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

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

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

The cross product of the two vectors is:

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

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

$C_x=A_yB_z-A_zB_y$

$C_y=A_zB_x-A_xB_z$

$C_z=A_xB_y-A_yB_x$

Comparing eq59a and eq69,

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

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 $\boldsymbol{\mathit{v}}=x\boldsymbol{\mathit{i}}+y\boldsymbol{\mathit{j}}+z\boldsymbol{\mathit{k}}$ is $\vert\boldsymbol{\mathit{v}}\vert=\sqrt{x^{2}+y^{2}+z^{2}}$ , we have $\vert\boldsymbol{\mathit{L}}\vert^{2}=L_{x}^{\, \, 2}+L_{y}^{\, \, 2}+L_{z}^{\, \, 2}$ 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 $\boldsymbol{\mathit{L}}$ . The significance of $L^{2}$ will be explored in subsequent articles.

Question

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

Answer

From eq59a,

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

Hence,

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

Classical orbital angular momentum

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Answer

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