One generally first encounters classical angular momentum in Cartesian coordinates for the sheer simplicity of calculations. Indeed, given the three-dimensional Lagrangian for a general central force potential

$L=\frac{m}{2}({\stackrel{·}{x}}^2+{\stackrel{·}{y}}^2+{\stackrel{·}{z}}^2)-V\left(r\right)$

the components of the angular momentum vector in Cartesian coordinates

$\begin{array}{l}\mathbf{L}=\mathbf{r}\times \mathbf{p}=m\mathbf{r}\times \stackrel{·}{\mathbf{v}}\\ \phantom{\mathbf{L}}=m(y\stackrel{·}{z}-z\stackrel{·}{y})\hat{\mathbf{x}}+m(z\stackrel{·}{x}-x\stackrel{·}{z})\hat{\mathbf{y}}+m(x\stackrel{·}{y}-y\stackrel{·}{x})\hat{\mathbf{z}}\end{array}$

are constant almost trivially. First temporal derivatives of spatial coordinates cancel immediately in pairs, while the second derivatives do as well by the Lagrange equations

$\frac{d}{dt}\frac{\partial L}{\partial {\stackrel{·}{q}}_i}=\frac{\partial L}{\partial q_i}\to {\stackrel{··}{q}}_i=-\frac{\partial V}{\partial q_i}=-\frac{V^{\prime }}{r}{q}_i$

Establishing this constancy is dead simple in Cartesian coordinates. So how about spherical coordinates?

This presentation employs Langrangians rather than Hamiltonians to display a more direct connection to coordinates for spatial position and momentum. This could potentially lead to confusion as to which quantity is indicated by a capital L, but the distinction is simple: angular momenum is a vector in bold, and while its square need not be bolded because a Lagrangian is never squared.

Axes in spherical coordinates are obtained from Cartesian axes in two stages. First perform a passive rotation around the z-axis through an angle φ, transforming the x- and y-axes into cylindrical ρ- and φ-axes. Then follow this with a second passive rotation around the φ-axis through an angle θ, which lying in the plane behaves like a y-axis. Keeping in mind that a rotation around a y-axis has a difference in sign from rotations around the other two Cartesian axes, the total transformation sought is

$\left[\begin{array}{ccc}cos\theta & 0& -sin\theta \\ 0& 1& 0\\ sin\theta & 0& cos\theta \end{array}\right]\left[\begin{array}{ccc}cos\phi & sin\phi & 0\\ -sin\phi & cos\phi & 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{cccc}cos\theta cos\phi & cos\theta sin\phi & -sin\theta & \\ -sin\phi & cos\phi & 0\\ sin\theta cos\phi & sin\theta sin\phi & cos\theta \end{array}\right]$

which produces a right-hand system in the order $[\hat{\mathbf{\theta }},\hat{\mathbf{\phi }},\hat{\mathbf{r}}]$ For a more familiar ordering, permute the rows of the transformation matrix to reorder the resulting vectors:

$\left[\begin{array}{c}\hat{\mathbf{r}}\\ \hat{\mathbf{\theta }}\\ \hat{\mathbf{\phi }}\end{array}\right]=\left[\begin{array}{ccc}sin\theta cos\phi & sin\theta sin\phi & cos\theta \\ cos\theta cos\phi & cos\theta sin\phi & -sin\theta & \\ -sin\phi & cos\phi & 0\end{array}\right]\left[\begin{array}{c}\hat{\mathbf{x}}\\ \hat{\mathbf{y}}\\ \hat{\mathbf{z}}\end{array}\right]$

To determine temporal derivatives of the unit vectors in spherical coordinates, designate this transformation matrix as T. Being a product of two orthogonal matrices, it is itself orthogonal. That means its inverse is its transpose, which can be easily verified.

The first temporal derivatives of these unit vectors can thus be written

$\frac{d}{dt}\left[\begin{array}{c}\hat{\mathbf{r}}\\ \hat{\mathbf{\theta }}\\ \hat{\mathbf{\phi }}\end{array}\right]=\frac{dT}{dt}\left[\begin{array}{c}\hat{\mathbf{x}}\\ \hat{\mathbf{y}}\\ \hat{\mathbf{z}}\end{array}\right]=\frac{dT}{dt}{T}^{-1}\left[\begin{array}{c}\hat{\mathbf{r}}\\ \hat{\mathbf{\theta }}\\ \hat{\mathbf{\phi }}\end{array}\right]$

since the unit vectors in Cartesian coordinates are constant. Evaluting the product of the derivative of a matrix and its inverse takes a bit of algebra, with the end result

$\frac{d}{dt}\left[\begin{array}{c}\hat{\mathbf{r}}\\ \hat{\mathbf{\theta }}\\ \hat{\mathbf{\phi }}\end{array}\right]=\left[\begin{array}{ccc}0& \stackrel{·}{\theta }& \stackrel{·}{\phi }sin\theta \\ -\stackrel{·}{\theta }& 0& \stackrel{·}{\phi }cos\theta \\ -\stackrel{·}{\phi }sin\theta & -\stackrel{·}{\phi }cos\theta & 0\end{array}\right]\left[\begin{array}{c}\hat{\mathbf{r}}\\ \hat{\mathbf{\theta }}\\ \hat{\mathbf{\phi }}\end{array}\right]$

To formulate the angular momentum vector in spherical coordinates, first note that $\mathbf{r}=r\hat{\mathbf{r}}$ by the very definition of a radial axis. The explicit expression for the angular momentum is then simple:

$\begin{array}{l}\mathbf{L}=m\mathbf{r}\times \stackrel{·}{\mathbf{v}}=mr\hat{\mathbf{r}}\times \frac{d}{dt}\left(r\hat{\mathbf{r}}\right)=m{r}^2\hat{\mathbf{r}}\times \frac{d}{dt}\hat{\mathbf{r}}\\ \phantom{\mathbf{L}}=m{r}^2\hat{\mathbf{r}}\times (\stackrel{·}{\theta }\hat{\mathbf{\theta }}+\stackrel{·}{\phi }sin\theta \hat{\mathbf{\phi }})\\ \mathbf{L}=m{r}^2(\stackrel{·}{\theta }\hat{\mathbf{\phi }}-\stackrel{·}{\phi }sin\theta \hat{\mathbf{\theta }})\end{array}$

The square of this vector is immediately

$L^2=m^2{r}^4({\stackrel{·}{\theta }}^2+{\stackrel{·}{\phi }}^2{sin}^2\theta )$

To establish the constancy of these two expression, one needs the equations of motion of combinations of spherical variables. Transforming the Lagrangian above into spherical coordinates is a simple matter of evaluating temporal derivatives of Cartesian variables in those coordinates. The standard result is

$L=\frac{m}{2}({\stackrel{·}{r}}^2+r^2{\stackrel{·}{\theta }}^2+r^2{sin}^2\theta {\stackrel{·}{\phi }}^2)-V\left(r\right)$

The equations of motion for the combinations are

$\frac{d}{dt}\left(r^2\stackrel{·}{\theta }\right)=r^2{\stackrel{·}{\phi }}^{2}sin\theta cos\theta \begin{array}{c}\frac{d}{dt}\left(r^2{sin}^2\theta \stackrel{·}{\phi }\right)=0\\ \frac{d}{dt}(r^{2}sin\theta \stackrel{·}{\phi })=-r^2\stackrel{·}{\theta }\stackrel{·}{\phi }cos\theta \end{array}$

Assembling expressions above, the temporal derivative of angular momentum is

$\begin{array}{l}\frac{1}{m}\frac{d\mathbf{L}}{dt}=r^2{\stackrel{·}{\phi }}^{2}sin\theta cos\theta \hat{\mathbf{\phi }}-r^2\stackrel{·}{\theta }\stackrel{·}{\phi }(sin\theta \hat{\mathbf{r}}+cose\theta \hat{\mathbf{\theta }})\\ +r^2\stackrel{·}{\theta }\stackrel{·}{\phi }cos\theta \hat{\mathbf{\theta }}+r^2\stackrel{·}{\phi }sin\theta (\stackrel{·}{\theta }\hat{\mathbf{r}}-\stackrel{·}{\phi }cos\theta \hat{\mathbf{\phi }})=0\end{array}$

as expected. Its square is then automatically constant as well, but just as easy to show explicity:

$\frac{1}{2m^2}\frac{d{L}^2}{dt}=r^2\stackrel{·}{\theta }·r^2{\stackrel{·}{\phi }}^{2}sin\theta cos\theta -r^2\stackrel{·}{\phi }sin\theta ·r^2\stackrel{·}{\theta }\stackrel{·}{\phi }cos\theta =0$

Note that establishing these results does not need the full set of equations of the dynamic system, just the behavior of the relevant combinations. This is in a sense more compact than using Cartesian coordinates.

Such a simple classical result, but try searching for it on the web. Not easy to find all in one place, but handy to have.

This short presentation has application to motion in the presence of Schwarzschild black holes, as well as any other spherically symmetric system in classical variables.

Uploaded 2026.09.17 analyticphysics.com