okay as I read somewhere "angular momentum is not defined about an axis instead it is defined about a point but while dealing with fixed axis rotation or rotation about axis in translation we need angular momentum about an axis".

Now three questions derive from this statement.

  1. Why it is inconvenient to define angular momentum about an AXIS (fix and free)?

  2. Even if we define it , how to calculate its magnitude? (answer separately for a particle and for a rigid body)

  3. how to determine its direction?


2 Answers 2


An "axis" is an infinite line. This immediately raises an issue in defining the position vector, $\mathbf{r}$, from which one computes the cross-product with the linear momentum, $\mathbf{p}$. That is, angular momentum is defined as: $$ \mathbf{L} = \mathbf{r} \times \mathbf{p} \tag{0} $$ which in the case of 3-vectors can be reduced to: $$ \mathbf{L} = r \ p \ \sin{\theta} \ (\hat{r} \times \hat{p}) \tag{1} $$ where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{p}$.

So how would you define $\mathbf{r}$ in reference to an infinite line?

  • $\begingroup$ thats what I'm asking? still it is used in calculations. $\endgroup$ Commented Mar 13, 2018 at 16:13
  • $\begingroup$ No, no one calculates this with respect to an axis. It is a vector so they perform the calculation in a specific coordinate basis, but that does not mean it is defined with respect to an axis. $\endgroup$ Commented Mar 13, 2018 at 16:18
  • 1
    $\begingroup$ @honeste_vivere Are you kidding? Calculating angular momentum with respecto to an axis is a common calculation, especially for rigid bodies, where $r$ is the distance from the straight line to the point. This is usually done using the moment of intertia $\endgroup$
    – FGSUZ
    Commented Jul 31, 2019 at 21:15
  • $\begingroup$ @FGSUZ - If by "axis" you mean the $\mathbf{r}$ vector, then fine, yes. But that is not what I meant. If you choose your coordinate basis such that $\mathbf{r}$ is aligned with one of the basis vectors, then again yes. However, this does not mean that the angular momentum is with respect to an axis, it's a vector defined in any coordinate basis you choose. For rigid bodies, it is simpler to choose the coordinate basis such that it aligns with axes of the rigid body but again, this is not what you are implying. $\endgroup$ Commented Aug 1, 2019 at 14:44

I think the point being made is that angular momentum is a vector field, meaning it has different values depending on where in the space it is expressed at. So angular momentum must be defined at a specific location.

The information described by the angular momentum vector is only that of the location of the momentum axis in space. The momentum axis is also known as the axis of percussion also. In contrast to linear momentum which describes the direction of this axis. Remember a location and a direction define a line in space.

Try to visualize the following. A rigid body of mass $m$ has a velocity vector of the center of mass $\boldsymbol{v}_C$. At some instant, the center of mass is located at the point $\boldsymbol{r}_C$ and at that location, the mass moment of inertia is defined as $\mathrm{I}_C$. Also, the body is rotating with $\boldsymbol{\omega}$.

The following can be deduced from this situation

  1. Linear momentum is $$\boldsymbol{p} = m \boldsymbol{v}_C \tag{1}$$
  2. Angular momentum at the center of mass C is $$\boldsymbol{L}_C = \mathrm{I}_C \,\boldsymbol{\omega} \tag{2}$$
  3. There exists a special axis in space that makes angular momentum zero, or at least minimum (by being parallel to this axis). This is a found as follows:

  4. The axis of percussion has direction parallel to linear momentum $$\hat{\boldsymbol{z}} = \frac{ \boldsymbol{p} }{\| \boldsymbol{p} \|} \tag{3}$$

  5. Find a point A on this axis closest to the origin with $$ \boldsymbol{r}_A = \boldsymbol{r}_C + \frac{ \boldsymbol{p} \times \boldsymbol{L}_C}{\| \boldsymbol{p} \|^2} \label{eqn:location} \tag{4} $$


Transfer angular momentum from point C to point A using the standard transformation rules $$ \boldsymbol{L}_A = \boldsymbol{L}_C + (\boldsymbol{r}_C - \boldsymbol{r}_A) \times \boldsymbol{p} \tag{5}$$

Substitute $\eqref{eqn:location}$ into the above for $$ \boldsymbol{L}_A = \boldsymbol{L}_C - \left( \frac{ \boldsymbol{p} \times \boldsymbol{L}_C}{\| \boldsymbol{p} \|^2} \right) \times \boldsymbol{p} \tag{6}$$ and use the vector triple product identity $(a \times b) \times c = b (c \cdot a) - a ( c \cdot b)$ to simplify the above $$ \boldsymbol{L}_A = \boldsymbol{L}_C - \frac{ \boldsymbol{L}_C ( \boldsymbol{p} \cdot \boldsymbol{p}) - \boldsymbol{p} ( \boldsymbol{p} \cdot \boldsymbol{L}_C) }{ \| \boldsymbol{p} \|^2 } = h\, \boldsymbol{p} \tag{7}$$ which is parallel to $\boldsymbol{p}$ with the scalar $h=\frac{ \boldsymbol{p} \cdot \boldsymbol{L}_C}{ \| \boldsymbol{p} \|^2}$ value called the pitch.


Linear momentum is a vector not associated with a point in space (as it is shared by the entire rigid body) and is used to establish the direction of the axis of percussion. Angular momentum is just the moment of momentum and it is used to establish the location of the axis of percussion, and thus it must be defined at a specific location.


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