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Exercises · 6.6

Q.Find the components along the xx, yy, zz axes of the angular momentum l⃗\vec{l} of a particle, whose position vector is r⃗\vec{r} with components xx, yy, zz and momentum is p⃗\vec{p} with components pxp_x, pyp_y and pzp_z. Show that if the particle moves only in the xx-yy plane the angular momentum has only a zz-component.

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Angular momentum l⃗=r⃗×p⃗\vec{l} = \vec{r} \times \vec{p} expands into three components using the cross product determinant. When motion is confined to the xx-yy plane, both zz and pzp_z are zero, leaving only lz=xpy−ypxl_z = x p_y - y p_x.

The definition of angular momentum for a particle is the cross product of its position vector and its linear momentum:

l⃗=r⃗×p⃗\vec{l} = \vec{r} \times \vec{p}

This is a vector quantity. Its direction is perpendicular to the plane containing r⃗\vec{r} and p⃗\vec{p}, following the right-hand rule. The magnitude tells us how much "rotational oomph" the particle has about the origin.

To find the components, we write the cross product in determinant form:

l⃗=∣i^j^k^xyzpxpypz∣\vec{l} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix}

Expanding this determinant gives us the three components directly.

  1. The xx-component comes from the i^\hat{i} term. We take the determinant of the submatrix formed by deleting the row and column of i^\hat{i}:

lx=ypz−zpyl_x = y p_z - z p_y

  1. The yy-component comes from the j^\hat{j} term. Remember the alternating sign pattern (+,−,++,-,+ for i^,j^,k^\hat{i}, \hat{j}, \hat{k}):

ly=zpx−xpzl_y = z p_x - x p_z

  1. The zz-component comes from the k^\hat{k} term:

lz=xpy−ypxl_z = x p_y - y p_x

So the full angular momentum vector is:

l⃗=(ypz−zpy) i^+(zpx−xpz) j^+(xpy−ypx) k^\vec{l} = (y p_z - z p_y)\,\hat{i} + (z p_x - x p_z)\,\hat{j} + (x p_y - y p_x)\,\hat{k}

Now for the second part: the particle moves only in the xx-yy plane.

Watch out

A common mistake is to think that "moving in the xx-yy plane" only means z=0z = 0. But it also means the velocity (and hence momentum) has no zz-component — otherwise the particle would leave the plane. …

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