# What actually is the vector of angular momentum?

If an object spins around a central point, it gets angular momentum which is a vector with an orientation dependent on whether its clockwise rotation or anticlockwise, i get that. But what the vector part actually is, is confusing me. Does it mean that the object spinning gets a force going upwards or downwards? In that case, how do you explain bicycles? Shouldn't the spinning result in angular momentum pointing in one direction thus forcing your bike to fall over the same way someone would if they pushed you from either side of the bike?

• Your question seems to be not clear. Try to improve your question – Tojrah Apr 30 at 13:54
• The angular momentum is defined as $\mathbf{L}=\mathbf{r}\times\mathbf{p}$. If you calculate its derivative with respect to time you will see it can be expressed in terms of angular velocity. No more than that, don't overengineer it, don't try to "interpret" it. – gented Apr 30 at 14:02
• Why would there be a force upwards just because angular momentum points up? – jacob1729 Apr 30 at 14:55
• Angular momentum isn't really a vector, it's a pseudo vector (aka axial vector) which is the antisymmetric part of a rank-2 tensor, so thinking of it like a (vector) force will most likely lead to confusion. They are different geometrical objects. – JEB Apr 30 at 15:39
• @JEB - angular momentum and toque are vectors, but they are free vectors as opposed to linear momentum and force which are line vectors. Line vectors operating along a line in space and free vectors don't have a location, just a direction, and magnitude. – ja72 Apr 30 at 18:41

## 8 Answers

Angular momentum has a magnitude, and is about some axis (thus has some sense of "direction). So vectors are used to represent the quantity, and much of the machinery of vectors applies to it. However, there are ways in which "vector" is a misleading term when applied to it, and some people characterize them as being in other mathematical structures, such as a Clifford algebra. Regardless of the exact mathematical formulation, they do not live in the same "space" as the physical world. The vector that represents angular momentum has physical meaning, but it doesn't physically exist. If an angular momentum is represented by a vector pointing east, that doesn't mean that there is any physical thing pointing east. It just means that the angular momentum is around an east-west axis, and that it is "positive" in the eastern direction, where "positive" is an arbitrary convention that we use to keep track of direction.

And apart from all the above, angular momentum is not a force, any more than linear momentum is. It is rotation around an axis.

1. Angular momentum of a particle only describes the location of the linear momentum vector $$\boldsymbol{p} = m \boldsymbol{v}$$ where $$\boldsymbol{v}$$ is the linear velocity and $$m$$ the mass. Momentum acts along a line in space called the axis of percussion.

2. Angular momentum $$\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$$ as measured at the origin, when a particle is located at $$\boldsymbol{r}$$. It measures the moment of momentum in that the location of momentum can be recovered by $$\boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L}}{ \| \boldsymbol{p} \|^2}$$

3. This is entirely analogous to torque, which is the moment of force. The torque vector tells us the location of the line of action of a force with $$\boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau} }{ \| \boldsymbol{F} \|^2 }$$

4. It is also entirely analogous to linear velocity being the moment of rotation. The linear velocity of a rotating rigid body measured at the origin is $$\boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega}$$ were $$\boldsymbol{r}$$ is the location of the rotation axis.

5. The location of the rotation axis is recovered similarly with $$\boldsymbol{r} = \frac{ \boldsymbol{\omega} \times \boldsymbol{v} } {\| \boldsymbol{\omega} \|^2}$$

6. For rigid bodies, angular momentum can exist without linear momentum when a body is rotating about its center of mass. The interpretation here is the there are particles that are rotating about an axis and the direction of angular momentum vector is somewhat related (but not exactly) to the direction of rotation $$\boldsymbol{L} = \mathrm{I} \boldsymbol{\omega}$$ It represents the direction you need to apply a torque at in order to instantaneously stop a rotating body.

7. The full expression of angular momentum of a rigid body includes the motion of the center of mass, as well as the motion about the center of mass

\begin{aligned} \boldsymbol{p} & = m \, \boldsymbol{v}_{\rm com} \\ \boldsymbol{L} & = \mathrm{I}_{\rm com} \boldsymbol{\omega} + \boldsymbol{r}_{\rm com} \times \boldsymbol{p} \end{aligned}

See this similar post as well as this more extensive post for more details.

• It looks like you have a typo in 1. Angular momentum of a particle only describes the location of the linear momentum vector p=mv – PM 2Ring Apr 30 at 20:16
• @PM2Ring - I still don't see it. Which word has the typo? Thanks. – ja72 May 1 at 11:59
• Ok, maybe I just don't get what you're trying to say. How does the angular momentum of a particle describe the location of the linear momentum vector? – PM 2Ring May 1 at 12:02
• @PM2Ring Use the first equation above to get the location of momentum. Angular momentum is just the moment of linear momentum, just as torque is the moment of force. This is just saying that angular momentum describes linear momentum at a distance, just as toque describes a force at a distance. The term moment of mathematically means $${\rm quantity} = {\rm position} \times {\rm something}$$ – ja72 May 1 at 18:27
• The vector form of a linear property (force, velocity, momentum...) is a mathematical tool telling direction.

• The vector form of a rotational property (torque, angular velocity, angular momentum...) is a mathematical tool telling rotational direction.

By rotational direction we just mean the right-hand-rule, so, a way to determine whether the rotation is clockwise or counterclockwise.

Don't mix the two. Vector forms are mathematical inventions that can represent physical properties depending on the situation. They mean different things for different types of properties. They sometimes feel fairly "intuitive" (when they define direction) whereas they other times do not make much physical sense but are "pseudo" vectors (when they define an imaginary axis about which we rotate).

The angular momentum vector has a magnitude and a direction, just like any other vector. However, there is an issue with angular momentum that doesn't apply for objects moving in a linear direction.

If you apply torque to the outside rim of a bicycle wheel, it spins in the direction of the applied force that is producing that torque. That bicycle wheel acquires angular momentum as a result. If you look at one side of the bicycle wheel, you would say that it's spinning in the clockwise direction. Somebody on the other side of the bicycle wheel would say that it's spinning in the counter-clockwise direction. Thus, the two of you will never agree because you are in different reference frames. This means that something must be defined for the angular momentum vector that doesn't change depending on the reference frame.

By convention, if you use your right hand, and let your fingers move in the direction of the rotating bicycle wheel as you close your hand, with your thumb pointing "up" relative to your hand, your thumb will point in a direction that is perpendicular to the plane of rotation of the bicycle wheel and it will also point in the direction of the angular momentum vector. A person on the other side of the rotating bicycle wheel, performing the same right-hand rule, would see his thumb pointing in the same direction as your thumb, meaning that you would both now agree on the direction of the angular momentum vector.

Regarding the bicycle not falling over as you are riding down the road, conservation of angular momentum means that the rotating bicycle wheels will "try" to keep their orientation in space as you ride down the road. If the wheels are spinning in a vertical direction, they will keep doing that until a torque is applied to them to change that orientation. Thus, angular momentum of the bicycle wheels stabilizes the bicycle, and the higher the speed of the wheels the more angular momentum that they have, which is why bicycles become easier to balance as the speed of rotation of the wheels increases.

This is one of those times where simplifications can get in the way.

In more formal contexts, angular momentum is not actually a vector. It's something called a bi-vector. It is the bivector constructed by taking the exerior product of the vector from the axis of rotation to the object and its momentum, $$L=x\land p$$, rather than their cross product $$x\times p$$.

However, there's a catch. Exterior products take a while to learn, and the reasons why you would want to use the are rather nuanced. And, very conveniently, in 3 dimensions, for every bivector $$x\land p$$ there is a corresponding pseudovector $$x \times p$$ called an "axial vector" for which the equations are identical. You'll do the same operations to manipulate the numbers because the same $$|x||p|sin\theta$$ term will show up in both sets of math. (this stops being a universal thing that can be done as we go into higher dimensions)

So we teach momentum as a cross product. The price of this is that the meaning of the direction of that vector is kind of murky, as you noticed. It's a whole lot easier to teach, but we lost a little intuition along the way. If you look at the way one thinks of a bivector and how we visualize them, the momentum bivector is all within the plane of rotation. You have the exterior product between the displacement between the axis of rotation (clearly in plane) and the momentum of the object (clearly in plane). One thinks entirely in the plane of rotation. The only reason the out of plane vectors ever appeared was because we used the cross product to create an axial vector to represent the deeper bi-vector. If you avoid that extra step, then we never see anything out of plane.

So we teach angular momentum using axial vectors, because it makes it easier to teach. We don't need to teach new operators. However, at a deeper conceptual level, they are more correctly bivectors, and that should help explain why that non-intuitive out-of-plane angular momentum feels out of place... because it is. It's not wrong, just a little troublesome for intuition.

What actually is the vector of angular momentum?

It's the mathematical representation of physical rotation. That is, rotation can be quantified by a vector quantity. That's it. There's nothing "real" about the "vector of angular momentum." It's kind of like using complex numbers to represent wave phases. It's just a nice way to encode a real-world relationship into a mathematical model.

The direction of the vector describes direction of planar rotation (right-hand rule) while the magnitude of the vector describes the magnitude of momentum. To speak of angular momentum, imagine two disks of equal size, one made of styrofoam, the other of granite. It will be much harder to accelerate the stone disk than the styrofoam disk. Similarly, once spinning it will be harder to decelerate the stone disk because it has higher angular momentum (due to more mass).

You are encountering here an example of the subtle but significant difference between the physical quantity and its mathematical representation. The physical quantity is a measure of the degree to which mass is moving around some axis. There are many ways one could try to quantity that idea. The way that is adopted in physics is to define angular momentum so that it will be a conserved quantity in an isolated system. That is, while the internal parts of the system may move and push each in complicated ways, the total angular momentum of the system will not be affected by all such internal goings-on. This how we arrive at the definition $${\bf L} = {\bf r} \times {\bf v}$$ for the angular momentum of a particle about the origin. I am omitting the proof that this is indeed a nice conserved quantity, when added up over all the parts of an isolated system. For a rigidly rotating object, the resulting angular momentum vector is along the direction you know (i.e. along the axis of rotation, and pointing in the right-handed sense).

What happens next is the mathematics says: "ok; you now have a useful quantity. And according to the way we defined it, it turns out to be a vector which points in the direction at right angles to both $$\bf r$$ and $$\bf p$$. If that direction seems puzzling to you, don't worry---it will become familiar as you see it working correctly in lots of examples. It does not mean anything is moving along the direction of $$\bf L$$. Rather, the direction of $$\bf L$$ indicates the direction of an axis about which some rotation or orbital motion is happening."

That's the end of the answer to your question, but since you mention the example of a bicycle I'll say a little about that too.

In the case of a bicycle the angular momentum associated with rotation of the wheels does not contribute much to the stability. What is much more important is the angle between the front fork and the axle, which determines how the weight of the bike tends to align the front wheel. However, just for interest, let's see what happens to the angular momentum of the wheels when you ride a bike.

As you sit on the bike rolling forward, the angular momentum vector of either wheel is pointing to the left. (Use your right hand to check this). If we wish to turn the bike towards the right, we require, among other things, a torque to be applied to the front wheel so as to rotate its angular momentum vector. To make the correct change for a right turn the torque vector should be directed forwards, because we want the change $$\Delta {\bf L}$$ to be in the forwards direction, in order to swing $$\bf L$$ around in the correct sense. One way to provide such a torque vector is to tip the wheel over, and the interesting point is that you should tip the wheel in the very sense that in any case you require in order to turn the bike, which is to say tipping into the corner, towards the right as you take a right hand bend (and towards the left as you take a left hand bend). (This tipping of the bike also ensures that the combination of friction and normal reaction from the road surface does not flip the bike over as you take the turn, and in fact this is its main role.)

Whenever you see the cross product, know you’re dealing with a second rank antisymmetric tensor in disguise. Second rank tensors acting on 3D vectors can be represented by $$3\times3$$ matrices, and if that matrix is required to be antisymmetric, that puts six constraints on it, leaving three degrees of freedom to define it. Thus the action of the antisymmetric tensor

$$A=\begin{pmatrix} 0 & a_1 & a_2 \\ -a_1 & 0 & a_3 \\ -a_2 & -a_3 & 0 \end{pmatrix}$$

on a vector $$\mathbf{v},$$ that is, $$A\mathbf{v},$$ can be conveniently denoted as $$\mathbf{a} \times \mathbf{v},$$ where $$\mathbf{a} = (-a_3, a_2, -a_1)^{\mathrm{T}}.$$ Let’s check:

$$\begin{pmatrix}a_x \\ a_y \\ a_z\end{pmatrix} \times \begin{pmatrix}v_x \\ v_y \\ v_z\end{pmatrix} = \begin{pmatrix} 0 & -a_z & a_y \\ a_z & 0 & -a_x \\ -a_y & a_x & 0 \end{pmatrix} \begin{pmatrix}v_x \\ v_y \\ v_z\end{pmatrix} = \begin{pmatrix} -a_z v_y+a_yv_z \\ \phantom{-}a_zv_x - a_xv_z \\ -a_yv_x+a_xv_y \end{pmatrix}$$ which matches the definition of the cross product in terms of the Cartesian coordinates.

So it’s just a notational trick based on the fact that in 3D space (and in no other number of dimensions), rank 2 antisymmetric tensors have exactly as many degrees of freedom as vectors and can be put into correspondence. These pseudo-vectors end up pointing in directions which don’t necessarily mean anything is moving in that direction, and their magnitudes might be devoid of physical meaning as well.

A similar approach can be used with other kinds of tensors, possibly requiring more than one vector to match the amount of information contained in the tensor. This is one way to define the $$\mathbf{E}$$ and $$\mathbf{B}$$ vectors in electromagnetism theory, while tensor forms of the laws combine those quantities in a single entity.