is, it gives us a velocity, as illustrated below:

enter image description here

The Lie algebra of a Lie group is simply the tangent space at the identity element of the group. For $\mathrm{SO}(n)$, the identity element is the rotation that does nothing, i.e. no rotation.

Therefore, an angular velocity is an element of the Lie algebra of $\mathrm{SO}(n)$, which is denoted by $\mathfrak{so}(n)$.

Side note: In terms of matrices, $\mathrm{SO}(n)$ can be represented as the set of $n \times n$ orthogonal matrices with determinant 1, while $\mathfrak{so}(n)$ can be represented as the set of $n \times n$ antisymmetric matrices. The matrix exponential gives us the exponential map from the latter to the former.

So what is $\mathfrak{so}(n)$ like? Intuitively, we can specify any angular velocity $\omega$ as follows:

  • Rotate this fast ($a_1$) on this plane ($p_1$) through the origin.
  • Rotate this fast ($a_2$) on this plane ($p_2$) through the origin.
  • etc.

Each plane $p_i$ also carries an orientation which tells us which way we intend to rotate.

In short, we can think of $\omega$ as a weighted sum $a_1 p_1 + a_2 p_2 + \dots$. But what is $p_i$, mathematically? To specify a plane, we only need 2 unit vectors (say $\mathbf{u}$ and $\mathbf{v}$), as illustrated below:

enter image description here

The resulting plane is the wedge product of $\mathbf{u}$ and $\mathbf{v}$, which is denoted by $\mathbf{u} \wedge \mathbf{v}$. Switching the order of $\mathbf{u}$ and $\mathbf{v}$ switches the orientation of the plane. When combined, they cancel out: \begin{align} \mathbf{u} \wedge \mathbf{v} + \mathbf{v} \wedge \mathbf{u} = 0 \end{align}

This corresponds to the fact that if we rotate this fast in one direction and equally fast in the opposite direction, we get nothing. Scaling either vector by a scalar $a$ simply scales the resulting angular velocity:

\begin{align} a \mathbf{u} \wedge \mathbf{v} = \mathbf{u} \wedge a \mathbf{v} = a (\mathbf{u} \wedge \mathbf{v}) \end{align}

Thus each summand $a_i p_i$ of our angular velocity is the wedge product of 2 vectors, i.e. a blade. Thus our angular velocity $\omega$ is a sum of blades, i.e. a bivector. The set of bivectors is denoted by $\wedge^2 \mathbb{R}^n$.

In 2 and 3 dimensions, something special happens: Any sum of blades is a blade. Thus we only need a single blade to specify an angular velocity. Consequently, every rotation is a simple rotation.

Furthermore, in 3 dimensions, the dual of a bivector is a vector, i.e. $\star \left( \wedge^2 \mathbb{R}^3 \right) = \mathbb{R}^3$. This is why in 3D we typically describe planes using "normal vectors": \begin{align} \mathbf{u} \times \mathbf{v} &\stackrel{\text{def}}{=} \star (\mathbf{u} \wedge \mathbf{v}) \end{align}

and rotations using "rotation axes" (see Euler's rotation theorem).

This trick doesn't work in other dimensions. For example, in 2 dimensions, the dual of a bivector is a scalar, which is why we typically describe 2D rotations using scalars.

In 4-dimensional space, something even stranger happens: Not only is the dual of a bivector not a vector, but there are bivectors that are not blades. Consequently, there are rotations in 4-dimensional space that cannot be described as rotations on a single plane. These are called double rotations. An example is the rotation given by

\begin{align} \mathbf{u} \wedge \mathbf{v} + \mathbf{w} \wedge \mathbf{x} \end{align}

where $\mathbf{u},\mathbf{v},\mathbf{w},\mathbf{x}$ are all mutually orthogonal. The animation below shows a double rotation acting on the 4D unit cube (stereographically projected to 3D, of course):

enter image description here

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    $\begingroup$ While a bit dense for a high schooler, this one is the right answer. The wedge product is the thing that is "meaningful," including in higher dimensions. It just happens that in 2 or 3 dimensions, the math of these wedge products is identical to the math required for cross products. So the direction of the cross product itself isn't all that meaningful, but it's really close to something which is. $\endgroup$
    – Cort Ammon
    Oct 12, 2020 at 0:21
  • $\begingroup$ I loved this answer! Could you recommend a book about this? I'm familiar with alternating tensors but not too advanced. $\endgroup$
    – Suriya
    Feb 8, 2021 at 13:12
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There is in fact a way of expressing angular velocity in such a way that there is no ambiguity of what-part-of-this-is-convention.

Angular velocity occurs in a plane, and it has direction and magnitude. To specify a plane two vectors that lie in that plane are specified, with the order of the two vectors giving the direction of the angular velocity. The magnitude of the angular velocity can for example be specified with a separate number.

If you would be working with a space with four spatial dimensions the above way to specify would be the only possible way; with four spatial dimensions each plane has two vectors that are perpendicular to it. (And with a higher number of spatial dimensions there are more perpendicular vectors.)


Our space has three spatial dimensions, and with three spatial dimensions a shorthand notation of angular velocity is available.

In a space with three spatial dimensions every plane has a single vector that is perpendicular to it. So: to specify a particular plane in a space with three spatial dimensions it is sufficient to specify the vector perpendicular to that plane. And then you can make the magnitude of that single vector represent the magnitude of the angular velocity.

That notation is so much shorter, so much more compact, using that notation is a no-brainer. (It should be kept in mind thoug though that it is something of a fluke; it works only with a space with three spatial dimensions.)

Direction of the rotation
There is one thing, of course. The notation is so compact that there is no room to specify the direction of the rotation. It's literally one bit of information: this-way-around or the-other-way-around. But the shorthand notation has no room to spare; it cannot express that bit.

That is why the shorthand notation is supplemented with the right-hand-rule, the right hand rule fills in that one necessary bit of information.

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  • $\begingroup$ So let's say that if there was a property having two types of values A and B of which one is applied when a body is rotating clockwise and the other applies the other way around then would we have assigned directions in direction of A and B? $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 11:29
  • $\begingroup$ I just want to know that is it just a mathematical fitting to make things work? $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 11:32
  • $\begingroup$ I got my query solved by combining the answer from J Thomas and Cleonis. So thanks for helping me and I respect your effort to help people like us. $\endgroup$
    – Ritanshu
    Oct 12, 2020 at 7:42
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I am learning about Angular velocity. And I am confused that is the direction of angular velocity just a definition or has a physical significance.

You are going to get confusing answers, because your question as stated doesn't mean much. But it means something....

There are things in mathematical notation that are basicly arbitrary. Somebody chose to write them that way, and they worked, and now everybody does it that way. Like multiplication distributes over addition, and we write $a(b+c)$. We could have used any other symbol in place of (). $a:b,c:$ would have worked as well. For $a(b-c)$ we could do $a:b,-c:$.

We could have a convention that each time you have a newline.

$a(b(d+e-f)+c)$ becomes

a:  
   b:
     d,e,-f
   ,c

That would some ways work better though it would take more space on the page. It's basicly arbitrary which way we use.

But the fact that $a(b+c)=ab+ac$ is not arbitrary. It's important.

It looks to me like you're asking what's the important part, and what's just convention.

Could we have defined the direction of Angular Velocity to any other direction if we had more options or let's say we had 4-dimensions reality?

It would have to amount to the same thing -- if it gave a different answer then it would be a wrong answer. Unless we changed the concepts around somehow so they combined differently to get the same end result.

But yes, instead of defining a vector axis as the defining direction, we could have two vectors to define the plane the rotation is in. And then at any one moment the velocity would be something in that plane. That wouldn't make any practical difference in 3D but it might be clearer.

A rotation is in some particular plane. If you use polar notation, rotation changes the angle but not the length. Polar coordinates (or for that matter cartesian coordinates) gives you an arbitrary zero point, and whatever point you rotate around, you arbitrarily subtract its displacement from all the locations so it will be at zero to do the rotation. You can add the displacement back later with no loss.

Using the normal vector is only one possible way to describe which plane the rotation is in. That's arbitrary notation.

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  • $\begingroup$ I got my query solved by combining the answer from J Thomas and Cleonis. So thanks for helping me and I respect your effort to help people like us. $\endgroup$
    – Ritanshu
    Oct 12, 2020 at 7:42
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First lets be on the same page regarding what is angular velocity ?

Angular velocity often, denoted as $\omega$ is the rate of angular displacement, denoted as $\theta$ with respect to time i.e. you might have seen this equation a lot $$\displaystyle{\vec{\omega} = \frac{\vec{\theta}}{t}}$$ and if we are talking about instantaneous angular velocity then : $$\displaystyle{d\vec{\omega} = \frac{d\vec{\theta}} {dt}}$$ The angular displacement is the change in the plane angle, subtended by the body which is performing the motion, at some reference point.

Direction of $\vec\omega$

The direction of angular velocity, basically tells you in which direction is a body rotating or performing circular motion with respect to a reference point i.e. it tells you the direction of the angular displacement.

Just like you can assume what direction you want to take for positive $x$ axis and which one for positive $y$ axis, you can also assume which direction to take positive and which one to take negative. Say, you took counterclockwise motion as positive then you have to take clockwise motion as negative.

And of course you can use the Right Hand thumb rule, also referred to as Maxwell's Corkscrew Rule to find the direction of angular velocity. Some teachers in Highschools, while teaching mechanics to students, say that angular velocity is a vector but they treat it like a scalar, which is wrong.

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  • $\begingroup$ So let's say that if there was a property having two types of values A and B of which one is applied when a body is rotating clockwise and the other applies the other way around then would we have assigned directions in direction of A and B? $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 11:29
  • $\begingroup$ I just want to know that is it just a mathematical fitting to make things work? $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 11:32
  • $\begingroup$ I couldn't clearly understand your question properly. If the property you are talking about in your first question is angular velocity, then it has 1 value just different signs. Kinda like how we teach counting to children, if you go to the left of zero, you have negative numbers, we told them that if they count to the left of 0 it is considered negative. That is a general convention. But, If you want to you can take going left as positive. $\endgroup$ Oct 3, 2020 at 14:16
  • $\begingroup$ I don't know if you have been taught about/ learned about torque or not ? But, after seeing your edit, I would like to make an analogy to how you said force has a physical significance. If a body has angular velocity in say +x direction and a torque is being applied to it in that direction that it will start spinning faster in that direction. But if the direction of both is opposite then the body will slow down, eventually. $\endgroup$ Oct 3, 2020 at 14:22
  • $\begingroup$ Or for example, there exists a sorta definition of angular velocity that the cross product of the position vector and angular velocity of a body is its linear velocity with respect to a reference point. So you can see that if the direction of angular velocity changes then so does linear velocity $\endgroup$ Oct 3, 2020 at 14:24
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Defining angular velocity as a vector that is perpendicular to the plane of rotation is useful in three dimensional scenarios because it allows angular velocities to be added using the rules of vector addition (the parallelogram rule). If an object is rotating with angular velocity vector $\vec \omega_1$ relative to a reference frame $F_1$, and $F_1$ is rotating about the same centre with angular velocity vector $\vec \omega_2$ relative to reference frame $F_2$ then the angular velocity of the object relative to $F_2$ is the vector sum $\vec \omega_1 + \vec \omega_2$. So, yes, the angular velocity vector does have physical significance.

However, the additional of angular velocity vectors would still work if we replaced $\vec \omega_1$ with $-\vec \omega_1$ and $\vec \omega_2$ with $-\vec \omega_2$ i.e. if we used a left hand rule instead of a right hand rule to find the direction of the angular velocity vector. So the use of a right hand rule to determine the direction of the angular velocity vector is the part that is convention.

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  • $\begingroup$ So let's say that if there was a property having two types of values A and B of which one is applied when a body is rotating clockwise and the other applies the other way around then would we have assigned directions in direction of A and B? $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 11:28
  • $\begingroup$ I just want to know that is it just a mathematical fitting to make things work? $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 11:32
  • $\begingroup$ @user921307 I don't understand what you mean by "two types of values A and B". If two objects are rotating in the same plane, one clockwise and the other anti-clockwise, then their angular velocity vectors point in opposite directions. $\endgroup$
    – gandalf61
    Oct 3, 2020 at 11:33
  • $\begingroup$ I know, but in short, I want to ask that is it just a mathematical trick to help resolve ambiguities and make things work because it seems the same for the definition of the cross product too $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 11:34
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The vectors representing rotation are chosen along the axis if rotation because that is the only direction in the system which is usually not continuously changing direction. That said, such vectors can accurately represent the direction and magnitude of rotational quantities.

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  • $\begingroup$ Thanks, I got the point $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 17:18
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The magnitude of $\vec{\omega}$ gives you the speed of rotation. But the direction of rotation gives you the orientation of the rotation axis.

The general motion of a 3D solid body is a rotation about an instantaneous axis (line in space) and a parallel translation along the axis. This is the so called Chasles's theorem.

To describe a line in space you need the direction of the line, as well as any point along the line.

The motion of a solid body is described by the following properties, derived from the rotation vector $\vec{\omega}$, as well as the velocity vector $\vec{v}$ of some point on the body (the reference point).

  • Magnitude of rotation $$\omega = \| \vec{\omega} \| \tag{1}$$
  • Direction of rotation $$\hat{z} = \frac{ \vec{\omega}}{\omega} \tag{2}$$
  • Point on rotation axis closest to reference point $$ \vec{r} = \frac{ \vec{\omega} \times \vec{v} }{\omega^2} \tag{3}$$ this position is measured from the reference point.
  • Parallel motion along the rotation axis $$ \vec{v}_\parallel = \left( \frac{\vec{\omega} \cdot \vec{v}}{\omega^2} \right) \vec{\omega} \tag{4} $$

So the direction of $\vec{\omega}$ not only gives you (2), but contributes to the location of the rotation axis via (3).

In reverse, you can transform the rotation about the axis plus the parallel motion to the velocity of the reference point with the following expression

$$ \vec{v} = \vec{v}_\parallel + \vec{\omega} \times (-\vec{r}) \tag{5}$$


PS> Something similar happens with force vector $\vec{F}$ given the torque of the force at some reference point $\vec{\tau}$. You can find similarly the position $\vec{r}$ and the parallel torque of the line of action of the force.

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