Classical orbital angular momentum

The classical definition of angular momentum is a pseudo-vector, which can be separated into its 3 components in Cartesian coordinates as follows:

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are two vectors

$\boldsymbol{A}=\boldsymbol{i}A_x+\boldsymbol{j}A_y+\boldsymbol{k}A_z\; \; \; \; \; \;\boldsymbol{B}=\boldsymbol{i}B_x+\boldsymbol{j}B_y+\boldsymbol{k}B_z$

where $A_x$ is the component of $A$ in the $x$ -direction and $\boldsymbol{i},\boldsymbol{j},\boldsymbol{k}$ are unit vectors in the $x,y,z$ directions.

The cross product of the two vectors is:

$\boldsymbol{C}=\boldsymbol{A}\times\boldsymbol{B}=\boldsymbol{i}(A_yB_z-A_zB_y)+\boldsymbol{j}(A_zB_x-A_xB_z)+\boldsymbol{k}(A_xB_y-A_yB_x)\; \; \; \; \; \; \; \; 69$

Since $\boldsymbol{C}=\boldsymbol{i}C_x+\boldsymbol{j}C_y+\boldsymbol{k}C_z$

$C_x=A_yB_z-A_zB_y$

$C_y=A_zB_x-A_xB_z$

$C_z=A_xB_y-A_yB_x$

Comparing eq59a and eq69,

$\boldsymbol{L}=\boldsymbol{r}\times\boldsymbol{p}=\boldsymbol{i}(r_yp_z-r_zp_y)+\boldsymbol{j}(r_zp_x-r_xp_z)+\boldsymbol{k}(r_xp_y-r_yp_x)$

and

$L_x=r_yp_z-r_zp_y$

$L_y=r_zp_x-r_xp_z$

$L_z=r_xp_y-r_yp_x$

$L_x$ , $L_y$ and $L_z$ are the classical orbital angular momenta about the $x$ -axis, $y$ -axis and $z$ -axis respectively. Since the magnitude of a vector $\boldsymbol{v}=x\boldsymbol{i}+y\boldsymbol{j}+z\boldsymbol{k}$ is $\vert\boldsymbol{v}\vert=\sqrt{x^{2}+y^{2}+z^{2}}$ , we have $\vert\boldsymbol{L}\vert^{2}=L_{x}^{\, \, 2}+L_{y}^{\, \, 2}+L_{z}^{\, \, 2}$ or simply

$L^{2}=L_{x}^{\, \, 2}+L_{y}^{\, \, 2}+L_{z}^{\, \, 2}\; \; \; \; \; \; \; \; 70$

In other words, $L^{2}$ is square of the magnitude of the vector $\boldsymbol{L}$ . The significance of $L^{2}$ will be explored in subsequent articles.

Question

Show that $\boldsymbol{\tau}=\frac{d\boldsymbol{L}}{dt}$ .

Answer

From eq59a,

$\frac{d\boldsymbol{L}}{dt}=\boldsymbol{r}\times\frac{d\boldsymbol{p}}{dt}+\boldsymbol{p}\times\frac{d\boldsymbol{r}}{dt}=\boldsymbol{r}\times m\frac{d\boldsymbol{v}}{dt}+m\boldsymbol{v}\times\boldsymbol{v}=\boldsymbol{r}\times\boldsymbol{F}=\boldsymbol{\tau}$

Hence,

$\boldsymbol{\tau}=\frac{d\boldsymbol{L}}{dt}\; \; \; \; \; \; \; \; 71$

Classical orbital angular momentum

Question

Answer

Next article: quantum orbital angular momentum operators (Cartesian coordinates)

Previous article: relation between magnetic dipole moment and angular momentum

Content page of quantum mechanics

Content page of advanced chemistry

Main content page

Leave a Reply Cancel reply