Angular momentum (classical)

In physics, angular momentum is a kinematic property of one or more point masses. Its importance derives from the fact that it is a conserved quantity in several physical circumstances. (Recall that a conserved quantity of a physical system is a property that does not change over time.)

Point mass m with velocity v has angular momentum L ≡ m r × v with respect to O. L is a vector pointing towards the reader

The angular momentum of a single point mass m is defined with respect to a point O. Denote the vector from O to m by r (see the figure). Let the mass have velocity v, then the angular momentum L of the point mass is the cross product,

$\mathbf {L} \equiv m\mathbf {r} \times \mathbf {v} =\mathbf {r} \times \mathbf {p} .$

It follows from the definition of cross product that the vector L is perpendicular to the plane of the figure and points towards the reader. Note that the definition p ≡ m v for the linear momentum is introduced here.

The definition of the angular momentum of a system of n point masses is the following simple generalization:

$\mathbf {L} _{\mathrm {sys} }\equiv \mathbf {r} _{1}\times \mathbf {p} _{1}+\mathbf {r} _{2}\times \mathbf {p} _{2}+\cdots +\mathbf {r} _{n}\times \mathbf {p} _{n}=\sum _{i=1}^{n}\mathbf {r} _{i}\times \mathbf {p} _{i}.$

Conservation of angular momentum

For simplicity we consider the case of one point mass, generalization to n point masses is straightforward. Using Newton's dot notation for time derivatives, we find

$\mathbf {\dot {L}} =m\mathbf {\dot {r}} \times \mathbf {v} +m\mathbf {r} \times \mathbf {\dot {v}} .$

Use

$\mathbf {\dot {r}} \times \mathbf {v} \equiv \mathbf {v} \times \mathbf {v} =0\quad {\hbox{and}}\quad m\mathbf {r} \times \mathbf {\dot {v}} =m\mathbf {r} \times \mathbf {a} =\mathbf {r} \times \mathbf {F} ,$

where we invoked Newton's second law $\scriptstyle m\mathbf {\dot {v}} \equiv m\mathbf {a} =\mathbf {F} .$ Hence the time derivative of the angular momentum is equal to the torque N,

$\mathbf {\dot {L}} =\mathbf {r} \times \mathbf {F} \equiv \mathbf {N} .$

It follows that angular momentum is conserved—its time derivative is zero—if the applied torque N is zero, which means that either F is zero, or r and F are parallel, because in that case the cross product vanishes.

The former case occurs when the mass moves uniformly in homogeneous space (no external forces). Then also the linear momentum of the point mass is conserved, and therefore one rarely considers the angular momentum in the case of uniform rectilinear motion.

The second case is more interesting. Let us assume that F is conservative, i.e., its curl vanishes, then F is the gradient of a potential

$\mathbf {\nabla } \times \mathbf {F} =0\Longrightarrow \mathbf {F} =-\mathbf {\nabla } V(x,y,z),$

so that for a conservative force the time derivative of L is

$\mathbf {\dot {L}} =-\left(\mathbf {r} \times \mathbf {\nabla } \right)V(x,y,z).$

The quantity in brackets is known in group theory as a Lie derivative of the full rotation group SO(3). In quantum mechanics it is a well-known operator, namely the orbital angular momentum operator (except for the factor $\scriptstyle i\hbar$).

From the general properties of Lie derivatives—or equivalently orbital angular momenta—it follows that the time derivative of L vanishes whenever the potential V is rotationally invariant. This means that the function V(x,y,z) depends on r only, when expressed in spherical polar coordinates (r, θ, φ). By expressing the gradient in spherical polar coordinates and writing the cross product as a determinant, we can prove this explicitly