Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Exercise
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$\text{l}_{\text{x}}=\text{yp}_{\text{z}}-\text{zp}_{\text{y}}$ $\text{l}_{\text{y}}=\text{zp}_{\text{x}}-\text{zp}_{\text{z}}$ $\text{l}_{\text{z}}=\text{xp}_{\text{y}}-\text{yp}_{\text{x}}$ Linear momentum of the particle, $\vec{\text{p}}=\text{p}_{\text{x}}\hat{\text{i}}+\text{p}_{\text{y}}\hat{\text{j}}+\text{p}_{\text{z}}\hat{\text{k}}$ Position vector of the particle, $\vec{\text{r}}={\text{x}}\hat{\text{i}}+{\text{y}}\hat{\text{j}}+{\text{z}}\hat{\text{k}}$ Angular momentum, $\vec{\text{l}}=\vec{\text{r}}\times\vec{\text{p}}$ $=({\text{x}}\hat{\text{i}}+{\text{y}}\hat{\text{j}}+{\text{z}}\hat{\text{k}})\times(\text{p}_{\text{x}}\hat{\text{i}}+\text{p}_{\text{y}}\hat{\text{j}}+\text{p}_{\text{z}}\hat{\text{k}})$ $=\begin{vmatrix}\vec{\text{i}}&\vec{\text{j}}&\vec{\text{k}}\\\text{x}&\text{y}&\text{z}\\\text{p}_{\text{x}}&\text{p}_{\text{y}}&\text{p}_{\text{z}}\end{vmatrix}$ $=\text{l}_{\text{x}}\hat{\text{i}}+\text{l}_{\text{y}}\hat{\text{j}}+\text{l}_{\text{z}}\hat{\text{k}}$ $=\hat{\text{i}}(\text{yp}_{\text{z}}-\text{zp}_{\text{y}})-\hat{\text{j}}(\text{zp}_{\text{x}}-\text{zp}_{\text{z}})+\hat{\text{k}}(\text{xp}_{\text{y}}-\text{yp}_{\text{x}})$ Comparing the coefficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get $ \text{l}_{\text{x}}=\text{yp}_{\text{z}}-\text{zp}_{\text{y}},\text{ l}_{\text{y}}=\text{zp}_{\text{x}}-\text{zp}_{\text{z}},\text{ l}_{\text{z}}=\text{xp}_{\text{y}}-\text{yp}_{\text{x}}\ ...(\text{i})$ The particle moves in the x-y plane. Hence, the z-component of the position vector and linear momentum vector becomes zero, i.e., Z = Pz = 0 Thus, equation (i) reduces to, $\text{l}_{\text{x}}=0,\text{ l}_{\text{y}}=0,\text{ l}_{\text{y}}=\text{x}\text{p}_{\text{y}}-\text{y}\text{p}_{\text{x}}\ ...(\text{ii})$ Therefore, when the particle is confined to move in the x-y plane, the direction of angular momentum is along the z-direction.
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