Angular momentum is defined as $L = r \times p$. By the definition of the cross product, $L$ is going to be orthogonal to both $r$ and $p$, and the magnitude $|L|$ tells us how much angular momentum the object has. What I am unclear of is what does the vector $L$ itself tell us? For example, if $r$ and $p$ are confined in the $xy$ plane in $\mathbb{R}^3$, then clearly $L$ must parallel to the $z$ axis. Does the vector $L$ tell us anything more than the magnitude of the angular momentum and the direction of its spin, or is there some real physical meaning to it, for example is there really momentum in the $z$ direction? To put it differently, what does the direction of $L$ really tell us?

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    $\begingroup$ Is there really momentum in the 𝑧 direction? Are you asking if there is linear momentum in the $z$-direction? $\endgroup$
    – Ghoster
    Nov 25, 2022 at 1:39
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    $\begingroup$ @Ghoster I phrased it like that for like of better wording, but what I meant is we say the vector $L$ is pointing in the $z$ direction (in my example). Does that really tell us anything other than the direction of rotation about $L$? $\endgroup$
    – CBBAM
    Nov 25, 2022 at 1:42
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    $\begingroup$ I don't quite understand the question "or is there some real physical meaning to it." Angular momentum is just as physically meaningful as linear momentum (and their physical meanings are precisely analogous). Is that the subject of your question? $\endgroup$
    – J. Murray
    Nov 25, 2022 at 12:56
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    $\begingroup$ Pretty sure that it's technically a pseudovector and not a real vector, which is one reasons that it requires interpretation and convention for its application. $\endgroup$ Nov 25, 2022 at 14:42
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    $\begingroup$ This pseudovector/"real" vector stuff is really more mathematical nuance, a statement about how it behaves when certain transformations are applied, rather than a statement about the reality of things. What's really going to blow your mind is that all of these are just models (ways of describing) various aspects of real physical phenomena. E.g. consider force: we think of it as of a vector describing some influence acting at a distance. But there isn't actually an invisible arrow attached to an object, pulling it or pushing it away. It's no more real (or unreal) than angular momentum is. $\endgroup$ Nov 27, 2022 at 12:50

3 Answers 3


Angular momentum is the generator of rotations. Linear momentum is the generator of linear translations.

This can be given a precise mathematical meaning, but intuitively it means something like the following: if an object has angular momentum, this causes the object to rotate. If an object has linear momentum, this causes the object to move (“translate”).

Technically speaking, angular momentum is best thought of as a “2-form.” In other words, you can think of the angular momentum as being defined in a plane. However in 3 dimensions there is a coincidence: every plane can be uniquely labeled (up to scaling) by a vector. The vector which labels the plane is taken to be the vector perpendicular to the plane. This coincidence is because a 3D space has 3 dimensions, and $3-2=1$, so the plane with 2 dimensions can be labeled with the leftover 1 dimension.

If we lived in a 4-dimensional spatial universe, this coincidence would no longer apply to us. So we wouldn't represent angular momentum with a vector, we would have to use a 2-form.

The takeaway here is that if you find it helpful, you should think about angular momentum as properly being defined by a plane, together with a sign which gives the direction of rotation in the plane. In 3D, notice that a plane can just be labeled by the vector normal to it.

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    $\begingroup$ So the vector $L$ really defines the normal to the surface of rotation and that's it? $\endgroup$
    – CBBAM
    Nov 25, 2022 at 7:48
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    $\begingroup$ @CBBAM That is the first thing you should have been told when being taught it. Right hand rule convention. Were you not? Did they just throw a math expression at you? I think most people think of it more as the axis of rotation than "normal to the surface of rotation". $\endgroup$
    – DKNguyen
    Nov 25, 2022 at 15:14
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    $\begingroup$ What's up with all this stuff about momentum generating/causing movements? How is that any more accurate than saying movement causes momentum? I could say if an object is moving, it causes it to have momentum. I'm not sure why you started with those two paragraphs when cause and effect isnt really a relevant notion between momentum and movement. $\endgroup$
    – JMac
    Nov 25, 2022 at 19:03
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    $\begingroup$ @JMac Momentum is often considered more fundamental than movement, because momentum is a conserved quantity and movement isn't. $\endgroup$
    – zwol
    Nov 25, 2022 at 21:58
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    $\begingroup$ @CBRAM: It's a "bivector". You can think of it as a plane parallelogrammic figure of indeterminate shape but determinate area, with the edges oriented either clockwise or counterclockwise. Physically, the plane of the figure then stands, in a sense, for the "equatorial plane" of the rotating object. $\endgroup$ Nov 25, 2022 at 22:23

There is a field of math called geometric algebra, which is often praised because it captures many geometrical quantities in more natural mathematical objects.

In geometric algebra we can form an area element from two vectors using what is called the wedge product. For two vectors $u,v$ it is defined as $u\wedge v$ and it represents a parallelogram that is spanned by the two vectors (see the image below). It is a 'signed' area element, which means that if you switch $u$ and $v$ the area element picks up a minus sign $ u\wedge v=-v\wedge u$. Switching $u$ and $v$ is the same as rotating the area element by $180^\circ$ like a pancake.

The reason this is nice is that we can define angular momentum and torque as bivectors $$L=r\wedge p,\quad\tau=r\wedge F\,;$$ and by doing this we never have to reference dimensions we don't use. The objects that represent rotations never leave the plane that they rotate in. As an example consider angular momentum in 2D. \begin{align} L&=(x\,e_1+y\,e_2)\wedge(p_x\,e_1+p_y\,e_2)\\ &=xp_x\,e_1\wedge e_1+yp_x\, e_2\wedge e_1+xp_y \, e_1\wedge e_2+yp_y\, e_2\wedge e_2\\ &=(xp_y-yp_x)e_1\wedge e_2 \end{align} Here I used $u\wedge u=0$, which is not hard to prove using $ u\wedge v=-v\wedge u$. We can notice two things from this computation. Firstly, the numerical computation is the same as you would do using a cross product. Secondly, the third dimension is never referenced here. Normally to calculate $\vec L$ in 2D you would have to imagine the z-dimension and then calculate $L_z$. The geometric algebra approach is more general and works in any dimension. Three dimensions are special because we can map bivectors uniquely to vectors, for example $e_1\wedge e_2\leftrightarrow e_3$. This is the reason that the cross product works in 3D.

To come back to your question: the angular momentum vector is just a way to encode an axis and a magnitude in a convenient manner. As I showed you above, this is just one choice you could make to represent the physical quantities you are interested in.

enter image description here

source of image: https://favpng.com/png_view/positivity-geometric-algebra-exterior-algebra-geometry-png/pUEKctM4

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    $\begingroup$ This is very nice! I believe it captures the idea of angular momentum better than the usual cross product, even if the two approaches are equivalent. $\endgroup$
    – CBBAM
    Nov 25, 2022 at 23:08

To put it simply: the direction of $\vec{L}$ has no physical meaning. It's just a convention. In fact, the direction of $\vec{L}$ acts quite differently from the direction of $\vec{p}$ or most other vectors you'll work with in mechanics.

Imagine something has linear momentum upward (positive Z direction). If you look in a mirror, it will still have linear momentum upward. Your coordinate system has flipped around, but positive Z is still positive Z.

Now imagine something has angular momentum upward (positive Z direction). If you look in a mirror, its angular momentum is now pointing downward! The angular momentum has actually changed with respect to its coordinate system, positive Z becoming negative Z!

Furthermore, there are plenty of ways you can measure $\vec{p}$. But there's no way to directly measure $\vec{L}$. Any actual measurement will involve taking a cross product somewhere, which cancels out the weird mirror behavior.

Why does this happen? Well, it's common, in three dimensions, to talk about rotations as happening around a line. This is where you get concepts like the "axis of rotation", and the angular momentum vector pointing along this axis. But how does this work when you go down to two dimensions? Can you have rotations in a two-dimensional world?

You can indeed. They just happen around a point rather than a line: you have a "center of rotation" rather than an "axis of rotation". Suddenly you need a scalar rather than a vector for angular momentum.

This is because rotations* don't actually happen around a line: in general, they happen in a plane, a two-dimensional space. The maximum number of linearly-independent planes in N-dimensional space is $N \choose 2$ (i.e. the number of unique combinations of two of those dimensions), so this is the dimensionality** of our space of rotations.

So let's look at those spaces. In two dimensions, there's only one possible plane, so the space of rotations is one-dimensional. In four dimensions, you can have up to six linearly-independent planes (XY, XZ, XW, YZ, YW, ZW), so the space of rotations is six-dimensional.

But in three dimensions, you can only have three linearly-independent planes (XY, XZ, YZ). Which means the space of rotations is three-dimensional. This gives an isomorphism between rotations and vectors in space—and thus, you can pretend rotations are the same as standard spatial vectors, if you really want to. And physicists really want to.

But, these aren't actually vectors in space, not the way $\vec{p}$ is. They don't behave the same as "standard" vectors, like $\vec{p}$: they flip upside down if you look in a mirror, for example. Treating rotations as vectors like this requires imposing a handedness on the entire universe (hence "the right hand rule" instead of "the left hand rule", and everything going weird when you look in a mirror, which flips handedness), which isn't required to work with "standard" vectors ($\vec{p}$ doesn't care if you look in a mirror or not).

So if you don't want to pretend your rotations are vectors, you can treat them as something else instead: as directed plane segments, or "bivectors"***. AccidentalTaylorExpansion's answer goes into this further. This lets you deal with angular momentum in a way that generalizes to any number of spatial dimensions, instead of crashing and burning the moment you try to work in a plane instead of a three-space.

* Technically, this is only true for simple rotations. In three dimensions and below, all rotations are simple. In four dimensions and above, you can also have a rotation that's a linear combination of simple rotations: a tesseract can rotate simultaneously in the XY and ZW planes, for example, and those rotations just have nothing to do with each other.

** That is, the minimum number of elements you need to span the space. The examples I'm giving here aren't the only possible bases for these spaces, they're just the most convenient ones to write out.

*** Or more technically, members of the even subalgebra of the Clifford algebra, which is important when you're dealing with non-simple rotations.

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    $\begingroup$ This was a very nice answer and precisely what I was looking for, thank you. After reading your answer, as well as all the others, it seems the vector $L$ is merely a convenience (and hence has no physical interpretation) to express the very real concept of angular momentum. $\endgroup$
    – CBBAM
    Nov 25, 2022 at 23:11
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    $\begingroup$ @CBBAM Yep! One other interesting point: fake vectors like this (properly axial vectors or pseudovectors) can never actually be measured directly. Any measurement of a fake vector requires taking a cross product first, to turn it back into a real vector. $\endgroup$
    – Draconis
    Nov 25, 2022 at 23:21

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