Phy5645/AngularMomentumExercise

If the earth (classically) revolves around the sun counter-clockwise in the x-y plane with the sun at the origin, what is the minimum angle between the angular momentum vector of the earth and the z axis? Ignore the intrinsic spin of the earth. The angular momentum of the earth around the sun is $\ 4.83*10^{31}$ ${\frac {kg*m^{2}}{s}}$. Compare the minimum angle to that of a quantum particle with $\ l=4$.

Solution:

Recall that in QM: $\ L=\hbar {\sqrt {l(l+1)}}$; $\ L_{z}=m\hbar$.

The angle $\ \theta$ between $\mathbf {L}$ and the z-axis fulfills: $\ \cos \theta ={\frac {L_{z}}{L}}={\frac {m}{\sqrt {l(l+1)}}}$.

To make $\ \theta$ as small as possible, $\ m$ must be maximum. This is when $\ m=l$. Therefore, the minimum angle $\ \alpha$ obeys: $\ \cos \alpha ={\frac {l}{\sqrt {l(l+1)}}}$

We solve $\ L=\hbar {\sqrt {l(l+1)}}$ to find $\ l$. Since $\ l$ will be very large we invoke the approximation: $\ {\sqrt {l(l+1)}}\approx l{\sqrt {(1+{\frac {1}{l}})}}=l+{\frac {1}{2}}$. We resist the urge to discard the ${\frac {1}{2}}$ because without it our result will be trivial. $\ L=\hbar (l+{\frac {1}{2}})$, therefore <math>\ l = \frac{L}{\hbar}-\frac{1}{2}.