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I have read that it's not necessary for angular momentum and angular velocity to be parallel, but it is necessary for linear momentum and linear velocity to be parallel. How is this correct?

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    $\begingroup$ Possible duplicate of Can the direction of angular momentum and angular velocity differ? $\endgroup$ – Yuvraj Singh... Jul 24 '19 at 13:59
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    $\begingroup$ @yuvrajsingh I disagree. That is more of a related question. This question is asking for more of a comparison between "linear" and "angular", and why a difference arises. $\endgroup$ – Aaron Stevens Jul 24 '19 at 14:50
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    $\begingroup$ FWIW, canonical/conjugate momentum does not have to be parallel to velocity, e.g. in an E&M background. $\endgroup$ – Qmechanic Jul 24 '19 at 21:12
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    $\begingroup$ Note that the angular momentum pseudovector is a different geometric object from the linear momentum vector. This distinction becomes apparent in Special Relativity where there is a linear momentum four-vector but an antisymmetric angular momentum tensor. $\endgroup$ – Hal Hollis Jul 24 '19 at 21:33
  • $\begingroup$ Linear momentum and linear velocity are not necessarily parallel either, c.f. the Shockley-James paradox. journals.aps.org/prl/abstract/10.1103/PhysRevLett.18.876 Although for a closed system, the total momentum (the Noether charge of translations) is proportional to the velocity of the center of energy (relativistic generalization of center of mass), for the parts of a system it's not the case. journals.aps.org/pr/abstract/10.1103/PhysRev.171.1370 aapt.scitation.org/doi/10.1119/1.3152712 $\endgroup$ – Robin Ekman Jul 26 '19 at 15:42
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Note that this answer, like the tags of the question, is only focusing on Newtonian mechanics.

The definition of linear momentum $\mathbf p$ is expressed in terms of the linear velocity $\mathbf v$ using $$\mathbf p=m\mathbf v$$ Since $m$ is just a scalar quantity, the vectors $\mathbf p$ and $\mathbf v$ are obviously parallel. i.e. $$p_x=mv_x$$ $$p_y=mv_y$$ $$p_z=mv_z$$

However, the angular momentum $\mathbf L$ is related to the angular velocity $\boldsymbol \omega$ by $$\mathbf L=\mathbf I\boldsymbol\omega$$

where $\mathbf I$ is the moment of inertia tensor. This is explicitly written out as

$$ \begin{bmatrix} L_x\\L_y\\L_z \end{bmatrix}= \begin{bmatrix} I_{xx}&I_{xy}&I_{xz}\\ I_{yx}&I_{yy}&I_{yz}\\ I_{zx}&I_{zy}&I_{zz} \end{bmatrix} \begin{bmatrix} \omega_x\\\omega_y\\\omega_z \end{bmatrix} $$ Or each component written out: $$L_x=I_{xx}\omega_x+I_{xy}\omega_y+I_{xz}\omega_z$$ $$L_y=I_{yx}\omega_x+I_{yy}\omega_y+I_{yz}\omega_z$$ $$L_z=I_{zx}\omega_x+I_{zy}\omega_y+I_{zz}\omega_z$$

This shows that, in general, $\mathbf L$ and $\boldsymbol\omega$ are not parallel. We have something more complicated than the linear case because each component of the angular momentum vector depends on all of the angular velocity components.

However, this complexity does not prevent our two vectors from being parallel. We can determine when they are parallel by looking at the equation $$\mathbf L=\mathbf I\boldsymbol\omega=I\boldsymbol\omega$$ where $I$ is a scalar quantity. What this means is that $\boldsymbol\omega$ needs to be an eigenvector of $\mathbf I$ in order for $\mathbf L$ and $\boldsymbol\omega$ to be parallel. Note that this is usually what you encounter in your introductory physics classes. You, without knowing it, pick axes in such a way that the moment of inertia tensor is diagonal and your angular velocity is only along a single eigenvector (usually taken to be along the z-axis). For example, for a cylinder of radius $R$, height $H$, and mass $M$ rotating about its central axis aligned with the z-axis, we have $$ \begin{bmatrix} L_x\\L_y\\L_z \end{bmatrix}= \begin{bmatrix} \frac{1}{12}M\left(3R^2+H^2\right)&0&0\\ 0&\frac{1}{12}M\left(3R^2+H^2\right)&0\\ 0&0&\frac12MR^2 \end{bmatrix} \begin{bmatrix} 0\\0\\\omega_z \end{bmatrix} $$ Therefore we end up with the simple $$\mathbf L=L_z\hat z=I_{zz}\omega_z\hat z=\frac12MR^2\omega_z\hat z$$ or you might even see in an introductory physics class just $$L=\frac12MR^2\omega$$


As a small note, the reason things become so much more complicated with the rotations is that the angular momentum not only depends on the mass of the object, but also how that mass is distributed. This is evident from the definition of angular momentum for a point particle $\mathbf L=\mathbf r\times\mathbf p$

However, if we look at an extended body's momentum we just get $$\mathbf p_T=m_T\mathbf v_{\text{com}}$$ where the $T$ subscript stands for "total" and "com$ is center of mass. Therefore, we still end up with a scalar mass instead of a tensor.

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    $\begingroup$ That's nice, but I think the OP want to know WHY mass can't become a tensor as well. $\endgroup$ – FGSUZ Jul 24 '19 at 15:45
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    $\begingroup$ I'll try to explain. Most answers are explaining that $m$ is a scalar, and therefore $\vec{p}$ is parallel to $\vec{v}$, as $\vec{p}=m\vec{v}$. On the other hand, since $I$ is a tensor, then $\vec{L}=I\vec{\omega}$ is different. This can be an answer, but it would be great if someone went deeper. Why can't mass be a tensor too, so that $p$ needent be parallel to $v$ anymore? $\endgroup$ – FGSUZ Jul 24 '19 at 19:10
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    $\begingroup$ @FGSUZ The answer by WetSavannahAnimal goes pretty deep. $\endgroup$ – Aaron Stevens Jul 24 '19 at 21:20
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    $\begingroup$ @FGSUZ it can, see e.g. effective mass in semiconductors. But that's somewhat outside of Newtonian mechanics, although you could apply Newton's laws to (well-spread to ignore dispersion but still small enough to treat them as points) wave packets. $\endgroup$ – Ruslan Jul 25 '19 at 10:32
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    $\begingroup$ In “What this means in that $\omega$ needs to be an eigenvalue of I...” — do you mean eigenvector here? $\endgroup$ – Santana Afton Jul 26 '19 at 12:48
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The other answers that say the difference arises because there is an inertia tensor in the rotation case are all perfectly correct, but i think we can go deeper and more intuitive that this and say:

Translations commute. Rotations don’t. That’s the fundamental reason for the difference. And it’s that way that these two different behaviors bear upon Noether’s theorem gives root to the difference between linear momentum, with its simple, scalar multiple relationship with velocity and angular momentum, with its more general linear transformation from the angular velocity vector.

Actually, the statement "Translations commute. Rotations don’t“ is a little glib and imprecise. But it's very close to being accurate. To be more precise, one would say rotations form a more complicated, noncommutative Lie group, whereas translations form „the" three dimensional commutative Lie group (commutative Lie groups of a given dimension are essentially all the same - there can be a topological difference in that the mutually commuting one parameter groups can be compact or not, but from the Lie algebra standpoint they are all exactly the same, and Noether’s theorem is only influenced by the Lie algebra, not by the group topology).

Recall that the reason for being for angular and linear momentum - what makes them useful concepts - is that they are conserved quantities of a physical system (in the absence of external forces). And, although Noether’s theorem is not the only mechanism that gives rise to conserved quantities in physics, it is in this case. So, if we want fundamental, intuitive insight into this question, we must look at how Noether’s theorem plays out in the two cases.

Let’s grab the expression from Wikipedia for the conserved Noether current:

$$\left(\frac{\partial L}{\partial \dot{\mathbf{q}}} \cdot \dot{\mathbf{q}} - L \right) T_r - \frac{\partial L}{\partial \dot{\mathbf{q}}} \cdot \mathbf{Q}_r\tag{1}$$

This is a very physicist equation and needs explanation if you’re not familiar with its notation and jargon. Noether’s theorem states that if the Lagrangian of a system is time-translation-invariant and is invariant with respect to a continuous, one-parameter group of transformations on the system’s generalized co-ordinates, then there is a conserved quantity for each such one parameter group, called to Noether charge. For example: rotations of the co-ordinate system about an axis - the rotation’s magnitude can be smoothly varied with the rotation angle - or translations along a given direction of the co-ordinate origin, which are parameterized by a 1continuous signed distance value. In (1), $\mathbf{Q}_r$ is the „generator" of the continuous transformation in question (explained below). The first part of (1) - the bit to the left of $T_r$, has to do with the continuous transformation of translating the time co-ordinate, and is not important for the present considerations. It gives rise to the energy as a c1onserved quantity. So we look at the part that is nonzero for the transformations (on the non-time co-ordinates) that we’re interested in:

$$\frac{\partial L}{\partial \dot{\mathbf{q}}} \cdot \mathbf{Q}_r\tag{2}$$

As someone versed in a more Lie theoretical understanding of these matters, i would write the above equation more to my taste as follows:

$$\mathbf{Q}_r \frac{\partial L}{\partial \dot{\mathbf{q}}} = \mathbf{Q}_r \, \mathbf{p}\tag{3}$$

where $\frac{\partial L}{\partial \dot{\mathbf{q}}}$ is a vector whose components are the generalized momenta $p_i = \frac{\partial L}{\partial \dot{q}_i}$ corresponding to the generalized co-ordinates $p_i$ of the Lagrangian description. In the above, $\mathbf{Q}_r$ is the matrix of the Lie algebra member when the Lie group in question acts on the generalized co-ordinates.

So, let’s see how (3) pans out. If our Lagrangian is independent of the generalized co-ordinates themselves, then the matrix of the translation in the $i^{th}$ direction is simply diagonal with noughts along its leading diagonal except at the $i^{th}$ position. So (3) just says that the $i^{th}$ component of the generalized momentum $\mathbf{p}$ is conserved. Or, repeating the argument for each $i$, the $\mathbf{p}$ itself is conserved. So the situation is very simple for translations in the co-ordinates themselves.

However, if the concept of rotation of generalized co-ordinates is meaningful, and, if further, the Lagrangian is invariant with respect to rotations in the generalized co-ordinates, then the matrix $\mathbf{Q}_r$ in (3) is the Lie algebra member of $SO(N)$ corresponding to rotation about the axis in question, times the distance $r$ from the co-ordinate origin (this is the $N$-dimensional generalization corresponding to the vector calculus operator $\mathbf{\omega}\times\mathbf{r}$ in 3 dimensions). So already, without assuming anything about the expressions for angular or linear momentum aside from that our Lagrangian is invariant to translations and rotations, we can see that the expressions for the conserved linear momentums are simply the generalized momentums themselves, whereas the conserved angular momentums, by dent of the more complicated Lie algebra for $SO(N)$ are matrix operations on the generalized momentums.

If we further assume that the generalized momentum is $m_i\,r\, \mathbf{Q}_r \dot{\theta}_r$ (the $N$-dimensional generalization corresponding to the vector calculus expression $\mathbf{\omega}\times \mathbf{r}$ in 3 dimensions), then our Noether charge is:

$$m \,r^2 \,\mathbf{Q}_r^2 \dot{\theta}_r\tag{5}$$

and this quantity is conserved for each co-ordinate $r$. This is readily shown to mean the same thing as the conservation of $\mathbf{L} = \mathbf{I}\,\mathbf{\omega}$ when summed over all particles in a rigid body.


Actually, i believe my answer above is the fundamental grounds for Aaron Stevens' excellent physical intuition::

As a small note, the reason things become so much more complicated with the rotations is that the angular momentum not only depends on the mass of the object, but also how that mass is distributed.

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Mass is a scalar but mass moment of inertia is a tensor. As a result, a scalar can only change the magnitude of a vector, but a tensor can change both the magnitude and the direction.

From linear algebra:

  • $\boldsymbol{p} = m \boldsymbol{v}$ is always parallel to $\boldsymbol{v}$ for $m \neq 0$
  • $\boldsymbol{L} = \rm{I}\, \boldsymbol{\omega}$ is only parallel when $\boldsymbol{\omega}$ is an eigenvector of $\mathrm{I}$. A special case exists for symmetric objects like spheres and cubes where $\mathrm{I}$ is a scalar multiple of the identity matrix.
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    $\begingroup$ @AaronStevens - I agree. I edited the answer. $\endgroup$ – ja72 Jul 24 '19 at 15:44
  • $\begingroup$ @AaronStevens - Dang it!. Right again, a cube for example. Thank you for the corrections. $\endgroup$ – ja72 Jul 24 '19 at 17:57
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    $\begingroup$ I think the question is actually asking WHY mass is a scalar, but its rotational analog is a tensor. $\endgroup$ – Dawood says reinstate Monica Jul 25 '19 at 20:33
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Linear momentum is defined as

$$\vec{p}=m\vec{v}$$

where $m$ is a scalar (i.e. just a number). Multiplying a vector by a scalar does not change its direcction.

In contrast, angular momentum is defined as

$$\vec{L}=\overset{\leftrightarrow}I\vec{\omega}$$

where $\overset{\leftrightarrow}I$ is a tensor (i.e. something like, but not quite exactly, a matrix). Multiplying a vector by a tensor can, and often does, change the direction of the vector.

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