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Suppose the earth revolves around the sun counter-clockwise in the x-y plane with the sun at the origin. Quantum mechanically, what is the minimum angle of the angular momentum vector of the earth with the z axis? Ignore the intrinsic spin of the earth. The angular momentum of the earth around the sun is <math>\ 4.83 \cdot 10^{31} J \cdot s</math>. | |||
Compare the minimum angle to that of a quantum particle with <math>\ l=4 </math>. | Compare the minimum angle to that of a quantum particle with <math>\ l=4 </math>. | ||
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We solve <math>\ L = \hbar\sqrt{l(l+1)}</math> to find <math>\ l </math>. | We solve <math>\ L = \hbar\sqrt{l(l+1)}</math> to find <math>\ l </math>. | ||
Since <math>\ l </math> will be very large we invoke the approximation: <math>\ \sqrt{l(l+1)} \approx l \sqrt{(1+\frac{1}{l})} = l+\frac{1}{2}</math>. We resist the urge to discard the <math> \frac{1}{2} </math> because without it our result will be trivial. | Since <math>\ l </math> will be very large we invoke the approximation: <math>\ \sqrt{l(l+1)} \approx l \sqrt{(1+\frac{1}{l})} = l+\frac{1}{2}</math>. We resist the urge to discard the <math> \frac{1}{2} </math> because without it our result will be trivial. | ||
<math>\ L= \hbar(l+\frac{1}{2})</math>, therefore <math>\ l = \frac{L}{\hbar}-\frac{1}{2}. | <math>\ L= \hbar(l+\frac{1}{2})</math>, therefore <math>\ l = \frac{L}{\hbar}-\frac{1}{2}</math>. Plugging this expression into the equation for <math>\ \alpha </math> and using the previous approximation again, we have: <math>\ \cos \alpha \approx \frac{l}{l+\frac{1}{2}} = \frac{\frac{L}{\hbar}-\frac{1}{2}}{\frac{L}{\hbar}} = \frac{2L-\hbar}{2L}</math>. | ||
<math>\ \alpha \approx cos^{-1}\frac{2L-\hbar}{2L}</math>. | |||
Plugging in <math>\ L = 4.83 \cdot 10^{31} J \cdot s </math> and <math>\ \hbar = 1.055 \cdot 10^{-34} J \cdot s</math> we obtain: | |||
<math>\ \alpha \approx cos^{-1}(0.999999999999999999999999999999999999999999999999999999999999999998908) \approx 1.48 \cdot 10^{-33} rad</math>. | |||
This is the smallest angle that <math>\ \mathbf{L} </math> makes with the z-axis in the case of the earth going around the sun. | |||
Revision as of 18:13, 6 December 2009
Suppose the earth revolves around the sun counter-clockwise in the x-y plane with the sun at the origin. Quantum mechanically, what is the minimum angle of the angular momentum vector of the earth with the z axis? Ignore the intrinsic spin of the earth. The angular momentum of the earth around the sun is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ 4.83 \cdot 10^{31} J \cdot s} . Compare the minimum angle to that of a quantum particle with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ l=4 } .
Solution:
Recall that in QM: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ L = \hbar\sqrt{l(l+1)}}
; Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ L_z = m\hbar}
.
The angle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \theta } between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{L} } and the z-axis fulfills: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \cos \theta = \frac{L_z}{L} = \frac{m}{\sqrt{l(l+1)}}} .
To make Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \theta } as small as possible, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ m } must be maximum. This is when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ m=l } . Therefore, the minimum angle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \alpha } obeys: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \cos \alpha = \frac{l}{\sqrt{l(l+1)}} }
We solve Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ L = \hbar\sqrt{l(l+1)}} to find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ l } . Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ l } will be very large we invoke the approximation: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \sqrt{l(l+1)} \approx l \sqrt{(1+\frac{1}{l})} = l+\frac{1}{2}} . We resist the urge to discard the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2} } because without it our result will be trivial. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ L= \hbar(l+\frac{1}{2})} , therefore Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ l = \frac{L}{\hbar}-\frac{1}{2}} . Plugging this expression into the equation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \alpha } and using the previous approximation again, we have: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \cos \alpha \approx \frac{l}{l+\frac{1}{2}} = \frac{\frac{L}{\hbar}-\frac{1}{2}}{\frac{L}{\hbar}} = \frac{2L-\hbar}{2L}} .
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \alpha \approx cos^{-1}\frac{2L-\hbar}{2L}} .
Plugging in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ L = 4.83 \cdot 10^{31} J \cdot s } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \hbar = 1.055 \cdot 10^{-34} J \cdot s} we obtain:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \alpha \approx cos^{-1}(0.999999999999999999999999999999999999999999999999999999999999999998908) \approx 1.48 \cdot 10^{-33} rad} .
This is the smallest angle that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \mathbf{L} } makes with the z-axis in the case of the earth going around the sun.